9 Electric Cells Electric Cells

9 Electric Cells Electric Cells
chapter
9
In this chapter,
you will be able to
•
demonstrate an
understanding of oxidation
and reduction in terms of the
transfer of electrons or
change in oxidation number;
•
identify and describe the
functioning of the
components in electric cells;
•
describe galvanic cells in
terms of oxidation and
reduction half-cells and
electric potential differences;
•
describe the function of the
hydrogen reference half-cell
in assigning reduction
potential values;
•
explain corrosion as an
electrochemical process, and
describe corrosion-inhibiting
techniques;
•
demonstrate
oxidation–reduction
reactions through
experiments and analyze
these reactions;
•
write balanced chemical
equations for redox reactions;
•
determine oxidation and
reduction half-cell reactions,
current and ion flow, electrode
polarity and cell potentials of
typical galvanic cells;
•
predict the spontaneity of
redox reactions and cell
potentials using a table of
half-cell reduction potentials;
•
describe examples of
common galvanic cells and
evaluate their environmental
and social impact;
•
research and assess
environmental, health, and
safety issues involving
electrochemistry.
650 Chapter 9
Electric
Cells
Since their invention in 1888, vehicles powered entirely by electricity have drifted in and
out of fashion. Many experts predict that will end in the next decade, when electric vehicles finally make a breakthrough. A combination of political, economic, and environmental
factors are slowly making electric power a viable alternative to gasoline power. One
advantage of electric cars over gasoline-fuelled cars is environmental. There is real
promise of a significant reduction in pollution from the vehicles we drive. Also, cars
powered by gasoline engines are about 15% efficient, but many electric cars are 90%
efficient. (Of course, overall efficiency depends on how the electricity and gasoline are
produced in the first place.) Other attractive features of electric vehicles are that they are
nearly silent and require minimal maintenance.
The most serious obstacle to the widespread use of electric cars is the lack of a powerful, lightweight, inexpensive battery to increase range and usefulness. Scientists and engineers are researching alternatives to the common lead-acid battery. Perhaps the most
promising alternative is not a battery that needs to be periodically recharged, but one that
runs continuously as fuel is supplied. One such alternative is the aluminum-air fuel cell,
which uses aluminum metal as the fuel and oxygen from the air to produce electricity.
Another possibility is the solid polymer hydrogen (or methanol) fuel cell, in which
hydrogen (or a hydrogen-rich fuel) and oxygen from the air produce electricity. The
discovery of the hydrogen fuel cell led to a four-fold improvement in power, so that
liquid and gas fuel cells are now feasible batteries for electric cars.
In this chapter, you will study the technological development and scientific understanding of cells and batteries that produce electricity for many uses, including,
potentially, electric cars.
REFLECT on your learning
1. How do cells and batteries work?
2. What are the key scientific concepts or principles that can be used to understand and
explain the internal components and processes of a cell that produces electricity?
3. List the types and uses of a variety of common electric cells. Include an assessment
of the impact of each one on our lives.
NEL
TRY THIS activity
A Simple Electric Cell
A cell that produces electricity can be amazingly simple because it uses very
common materials and requires no technical expertise to construct. Anyone
can make one and then improve its efficiency without much understanding
of the scientific principles involved. That is why the electric cell was used for
more than 100 years before scientists understood how it worked.
Materials: copper and zinc metal strips (or any two different metals); steel
wool, orange, apple, potato (and other fruits or vegetables); LCD clock;
voltmeter (or multimeter) with leads
• Clean the metals with steel wool to remove any coating or oxides.
• Stick both metal strips into the orange. Make sure that the metal strips are
not in contact inside the orange.
• Momentarily touch the leads (red—positive; black—negative) from the
voltmeter, one to each metal strip. Now reverse the leads and test again.
(a) Record and describe what happened in each case.
• Connect the leads to the LCD clock, paying attention to positive and
negative connections.
(b) Does the clock work? If it does not, suggest a solution to make it
work. Try it.
(c) Explain, in your own words, what you think happened in (b).
• Repeat the above process using other fruits and vegetables.
(d) Did you produce electricity in each case?
(e) What do all fruits and vegetables have in common?
(f) How could you improve upon your electric cell?
Wash your hands after completing this activity.
NEL
Electric Cells 651
9.1
7.1
Figure 1
The technology of metallurgy has a
long history, preceding by thousands of years the scientific understanding of the processes.
In prehistoric times, people learned to extract metals from rocks and minerals (Figure 1).
This discovery initiated both the technology of metallurgy and humanity’s progression
from the Stone Age, through the Bronze Age and the Iron Age, to our increasingly technological modern age. Only a few metals, such as gold and silver, exist naturally in the form
of a pure element. Most metals exist in a variety of compounds mixed with other substances
in rocks called ores. The pure metals must be extracted, or refined, from the ores. For
some metals, the basic procedures are quite simple and were developed early in human
history; for others, more complex procedures have been developed more recently.
bronze
(copper and tin)
copper
5000
Oxidation and Reduction
4000
3000
brass
(copper and zinc)
zinc
aluminum
iron
2000
1000
B.C.
A.D.
1000
2000
From metallurgy, the term reduction came to be associated with producing metals
from their compounds. Red iron(III) oxide “reduced” by carbon monoxide gas to iron
metal is a typical example of this process. The production of tin and copper metals are
other examples where a metal compound is reduced to the metal.
3 CO(g) → 2 Fe(s) 3 CO2(g)
SnO2(s) C(s) → Sn(s) CO2(g)
CuS(s) H2(g) → Cu(s) H2S(g)
Fe2O3(s)
Figure 2
Making steel requires higher temperatures than those provided by a
simple wood fire. Only a few cultures developed the technology to
make steel early in their history. In
Japan, steel was used in the crafting
of samurai swords. At a time when
there was no written language, the
process of sword-making was made
into a ritual so that it could be more
accurately passed on from one generation to the next.
652 Chapter 9
As you can see from these chemical equations, another substance, called a reducing agent,
causes or promotes the reduction of a metal compound to an elemental metal. In the preceding examples, carbon monoxide is the reducing agent for the production of iron
from iron(III) oxide. Carbon (charcoal) is the reducing agent for the production of tin
from tin(IV) oxide, and hydrogen is the reducing agent for the production of copper
from copper(II) sulfide. These are three of the most common reducing agents used in
metallurgical processes.
Although the discovery of fire occurred much earlier than that of metal refining, both
discoveries advanced the development of civilization significantly. There are also important similarities in the chemistry behind these technological developments. Of course, it
does not require a detailed scientific understanding of the processes to use fire (Figure 2).
Only relatively recently, in the 18th century, have we come to realize the role of oxygen in
burning. Understanding the connection between corrosion and burning is an even more
recent development. Corrosion, including the rusting of metals, is now understood to
NEL
Section 9.1
be similar to combustion, although corrosion reactions occur more slowly. Reactions of
substances with oxygen, whether they were the explosive combustion of gunpowder, the
burning of wood, or the slow corrosion of iron, came to be called oxidation. As the study
of chemistry developed, it became apparent that oxygen was not the only substance that
could cause reactions similar to oxidation reactions. For example, metals can be converted to compounds by most nonmetals and by some other substances as well. The rapid
reaction process we call burning may even take place with gases other than oxygen, such
as chlorine or bromine (Figure 3). The term “oxidation” has been extended beyond reactions with oxygen to include a wide range of combustion and corrosion reactions, such
as the following:
O2(g) → 2 MgO(s)
2 Al(s) 3 Cl2(g) → 2 AlCl3(s)
Cu(s) Br2(g) → CuBr2(s)
2 Mg(s)
A substance that causes or promotes the oxidation of a metal to produce a metal compound is called an oxidizing agent. In the reactions shown above, the oxidizing agents are
oxygen, chlorine, and bromine. As you will see, an understanding of reduction and
oxidation is necessary to explain how electric cells (batteries) work.
Figure 3
Copper metal is oxidized by reactive
nonmetals such as bromine.
DID YOU
Practice
Understanding Concepts
1. Write an empirical definition for each of the following terms:
(a) reduction
(d) reducing agent
(b) oxidation
(e) metallurgy
(c) oxidizing agent
(f) corrosion
2. For each of the following, classify the reaction of the metal or metal compound as
reduction or oxidation, and identify the oxidizing agent or the reducing agent.
(a) 4 Fe(s) 3 O2(g) → 2 Fe2O3(s)
(b) 2 PbO(s) C(s) → 2 Pb(s) CO2(g)
(c) NiO(s) H2(g) → Ni(s) H2O(l)
(d) Sn(s) Br2(l) → SnBr2(s)
(e) Fe2O3(s) 3 CO(g) → 2 Fe(s) 3 CO2(g)
(f) Cu(s) 4 HNO3(aq) → Cu(NO3 )2(aq) 2 H2O(l) 2 NO2(g)
KNOW
?
Putting Out Class D Fires
A Class D fire includes combustible metals such as magnesium and the alkali metals. These
metals burn at very high temperatures and water can make the fire
much worse, because of violent
reactions. Carbon dioxide extinguishers don’t help either—magnesium burns very well in carbon
dioxide. So how can you put out
such a fire? Sand is a simple
option, but a special fire extinguisher such as MET-L-X, which
contains sodium chloride, is the
preferred method.
3. List three reducing agents used in metallurgy.
4. What class of elements behaves as oxidizing agents for metals?
Making Connections
5. In the history of metallurgy, which came first, technological applications or scientific
understanding? Elaborate on your answer.
Extension
6. Archaeometallurgy is the study of ancient metallurgy using modern analytical techniques (Figure 4). Give some examples of research in the field. What metals and time
periods have been studied? Can a metal from one mine be distinguished from the
same metal from another mine? How is this information used in archaeological
studies?
GO
NEL
www.science.nelson.com
Figure 4
Aslihan Yener pioneered the use of
modern X-ray techniques to identify
metals from as early as 8000 B.C.E.
Electric Cells 653
Electron Transfer Theory
Figure 5
A common single displacement
reaction is the reaction of active
metals with an acid, such as zinc
with hydrochloric acid.
INVESTIGATION 9.1.1
Single Displacement Reactions
(p. 715)
Some familiar single displacement
reactions provide evidence for
reduction and oxidation processes.
Single displacement reactions are familiar reactions in which one element replaces
another similar element in a compound. These reactions are useful to investigate first
because they provide a relatively simple introduction to the modern theoretical definitions of oxidation and reduction. A useful idea is to imagine that a reaction is a combination of two parts, called half-reactions. A half-reaction represents what is happening
to only one reactant in an overall reaction. It tells only part of the story. Another halfreaction is required to complete the description of the reaction. Splitting a chemical
reaction equation into two parts not only makes the explanations simpler but also leads
to some important applications that are discussed later in this unit.
For example, when zinc metal is placed into a hydrochloric acid solution, gas bubbles form as the zinc slowly dissolves (Figure 5). Diagnostic tests show that the gas is
hydrogen and that zinc ions are present in the solution. Notice that zinc metal is oxidized
to zinc chloride. This is a corrosion of zinc caused by the hydrochloric acid.
Zn(s)
2 HCl(aq) → ZnCl2(aq) H2( g)
What happens to the zinc and what happens to the hydrochloric acid? The half-reactions help to answer these questions. The zinc atoms in the solid, Zn(s), are converted to
zinc ions in solution, Zn2
(aq) . Atomic theory requires that the zinc atoms lose two
electrons, as shown by the following half-reaction equation:
Zn(s) → Zn2
(aq)
2 e
Simultaneously, hydrogen ions in the solution gain electrons and are converted into
hydrogen gas, as shown below.
2 H
(aq)
2 e → H2(g)
Notice that both of these half-reaction equations, or half-reactions, are balanced by mass
(same number of element symbols on both sides) and by charge (same total charge on
both sides).
Figure 6
A piece of copper before it is placed
into a beaker of silver nitrate solution (left). Note the changes after
the reaction has occurred (right).
The blue colour of the solution
2 ions are present.
indicates Cu(aq)
654 Chapter 9
NEL
Section 9.1
In a laboratory, single displacement reactions in aqueous solution are easier to study
than the metallurgy or corrosion reactions discussed earlier in this chapter. However,
all of these reactions share a common feature—ions are converted to atoms and atoms
are converted to ions. For example, consider the reduction of aqueous silver nitrate to
silver metal in the presence of solid copper (Figure 6). According to atomic theory, silver
atoms are electrically neutral particles (47p, 47e) and silver ions are charged particles
(47p, 46e). In this reaction, an electron is required to convert a silver ion into a silver
atom. The following half-reaction equation explains the reduction of silver ions using
the theoretical rules for atoms and ions.
Ag+(aq)
e → Ag0(s)
(reduction)
According to modern theory, the gain of electrons is called
reduction.
Although this theoretical definition of reduction is in agreement with current atomic
theory, it does not explain where the electrons come from. As crystals of silver metal are
produced, the solution becomes blue, indicating that copper atoms are being converted
to copper(II) ions. According to atomic theory, copper atoms (29p, 29e) must be
losing electrons as they form copper(II) ions (29p, 27e).
Cu0(s) → Cu2
(aq)
2 e
(oxidation)
LEARNING
TIP
Balancing Half-Reaction
Equations
A common difficulty in writing halfreaction equations is deciding on
which side of the equation the electrons should appear. There are two
ways to approach this. You can use
your theoretical knowledge of
atoms and ions to determine
whether the species is losing or
gaining electrons. The other alternative is to add up the electric charges
on both sides of the equation arrow.
The total must be the same on both
sides. Try this for the half-reaction
equations on this page.
reduction a process in which electrons are gained
oxidation a process in which electrons are lost
According to modern theory, the loss of electrons is called
oxidation.
DID YOU
Evidence shows that the silver-coloured solid and the blue colour of the solution are
simultaneously formed near the surface of the copper metal. Therefore, scientists believe
that the electrons required by the silver ions are supplied when silver ions collide with
copper atoms on the metal surface.
Theory suggests that the total number of electrons gained in a reaction must equal the
total number of electrons lost. Also, oxidation and reduction are separate processes. This
theoretical description requires oxidation and reduction to occur simultaneously rather
than sequentially. Oxidation–reduction reactions are often simply called “redox” reactions.
The equations for reduction and oxidation half-reaction and the overall (net) ionic
equation summarize the electron transfer that is believed to take place during a redox
reaction. As in other chemical reactions, the net equation must be balanced.
Writing and Balancing Half-Reaction Equations
1.
KNOW
?
Redox Mnemonics
A mnemonic is a word or group of
words used to help remember
some information. “LEO the lion
says GER” is a mnemonic to help
remember that “Loss of Electrons
is Oxidation” and “Gain of
Electrons is Reduction.” Another
mnemonic is “OIL RIG” which
translates as “Oxidation Is Loss,
Reduction Is Gain.” Sometimes
mnemonics are better if you invent
them yourself.
SAMPLE problem
Write a balanced net equation for the reaction of copper metal with aqueous
silver nitrate.
To show that the number of electrons gained equals the number lost in two half-reaction
equations, it may be necessary to multiply one or both half-reaction equations by an
integer to balance the electrons. In this example, the silver half-reaction equation must be
multiplied by 2.
→ Cu2
(aq)
Cu(s)
2 [Ag
(aq)
NEL
e
→ Ag(s)]
2 e
(two electrons lost by one atom)
(two electrons gained by two ions)
Electric Cells 655
Ag
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Now, add the half-reaction equations and cancel terms that appear on both sides of the
equation to obtain the net ionic equation.
2 Ag
(aq)
2 Ag
(aq)
Ag
2
e Cu(s) → 2 Ag(s) Cu2
e
(aq) 2
Cu(s) → 2 Ag(s) Cu2
(aq)
Silver ions are reduced to silver metal by reaction with copper metal. Simultaneously,
copper metal is oxidized to copper(II) ions by reaction with silver ions (Figure 7).
oxidized to metal ion
+
2 Ag(aq)
Cu(s)
→
+
2 Ag(s) Cu2(aq)
reduced to metal
Cu
Cu
Cu
Cu
Ag
Ag
Cu
Cu2
Cu
Cu
Cu
Cu
To evaluate this theory of oxidation and reduction you should look to see if it is
consistent with other accepted theories and definitions. The theoretical definitions
of oxidation and reduction are consistent with the historical, empirical definitions
presented earlier in this chapter; for example, a metal-containing compound is
reduced to a metal and a metal is oxidized to form a compound. Redox theory is
also consistent with accepted atomic theory and the collision–reaction theory. Most
importantly, redox theory explains the observations made by scientists.
Example 1
Figure 7
A model of the reaction of copper
metal and silver nitrate solution illustrates aqueous silver ions reacting at
the surface of a copper strip.
Write and label two balanced half-reaction equations to describe the reaction of zinc
metal with aqueous lead(II) nitrate, as given by the following chemical equation.
Zn(s)
Solution
Zn(s) → Zn2
(aq)
Pb2
(aq)
LEARNING
Pb(NO3 )2(aq) → Pb(s) Zn(NO3 )2(aq)
2 e
TIP
Cancelling Terms for a Net
Ionic Equation
The terms you can cancel must
be identical, including their
states of matter. Electrons must
always cancel completely. This
is because the electrons that
appear in each half-reaction
equation are the same electrons. They are the electrons
that transfer from one particle to
another.
SUMMARY
2 e
(oxidation)
→ Pb(s)
(reduction)
Electron Transfer Theory
• A redox reaction is a chemical reaction in which electrons are transferred
between particles.
• The total number of electrons gained in the reduction equals the total number
of electrons lost in the oxidation.
• Reduction is a process in which electrons are gained.
• Oxidation is a process in which electrons are lost.
Practice
Understanding Concepts
7. Write a theoretical definition for each of the following terms:
(a) redox reaction
(b) reduction
(c) oxidation
8. Write a pair of balanced half-reaction equations—one showing a gain of electrons
and one showing a loss of electrons—for each of the following reactions:
2
(a) Zn(s) Cu2
(aq) → Zn(aq) Cu(s)
(b) Mg(s) 2 H(aq) → Mg2
(aq) H2( g)
656 Chapter 9
NEL
Section 9.1
9. For each of the following, write and label the oxidation and reduction half-reaction
equations. Ignore spectator ions.
(a) Ni(s) Cu(NO3 )2(aq) → Cu(s) Ni(NO3 )2(aq)
(b) Pb(s) Cu(NO3 )2(aq) → Cu(s) Pb(NO3 )2(aq)
(c) Ca(s) 2 HNO3(aq) → H2( g) Ca(NO3 )2(aq)
(d) 2 Al(s) Fe2O3(s) → 2 Fe(l) Al2O3(s) (Figure 8)
10. We have only looked at one type of single displacement reaction—a metal displacing another metal from an ionic compound. A nonmetal can also displace
another nonmetal from an ionic compound. For example,
Cl2(aq)
2 NaI(aq) → I2(s) 2 NaCl(aq)
Using your knowledge of atoms and ions and the ideas presented in this chapter,
write a pair of balanced half-reaction equations for this reaction—one showing a
gain of electrons and one showing a loss of electrons
11. Ionic compounds can react in double displacement reactions. For example,
FeCl3(aq)
3 NaOH(aq) → Fe(OH)3(s) 3 NaCl(aq)
According to ideas discussed in this chapter, has a redox reaction taken place in
the reaction above? Explain your answer.
Oxidation States
Historically, oxidation and reduction were considered separate processes, more of interest
for technology than for science. With modern atomic theory came the idea of an electron
transfer involving both a gain of electrons by one particle and a loss of electrons by
another particle. This theory of redox reactions is most easily understood for atoms or
monatomic ions. Metals and monatomic anions tend to lose electrons (become oxidized),
whereas nonmetals and monatomic cations tend to gain electrons (become reduced).
More complex redox reactions, such as the reduction of iron(III) oxide by carbon
monoxide in the process of iron production, the oxidation of glucose in the biological
process of respiration, and the use of dichromate ions as a strong oxidizing agent in
chemical analysis are not adequately described or explained with simple redox theory.
In order to describe oxidation and reduction of molecules and polyatomic ions,
chemists have developed a method of “electron bookkeeping” to keep track of the loss
and gain of electrons. The method is arbitrary but it works well. In this system, the oxidation state of an atom in an entity is defined as the apparent net electric charge that an
atom would have if electron pairs in covalent bonds belonged entirely to the more electronegative atom. An oxidation state is a useful idea for keeping track of electrons but
it does not usually represent an actual charge on an atom—oxidation states are arbitrary charges.
An oxidation number is a positive or negative number corresponding to the oxidation
state assigned to an atom. In a covalently bonded molecule or polyatomic ion, the more
electronegative atoms are considered to be negative and the less electronegative atoms
are considered to be positive. For example, in a water molecule the oxygen atom (electronegativity 3.5) is assigned the bonding electron from each hydrogen atom (electronegativity 2.1); that is, the oxidation number of the oxygen atom is 2 and the
oxidation number of each hydrogen atom is 1 (Figure 9). In order to distinguish these
numbers from actual electrical charges, oxidation numbers are written in this book as
positive or negative numbers, that is, with the sign preceding the number. Oxidation
NEL
Figure 8
The reduction of iron(III) oxide by
aluminum is called the "thermite"
reaction. Because this reaction is
rapid and very exothermic, molten
white-hot iron is produced. Here a
falling aluminum wrench momentarily sparks a thermite reaction
when it strikes a rusted iron block.
LEARNING
TIP
Although the meaning of the terms
oxidation state and oxidation
number are slightly different, some
people use these terms interchangeably.
O H
H
Figure 9
An oxygen atom has 8p and 8e. If
the oxygen atom gets to count the
two hydrogen electrons in the two
shared pairs of electrons, then 8p
and 10e produces an apparent net
charge of 2. Each hydrogen atom
with 1p has no additional electrons. Its one electron has already
been counted by the oxygen atom.
Therefore, the hydrogen has an
apparent net charge of 1 .
oxidation number a positive or
negative number corresponding to
the apparent charge that an atom in
a molecule or ion would have if the
electron pairs in covalent bonds
belonged entirely to the more
electronegative atom
Electric Cells 657
LEARNING
TIP
Oxidation Number Format
Oxidation numbers are simply
positive or negative numbers
assigned on the basis of a set of
arbitrary rules. It is important for
you to realize that these are not
electric charges. For this reason,
we write an oxidation with the
sign preceding the number, as
in +2 or “positive two.” This differs from an ion charge of 2+ or
“two positive” units of electric
charge.
numbers can be assigned to many common atoms and ions (Table 1) and they can then
be used to determine the oxidation numbers of other atoms.
Oxidation numbers are simply a systematic way of counting electrons. Therefore, the
sum of the oxidation numbers in a compound or ion must equal the total charge—zero
for neutral compounds, and the ion charge for ions.
Table 1 Common Oxidation Numbers
Atom or ion
Oxidation number
all atoms in elements
hydrogen in all compounds
except hydrogen in hydrides
oxygen in all compounds
except oxygen in peroxides
all monatomic ions
SAMPLE problem
0
1
1
2
1
charge on ion
Examples
Na is 0, Cl in Cl2 is 0
H in HCl is 1
H in LiH is 1
O in H2O is 2
O in H2O2 is 1
Na+ is 1, S2 is 2
Oxidation Number in a Molecular Compound
1.
What is the oxidation number of carbon in methane, CH4?
This is determined by assigning an oxidation number of +1 to hydrogen (Table 1).
x 1
CH4
Now solve for x. Since a methane molecule is electrically neutral, the oxidation numbers
of the one carbon atom and the four hydrogen atoms (4 times +1) must equal zero.
x 4(1) 0
x 4
x 1
4 1
becomes
CH4
CH4
Carbon in methane has an oxidation number of 4.
2.
What is the oxidation number of manganese in a permanganate ion, MnO4.
In a polyatomic ion, like a neutral compound, the total of the oxidation numbers of all
atoms must equal the charge. The oxidation number of manganese in the permanganate
ion, MnO4, is determined using the oxidation number of oxygen as 2 (Table 1) and the
knowledge that the charge on the ion is 1. The total of the oxidation numbers of the one
manganese atom (x) and the four oxygen atoms (4 times 2) must equal the charge on
the ion (1).
x 4(2) = 1
x = 7
x 2
MnO4
7 2
becomes
MnO4
The oxidation number of manganese in MnO4 is 7.
658 Chapter 9
NEL
Section 9.1
TIP
Example
LEARNING
What is the oxidation number of sulfur in sodium sulfate?
Oxidation Numbers
The summary of the method using
Table 1 to determine oxidation
numbers is restricted to molecules
or polyatomic ions in which there is
only one unknown, i.e., only one
kind of atom other than hydrogen
or oxygen. Situations not covered
by Table 1 can still be done using
electronegativities to assign electrons to atoms in a Lewis structure.
Solution
1 x 2
2(1) x 4(2) 0
x 6
Na2SO4
The oxidation number of sulfur in sodium sulfate is 6.
SUMMARY
Determining Oxidation Numbers
Step 1 Assign common oxidation numbers (Table 1).
Step 2 The total of the oxidation numbers of atoms in a molecule or ion equals
the value of the net electric charge on the molecule or ion.
(a) The sum of the oxidation numbers for a compound is zero.
(b) The sum of the oxidation numbers for a polyatomic ion equals the
charge on the ion.
Step 3 Any unknown oxidation number is determined algebraically from the
sum of the known oxidation numbers and the net charge on the entity.
Practice
Understanding Concepts
12. Determine the oxidation number of
(c) S in SO42
(a) S in SO2
(b) Cl in HClO4
(d) Cr in Cr2O72
13. Determine the oxidation number of nitrogen in
(a) N2O( g)
(d) NH3( g)
(b) NO( g)
(e) N2H4( g)
(c) NO2( g)
(f) NaNO3(s)
(e) I in MgI2
(f) H in CaH2
(g) N2( g)
(h) NH4Cl(s)
14. Determine the oxidation number of carbon in
(a) graphite (elemental carbon)
(c) sodium carbonate
(b) glucose
(d) carbon monoxide
15. In a breathalyzer (Figure 10), ethanol is oxidized by an acidic dichromate solution.
(a) Determine the oxidation number of every atom or ion in the following breathalyzer reaction:
3
2 + 3 C H OH
16 H(aq)
+ 2 Cr2O7(aq)
2 5
(aq) → 4 Cr(aq) 3 CH3COOH(aq) H2O( l)
(b) What colour change would you expect in this reaction?
16. Carbon can be progressively oxidized in a series of organic reactions. Determine
the oxidation number of carbon in each of the compounds in the following series
of oxidations:
methane→ methanol→ methanal→ methanoic acid→ carbon dioxide
Extension
17. Assigning oxidation numbers using the rules we have established may occasionally produce some unusual results. For example, consider Fe3O4.
(a) Determine the oxidation number of iron in Fe3O4.
(b) What is unusual about your answer?
(c) Suggest a reason why this might occur.
NEL
Figure 10
It is a criminal offence to drive a
vehicle when you have a blood
alcohol content greater than
0.08 g/100 mL of blood. The breathalyzer measures the alcohol content
of exhaled breath, which is assumed
to be proportional to the blood
alcohol content. Inside the device,
the alcohol in the breath sample is
oxidized by acidic potassium
dichromate in an acidic solution.
Answers
12. (a) 4
(d) 6
(b) 7
(e) 1
(c) 6
(f) 1
13. (a) 1
(e) 2
(b) 2
(f) 5
(c) 4
(g) 0
(d) 3
(h) 3
14. (a) 0
(c) 4
(b) 0
(d) 2
15. 1; 6,2; 2, 1, 2, 1;
3; 0,1, 0, 2, 1; 1; 2
16. 4; 2; 0; 2; 4
Electric Cells 659
oxidation an increase in oxidation
number
reduction a decrease in oxidation
number
oxidation
number
1
0
1
reduction
oxidation
2
Oxidation Numbers and Redox Reactions
Although the concept of oxidation states is somewhat arbitrary, because it is based on
assigned charges, it is self-consistent and allows predictions of electron transfer. If the oxidation number of an atom or ion changes during a chemical reaction, then an electron
transfer (that is, an oxidation–reduction reaction) has occurred. An increase in the oxidation number is defined as an oxidation and a decrease in the oxidation number is a
reduction. If oxidation numbers are listed as positive and negative numbers on a number
line as they are in Figure 11, then the process of oxidation involves a change to a more
positive value (“up” on the number line) and reduction is a change to a more negative
value (“down” on the number line). If the oxidation numbers do not change, this is
interpreted as no transfer of electrons. A reaction in which all oxidation numbers remain
the same is not a redox reaction.
• An oxidation is an increase in oxidation number.
• A reduction is a decrease in oxidation number.
• In a redox reaction, oxidation numbers change.
2
Figure 11
In a redox reaction, both oxidation
and reduction occur.
Coal is a fossil fuel that is burned in huge quantities in some electrical power generating stations. If we assume pure carbon and a complete combustion, carbon is converted to carbon dioxide. In this reaction, the oxidation number of carbon changes from
0 in C(s) to 4 in CO2(g). Simultaneously, oxygen is reduced from 0 in O2(g) to 2 in
CO2(g).
oxidation
LAB EXERCISE 9.1.1
Oxidation States of Vanadium
(p. 716)
Vanadium is an interesting element
because it forms many different ions
with different colours.
0
0
+4 –2
C(s) + O2(g) → CO2(g)
reduction
The main purpose of assigning oxidation numbers is to see how these numbers change
as a result of a chemical reaction. In any redox reaction, like the combustion of carbon, there
will always be both an oxidation and a reduction. We will use these changes in the next section to balance redox equations, but first we will look at some additional examples.
Sample
Oxidation Number Changes in a Reaction
SAMPLE problem
1.
You have seen the reaction of active metals like zinc with an acid (Figure 5).
Identify the oxidation and reduction in the reaction of zinc metal with
hydrochloric acid.
First, you need to write the chemical equation, as it is not provided. Net ionic equations
are best, but the procedure will still work if you write a non-ionic equation.
2
Zn(s) 2 H
(aq) → Zn(aq) H2(g)
After writing the equation, determine all oxidation numbers.
0
1
2
0
2
Zn(s) 2 H
(aq) → Zn(aq) H2( g)
Now look for the oxidation number of an atom/ion that increases as a result of the reaction and label the change as oxidation. There must also be an atom/ion whose oxidation
number decreases. Label this change as reduction.
660 Chapter 9
NEL
Section 9.1
oxidation
0
+1
Zn(s) + 2
+
H(aq)
+2
→
DID YOU
0
2+
Zn(aq)
+ H2(g)
reduction
2.
When natural gas burns in a furnace, carbon dioxide and water form. Identify
oxidation and reduction in this reaction.
First, write the equation.
CH4(g) 2 O2(g) → CO2(g) 2 H2O(g)
• Removal of oxygen decreases
the oxidation number of carbon
(i.e., a reduction),
e.g.,
+4 –2
+1 –2
e.g.,
Example
The determination of blood alcohol content from a sample of breath or blood involves the
reaction of the sample with acidic potassium dichromate solution. If ethanol is present,
chromium(III) ions, water, and acetic acid are produced. Identify the oxidation and reduction in the following chemical reaction:
2–
+
3+
2Cr2O7(ag)
+ 16H(ag)
+ 3C2H5OH(ag) → 4Cr(ag)
+ 11H2O(g) + 3CH3COOH(ag)
Solution
oxidation
–2 +1 –2 +1
C2H6 → C2H4
C is 3
reduction
Carbon is oxidized from 4 in methane to 4 in carbon dioxide as it reacts with
oxygen. Simultaneously, oxygen is reduced from 0 in oxygen gas to 2 in both products.
Notice that the oxygen atoms in the reactant are distributed between the two products.
This does not change our procedure because we are only looking for the change from
reactant to product. We say that “oxygen is reduced” in this reaction and it does not
matter where the reduced oxygen appears in the products.
+1
C is 0
• Removal of hydrogen increases
the oxidation number of carbon
(i.e., an oxidation),
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
+6 –2
HCOOH → HCHO
C is 2
oxidation
0
?
Redox in Biological Systems
Biologists often classify oxidation
and reduction in terms of the addition or removal of oxygen or
hydrogen.
Now we can insert the oxidation numbers and arrows.
–4 +1
KNOW
+3
+1 –2
DID YOU
C is 2
KNOW
?
Redox Reactions in Living
Organisms
The ability of carbon to take on
different oxidation states is essential to life on Earth. Photosynthesis
involves a series of reduction reactions in which the oxidation
number of carbon changes from
4 in carbon dioxide to an
average of 0 in sugars such as
glucose. In cellular respiration,
carbon undergoes a series of oxidations, after which the oxidation
number of carbon is again 4 in
carbon dioxide.
0+1 0–2–2 +1
2–
3+
2 Cr2O7(aq)
+ 16 H+(aq) + 3 C2H5OH(aq) → 4 Cr(aq)
+ 11 H2O(g) + 3 CH3COOH(aq)
reduction
Chromium atoms in Cr2O72 are reduced (6 to 3). Carbon atoms in C2H5OH are
oxidized (2 to 0).
SUMMARY
Oxidation States
• Oxidation is an increase in oxidation number.
• Reduction is a decrease in oxidation number.
• A redox reaction involves both an oxidation and a reduction.
NEL
Electric Cells 661
Practice
Understanding Concepts
18. Methanol reacts with acidic permanganate ions as shown below:
Answers
18. (a) 2, 1, 2, 1; 7,
2; 1; 0, 1, 2; 2;
1, 2
19. (a) 0; 1,5, 2; 0; 2,
5, 2
(b) 2,5, 2; 1, 1; 2,
1; 1, 5, 2
(c) 0;1, 1; 0; 1, 1
(d) 1, 1; 0; 0
(e) 1,1; 1, 2, 1; 1,
2, 1; 1, 1
(f) 0; 0; 3, 1
1
(g) 22, 1; 0; 4, 2;1,
2
(h) 1, 1; 1, 2; 0
2
5 CH3OH(l) 2 MnO4
(aq) 6 H(aq) → 5 CH2O(l) 2 Mn(aq) 8 H2O(l)
(a) Assign oxidation numbers to all atoms/ions.
(b) Which atom/ion is oxidized? Label the oxidation above the equation.
(c) Which atom/ion is reduced? Label the reduction below the equation.
19. For each of the following chemical reactions, assign oxidation numbers to each
atom/ion and indicate whether the equation represents a redox reaction. If it does,
identify the oxidation and reduction.
(a) Cu(s) 2 AgNO3(aq) → 2 Ag(s) Cu(NO3 )2(aq)
(b) Pb(NO3 )2(aq) 2 KI(aq) → PbI2(s) 2 KNO3(aq)
(c) Cl2(aq) 2 KI(aq) → I2(s) 2 KCl(aq)
(d) 2 NaCl(l) → 2 Na(l) Cl2(g)
(e) HCl(aq) NaOH(aq) → HOH(l) NaCl(aq)
(f) 2 Al(s) 3 Cl2(g) → 2 AlCl3(s)
(g) 2 C4H10(g) 13 O2(g) → 8 CO2(g) 10 H2O(l)
(h) 2 H2O2(l) → 2 H2O(l) O2(g)
20. Classify the chemical equations in question 19 (ah) using the five reaction
types—formation, decomposition, single displacement, double displacement, and
combustion. Which reaction type does not appear to be a redox reaction?
21. Hydrogen peroxide, H2O2(l), can either be oxidized or reduced depending on the substance with which it reacts. Use oxidation numbers to explain why this is possible.
Making Connections
22. Earth has an oxidizing atmosphere of oxygen. The planet Saturn has a reducing
atmosphere of hydrogen and methane. Describe the two types of atmospheres in
terms of changes in oxidation numbers of carbon and of the likely reactions.
SUMMARY
Electron Transfer and Oxidation States
• According to current theory, a redox reaction is a chemical reaction in which
electrons are transferred and the oxidation numbers change.
• Oxidation is the increase in oxidation number and corresponds to a loss of
electrons.
• Reduction is the decrease in oxidation number and corresponds to a gain of
electrons.
662 Chapter 9
NEL
Section 9.1
Section 9.1 Questions
Understanding Concepts
1. Copy and complete the following table to distinguish
between oxidation and reduction:
Electron transfer
Oxidation states
oxidation
8. When carbon dioxide is released into the atmosphere from
natural or human activities, some of it reacts with water to
form carbonic acid. This accounts for the natural acidity of
rainwater and may also contribute to acid rain.
(a) Write the balanced chemical equation for the reaction
of carbon dioxide with water to form carbonic acid.
(b) Is this a redox reaction? Justify your answer.
reduction
Applying Inquiry Skills
2. Write and label a pair of balanced half-reaction equations
for each of the following reactions.
2
(a) Pb(s) Cu2
(aq) → Pb(aq) Cu(s)
(b) Cl2(aq) 2 Br(aq) → 2 Cl
(aq) Br2(l)
3. What is an oxidation number?
4. State two ways you can recognize a redox reaction, using a
chemical reaction equation.
5. In a redox reaction, what happens to the oxidation number of
(a) an atom that is oxidized?
(b) an atom that is reduced?
6. Write the oxidation number of each atom/ion in the following substances:
(a) carbon monoxide, CO(g), a toxic gas
(b) ozone, O3(g), ozone layer
(c) ammonium chloride, NH4Cl(s), used in dry cells (batteries)
(d) phosphoric acid, H3PO4(aq), in cola soft drinks
(e) sodium thiosulfate, Na2S2O3(s), antidote for cyanide
poisoning
(f) sodium tripolyphosphate, Na5P3O10(s), in laundry detergents
7. Redox reactions are common in organic chemistry. For
example, carboxyl groups can be oxidized to form carbon
dioxide. In the following chemical equation, permanganate
ions convert oxalate ions to carbon dioxide in an acidic
solution.
2
2 MnO4
(aq) 5 C2O4(aq) 16 H(aq) →
2 8 H O
2 Mn(aq)
2 (l) 10 CO2(g)
(a) Assign oxidation numbers to all atoms/ions.
(b) Which atom is oxidized? State the change.
(c) Which atom is reduced? State the change.
9. In the Try This Activity at the beginning of this chapter, you
put copper and zinc strips into an orange (or other fruit)
and a reaction occurred to produce electricity. Write a brief
design (plan) to determine if either of the copper or zinc
metals is oxidized or reduced.
Making Connections
10. The dark tarnish that forms on silver objects is silver sulfide.
A common home remedy is to restore the silver with baking
soda and aluminum foil (Figure 12).
(a) Write the single displacement reaction of silver sulfide
and aluminum metal.
(b) Identify what is oxidized and what is reduced.
(c) Do you think this is a better method of cleaning than
polishing or scrubbing the silver? Why or why not?
Figure 12
If the tarnished silver is placed in a hot solution of
baking soda on aluminum foil in a nonmetallic dish
(e.g., glass), a redox reaction converts the tarnish back
into silver metal.
11. Police forces use various instruments and processes to
determine whether a person is legally impaired. What is the
difference in operation between the Breathalyzer and the
Intoxilyzer? Describe briefly how redox reactions are
involved.
GO
NEL
www.science.nelson.com
Electric Cells 663
9.2
LEARNING
TIP
Electron Transfer
All redox reactions are electron
transfer reactions. This means
that electrons that are lost by
one particle are the same electrons that are gained by
another. This is like you giving
five dollars to a friend. You lose
five dollars and your friend
gains five dollars—five dollars
has been exchanged. Obviously,
the money lost must equal the
money gained. The same is true
for electrons in a redox reaction.
Balancing Redox Equations
Knowing the ratio of reacting chemicals is necessary in many applications. Chemists
use the mole ratio from a balanced chemical equation to study the nature of the reaction. The stoichiometry of a reaction is essential in many types of chemical analysis such
as the breathalyzer. Finally, chemical industries need to know the quantities of reactants
to mix and the yield of a desired product. In this and previous courses, you have already
seen many examples of the use of balanced chemical equations.
Simple redox reaction equations can be balanced by inspection or by a trial-and-error
method. You have done this often in previous courses for reactions such as single displacement and combustion. More complex redox reactions may be very difficult to balance this way due to the number and complexity of the reactants and products. As you
will see in this section, oxidation numbers and half-reaction equations can be used to
balance any redox equation.
Oxidation Number Method
One way of recognizing a redox reaction is to assign oxidation numbers to each atom or
ion and then look for any changes in the numbers. Any change in the oxidation number
of a particular atom or ion is believed to be related to a change in the number of electrons.
Because electrons are transferred in a redox reaction, the total number of electrons lost
by one atom/ion must equal the total number of electrons gained by another atom/ion.
In terms of oxidation numbers, this means that the changes in oxidation numbers must
also be balanced.
The total increase in oxidation number for a particular
atom/ion must equal the total decrease in oxidation number
of another atom/ion.
Let’s look at a simple example first. You could easily balance this equation by inspection, but we will use it to illustrate the main points of the oxidation number method.
SAMPLE problem
Balancing Redox Equations
Hydrogen sulfide is an unpleasant constituent of “sour” natural gas. Hydrogen sulfide is not only very toxic but it also smells terrible, like rotten eggs. It is common
practice to “flare” or burn small quantities of sour natural gas that occur with oil
deposits (Figure 1). The gas is burned because it is not worth recovering and
treating a small quantity of gas. When this gas is burned, hydrogen sulfide is
converted to sulfur dioxide.
H2S(g) O2(g) → SO2(g) H2O(g)
Balance this equation.
If you are using oxidation numbers to balance a redox equation, the first step is to assign
oxidation numbers to all atoms/ions and look for the numbers that change. Circle or
highlight the oxidation numbers that change.
664 Chapter 9
NEL
Section 9.2
oxidation
+1 −2
+4 −2
0
+1 −2
SO2(g) + H2O(g)
H2S(g) + O2(g)
reduction
Notice that a sulfur atom is oxidized from 2 to 4. This is a change of 6 and means
6e have been transferred. An oxygen atom is reduced from 0 to 2, a change of 2 or 2e
transferred. Because the substances in the equation are molecules, not atoms, we need to
specify the change in the number of electrons per molecule.
1 2
H2S(g)
4 2
0
6e/S atom
6e/H2S
O2(g)
→
SO2(g)
+1 2
H2O(g
2e/O atom
4e/O2
An H2S molecule contains one sulfur atom. Therefore, the number of electrons transferred per sulfur atom is the same number per H2S molecule. An O2 molecule contains two
O atoms. Therefore, when one O2 molecule reacts, two oxygen atoms transfer 2e each for
a total of 4e.
The next step is to determine the simplest whole numbers that will balance the number
of electrons transferred for each reactant. The numbers become the coefficients for the
reactants.
1 2
H2S(g)
4 2
0
O2(g)
→
SO2(g)
Figure 1
Oil deposits often contain small
quantities of natural gas. The gas
is simply burned (“flared”). Since
the gas often contains hydrogen
sulfide, this practice can be a significant source of pollutants. It is
also a waste of energy when many
of these flares operate in a large
oil field.
1 2
H2O(g)
TIP
6e/S atom
2e/O atom
LEARNING
6e/H2S
2
12
4e/O2
3
12
You can adjust the number of electrons per atom to the number per
molecule by multiplying the number
per atom by the subscript of the
atom in the chemical formula.
Now you have the coefficients for the reactants.
2 H2S(g) 3 O2(g) → SO2(g) H2O(g)
The coefficients of the products can easily be obtained by balancing the atoms whose oxidation numbers have changed and then any other atoms. The final balanced equation is
shown below.
2 H2S(g)
3 O2(g)
→
2 SO2(g)
2 H2O(g)
Sometimes you may not know all of the reactants and products of a redox reaction.
The main reactants and oxidized/reduced products will always be given and you will know
if the reaction took place in an acidic or basic solution. Experimental evidence shows that
water molecules, hydrogen ions, and hydroxide ions play important roles in reactions in
such solutions. The procedure for balancing such equations is initially the same as the one
we have discussed, but you will need to add water molecules, hydrogen ions, and/or
hydroxide ions to finish the balancing of the overall equation. The following two sample
problems illustrate this procedure.
NEL
Electric Cells 665
SAMPLE problem
Balancing Redox Equations in Acidic and Basic Solutions
1.
Chlorate ions and iodine react in an acidic solution to produce chloride ions
and iodate ions.
ClO3 (aq) I2(aq) → Cl(aq) IO3 (aq)
Balance the equation for this reaction.
Assign oxidation numbers to each atom/ion and note which numbers change.
5 2
1
0
ClO 3
(aq)
5 2
Cl
(aq)
→
I 2(aq)
IO 3 (aq)
A chlorine atom is reduced from 5 to 1, a change of 6. Simultaneously, an iodine
atom is oxidized from 0 to 5, a change of 5. Record the change in the number of electrons per atom and per molecule or polyatomic ion.
5 2
1
0
ClO 3
(aq)
6e/Cl
5e/I
6e/ClO3
10e/I2
5 2
Cl
(aq)
→
I 2(aq)
IO3
(aq)
The total number of electrons transferred by each reactant must be the same. Multiply
the numbers of electrons by the simplest whole numbers to make the totals equal, in this
case, 30e. You can now write the coefficients for the reactants and the products.
5 2
5 ClO 3
(aq)
6e/Cl
6e/ClO3
5
0
3 I2(aq)
→
1
5 Cl
(aq)
5 2
6 IO3 (aq)
5e/I
10e/I2
3
Although the chlorine and iodine atoms are now balanced, notice that the oxygen
atoms are not; 15 on the left versus 18 on the right. Because this reaction occurs in an
aqueous solution, we can add H2O molecules to balance the O atoms. The reactant side
requires 3 oxygen atoms (from 3 water molecules) to equal the total of 18 oxygen atoms
on the product side.
3 H2O(l) 5 ClO3
(aq) 3 I2(aq) → 5 Cl (aq) 6 IO 3(aq)
LEARNING
TIP
A balanced chemical reaction
equation includes both a mass
and charge balance. Mass is
balanced using the atomic symbols. If the symbols balance but
not the charge, the equation is
not balanced. Be sure to check
both symbol and charge.
In adding water molecules, we are also adding H atoms. Because this reaction occurs in
an acidic solution, we will add H+
(aq) to balance the hydrogen.
3 H2O( l ) 5 ClO3 (aq) 3 I2(aq) → 5 Cl (aq) 6 IO3 (aq) 6 H (aq)
The redox equation should now be completely balanced. Check your work by checking the
total number of each atom/ion on each side and checking the total electric charge, which
should also be balanced.
2.
Methanol reacts with permanganate ions in a basic solution. The main reactants and products are shown below.
2
2
CH3OH(aq) MnO4
(aq) → CO3(aq) MnO4(aq)
Balance the equation for this reaction.
666 Chapter 9
NEL
Section 9.2
We will follow the same procedure as in the previous problem, adjusting for a basic solution at the end: assign oxidation numbers; note which ones change and by how much per
reactant; and then balance the total number of electrons to obtain the coefficients for the
main reactants and products.
2 1 2 1
1 CH3OH(aq)
7 2
6 MnO4
(aq)
6e/C
1e/Mn
6e/CH3OH
1
1e/MnO4
6
→
4 2
6 2
1 CO32
(aq) 6 MnO42
(aq)
Just as before, add H2O(l) to balance the O atoms. The reactant side requires 2 oxygen
atoms (from 2 water molecules) to equal the 27 oxygen atoms on the product side. Next,
balance the H atoms using H
(aq). The product side requires 8 hydrogens to balance the 8
on the reactant side (4 in water and 4 in methanol).
2
2
2 H2O( l ) CH3OH(aq) 6 MnO4 (aq) → CO3(aq) 6 MnO4(aq) 8 H (aq)
If this reaction occurred in an acidic solution, you would now be finished. For a basic
solution, however, we add enough OH
(aq) to both sides to equal the number of H (aq)
present. The hydrogen and hydroxide ions on the same side of the equation are then combined to form water.
→ CO 2 6 MnO 2 8 H 8 OH
8 OH(aq)
2 H2O(l) CH3OH(aq) 6 MnO4(aq)
(aq)
(aq)
3(aq)
4(aq)
8 H20
Finally, cancel the same number of H2O molecules on both sides. In this case, the H2O
on the reactant side can be cancelled by also removing 2 H2O from the product side,
leaving the extra 6 H2O in the final equation.
2
2
8 OH (aq) CH3OH(aq) 6 MnO4 (aq) → CO3(aq) 6 MnO4(aq) 6 H2O(l)
SUMMARY
Procedure for Balancing Redox
Equations Using Oxidation Numbers
Step 1 Assign oxidation numbers and identify the atoms/ions whose oxidation
numbers change.
Step 2 Using the change in oxidation numbers, write the number of electrons
transferred per atom.
Step 3 Using the chemical formulas, determine the number of electrons
transferred per reactant. (Use the formula subscripts to do this.)
Step 4 Calculate the simplest whole number coefficients for the reactants that
will balance the total number of electrons transferred. Balance the reactants and products.
.
Step 5 Balance the O atoms using H2O(l), and then balance the H atoms using H(aq)
For basic solutions only,
Step 6 Add OH
(aq) to both sides equal in number to the number of H(aq) present.
Step 7 Combine H
(aq) and OH (aq) on the same side to form H2O(l), and cancel
the same number of H2O(l) on both sides.
NEL
Electric Cells 667
Check the balancing of the final equation. Make sure that both symbols and charge are
balanced.
Example 1
Balance the chemical equation for the oxidation of ethanol by dichromate ions in a
breathalyzer to form chromium(III) ions and acetic acid in an acidic solution.
Solution
6 2
21 2 1
3
3e/Cr
2e/C
6e/Cr2O72
4e/C2H5OH
2
3
0 1 0 2 1
3
2
16 H (aq) 2 Cr2O7(aq) 3 C2H5OH(aq) → 4 Cr (aq) 3 CH3COOH(aq) 11 H2O(l)
Example 2
Balance the following chemical equation, assuming the reaction occurs in a basic solution.
NO2 (aq) I2(aq) → NO3 (aq) I(aq)
Solution
3 2
0
5 2
1
2 OH
(aq) NO2 (aq) I2(aq) → NO3(aq) 2 I (aq) H2O(l)
2e/N
1e/I
Practice
Understanding Concepts
Answers
2. (a) 1 Cr2O72, 6 Cl, 14 H;
2 Cr 3, 3 Cl2, 7 H2O
(b) 2 IO3, 5 HSO3; 5 SO42,
1 I2, 3 H, 1 H2O
(c) 2 HBr, 1 H2SO4; 1 Br2,
2 H2O
3. (a) 2 MnO4, 3 SO32, 1 H2O;
3 SO42, 2 MnO2, 2 OH
(b) 4 ClO3, 3 N2H4;
6 NO, 4 Cl, 6 H2O
1. Why is the change in oxidation number of an atom the same as the number of
electrons transferred?
2. Balance the following chemical equations for reactions in an acidic solution:
3
(a) Cr2O72
(aq) Cl (aq)→ Cr (aq) Cl2(aq)
(b) IO3(aq) HSO3(aq) → SO42
(aq) I2(s)
(c) HBr(aq) H2SO4(aq) → SO2(g) Br2(l)
3. Balance the following chemical equations for reactions in a basic solution:
2
2
(a) MnO4
(aq) SO3(aq) → SO4(aq) MnO2(s)
(b) ClO3 (aq) N2H4(aq) → NO(g) Cl
(aq)
4. Ammonia gas undergoes a combustion to produce nitrogen dioxide gas and water
vapour. Write and balance the reaction equation.
Half-Reaction Method
An alternative to the oxidation number method for balancing redox equations is to write
balanced oxidation and reduction half-reaction equations. Once these half-reaction
equations are obtained, it is a simple matter to balance electrons and obtain the final
balanced redox equation. We will first address the writing of an individual halfreaction equation.
Although most metals and nonmetals have relatively simple half-reaction equations,
polyatomic ions and molecular compounds undergo more complicated oxidation and
reduction processes. In most of these processes, the reaction takes place in an aqueous
668 Chapter 9
NEL
Section 9.2
solution that is very often acidic or basic. As before, we must consider the important
role that water molecules, hydrogen ions, and hydroxide ions play an important role in
these half-reactions. A method of writing half-reactions for polyatomic ions and molecular compounds requires that water molecules and hydrogen or hydroxide ions be
included. This method, illustrated in the following sample problem, is sometimes called
the “half-reaction” or “ion-electron” method.
Writing Half-Reaction Equations
1.
SAMPLE problem
Nitrous acid can be reduced in an acidic solution to form nitrogen monoxide
gas. What is the reduction half-reaction for nitrous acid?
The first step is to write the reactants and products.
HNO2(aq) → NO(g)
If necessary, you should balance all atoms other than oxygen and hydrogen in this
partial equation. In this example, there is only one nitrogen atom on each side.
Next, add water molecules, present in an aqueous solution, to balance the oxygen
atoms, just as we did in the oxidation number method.
HNO2(aq) → NO(g) H2O(l)
Because the reaction takes place in an acidic solution, hydrogen ions are present, and
these are used to balance the hydrogen on both sides of the equation.
H
(aq) HNO2(aq) → NO(g) H2O(l)
At this stage, all of the atoms should be balanced, but the charge on both sides will not
be balanced. Add an appropriate number of electrons to balance the charge. Because
electrons carry a negative charge, they are always added to the less negative, or more
positive, side of the half-reaction.
e H
(aq) HNO2(aq) → NO(g) H2O(l)
This balanced half-reaction equation represents a gain of electrons—in other words, a
reduction of the nitrous acid. Check to make sure that both the atom symbols and the
charge are balanced.
In a basic solution, the concentration of hydroxide ions greatly exceeds that of
hydrogen ions. For basic solutions, we will develop the half-reaction as if it occurred in an
acidic solution and then convert the hydrogen ions into water molecules using hydroxide
ions. This trick works because a hydrogen ion and a hydroxide ion react in a 1:1 ratio to
form a water molecule. The following problem illustrates the procedure for writing halfreaction equations that occur in basic solutions.
2.
LEARNING
TIP
Notice the similarity in balancing
O atoms and H atoms with what
you did in the oxidation number
method. Reactions in a basic solution are also treated in the same
way as you did for the oxidation
number method.
Copper metal can be oxidized in a basic solution to form copper(I) oxide.
What is the half-reaction for this process?
Following the same steps as before, we write the formula and balance the atoms, other
than oxygen and hydrogen. Here the copper atoms must be balanced.
2 Cu (s) → Cu2O(s)
Next, balance the oxygen using water molecules and balance the hydrogen using
hydrogen ions, assuming, for the moment, an acidic solution. The charge is balanced
using electrons.
H2O(l) 2 Cu(s) → Cu2O(s) 2 H
(aq) 2 e
NEL
Electric Cells 669
Because the half-reaction occurs in a basic solution, add the same number of hydroxide
ions as there are hydrogen ions, to both sides of the equation. This is done to maintain the
balance of mass and charge.
2 OH
2 OH
(aq) H2O(l) 2 Cu(s) → Cu2O(s) 2 H (aq) 2 e
(aq)
Combine equal numbers of hydrogen ions and hydroxide ions to form water molecules.
2 OH
(aq) H2O(l) 2 Cu(s) → Cu2O(s) 2 H2O(l) 2 e
Finally, cancel H2O and anything else that is the same from both sides of the equation.
Check that the atom symbols and charge are balanced.
2 OH
(aq) H
2O(l) 2 Cu(s) → Cu2O(s) 2 H
2O(l) 2 e
2 OH
(aq) 2 Cu(s) → Cu2O(s) H2O(l) 2 e
Example
Chlorine is converted to perchlorate ions in an acidic solution. Write the half-reaction
equation. Is this half-reaction an oxidation or a reduction?
LEARNING
TIP
LEO says GER
Recall that loss of electrons is
oxidation (LEO) and gain of
electrons is reduction (GER)
Solution
4 H2O(l) Cl2(aq) → 2 ClO4 (aq) 8 H(aq) 8 e
This half-reaction is an oxidation.
Example
Aqueous permanganate ions are reduced to solid manganese(IV) oxide in a basic
solution. Write the half-reaction equation. Is the half-reaction an oxidation or a reduction?
Solution
4 H
4 OH
(aq) 3 e
(aq) MnO4 (aq) → MnO2(s) 2 H2O(l) 4 OH(aq)
4 H2O(l) 3 e MnO4
(aq) → MnO2(s) 2 H2O(l) 4 OH(aq)
MnO4
(aq) 2 H2O(l) 3 e → MnO2(s) 4 OH (aq)
This half-reaction is a reduction.
SUMMARY
Writing Half-Reaction Equations
Step 1 Write the chemical formulas for the reactants and products.
Step 2 Balance all atoms, other than O and H.
Step 3 Balance O by adding H2O(l).
Step 4 Balance H by adding H
(aq).
Step 5 Balance the charge on each side by adding e and cancel anything that is
the same on both sides.
For basic solutions only,
to both sides to equal the number of H present.
Step 6 Add OH(aq)
(aq)
Step 7 Combine H
(aq) and OH(aq) on the same side to form H2O(l). Cancel
equal amounts of H2O(l) from both sides.
670 Chapter 9
NEL
Section 9.2
Practice
Understanding Concepts
5. For each of the following, complete the half-reaction equation and classify it as an
oxidation or a reduction.
(a) dinitrogen oxide to nitrogen gas in an acidic solution
(b) nitrite ions to nitrate ions in a basic solution
(c) silver(I) oxide to silver metal in a basic solution
(d) nitrate ions to nitrous acid in an acidic solution
(e) hydrogen gas to water in a basic solution
Balancing Redox Equations Using Half-Reaction Equations
A redox reaction includes both an oxidation and a reduction. In other words, one substance
has to lose electrons as another substance gains electrons. Now that you know how to
write half-reaction equations, we can combine an oxidation half-reaction equation and
a reduction half-reaction equation to obtain the overall balanced redox equation.
For a particular reaction, chemists know the main starting materials and the reaction
conditions (e.g., acidic or basic). A chemical analysis of the products determines the
oxidized and reduced species produced in the reaction. This provides a skeleton equation showing the main reactants and products. The details of the final redox equation
will be provided by the individual balanced half-reaction equations.
Balancing Redox Equations Using Half-Reactions
SAMPLE problem
In a chemical analysis, a solution of dichromate ions is reacted with an acidic
solution of iron(II) ions (Figure 2). The products formed are iron(III) and
chromium(III) ions as shown by the following skeleton equation.
3
3
2
Fe2
(aq) Cr2O7(aq) → Fe(aq) Cr (aq)
Balance the equation.
The first step is to separate the equation into two skeleton half-reaction equations,
keeping related entities together.
3
Fe2
(aq) → Fe (aq)
3
Cr2O72
(aq) → Cr (aq)
Now you can treat each half-reaction equation separately to obtain a balanced
equation. The iron(II) half-reaction requires only the addition of an electron to balance the
charge.
Figure 2
The concentration of dichromate
ions can be determined by a reaction of a standard iron(II) solution. A
redox indicator is usually added to
produce a sharp colour change at
the completion of the reaction.
3
Fe2
(aq) → Fe (aq) e
For the dichromate half-reaction, you need to follow the same procedure as you did
before: balance atoms other than O and H atoms; balance O atoms by adding H2O(l);
balance H atoms by adding H
(aq); and finally, balance the charge by adding electrons.
3
2
6 e 14 H
(aq) Cr2O7(aq) → 2 Cr (aq) 7 H2O(l)
NEL
Electric Cells 671
Recall that the total number of electrons lost must equal the total number of electrons
gained. Using simple whole numbers, multiply one or both half-reaction equations so that
the electrons will be balanced. In this example, the iron(II) half-reaction equation must be
multiplied by a factor of 6 to balance 6e in the dichromate half-reaction equation.
3
6 [Fe2
(aq) → Fe (aq) e ]
6
e
14
H
(aq)
3
Cr2O72
(aq) → 2 Cr (aq) 7 H2O(l)
Add the two half-reaction equations.
14 H
3
3
2
6 Fe 2
(aq) 6 e
(aq) Cr2O7(aq) → 2 Cr (aq) 7 H2O(l) 6 Fe (aq) 6 e
Cancel the electrons and anything else that is the same on both sides of the equation.
3
3
2
6 Fe2
e 14 H
e
(aq) 6
(aq) Cr2O7(aq) → 2 Cr (aq) 7 H2O(l) 6 Fe (aq) 6
3
3
2
6 Fe 2
(aq) 14 H(aq) Cr2O7(aq) → 2 Cr (aq) 7 H2O(l) 6 Fe (aq)
Check the final redox equation to make sure that both the atom symbols and the charge
are balanced.
For reactions that occur in basic solutions, it is easier to follow the same procedure outlined above and then convert to a basic solution. In other words, create the balanced
redox equation for an acidic solution, then add OH
(aq) to convert the H(aq) to water molecules. This is the same procedure you used for obtaining a redox equation using the oxidation number method. An example for a basic solution is shown below.
Example
Permanganate ions and oxalate ions react in a basic solution to produce carbon dioxide
and manganese(IV) oxide.
2
MnO4
(aq) C2O4(aq) → CO2(g) MnO2(s)
Write the balanced redox equation for this reaction.
Solution
2 [3 e 4 H
(aq) MnO4 (aq) → MnO2(s) 2 H2O(l)]
3 [C2O42
(aq) → 2 CO2(g) 2 e
2
8
2 MnO4 (aq) 3 C2O4(aq) → 2 MnO2(s) 4 H2O(l) 6 CO2(g)
8 H 2 MnO 3 C O 2 → 2 MnO
OH(aq)
(aq)
4(aq)
2 4(aq)
2(s) 4 H2O(l) 6 CO2(g) 8
2
4 H2O(l) 2 MnO4 (aq) 3 C2O4(aq) → 2 MnO2(s) 6 CO2(g) 8 OH
(aq)
]
H
(aq)
8
SUMMARY
OH(aq)
Balancing Redox Equations
Using Half-Reaction Equations
Step 1 Separate the skeleton equation into the start of two half-reaction
equations.
Step 2 Balance each half-reaction equation.
Step 3 Multiply each half-reaction equation by simple whole numbers to
balance the electrons lost and gained.
Step 4 Add the two half-reaction equations, cancelling the electrons and
anything else that is exactly the same on both sides of the equation.
672 Chapter 9
NEL
Section 9.2
For basic solutions only,
Step 5 Add OH
(aq) to both sides equal in number to the number of H(aq)
present.
Step 6 Combine H
(aq) and OH(aq) on the same side to form H2O(l), and cancel
the same number of H2O(l) on both sides.
Practice
Understanding Concepts
6. Balance the following skeleton redox equations using the half-reaction method. All
reactions occur in an acidic solution.
2
(a) Zn(s) NO3 (aq) → NH4(aq) Zn(aq)
2
(b) Cl2(aq) SO2(g) → Cl (aq) SO4(aq)
7. Balance the following skeleton redox equations using the half-reaction method. All
reactions occur in a basic solution.
(a) MnO4
(aq) I (aq) → MnO2(s) I2(s)
(b) CN
(aq) IO3(aq) → CNO(aq) I (aq)
(c) OCl (aq) → Cl (aq) ClO3(aq)
Answers
6. (a) 4 Zn, 1 NO3, 10 H;
1 NH4, 4 Zn2, 3 H2O
(b) 1 Cl2, 1 SO2, 2 H2O;
2 Cl, 1 SO42 ,4 H
7. (a) 2 MnO4, 6 I, 4 H2O;
2 MnO2, 3 I2, 8 OH
(b) 3 CN, 1 IO3; 1 I,
3 CNO
(c) 3 OCl; 2 Cl, 1 ClO3
Extension
8. Balance the following redox equation.
KMnO4(aq) H2S(aq) H2SO4(aq) → K2SO4(aq) MnSO4(aq) S(s)
8. 2 KMnO4, 5 H2S, 3 H2SO4;
1 K2SO4, 2 MnSO4, 5 S,
8 H2O
Section 9.2 Questions
Understanding Concepts
1. In what way are the two methods of balancing redox
equations similar?
2. Compare oxidation and reduction in terms of oxidation
numbers and electrons transferred.
3. Balance the following equations representing reactions that
occur in an acidic solution:
2
(a) Cu(s) NO3
(aq) → Cu(aq) NO2(g)
2
(b) Mn(aq) HBiO3(aq) → Bi3
(aq) MnO4(aq)
3
(c) H2O2(aq) Cr2O72
(aq)→ Cr (aq) O2(g) H2O(l)
4. Balance the following equations representing reactions that
occur in a basic solution:
2
(a) Cr(OH)3(s) IO3
(aq) → CrO4(aq) I (aq)
(b) Ag2O(s) CH2O(aq) → Ag(s) CHO2
(aq)
2
(c) S2O42
(aq) O2(g) → SO4(aq)
Applying Inquiry Skills
5. State two general experimental designs that could help determine the balancing of the main species in a redox reaction.
grease in the drains. Some solid drain cleaners contain solid
sodium hydroxide and finely divided aluminum metal. When
mixed with water this produces a very vigorous, exothermic
reaction shown by the following skeleton equation:
Al(s) H2O(l) → Al(OH)4 (aq) H2(g)
(a) Complete the balanced redox equation for this reaction.
(b) Describe and discuss some possible health and safety
issues associated with the use of solid drain cleaners.
Extension
7. The analysis of iron by an oxidation–reduction titration is a
common analytical method. A common titrant used in this
analyis is a solution of the cerium(IV) ion, which is reduced
to cerium(III) in the analysis. In one chemical analysis of
some iron ore, the sample is treated to convert all of the
iron to iron(II) ions. A 25.0-mL sample of iron(II) is titrated
with 0.125 mol/L cerium(IV) solution using a redox indicator. The average volume of cerium(IV) required to reach
the endpoint was 15.1 mL. Calculate the concentration of
the iron(II) ions in the sample.
Making Connections
6. Many commercially available drain cleaners contain a basic
solution of sodium hydroxide, which helps to remove any
NEL
Electric Cells 673
9.3
zinc strip
CuSO4(aq)
Figure 1
Copper(II) ions react spontaneously
with zinc metal. A copper(II) ion has
a stronger attraction for the valence
electrons of a zinc atom than zinc
does.
Predicting Redox Reactions
A redox reaction may be explained as a transfer of valence electrons from one substance
to another. Evidence indicates that the majority of atoms, molecules, and ions are stable
and do not readily release electrons. Since two particles must be involved in an electron
transfer, this transfer can be explained as a competition for electrons. Using a tug-of-war
analogy, each particle pulls on the same electrons. If one particle is able to pull electrons away from the other, a spontaneous reaction occurs (Figure 1). Otherwise, no
reaction occurs (Figure 2). In the spontaneous reaction of copper(II) ions and zinc
metal, the Cu2+ ion is electron deficient and pulls electrons from a Zn atom. The reaction occurs because Cu 2+ pulls harder on Zn’s electrons than Zn does. Cu2+ wins the
two valence electrons from a Zn atom. A successful electron transfer has occurred.
Without mixing all possible reactants and observing any evidence of reaction, how can
we predict if a reaction will occur? If a reaction occurs, what will be the products? The
answers to these questions cannot be obtained easily using redox theory. By observing
many successful and unsuccesful reactions, patterns emerge and empirical generalizations
can be made.
Oxidizing and Reducing Agents
Before we look at these patterns, some terms commonly used by chemists need to be
defined. When discussing possible reactants and comparing their reactivities, it is customary and convenient to classify the reactants in a redox reaction. This classification originated historically but is now defined in terms of an ability to lose or gain electrons. In
any redox reaction an electron transfer occurs, which means that one reactant is oxidized and one reactant is reduced.
Figure 2
The green nickel(II) ion colour
remains and the copper metal does
not react. Collisions between copper
atoms and nickel(II) ions apparently
do not result in the transfer of
electrons.
X
loses
increases
is oxidized
+
Y
gains
decreases
is reduced
→
products
electron change
oxidation number
Examples:
Zn(s)
+
Cu2
(aq)
→
Zn2
(aq) Cu(s)
2 Br
(aq)
Cl2(g)
→
Br2(l) 2 Cl
(aq)
3 CO(g)
Fe2O3(s)
→
3 CO2(g) 2 Fe(s)
Rather than saying “the reactant that is oxidized” and “the reactant that is reduced,”
chemists use the terms reducing agent and oxidizing agent. These terms originated in
the early history of metallurgy and corrosion. For example, to “reduce” a larger volume
of iron(III) oxide to a smaller volume of pure iron, a substance called a reducing agent
was required, e.g., CO(g).
+3
reducing agent a substance that
loses or gives up electrons to another
substance in a redox reaction
oxidizing agent a substance that
gains or removes electrons from
another substance in a redox reaction
674 Chapter 9
0
reducing agent + Fe2O3(s) → Fe + other products
reduction
Similarly, oxidation was originally associated with an oxidizing agent. For example, a
metal could be oxidized by certain substances called oxidizing agents. At first, oxygen was
the only known oxidizing agent, but others (e.g., halogens) can also oxidize or corrode
metals.
NEL
Section 9.3
e−
oxidation
0
OA + RA
2
oxidizing agent MgO(s) → Mg2+ other products
The terms oxidizing and reducing agents developed separately, long before any redox
theory of electron transfer emerged. Today, chemists routinely think in terms of electron
transfer to explain redox reactions. A redox reaction is recognized as an electron transfer
between an oxidizing agent and a reducing agent (Figure 3).
Figure 3
In all redox reactions, electrons are
transferred from a reducing agent to
an oxidizing agent.
Development of a Redox Table
Some redox reactions such as single displacement reactions are easy to study experimentally. The evidence of a reaction is immediately obvious and the interpretation of an
electron transfer is relatively simple. In the past, you have generally assumed that all
single displacement reactions are spontaneous. However, by testing several combinations of metals and metal ions, it can easily be shown that some combinations react
immediately, but many do not react at all. The question that arises is, “How do you know
when a chemical reaction will occur spontaneously without actually doing the reaction?”
Let’s look at some examples of combinations of metals and metal ions. Suppose copper,
lead, silver, and zinc metals were combined one at a time with each of copper(II), lead(II),
silver, and zinc ion solutions. We can rank the ability of the metal ions to react with the
metals (Table 1).
LEARNING
TIP
Oxidizing and Reducing Agents
If a positively charged metal ion
reacts, then it is usually converted
to a metal atom. According to
redox theory, this requires a gain
of electrons and hence the metal
ion is behaving as an oxidizing
agent. Similarly, if a metal atom
reacts, then it is always converted
to a positively charged ion by
losing electrons. Metals always
behave as reducing agents.
Table 1 Reactivities of Metal Ions with Metals
Ions
Ag+
(aq)
Cu2+
(aq)
Pb2+
(aq)
Zn2+
(aq)
Reacted with
Cu(s), Pb(s), Zn(s)
Pb(s), Zn(s)
Zn(s)
none
Number of
reactions
3
2
1
0
Reactivity order most
least
INVESTIGATION 9.3.1
Spontaneity of Redox Reactions
(p. 716)
How many reactions will occur?
The most reactive metal ion, Ag+(aq), has the greatest tendency to gain electrons. On
the other hand, Zn2+(aq) shows no tendency to gain electrons in the combinations tested.
Therefore, the order of reactivity is also the order of strengths as oxidizing agents.
strongest oxidizing agent
Ag (aq) e → Ag(s)
Cu 2
(aq) 2 e → Cu(s)
Pb 2
(aq) 2 e → Pb(s)
weakest oxidizing agent
Zn 2
(aq) 2 e → Zn(s)
The order of reactivity of the four metals can be obtained in a similar way (Table 2).
Table 2 Reactivities of Metals with Metal Ions
Metals
Zn(s)
Pb(s)
Cu(s)
Ag(s)
Reacted with
2
2
Ag
(aq), Cu(aq), Pb(aq)
2
Ag
(aq), Cu(aq)
Ag
(aq)
none
Number of
reactions
3
2
1
0
Reactivity order
most
NEL
least
Electric Cells 675
The most reactive metal, Zn(s), has the greatest tendency to lose electrons and Ag(s) shows
no tendency to lose electrons in the combinations tested. Metals behave as reducing
agents and so Zn(s) is the strongest reducing agent among those tested.
strongest reducing agent
LEARNING
TIP
Organization of Redox Tables
• By convention, half-reaction
equations are written as
reductions (i.e., with the electrons on the left). The
strongest oxidizing agent
(SOA) is at the top left in a
table of relative strengths of
oxidizing and reducing
agents and the strongest
reducing agent (SRA) is at
the bottom right of the table.
• The double arrows may indicate an equilibrium in some
situations, but here they
simply indicate that the halfreactions may be read from
left to right (top arrow) or
from right to left (bottom
arrow).
weakest reducing agent
Zn(s)
→ Zn 2
(aq) 2 e
Pb(s)
→ Pb 2
(aq) 2 e
Cu(s)
→ Cu2
(aq) 2 e
Ag(s)
→ Ag
(aq) e
In these reactions, the metal ions are the oxidizing agents and the silver ion is the
strongest oxidizing agent (SOA) of the four ions because it is the most reactive in our
group. The metals are the reducing agents and the zinc metal is the strongest reducing
agent (SRA). The two lists of reactivity can be summarized using a single set of halfreactions as shown in Table 3.
Table 3 Relative Strengths of Oxidizing and Reducing Agents
SOA
Decreasing
reactivity of
oxidizing
agents
–
OA
+
+ n e–
RA
–
e
Ag(s)
–
Cu2+
(aq) + 2 e
Cu(s)
–
Pb2+
(aq) + 2 e
Pb(s)
–
Zn2+
(aq) + 2 e
Zn(s)
Ag (aq) +
Decreasing
reactivity of
reducing
agents
SRA
In Table 3, the metal ions are on the left side of the equations and the metal atoms are
on the right side. For metal ions (the oxidizing agents), the half-reaction equations are
read from left to right in the table. For metal atoms (the reducing agents), the half-reaction equations are read from right to left.
Practice
Understanding Concepts
1. Oxidation and reduction are processes, and oxidizing agents and reducing agents are
substances. Explain this statement, using definitions of the terms.
2. If a substance is a very strong oxidizing agent, what does this mean in terms of
electrons?
3. If a substance is a very strong reducing agent, what does this mean in terms of
electrons?
Refer to Tables 1, 2, and 3 to answer questions 4 to 8.
4. List the metal(s) that react spontaneously with a copper(II) ion solution.
5. Which metal(s) did not appear to react with a copper(II) ion solution?
6. Start with the position of Cu2
(aq) in Table 3 and note the position of the metal(s) that
reacted and the metal(s) that did not react. For a metal that reacts spontaneously
with Cu2
(aq), where does the metal appear on a table of reduction half-reactions
(Table 3)?
7. Repeat questions 4 to 6 for the Pb2
(aq) ion.
Applying Inquiry Skills
8. Your answers to 6 and 7 form an empirical hypothesis that can be tested by making
predictions for the other metal ions. Use Table 3 to predict which of the reactions
should be spontaneous. Are the predictions correct? Is your hypothesis verified?
676 Chapter 9
NEL
Section 9.3
9. An experiment similar to the example of metals and metal ions was conducted using
halogens and halide ions.
Question
What is the table of relative strengths of oxidizing and reducing agents for the
halogens?
Evidence
Only three combinations produced evidence of a reaction (Figure 4, Table 4).
(a)
(b)
(c)
Analysis
(a) Prepare a table of half-reaction equations like Table 3 for the halogens.
Table 4 Reactions of Halogens with Solutions of Halides
Br
(aq)
Cl
(aq)
I
(aq)
Br2(aq)
Cl2(aq)
I2(aq)
no reaction
yellow-brown
no reaction
no reaction
no reaction
no reaction
yellow-brown
yellow-brown
no reaction
Figure 4
None of the combinations of
aqueous solutions of chlorine,
bromine, and iodine with their corresponding halides show any evidence
of reaction except for the reaction
between (a) bromine and iodide
ions, (b) chlorine and bromide ions,
and (c) chlorine and iodide ions.
The Spontaneity Rule
Evidence obtained from the study of many redox reactions has been used to establish a
generalization, called the redox spontaneity rule. Figure 5 illustrates how you can use
the rule, along with a table of oxidizing and reducing agents, to predict whether or not
a reaction is spontaneous.
OA
redox spontaneity rule a spontaneous redox reaction occurs only if
the oxidizing agent (OA) is above
the reducing agent (RA) in a table
of relative strengths of oxidizing and
reducing agents.
RA
spontaneous
reaction
+
RA
+
nonspontaneous
reaction
OA
Figure 5
The redox spontaneity rule
Another Method for Building Redox Tables
Once a spontaneity rule is developed from experimental evidence, the rule may be used
to generate half-reaction tables. The evidence to be analyzed in this case is a net ionic equation, accompanied by observations of spontaneity. In the following method, the spontaneity rule, rather than the number of reactions observed, is used to order the oxidizing
and reducing agents to produce a redox table. The procedure for this type of analysis
and synthesis is illustrated by the following example.
NEL
LAB EXERCISE 9.3.1
Building a Redox Table (p. 717)
Several groups of experimental evidence are combined to make one
larger table.
Electric Cells 677
SAMPLE problem
Creating a Redox Table
Three reactions among indium, cobalt, palladium, and copper were investigated.
The reaction equations below indicate that two spontaneous reactions occurred
and only one combination did not react. Using these equations, construct a redox
table of half-reaction equations showing the relative strengths of the oxidizing and
reducing agents.
→ 2 In3
(aq) 2
2
Cu(aq) Co(s) → Co(aq) Cu(s)
Cu2
(aq) Pd(s) → no evidence of
3 Co2
(aq) 2 In(s)
OA
RA
2+
(a) Co(aq)
In(s)
2+
(b) Cu(aq)
Co(s)
2+
Co(aq)
In(s)
Pd(s)
(c)
2+
Cu(aq)
Co(s)
–
Pd2+
(aq) + 2 e
2+
(aq)
Cu
2+
(aq)
Co
In(s)
Figure 6
The relative position of a pair of
oxidizing and reducing agents
indicates whether a reaction will
be spontaneous.
reaction
To construct a table from this information, work with one equation at a time. Identify the
oxidizing and reducing agents for the first reaction, and arrange them in two columns
using the spontaneity rule. For the first reaction, this step is shown in Figure 6(a). Co2
(aq)
is the oxidizing agent and In(s) is the reducing agent. Since the reaction is spontaneous,
the oxidizing agent is above the reducing agent in the list.
In the second reaction, Cu2
(aq) is the oxidizing agent and Co(s) is the reducing agent. This
reaction is also spontaneous; therefore, Cu2
(aq) is above Co(s) in the list. Since a metal
appears on the same line as its ion in a half-reaction table, add Co(s) and extend the list
as shown in Figure 6(b).
No reaction occurs for the third pair of reactants. If a reaction had occurred, Cu2
(aq)
would be the oxidizing agent and Pd(s) would be the reducing agent. As this reaction is
not spontaneous, the oxidizing agent appears below the reducing agent. Figure 6(c)
shows the list extended to include Pd(s). To complete the table, write balanced halfreaction equations for each oxidizing/reducing agent pair.
SOA
2+
Co(aq)
3 Co(s)
2+
(aq)
In
Pd(s)
–
Cu(s)
–
Co(s)
–
In(s)
+2e
+2e
+3e
SRA
Practice
Understanding Concepts
10. The following reactions were performed. Construct a table of relative strengths of
oxidizing and reducing agents.
2
Co2
(aq) Zn(s) → Co(s) Zn(aq)
LEARNING
TIP
Communication of
Nonspontaneous Reactions
Nonspontaneity of a reaction is
communicated in several ways:
with the phrases “no evidence
of reaction,” or “nonspontaneous,” or “no reaction”; or with
“nonspont.” or “ns” written over
the equation arrow.
Mg2
(aq) Zn(s) → no evidence of reaction
11. In a school laboratory four metals were combined with each of four solutions.
Construct a table of relative strengths of oxidizing and reducing agents.
2
Be(s) Cd2
(aq) → Be(aq) Cd(s)
2
Cd(s) 2 H
(aq) → Cd(aq) H2(g)
Ca2
(aq) Be(s) → no evidence of reaction
Cu(s) H
(aq)
→ no evidence of reaction
12. Is the redox spontaneity rule empirical or theoretical? Justify your answer.
13. Use the relative strengths of nonmetals and metals as oxidizing and reducing
agents, as indicated in the following unbalanced equations, to construct a table of
half-reactions.
Ag(s) Br2(l) → AgBr(s)
Ag(s) I2(s)
→ no evidence of reaction
Cu2
(aq) I (aq) → no redox reaction
Br2(l) Cl
(aq) → no evidence of reaction
678 Chapter 9
NEL
Section 9.3
An Extended Redox Table
Evidence collected in many experiments has been analyzed to produce an extended
redox table of oxidizing and reducing agents such as the one found in Appendix C11. A
table such as this represents the combined efforts of many people over many years. A redox
table is an important reference for chemists. You can use this table to compare oxidizing
and reducing agents, and to predict spontaneous redox reactions.
Practice
Understanding Concepts
Use the redox table in Appendix C11 or the CRC Handbook of Chemistry and Physics to
answer the following questions.
14. Arrange the following metal ions in order of decreasing strength as oxidizing agents:
lead(II) ions, silver ions, zinc ions, and copper(II) ions. How does this order compare
with the one in Table 3?
15. What classes of substances (e.g., metals, nonmetals, acidic, basic) usually behave as
(a) oxidizing agents?
(b) reducing agents?
16. Use atomic theory to explain why nonmetals behave as oxidizing agents and metals
behave as reducing agents. Is there logical consistency between atomic theory and
the empirically determined table of oxidizing and reducing agents?
17. Trends in the reactivity of elements show that fluorine is the most reactive nonmetal.
How does this relate to the position of fluorine in the redox table of oxidizing and
reducing agents? State one reason why this element is the most reactive nonmetal.
Why is your reason an explanation? (Keep asking a series of “why” questions until
your theoretical knowledge is expended. Does your theory pass the test of being able
to explain the empirically determined table?)
LEARNING
TIP
Recognizing OA and RA
Although you can consult the
table in Appendix C11, it is much
more efficient to memorize the
answers to question 15. This general pattern helps to speed up the
process of recognizing oxidizing
and reducing agents and is necessary to classify substances that
don’t appear in Appendix C11.
Fe
GER/OA
Fe2+
LEO/RA
Fe3+
Figure 7
Iron(II) ions can either lose or gain
electrons and, therefore, can act as
either reducing agents or oxidizing
agents.
18. Identify three oxidizing agents (other than Fe2
(aq), shown in Figure 7) from the table
that can also act as reducing agents. Try to explain this unique behaviour.
19. Use the redox spontaneity rule to predict whether the following mixtures will show
evidence of a reaction; that is, predict whether the reactions are spontaneous. (Do
not write the equations for the reaction.)
(a) nickel metal in a solution of silver ions
(b) zinc metal in a solution of aluminum ions
(c) an aqueous mixture of copper(II) ions and iodide ions
(d) chlorine gas bubbled into a bromide ion solution
(e) an aqueous mixture of copper(II) ions and tin(II) ions
(f) copper metal in nitric acid
Applying Inquiry Skills
20. Describe two experimental designs or methods to collect evidence from which halfreaction tables can be built.
DID YOU
KNOW
?
Getting Rid of Skunk Odour
The smell of a skunk is caused by
a thiol compound (R—SH). To
deodorize a pet sprayed by a
skunk, you need to convert the
smelly thiol to an odourless compound. Hydrogen peroxide in a
basic solution (usually from
sodium bicarbonate) acts as an
oxidizing agent to change the thiol
to a disulfide compound (RS—SR),
which is odourless.
Making Connections
21. From your own knowledge, list two metals that are found as elements and two that
are never found as elements in nature. Test your answer by referring to the position of
these metals in the table of oxidizing and reducing agents.
22. Of the two parallel ways of knowing, empirical and theoretical, which, to this point,
has been the most useful to you in predicting the spontaneity of redox reactions?
Explain.
NEL
Electric Cells 679
Table 5 Hints for Listing and
Labelling Entities
•
Aqueous solutions contain
H2O(l) molecules.
•
Acidic solutions contain H (aq)
ions.
•
Basic solutions contain OH
(aq)
ions.
•
Some oxidizing and reducing
agents are combinations, for
example, MnO4
(aq) and H(aq).
•
2
H2O(l), Fe2
(aq), Cu(aq), Sn(aq) and
2
Cr(aq) may act as either oxidizing
or reducing agents. Label both
possibilities in your list.
Predicting Redox Reactions in Solution
Arrhenius’s ideas about solutions provide an important starting point for predicting redox
reactions. In solutions, molecules and ions act approximately independently of each other.
A first step in predicting redox reactions is to list all entities that are present. (Some helpful
reminders are listed in Table 5.) For example, when copper metal is placed into an acidic
potassium permanganate solution, copper atoms, potassium ions, permanganate ions,
hydrogen ions, and water molecules are all present. Next, using your knowledge of oxidizing and reducing agents and Appendix C11, label all possible oxidizing and reducing
agents in the starting mixture. The permanganate ion is listed as an oxidizing agent only
in an acidic solution. To indicate this combination, draw an arc between the permanganate
and hydrogen ions as shown, and label the pair as an oxidizing agent. This procedure of listing
and identifying entities present is a crucial step in predicting redox reactions.
OA
OA
Cu(s)
K+(aq)
MnO4–(aq)
OA
+
H(aq)
OA
H2O(1)
RA
RA
Practice
Understanding Concepts
23. List all entities initially present in the following mixtures and identify all possible oxidizing and reducing agents.
(a) A lead strip is placed in a copper(II) sulfate solution.
(b) A gold coin is placed in a nitric acid solution.
(c) A potassium dichromate solution is added to an acidic iron(II) nitrate solution.
(d) An aqueous chlorine solution is added to a phosphorous acid solution.
(e) A potassium permanganate solution is mixed with an acidified tin(II) chloride
solution.
(f) Iodine solution is added to a basic mixture containing manganese(IV) oxide.
INVESTIGATION 9.3.2
The Reaction of Sodium
with Water (p. 718)
Test a prediction of a redox reaction.
We can use a redox table to identify the strongest oxidizing and reducing agents in a
mixture and then predict which reactions will occur. If we assume that collisions are
completely random, the strongest oxidizing agent and the strongest reducing agent will
react. (In some cases, further reactions may occur as well, but we will consider only the
initial reaction, unless otherwise specified.) When using the redox table in Appendix
C11 to predict redox reactions,
• Choose the strongest oxidizing agent present in your
mixture by starting at the top left corner of the redox
table and going down the list until you find the oxidizing
agent that is in your mixture.
• Choose the strongest reducing agent in your mixture by
starting at the bottom right corner of the redox table and
going up the list until you find the reducing agent that is
in your mixture.
• Reduction half-reaction equations are read from left to
right (following the forward arrow).
• Oxidation half-reaction equations are read from right to
left (following the reverse arrow).
• Any substances not present in the redox table will be
assumed to be spectator ions. You do not need to label
or consider these substances.
680 Chapter 9
NEL
Section 9.3
Using SOA and SRA to Predict Reactions
Suppose a solution of potassium permanganate is slowly poured into an acidified
iron(II) sulfate solution. Does a redox reaction occur and, if it does, what is the
reaction equation?
To make a prediction, the entities initially present are identified as oxidizing agents,
reducing agents, or both, as shown below.
OA
K+(aq)
OA
MnO4–(aq)
OA
+
H(aq)
OA
Fe2+
(aq)
OA
SO42–(aq)
OA
H2O(1)
RA
RA
Use the table in Appendix C11 to choose the strongest oxidizing agent and the
strongest reducing agent from your list and indicate them with SOA and SRA.
OA
K+(aq)
SOA
MnO4–(aq)
OA
+
H(aq)
OA
Fe2+
(aq)
OA
SO42–(aq)
SAMPLE problem
LEARNING
TIP
Identifying OA and RA
(1) If the half-reaction equation
shows two or more entities
present, then both must be in
your list. If there is only one,
then leave the entity unlabelled, as a spectator ion.
(2) Be careful with a few entities
that can act either as an OA or
a RA, for instance, the iron(II)
ion in this example.
OA
H2O(1)
SRA
RA
Now, write the half-reaction equation for the reduction of the SOA.
2
MnO4
(aq) 8 H (aq) 5 e → Mn(aq) 4 H2O(l)
Write the half-reaction equation for the oxidation of the SRA. Remember you are
reading from right to left on the table.
3
Fe2
(aq ) → Fe(aq) e
Before combining the half-reaction equations, balance the number of electrons transferred by multiplying one or both half-reaction equations by an integer so that the number
of electrons gained by the oxidizing agent equals the number of electrons lost by the
reducing agent.
In this case, the iron half-reaction must be multiplied by 5. Add the two equations, but
remember to cancel any common terms. You can cancel terms as you add (e.g., 5e) or
after you add the two half-reactions.
→ Mn2 4 H O
MnO4 (aq) 8 H (aq) 5 e
2 (l)
(aq)
3
5 [Fe2
(aq) → Fe (aq) e ]
2
3
2
MnO4 (aq) 8 H (aq) 5 Fe (aq) → 5 Fe (aq) Mn (aq) 4 H2O(l)
Finally, use the spontaneity rule to predict whether the net ionic equation represents a
spontaneous redox reaction. Indicate this by writing “spont.” or “non-spont.” over the
equation arrow.
spont.
2
3
2
MnO4
(aq) 8 H (aq) 5 Fe (aq) → 5 Fe(aq) Mn(aq) 4 H2O(l)
This prediction may be tested by mixing the solutions (Figure 8) and performing some
diagnostic tests. If the solutions are mixed and the purple colour of the permanganate ion
disappears, then it is likely that the permanganate ion reacted. If the pH of the solution is
tested before and after reaction, and the pH has increased, then the hydrogen ions likely
reacted.
NEL
Figure 8
A solution of potassium permanganate is being added to an acidic
solution of iron(II) ions. The dark
purple colour of MnO4
(aq) ions
instantly disappears. The interpretation is that MnO4
(aq) ions react with
Fe2
ions
to
produce
the yellow(aq)
2 ions.
brown Fe3
and
Mn
(aq)
(aq)
Electric Cells 681
Example 1
In a chemical industry, could copper pipe be used to transport a hydrochloric acid solution? To answer this question,
(a) predict the redox reaction and its spontaneity, and
(b) describe two diagnostic tests that could be done to test your prediction.
Solution
(a)
Cu(s)
SOA
+
H(aq)
OA
H2O(1)
Cl–(aq)
SRA
RA
RA
RA
+
2 H(aq)
+ 2 e– → H2(g)
2+
+ 2 e–
Cu(s) → Cu(aq)
non spont.
+
+ Cu(s)
2 H(aq)
Figure 9
Copper in hydrochloric acid does
not appear to react.
H2(g) + Cu2+
(aq)
Since the reaction is nonspontaneous, it should be possible to use a copper pipe to carry
hydrochloric acid.
(b) If no gas is produced when the mixture is observed, then it is likely that no hydrogen
gas was produced (Figure 9). If the colour of the solution did not change to blue,
then copper probably did not react to produce copper(II) ions. (If the solution is
tested for pH before and after adding the copper, and the pH did not increase, then
the hydrogen ions probably did not react.)
SUMMARY
Predicting Redox Reactions
Step 1 List all entities present and classify each as an oxidizing agent, reducing
agent, or both. Do not label spectator ions.
Step 2 Choose the strongest oxidizing agent as indicated in the table of relative
strengths of oxidizing and reducing agents, and write the equation for its
reduction.
Step 3 Choose the strongest reducing agent as indicated in the table, and write
the equation for its oxidation.
Step 4 Balance the number of electrons lost and gained in the half-reaction
equations by multiplying one or both equations by a number. Then add
the two balanced half-reaction equations to obtain a net ionic equation.
Step 5 Using the spontaneity rule, predict whether the net ionic equation
represents a spontaneous or nonspontaneous redox reaction.
Practice
Understanding Concepts
24. Predict the most likely redox reaction in each of the following situations. For any
spontaneous reaction, describe one diagnostic test to identify a primary product.
(a) During a demonstration, zinc metal is placed in a hydrochloric acid solution.
(b) A gold ring is placed into a hydrochloric acid solution.
(c) Nitric acid is painted onto a copper sheet to etch a design.
682 Chapter 9
NEL
Section 9.3
25. In your previous chemistry course, predictions of reactions were made according
to the single displacement generalization assuming the formation of the most
common ion.
(a) Use the generalization about single displacement reactions to predict the
reaction of iron metal with a copper(II) sulfate solution.
(b) Use redox theory and a table showing half-reactions to predict the most likely
redox reaction of iron metal with a copper(II) sulfate solution.
(c) Can both predictions be correct? Which do you think is likely correct and why?
26. Oxygen gas is bubbled into an aqueous solution of iron(II) iodide containing
excess hydrochloric acid. Predict all spontaneous reactions, in the order in which
they will occur.
Applying Inquiry Skills
27. Write one qualitative and one quantitative experimental design to test the two different predictions made for the reaction between iron metal and the copper(II)
sulfate solution in question 25.
DID YOU
KNOW
?
Aluminum Oxide Clouds
The solid rocket boosters of the
space shuttle contain the main
reactants ammonium perchlorate
and aluminum powder.
Ammonium perchlorate is a powerful oxidizing agent and aluminum is a relatively strong
reducing agent. Their very
exothermic reaction produces
finely divided aluminum oxide,
which forms the billows of white
smoke you see in the photograph.
28. Write a Prediction, with your reasoning, and an Experimental Design (including
diagnostic tests) to complete the investigation report.
Question
What are the products of the reaction of tin(II) chloride with an ammonium
dichromate solution acidified with hydrochloric acid?
Making Connections
29. When aluminum pots are used for cooking, small pits often develop in the metal.
Use your knowledge of redox reactions to explain the formation of these pits.
Suggest why this might be a slow process.
Extensions
30. Fluoride treatments of children’s teeth have been found to significantly reduce
tooth decay. When this was first discovered, toothpastes were produced containing
tin(II) fluoride.
Problem
What is the concentration of tin(II) ions in a solution prepared for research on
toothpaste?
Answer
30. (a) 0.258 mol/L
Experimental Design
An acidified tin(II) solution was titrated with a standardized potassium
permanganate solution.
Evidence
volume of tin(II) solution = 10.00 mL
concentration of permanganate solution = 0.0832 mol/L
average volume of permanganate reacted = 12.4 mL
Analysis
(a) Calculate the molar concentration of the tin(II) solution.
Section 9.3 Questions
Understanding Concepts
1. What is the key idea used to explain a redox reaction?
2. Write a theoretical definition of oxidation and reduction.
3. Distinguish between oxidation and oxidizing agent.
4. Distinguish between reduction and reducing agent.
NEL
5. Write and label two half-reaction equations to describe
each of the following reactions:
(a) Co(s) Cu(NO3)2(aq) → Cu(s) Co(NO3)2(aq)
(b) Cd(s) Zn(NO3)2(aq) → Zn(s) Cd(NO3)2(aq)
(c) Br2(l) 2 KI(aq) → I2(s) 2 KBr(aq)
Electric Cells 683
6. Using the redox table in Appendix C11, predict the spontaneity of each of the reactions shown in 5(a) to (c).
7. What is the relative strength of oxidizing and reducing
agents for strontium, cerium, nickel, hydrogen, platinum,
and their aqueous ions? Use the following information to
construct a table of relative strengths of oxidizing and
reducing agents.
2
3 Sr(s) 2 Ce3
(aq) → 3 Sr (aq) 2 Ce(s)
Ni(s) 2 H
(aq)
→
Ni 2
(aq)
H2(g)
2 Ce3
(aq) 3 Ni(s) → no evidence of reaction
→ no evidence of reaction (assume Pt 4 )
Pt (s) 2 H(aq)
(aq)
8. In the industrial production of iodine, chlorine gas is bubbled into seawater. Using only water and iodide ions in seawater as the possible reactants, predict the most likely
redox reaction, including appropriate equations for the halfreactions.
9. The steel of an automobile fender is exposed to acidic rain.
(Assume that steel is made mainly of iron.) Predict the most
likely redox reactions, including the equations for the relevant half-reactions.
10. A chemical technician prepares several solutions for use in
a chemical analysis. Will each of the solutions listed below
be stable if stored for a long period of time? Justify your
answer.
(a) acidic tin(II) chloride in an inert glass container
(b) copper(II) nitrate in a tin can
11. An excess of cobalt metal was left in an aqueous mixture
containing silver ions, iron(III) ions, and copper(II) ions for
an extended period of time. Write a balanced redox
equation for every reaction that occurs.
Applying Inquiry Skills
12. Prepare a redox table of half-reactions showing the relative
strengths of oxidizing and reducing agents in Table 6.
684 Chapter 9
Table 6 Reactions of Group 13 Elements and Ions
Al3+
(aq)
Tl+
(aq)
Ga3+
(aq)
In3+
(aq)
Al
X
√
√
√
Tl
X
X
X
X
Ga
X
√
X
√
In
X
√
X
X
X no evidence of a redox reaction
√ a spontaneous reaction occurred
Making Connections
13. Ursula Franklin (Figure 10) is an internationally recognized
scientist in her field. She has many interests and is outspoken on many topics. Briefly describe her pioneering scientific work and outline her views about science and
technology and funding for scientific research. What other
causes does she support?
GO
www.science.nelson.com
Figure 10
Ursula Franklin was born in
Germany in 1921 and received
her Ph.D. in 1948 from the
Technical University in Berlin.
She continued her studies at
the University of Toronto,
where she became a professor
in 1973. In 1984 she was
appointed a University
Professor at the University of
Toronto, an honour acknowledging that her academic and
scientific interests go far
beyond a single discipline.
NEL
Technology of Cells and Batteries
Before 1800, scientists knew that static electricity was produced by the friction created
by two moving objects in contact. They discovered ways of storing the charges temporarily, but when the energy was released in the form of an electrical spark, it could not
be put to practical use. Practical applications of electricity were developed only after
1800, the year in which Alessandro Volta announced his invention of the electric cell.
Volta invented the first electric cell but he got his inspiration from the work, almost
30 years earlier, of the Italian physician Luigi Galvani. Galvani noticed that the muscles
in a frog’s leg would twitch when a spark hit the leg. Galvani’s crucial observation was
that two different metals could make the muscle twitch. Unfortunately, Galvani thought
his discovery was due to some mysterious “animal electricity.” It was Volta who recognized that this effect had nothing to do with animals or muscle tissue, and everything to
do with conductors and electrolytes, as you have already observed in the activity at the
beginning of this chapter.
9.4
copper
metal
cell
paper
soaked
in a salt
solution
zinc
metal
battery
Cells and Batteries
The electric cells Volta invented produced very little electricity. Eventually, he came up
with a better design by joining several cells together. A battery is a group of two or more
electric cells connected to each other, in series, like railway cars in a train. Volta’s first battery consisted of several bowls of brine (aqueous sodium chloride) connected by metals
that dipped from one bowl into the next (Figure 1). This arrangement of metal strips and
electrolytes produced a steady flow of electric current. Volta improved the design of this
battery by replacing the strips of metal with flat sheets, and replacing the bowls with
paper or leather soaked in brine. This produced more electric current for a longer period
of time. As shown in Figure 2, Volta stacked cells on top of each other to form a battery,
known as a voltaic pile. When a loop of wire was attached to the top and bottom of this
voltaic pile, a steady electric current flowed. Volta assembled voltaic piles containing
more than 100 cells.
Volta’s invention was an immediate success because it produced an electric current more
simply and more reliably than methods that depended on static charges. It also produced a steady electric current—something no other device could do. The development
of this technology led to many advances in physics (for example, the theory and description of current electricity), in chemistry (for example, the discovery of Groups 1 and 2
metals), and in electrical and chemical engineering.
Zn
Cu
Figure 2
Volta’s revised cell design, simpler
than the first, consisted of a sandwich of two metals separated by
paper soaked in salt water (the electrolyte). A cell consisted of a layer of
zinc metal separated from a layer of
copper metal by the brine-soaked
paper. A large pile of cells could be
constructed to give more electrical
energy.
electric cell a device that continuously converts chemical energy into
electrical energy.
battery a group of two or more
electric cells connected in series
Figure 1
A version of Volta’s first battery. Each bowl contains two different metals, copper and zinc, in an
electrolyte, salt water. A series of bowls forms a series of cells (battery) whose total voltage is
the sum of the individual voltages of all cells.
NEL
Electric Cells 685
Basic Cell Design and Properties
+
–
voltmeter
cathode
(+)
Each electric cell is composed of two electrodes and one electrolyte (Figure 3). In the
cells we buy for home use, the electrolyte is usually a moist paste, containing only enough
conducting solution to make the cell function. The electrodes are usually two metals,
or graphite and a metal. In some designs, one of the electrodes is the container of the cell.
One of the electrodes is marked positive (+) and the other is marked negative (–).
anode
(–)
electrolyte
Figure 3
A cell always contains two electrodes—an anode and a cathode—and
an electrolyte. When testing the
voltage of a cell or battery, the red ()
lead of the voltmeter is connected to
the positive electrode (cathode), and
the black () lead is connected to the
negative electrode (anode).
electrode a solid electrical conductor
electrolyte an aqueous electrical
conductor
electric potential difference
(voltage) the potential energy difference per unit charge
In an electric cell or battery, the cathode is the positive
electrode and the anode is the negative electrode.
According to the theory that electricity is the flow of electrons, electrons move from
the anode of a battery through some conducting materials to the cathode. A battery
produces electricity only when there is an external conducting path, such as a wire,
through which electrons can move. Disconnecting the wire from the battery immediately
stops the electric current.
A voltmeter is a device that can be used to measure the energy difference, per unit
electric charge, between any two points in an electric circuit. The energy difference per
unit charge is called the electric potential difference or the voltage, and is measured in
volts (V). For example, the electrons transferred via a 1.5-V cell release only one-sixth
as much energy as the electrons from a 9-V battery.
Since the voltage is a ratio of energy to charge, it is not dependent on the size of the
cell. You may have noticed that you can buy the same type and brand of 1.5-V cells in a
variety of sizes, such as AA, B, C, and D. All are rated at 1.5 V. The larger cells can produce more energy at the same time as transferring more charge, but the ratio of energy
to charge is the same as the smaller cells. The voltage of a cell depends mainly on the chemical composition of the reactants in the cell.
Electric current, measured by an ammeter in amperes (A), is a measure of the rate
of flow of charge past a point in an electrical circuit (Figure 4). The larger the electric
cell of a particular kind, the greater the current that can be produced by the cell. The charge
transferred by a cell or battery is measured in coulombs (C) and expresses the total
charge transferred by the movement of charged particles. The power of a cell or battery
is the rate at which it produces electrical energy. Power is measured in watts (W), and
is calculated as the product of the current and the voltage of the battery. The energy
density, or specific energy of a battery, is a measure of the quantity of energy stored or supplied per unit mass. Energy density may be measured in joules per kilogram (J/kg).
Table 1 summarizes electrical quantities and their units of measurement.
Table 1 Electrical Quantities and SI Units
volt (V) the SI unit for electric
potential difference; 1 V 1 J/C
Quantity
charge
electric current the rate of flow of
charge past a point
current
potential difference
ampere (A) the SI unit for electric
current; 1 A 1 C/s
Symbol
Meter
Unit
Unit symbol
q
—
coulomb
C
I
ammeter
ampere
A (1 A 1 C/s)
V
voltmeter
volt
V (1 V 1 J/C)
power
P
—
watt
W (1 W 1 J/s)
energy density
—
—
joules per
kilogram
J/kg
coulomb (C) the SI unit for electric
charge
686 Chapter 9
NEL
Section 9.4
(a)
(b)
SUMMARY
Components of an Electric Cell
• An electric cell must have two electrodes and an electrolyte.
• An electrode is a solid conductor.
• An electrolyte is an aqueous conductor.
• The cathode is the electrode labelled positive.
• The anode is the electrode labelled negative.
• The electron flow is from the anode to the cathode.
Practice
Figure 4
(a) A dam built across a stream or
river stops the flow of water. Each
kilogram of water that backs up
behind the dam has a certain
quantity of potential energy relative to the bottom of the dam. In
other words, there is a potential
energy difference between a kilogram of water at the top of the
dam and a kilogram of water at
the bottom of the dam. A voltmeter can be used to measure
the height of the “dam” inside a
battery, that is, the potential
energy difference between a unit
number of electrons at the
cathode and a unit number of
electrons at the anode.
(b) If water is released from behind
the dam, it naturally flows from
the region of higher potential
energy (behind the dam) to a
lower potential energy below
the dam. Similarly, when the
circuit is connected to a cell or
battery, the electrons naturally
flow because there is a difference in potential energy.
Understanding Concepts
1. What are the parts of a simple electric cell?
2. Write an empirical definition of electrode and electrolyte, and a conventional definition of anode and cathode.
3. If a cassette player requires 6 V to operate, how many 1.5-V “dry” cells corrected in
series would it need?
4. Differentiate between electric current and voltage.
Making Connections
5. Why do manufacturers of battery-operated devices print a diagram showing the correct orientation of the batteries? (Supply two answers to this question—one from a
scientific perspective and one from a technological perspective.)
Technological Problem Solving
The initial development of cells and batteries preceded much of the current scientific
understanding of these devices. Cells and batteries existed almost 100 years before the
electron was discovered. The study of electric cells is a good illustration of tremendous
advances in technology based on very limited scientific knowledge. Technological development or problem solving is similar in some ways to scientific problem solving, but
its purpose differs. The purpose of technological problem solving is to find a realistic
way around a practical difficulty—to make something work—while the purpose of
scientific problem-solving is to describe, explain, and ultimately understand natural
and technological phenomena. Technology and science are dependent on each other.
Although scientific knowledge can be used to guide the creation of a technology, the technology may create new scientific understanding.
NEL
DID YOU
KNOW
?
Shocking Personal Experiments
“I introduced into my ears two
metal rods with rounded ends and
joined them to the terminals of the
apparatus. At the moment the circuit was completed, I received a
shock in the head—and began to
hear a noise—a crackling and
boiling. This disagreeable sensation, which I feared might be dangerous, has deterred me so that I
have not repeated the experiment.”
–Alessandro Volta (1745—1827)
Electric Cells 687
ACTIVITY 9.4.1
Developing an Electric Cell
(p. 719)
You are an inventor developing a
new electric cell.
DID YOU
KNOW
?
A “Not Quite Dry” Cell
The electrolyte in the “dry cell” is
actually a moist paste. If the cell
were completely dry it would not
work because the ions in the electrolyte must be able to carry the
electric current to complete the circuit. Just enough water is added so
that the ions can move, but not
enough to make the mixture liquid.
A systematic trial-and-error process, such as the following one, is often used in technological problem solving (Appendix A3):
• Develop a general design for problem-solving trials; for example, select which
variables to manipulate and which to control.
• Follow several prediction–procedure–evidence–analysis cycles, manipulating
and systematically studying one variable at a time.
• Complete an evaluation based on criteria such as efficiency, reliability, cost, and
simplicity.
This technological problem-solving model was important in the early development of
practical electric cells.
Consumer, Commercial, and Industrial Cells
Since Volta’s invention of the electric cell and battery, there have been many advances in
electrochemistry and technology. Invented in 1865, the zinc chloride cell is commonly
referred to as a dry cell because this design was the first to use a sealed container. These
1.5-V dry cells were used to make the first 9-V battery (Figure 5). Both the 1.5-V dry cell
and the 9-V battery are simple, reliable, and relatively inexpensive. Other cells, such as
the alkaline dry cell and the mercury cell (Table 2), were developed to improve the performance of the original dry cell. One problem with all of these cells is that the chemicals are eventually depleted and irreversible reactions prevent these cells from being
recharged. Cells that cannot be recharged are called primary cells. Later, we will discuss
two other types of cells that do not have this disadvantage.
Zinc Chloride Dry Cells
primary cell an electric cell that
cannot be recharged
carbon electrode
MnO2 and NH4Cl
electrolyte paste
Figure 5
Like a flashlight D cell, the dry cell
on the left has a voltage of 1.5 V.
The 9 V battery on the right is made
up of six 1.5-V dry cells in series.
zinc electrode
1.5 V cell
9 V battery
Secondary Cells
secondary cell an electric cell that
can be recharged
688 Chapter 9
Secondary cells can be recharged by using electricity to reverse the chemical reaction
that occurs when electricity is produced by the cell. Secondary cells and batteries include
the nickel-cadmium (Ni-Cad) cell and the lead-acid battery (Table 2 and Figure 6). A
relatively recently developed secondary cell with a unique design is the lithium-ion cell,
or Molicel (Figure 7, page 690).
NEL
Section 9.4
Table 2 Primary, Secondary, and Fuel Cells
Type
Name of Cell
Half-Reactions
Characteristics and Uses
primary
cells
dry cell (1.5 V)
2 MnO2(s) + 2 NH4+(aq) + 2 e– → Mn2O3(s) + 2 NH3(aq) + H2O(1)
–
Zn(s) → Zn2+
(aq) + 2 e
• inexpensive, portable, many
sizes
• flashlights, radios, many
other consumer items
secondary
cells
fuel cells
alkaline dry
cell (1.5 V)
–
2 MnO2(s) + H2O(1) + 2 e– → Mn2O3(s) + 2 OH(aq)
–
Zn(s) + 2 OH(aq) → ZnO(s) + H2O(1) + 2 e–
mercury cell
(1.35 V)
HgO(s) + H2O(1) + 2 e– → Hg(1) + 2 OH–
–
Zn(s) + 2 OH(aq)
→ ZnO(s) + H2O(1) + 2 e–
• small cell; constant voltage
during its active life
• hearing aids, watches
Ni-Cad cell
(1.25 V)
2 NiO(OH)(s) + 2 H2O(1) + 2 e– → 2 Ni(OH)2(s) + 2 OH–
–
Cd(s) + 2 OH(aq)
→ Cd(OH)2(s) + 2 e–
• can be completely sealed;
lightweight but expensive
• all normal dry cell uses, as
well as power tools, shavers,
portable computers
lead-acid cell
(2.0 V)
PbO2(s) + 4 H+(aq) + SO42–(aq) + 2 e– → PbSO4(s) + 2 H2O(1)
Pb(s) + SO42–(aq) → PbSO4(s) + 2 e–
• very large currents; reliable
for many recharges
• all vehicles
aluminum-air
cell (2 V)
–
3 O2(g) + 6 H2O(1) + 12 e– → 12 OH(aq)
3+
4 Al(s) → 4 Al(aq) + 12 e–
hydrogenoxygen cell (1.2 V)
–
O2(g) + 2 H2O(1) + 4 e– → 4 OH(aq)
2 H2(g) + 4 OH–(aq) → 4 H2O(1) + 4 e–
—
anode
• longer shelf life; higher
currents for longer
periods compared with
dry cell
• same uses as dry cell
• very high energy density;
made from readily available
aluminum alloys
• designed for electric cars
• lightweight; high efficiency;
can be adapted to use
hydrogen-rich fuels
• vehicles and space shuttle
cathode
+
+
—
—
+
cell spacer
H2SO4(aq)
electrolyte in
each cell
one cell
NEL
negative plates:
lead screen filled
with spongy lead
positive plates:
lead screen filled
with PbO2(s)
Figure 6
The anodes of a lead-acid car battery are composed of spongy lead
and the cathodes are composed of
lead(IV) oxide on a metal screen.
The large electrode surface area is
designed to deliver sufficient current
to start a car engine.
Electric Cells 689
TIP
LEARNING
Discharging and Charging
Discharging a cell or battery is
like letting the water spontaneously run out from the higher
level behind a dam. Charging (or
recharging) is like pumping the
water up behind the dam. This is
not a spontaneous process and
requires energy.
One of the most common and reliable secondary cells is the lead-acid cell in a typical
car battery. The discharging of this cell (see lead-acid cell, Table 2) produces approximately
2.0 V for the following net equation.
discharging
Pb(s) PbO2(s) 2 H2SO4(aq) → 2 PbSO4(s) 2 H2O(l)
To charge (or recharge) this cell requires the input (from the car’s alternator) of at least
2.0 V to force the products to change back to the reactants. The half-reactions for the
lead-acid cell listed in Table 2 both need to be reversed to obtain the following net equation.
charging
2 PbSO4(s) 2 H2O(l) → Pb(s) PbO2(s) 2 H2SO4(aq)
A battery can be recharged if the products are stable with no further reactions occurring
and if the products are able to travel through the electrolyte toward the appropriate electrode.
Practice
Understanding Concepts
6. What is the relationship between scientific knowledge and technological problem
solving?
7. What steps are involved in technological problem solving?
8. Suppose you decided to develop and market an aluminum-can cell. (See Activity 9.4.1.)
How and why would you alter the electrolyte?
9. Distinguish between primary and secondary cells, including a common example of
each.
10. What are some advantages and disadvantages of the zinc chloride dry cell?
Making Connections
safety header
separator
11. Find out how commercially available AA, C, and D cells differ. How do these
differences affect their performance?
12. What do the designs of the dry-cell container and the ice-cream cone have in common?
positive
electrode
negative
electrode
13. Portable electronic devices can be found everywhere. Laptop computers, cellular
telephones, mobile radios, cordless phones, portable disc and MP3 players, and
digital cameras all require an electric cell.
(a) What are some of the requirements for cells used in these applications?
(b) Why are some rechargeable batteries used in various portable devices supposed
to be totally “drained” (discharged) before recharging?
GO
Figure 7
Invented and manufactured in
British Columbia, the Molicel is a
high-energy, rechargeable cell in a
unique, jellyroll design.
690 Chapter 9
www.science.nelson.com
14. Moli Energy of Maple Ridge, BC was the first company in the world to develop a
commercial, rechargeable lithium cell, called a Molicel (Figure 7). Research the
characteristics and advantages of Molicels compared with other secondary cells.
GO
www.science.nelson.com
NEL
Section 9.4
Fuel Cells
A fuel cell is a different solution to the problem of the limited life of a primary cell.
Fuel cells produce electricity by the reaction of a fuel that is continuously supplied to keep
the cell operating. In principle, the fuel cell could be used forever, provided the fuel is continuously supplied. The fuel cell offers several advantages over methods that produce
electricity by the combustion of fossil fuels. For example, fuel cells generate electricity
more efficiently (Table 3), without the production of greenhouse gases or substances
that contribute to acid rain. The development of a cost-effective fuel cell is currently
the focus of much scientific study and technological research and development.
The first fuel cell was invented accidentally by William Grove in 1839 using platinum
electrodes, hydrogen and oxygen as fuels, and sulfuric acid as the electrolyte. Grove was
actually studying the reverse process—using electricity to convert water into hydrogen
and oxygen. After one experiment, he reconnected the two electrodes, without a power
supply attached, and found that a small current was produced spontaneously as hydrogen
and oxygen combined to form water. Grove continued to work on this cell but eventually decided it was not a practical device because the electric characteristics, such as
voltage, current, and energy capacity, were very low. Although many attempts to improve
this cell were made, including those by Nobel Prize winners Fritz Haber and Walther
Hermann Nernst, no significant progress was made. Many variables were manipulated,
such as different electrodes and electrolytes, but the reaction rates were too low and
corrosion of the electrodes was often a serious problem.
Finally, in 1955, Francis Bacon succeeded where many others had failed. He produced
a practical hydrogen-oxygen fuel cell using an alkaline electrolyte and electrodes constructed of porous nickel (Figure 8). Although the idea had been around for a long time,
Bacon’s cell was really the first practical fuel cell. NASA quickly adopted the hydrogenoxygen fuel cell as an electrical power source for space flights, because hydrogen and
oxygen are already available for propulsion systems and the product, water, can be purified for drinking. NASA’s fuel cell, a modification of the original Bacon cell, is an alkaline cell using potassium hydroxide as the electrolyte (Table 2). It produces 12 kW of
electricity and operates at 70% efficiency. Unfortunately, NASA’s fuel cell is expensive and
has a relatively short working life, primarily due to the corrosive electrolyte. As a result,
NASA’s cell is not economically viable for general or commercial applications.
The Ballard Fuel Cell
A variation of a hydrogen-oxygen fuel cell, also known more simply as the hydrogen
fuel cell, was developed for commercial applications by Ballard Power Systems in
Vancouver, BC. The Ballard fuel cell employs a proton exchange membrane (PEM) in place
of a liquid electrolyte. Normal electric cells use the ions in the liquid electrolyte to
transfer electric charge within the cell. In a hydrogen fuel cell, the PEM is a membrane
made from a solid proton-conducting polymer that transfers charge within the cell
(Figure 9). The PEM is simple, robust, eliminates corrosive liquids, and permits a high
energy density.
The Ballard fuel cell consists of an anode and a cathode separated by a polymer membrane electrolyte. Hydrogen fuel admitted through a porous anode is then converted
into hydrogen ions (protons) and free electrons in the presence of a catalyst at the anode.
An external circuit conducts the free electrons and produces the desired electrical current. Water and heat are produced when the protons, after migrating through the polymer
membrane to the cathode, react both with oxygen molecules from the air and with the
free electrons from the external circuit. Fuel cells can be connected in series (stacked) to
increase the voltage and power output (Figure 10). For example, an experimental
NEL
fuel cell an electric cell that produces electricity by a continually
supplied fuel
Table 3 Efficiencies of Different
Technologies*
Technology
Efficiency*
Fuel Cells
4070%
Electric power
plants
3040%
Automobile
engines
1723%
Gasoline lawn
mower
about 12%
*Efficiency is the fraction of the maximum
available energy that is actually usable.
cathode (+)
anode (–)
H2 gas
in
O2 gas
in
electrolyte
H2 gas
and
water
vapour
out
O2 gas
out
Figure 8
Hydrogen and oxygen gases are
continuously pumped into the cell,
and each reacts at a different electrode. Unused gases are removed,
filtered, and then recycled.
LEARNING
TIP
Inventions Require Patience
and Insight
Notice that the work done to try
and improve the original Grove cell
was very much a trial-and-error
development. Patience is required.
It took over 90 years to turn the
Grove cell into a technologically
useful cell.
Electric Cells 691
Vancouver transit bus uses an
electric motor powered by a
Ballard fuel cell that is capable
of 205 kW (or 250 hp).
fuel flow
oxidant flow
Ballard has development
field plate
field plate
agreements with most major car
manufacturers to use its cells in
exhaust
future electric cars. The zerofuel
water vapour
emission engines convert
recirculated
(no pollution)
hydrogen, or hydrogen-rich fuels
such as natural gas and even
methanol, into electricity, producing water and heat as the
main byproducts.
Although
the
Ballard
hydrogen fuel cell looks very
heat (90°C)
promising, there are several
water-cooled
problems yet to be solved. Cost
is a major factor, which may be
partially solved by mass production. The fuel is also under
air
fuel (hydrogen) debate. If hydrogen gas is used,
where does it come from?
Electrolysis of water uses a lot of
electrical energy and is an expensive way to obtain hydrogen. How would the hydrogen
gas be distributed and stored on board the vehicle? There are important safety concerns
associated with the handling and storage of hydrogen, which is flammable. Many scientists
and engineers believe that the solution is not to use hydrogen gas directly but to use
hydrogen-rich fuels. We have a lot of knowledge of reforming hydrocarbons to produce
hydrogen. If natural gas or even gasoline were reformed as needed on board the vehicle,
then we would have a familiar fuel source and an infrastructure in place to supply this
fuel. Not everyone agrees that this is a good solution.
proton exchange
membrane
Figure 9
The hydrogen fuel cell has the same
design as Volta’s original cell but the
electrolyte is a conducting solid.
LAB EXERCISE 9.4.1
Characteristics of a Hydrogen
Fuel Cell (p. 720)
Compare the electrical characteristics
of a hydrogen fuel cell and a dry cell.
Figure 10
Vehicles require high-power-density
fuel cells, i.e., ones that can produce
large quantities of energy per
second for every kilogram of fuel
cell. Fuel cell stacks developed by
Ballard have continued to improve.
The stack on the left (2001) has a
power density about 16 times
greater than the one on the rght
(1989), yet the size is similar.
692 Chapter 9
NEL
Section 9.4
Aluminum-Air Cell
Another type of fuel cell that has not received the media attention of hydrogen fuel cells
is the metal-air fuel cell, the most common of which is the aluminum-air cell (Table 2).
This is actually an aluminum-oxygen cell and has been developed for possible use in
electric cars. Air is pumped into the cell and oxygen reacts at the cathode while a replaceable mass of aluminum reacts at the anode (Figure 11). The fuel is solid aluminum
metal and the product, aluminum hydroxide, can be recycled back to aluminum metal.
The simple design means that this cell can be assembled in almost any size. The high
energy density of these cells results from the fact that three moles of electrons are released
from every mole of aluminum, and aluminum is a very lightweight metal. Unlike
hydrogen, storage and transportation of a solid fuel do not pose a problem. Estimates from
prototypes suggest that the aluminum anode will need replacement every 2500 km in an
electric car.
DID YOU
KNOW
aluminum anode
gas
diffusion
cathode
Large-Scale Commercial and Industrial Fuel Cells
The requirements for electrical power fuel cells for large-scale use in businesses and
industry are similar with regard to the fuel, but there is less concern about volume,
weight, or energy density. However, there is a need for cells with much longer lifetimes.
Fuel cells for large-scale commercial and industrial use are almost always co-generation
units. This means that they produce electricity as well as heat for space heating. Co-generation means that the overall efficiencies can be as high as 90%. Commercially viable
fuel cells today are usually acid electrolyte cells such as the phosphoric acid fuel cell,
which can produce 400 MW of power, sufficient for the electrical energy needs of a
small city (Figure 12). These cells usually use natural gas as a source of hydrogen for
the fuel cell and operate at temperatures of 200°C.
?
Methanol Production and Use
Methanol is produced in large
quantities from natural gas.
Methanol is widely used in
industry to produce a variety of
products, one of which is windshield washer fluid.
oxygen
(air)
oxygen
(air)
electrolyte
Practice
Understanding Concepts
15. Using several perspectives, state some advantages and disadvantages of a fuel cell.
16. List some potential uses of fuel cells.
17. For both the hydrogen-oxygen fuel cell and the aluminum-air fuel cell,
(a) write the two half-reaction equations (Table 2).
(b) label each equation from (a) as an oxidation or a reduction.
(c) write the net ionic equation for each cell.
Figure 11
Aluminum Power Inc., based in
Toronto, ON, has done extensive
development of the aluminum-air
solid fuel cell.
18. List some problems that must be solved before the Ballard cell sees widespread use.
19. Another Ballard-type fuel cell uses methanol as a fuel. What are some advantages of
methanol over hydrogen or natural gas?
Making Connections
20. One of the most successful batteries has been the lead-acid car battery.
(a) Identify the anode, cathode, and electrolyte.
(b) How are the large currents produced that are necessary to start a car?
(c) What has been the social impact of this battery?
(d) What are some possible environmental impacts of this battery?
21. Plastic batteries were the dream of the 1980s, the disappointment of the 1990s, and
the subject of the 2000 Nobel Prize for Chemistry. Now it appears that some commercial products will eventually result from the research and development invested in
plastic batteries. Briefly describe the electrodes and electrolyte for a plastic battery.
How is this battery similar to and different from an ordinary battery? What are some
advantages and disadvantages?
GO
NEL
www.science.nelson.com
Figure 12
The world’s first commercial phosphoric acid fuel cell, produced by
ONSI/International Fuel Cells. It has
been available since 1992 and uses
natural gas, waste methane,
propane, or hydrogen as fuels. This
unit produces 200 kW electricity and
200 kW heat at a total system
efficiency of 80%.
Electric Cells 693
EXPLORE an issue
Debate: Hydrogen Fuel Cells
No one doubts that internal-combustion vehicles are a major
source of pollution and environmental damage and all major
automobile manufacturers are racing to develop viable alternatives (e.g., an electric car). Judging from media reports and
automobile advertisements, fuel cells are seen as the “pollutionfree” alternative to the internal-combustion engine. Specifically,
hydrogen fuel cells are being widely promoted as the “green
alternative.” Often mentioned in the media, the product of a
hydrogen fuel cell is pure water. What could be better than
that?
(a) In small groups, prepare for a debate on the proposition,
“Hydrogen fuel cells are the ideal ‘green’ solution to the
internal-combustion engine.”
Define the Issue
Analyze the Issue
Identify Alternatives
Defend the Position
Research
Evaluate
In your research, consider:
•
where does the hydrogen come from?
•
from source to final end product, are hydrogen fuel
cells nonpolluting?
•
other perspectives, such as economic, social, political,
and technological.
•
alternatives to the hydrogen fuel cell.
(b) Develop your opinion, defending or opposing the proposition. Brainstorm and research arguments in support of your
position.
GO
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Section 9.4 Questions
Understanding Concepts
1. Draw a simple diagram of an electric cell and label: electrodes, electrolyte, cathode, anode, signs for cathode and
anode, and direction of electron flow through an external
wire.
2. What is the evidence that an electric cell involves a redox
reaction?
8. Criteria used to evaluate a battery include its reliability,
cost, simplicity of use, safety (leakage), size (volume), shelf
life, active life, energy density, power capacity, maintenance, disposal, environmental impact, and ability to be
recharged. Gather some information and analyze it to
determine what is the best cell or battery for a portable
radio, CD, or MP3 player.
3. What are the three types of electric cells used in consumer
and commercial operations? Briefly describe the main feature of each cell.
4. State two common examples of consumer cells and where
they may be used.
5. A silver oxide cell is often used when a miniature cell or
battery is required, as in watches, calculators, and cameras.
The following half-reaction equations occur in the cell:
Ag2O(s) H2O(l) 2 e → 2 Ag(s) 2 OH
(aq)
Zn(s) 2 OH
(aq) → Zn(OH)2(s) 2 e
(a) In which direction does the electric current flow—silver
to zinc or zinc to silver?
(b) Which is the anode and which is the cathode?
(c) Write the net redox equation for the discharging of the
silver oxide cell.
Making Connections
6. Suppose cells and batteries did not exist. What impact
would that have on your life?
7. (a) Why is there a great deal of interest in electric cars?
(b) Suggest some reasons why we don’t use lead-acid
batteries as the only power source for electric cars.
(c) How have advances in hydrogen fuel cells facilitated
the development of electric cars?
694 Chapter 9
Figure 13
A pacemaker includes the electronics and a built-in battery.
The whole unit is only a few centimetres in size and is
implanted under the skin near the collarbone.
9. People whose heart occasionally beats too slowly or too
quickly often have pacemakers to keep the heart beating
regularly (Figure 15). Pacemakers use a battery for electric
power. What kind of battery is commonly used today? How
long does it last? How does the doctor know when the battery is nearing the end of its life and needs to be replaced?
Why are rechargeable batteries generally not used?
GO
www.science.nelson.com
NEL
9.5
Galvanic Cells
Electric cells were invented about 1800 and were developed to serve practical purposes.
They were not explained scientifically until about 100 years after their invention. Their
use, however, contributed to scientific understanding of redox reactions and, later, this
knowledge helped explain reactions inside the cell itself. Electric cells adapted for scientific
study are often called galvanic cells (in recognition of Luigi Galvani) or voltaic cells (in
recognition of Alessandro Volta).
From a scientific perspective, the design of a cell “plays a trick” on oxidizing and reducing
agents, resulting in electrons passing through an external circuit rather than directly from
one substance to another. You have seen that the individual components of a cell—electrodes and electrolytes—determine electrical characteristics such as voltage and current.
Why is this so? What happens in different parts of a cell? To answer these questions,
chemists use a cell with a different design, with the parts of the cell separated so they can
be studied more easily. This is not a very practical arrangement but it greatly facilitates the
study of cells. Each electrode is in contact with an electrolyte, but the electrolytes surrounding each electrode are separated. This is accomplished by a porous boundary, a
barrier that separates electrolytes at least over a short time while still permitting ions to
move through tiny openings between the two solutions. Two common examples of porous
boundaries are the salt bridge and the porous cup, shown in Figure 1.
(a)
Figure 1
(a) A salt bridge is a U-shaped
tube containing an inert (unreactive) aqueous electrolyte
such as sodium sulfate. The
cotton plug allows ions to move
into or out of the ends of the
tube when the ends are
immersed in electrolytes.
(b) An unglazed porcelain (porous)
cup containing one electrolyte
sits in a container of a second
electrolyte. The two solutions are
separated, but ions can move in
and out of the cup through the
pores in the porcelain.
(b)
electrolyte
electrolytes
electrolyte
electrolyte
ions
ions
ions
cotton plugs
With this design modification, a cell can be split into two parts connected by a porous
boundary. Each part, called a half-cell, consists of one electrode and one electrolyte.
For example, the copper-zinc cell shown in Figure 2 has two half-cells, copper metal in
a solution of copper ions, and zinc metal in a solution of zinc ions. It can be represented
using the following abbreviated (“shorthand”) notation, called a cell notation:
half-cell an electrode and an electrolyte forming half of a complete cell
Cu(s) | Cu(NO3)2(aq) || Zn(NO3)2(aq) | Zn(s)
In this notation, a single line (|) indicates a phase boundary such as the interface of an
electrode and an electrolyte in a half-cell. A double line (||) represents a physical boundary
NEL
Electric Cells 695
wire
NaNO3(aq)
salt bridge
Figure 2
The essential parts of a cell are two
electrodes and an electrolyte. In this
design each electrode is in its own
electrolyte, forming a half-cell. The
two half-cells are connected by a
salt bridge (containing NaNO3(aq))
and by an external conductor to
make a complete circuit.
copper electrode
zinc electrode
copper (II) nitrate
electrolyte
zinc nitrate
electrolyte
Cu|Cu(NO3)2(aq)
half-cell
galvanic cell an arrangement of
two half-cells that can produce
electricity spontaneously
Zn(NO3)2(aq)|Zn(s)
half-cell
such as a porous boundary between half-cells. A galvanic cell is an arrangement of two
half-cells that can produce electricity spontaneously. Cells such as the one in Figure 2 are
especially suitable for scientific study.
ACTIVITY 9.5.1
Galvanic Cell Design (p. 721)
Examine the design and operation
of a galvanic cell.
A Theoretical Description of a Galvanic Cell
Observation of a galvanic cell as it operates provides evidence that explains what is happening inside the cell. For example, the study of a silver-copper cell in Activity 9.5.1 provides the evidence listed in Table 1. A theoretical interpretation of each point is included
in the table and is shown in Figure 3.
Table 1 Evidence and Interpretations of the Silver-Copper Cell
696 Chapter 9
Evidence
Interpretation
The copper electrode decreases in mass
and the intensity of the blue colour of the
electrolyte increases.
Oxidation of copper metal is occurring:
2 2 e
Cu(s) → Cu(aq)
blue
The silver electrode increases in mass as
long, silver-coloured crystals grow.
Reduction of silver ions is occurring:
Ag
(aq) e → Ag(s)
A blue colour slowly moves up the U-tube
from the copper half-cell to the silver half-cell
and the solution remains electrically neutral.
Copper(II) ions move toward the cathode.
Negative ions (anions) move toward the
anode.
A voltmeter indicates that the silver electrode
is the cathode (positive) and the copper
electrode is the anode (negative).
Electrons move from the copper electrode
to the silver electrode.
An ammeter shows that the electric current
flows between the copper electrode and the
silver electrode.
Electrons leave the copper half-cell and
enter the silver half-cell.
NEL
Section 9.5
LEARNING
e–
+
Na(aq)
Ag(s)
e–
–
NO3(aq)
Cu(s)
cathode (+)
anode (-)
+
Ag(aq)
2+
Cu(aq)
–
NO3(aq)
+
+ Na(aq)
Ag(aq)
–
NO3(aq)
cathode half-cell
+
Ag(aq) + e–
Ag(s)
(reduction)
–
NO3(aq)
–
NO3(aq)
2+
Cu(aq)
anode half-cell
Cu(s)
Cu(aq) + 2 e–
(oxidation)
According to the electron transfer theory and the concept of relative strengths of oxidizing and reducing agents, silver ions are the strongest oxidizing agents in the cell; they
undergo a reduction half-reaction at the cathode. The strongest oxidizing agent in the
cell always undergoes a reduction at the cathode. Copper atoms, which are the strongest
reducing agents in the cell, give up electrons in an oxidation half-reaction and enter the
solution at the anode. The strongest reducing agent in the cell always undergoes an oxidation at the anode. Therefore, the cathode is the electrode where reduction occurs and
the anode is the electrode where oxidation occurs.
• The strongest oxidizing agent present in the cell always
undergoes a reduction at the cathode.
• The strongest reducing agent present in the cell always
undergoes an oxidation at the anode.
Electrons released by the oxidation of copper atoms at the anode travel through the
connecting wire to the silver cathode. The direction of electron flow can be explained in
terms of competition for electrons. According to the table of relative strengths of oxidizing
and reducing agents in Appendix C11, silver ions are stronger oxidizing agents than
copper(II) ions. Silver ions win the tug of war for the electrons available from the conducting wire.
To write the net equation for the silver-copper galvanic cell, identify the strongest oxidizing and reducing agents present in the mixture. (This is the same procedure you followed
when predicting redox reactions in Section 9.3.) Then follow the same procedure for predicting half-reactions in which the two materials are in contact with each other.
NEL
TIP
Memory Devices
People often use acronyms or similar devices to help them remember
important information. “LEO says
GER” is an example. One way to
help you remember important
details of a cell is the expression
SOAC/GERC, loosely read as “soak
a jerk.” Translated, this means the
Strongest Oxidizing Agent at the
Cathode Gains Electrons and is
Reduced at the Cathode. Another
example is “An ox ate a red cat,”
which helps to recall Anode oxidation; reduction cathode.
Figure 3
A theoretical interpretation of the
silver-copper cell
cathode the electrode where
reduction occurs
anode the electrode where
oxidation occurs
DID YOU
KNOW
?
Electron Sources and Sinks
Chemists sometimes refer to the
anode as the electron source and
the cathode as the electron sink.
Can you see from the half-reactions why these terms apply?
Electric Cells 697
SOA
OA
+
Ag(s)  Ag(aq)
 Cu2+
(aq)  Cu(s)
RA
reduction at the cathode
+
2 [ Ag(aq)
+ e– → Ag(s) ]
2+
+ 2 e–
Cu(s) → Cu(aq)
oxidation at the anode
+
Cu(s) + 2 Ag(aq)
→ Cu2+
(aq) + 2 Ag(s)
net
LEARNING
TIP
Cell Notation
In this book, the cell notation is
written with the cathode first,
although the order may be
reversed. Writing the cathode
first is convenient later when we
calculate cell potentials.
SRA
The electrical neutrality in the half-cells and the salt bridge can be explained in terms
of the half-reactions and the movement of electrons and ions (Figure 4). If cations did
not move to the cathode, the removal of silver ions from the solution near the cathode
would create a net negative charge around the cathode and the buildup of negative
charge would prevent electrons from being transferred. Migration of cations toward the
cathode solution ensures that electrical neutrality is maintained. Likewise, the formation
of copper(II) ions at the anode would create a net positive charge, but this is balanced
by the movement of negative ions to the anode compartment through the salt bridge or
porous cup. The salt bridge permits the redistribution of charge that is needed to maintain electrical neutrality in the electrolyte solutions of the half-cells.
electrons
cathode (+)  electrolyte  electrolyte  anode (–)
(reduction)
(oxidation)
anions
cations
e–
e–
cations
cathode (+)
anode (–)
s
anion
Figure 4
In any operating cell, the electrical
circuit is completed by the electron
flow in the external part (wires) of
the cell and the ion flow in the
internal part (solutions) of the cell.
Galvanic Cells with Inert Electrodes
For cells containing metals and metal ions, the electrodes are usually the metals, and halfreactions take place on the surface of the metals. What happens if an oxidizing or a reducing
agent other than these is used? For example, an acidic dichromate solution is a strong oxidizing
agent that reacts spontaneously with copper metal. To construct this cell you can use a copper
half-cell, as in Figure 5, but an electrode is required for the dichromate half-cell. You cannot
use solid sodium dichromate as an electrode because solid ionic compounds do not
698 Chapter 9
NEL
Section 9.5
conduct electricity and solid sodium dichromate would also dissolve in the solution. You
need a solid conductor that will not react in the cell or interfere with the desired cell reaction.
In other words, you need an unreactive or inert electrode. Inert electrodes provide a location to connect a wire and a surface on which a half-reaction can occur. A carbon (graphite)
rod (Figure 5) or platinum metal foil are two inert electrodes that are commonly used.
Example
(a) Write equations for the half-reactions and the overall reaction that occur in the following cell:
2
C(s) | Cr2O72
(aq), H(aq) || Cu (aq) | Cu(s)
(b) Draw a diagram of the cell, labelling electrodes, electrolytes, the direction of electron
flow, and the direction of ion movement.
Solution
SOA OA
C(s)  Cr2O
2–
7 (aq)
+
(aq)
,H
OA
inert electrode a solid conductor
that will not react with any substances present in a cell (usually
carbon or platinum)
LEARNING
TIP
Cell Names
There are a variety of names used
for cells based upon spontaneous
redox reactions—electric, voltaic,
galvanic, and electrochemical. In
this book, electric cell is used for
consumer cells and galvanic cell
for scientific research cells.
2+
 Cu(aq)
 Cu(s)
SRA
cathode
anode
net
2–
7(aq)
Cr2O
+ 6 e → 2 Cr + 7 H2O(l)
3 [ Cu(s) → Cu + 2 e– ]
+
(aq)
+ 14 H
–
3+
(aq)
2+
(aq)
2–
+
2+
3+
Cr2O7(aq)
+ 14 H(aq)
+ 3 Cu(s) → 3 Cu(aq)
+ 2 Cr (aq)
+ 7 H2O(l)
e–
cations
Cu(s)
anode (-)
2+
Cu(aq)
NEL
anions
C(s)
cathode (+)
Figure 5
The copper electrode decreases in
mass, and the blue colour of the
electrolyte increases, indicating
oxidation at the anode. The
carbon electrode remains
unchanged, but the orange colour
of the dichromate solution
becomes less intense and more
yellow, evidence that reduction is
occurring in this half cell.
2Cr2O7(aq)
+
H(aq)
Electric Cells 699
SUMMARY
Galvanic Cells
• A galvanic cell consists of two half-cells separated by a porous boundary with
solid electrodes connected by an external circuit.
• The cathode is the positive electrode. Reduction of the strongest oxidizing agent
present in the cell occurs at the cathode.
• The anode is the negative electrode. Oxidation of the strongest reducing agent
present in the cell occurs at the anode.
• Electrons travel in the external circuit from the anode to the cathode.
• Internally, anions move toward the anode and cations move toward the cathode
as the cell operates. The solution remains electrically neutral.
Practice
Understanding Concepts
1. Write an empirical description of each of the following terms: galvanic cell, half-cell,
porous boundary, and inert electrode.
2. Write a theoretical definition of a cathode and an anode.
3. Indicate whether the following processes occur at the cathode or at the anode of a
galvanic cell.
(a) reduction half-reaction
(b) oxidation half-reaction
(c) reaction of the strongest reducing agent
(d) reaction of the strongest oxidizing agent
4. When is an inert electrode used?
5. What are the characteristics of the solution in a salt bridge? Provide an example.
6. For each of the following cells, use the given cell notation to identify the strongest
oxidizing and reducing agents. Write chemical equations to represent the cathode,
anode, and net cell reactions. Draw a diagram of each cell, labelling the electrodes,
electrolytes, direction of electron flow, and direction of ion movement.
2
(a) Ag(s) | Ag (aq) || Zn(aq) | Zn(s)
(b) Pt (s) | Na(aq), Cl(aq), O2(g), H2O(l) || Al3
(aq) | Al(s)
7. Ions move through a porous boundary between the two half-cells of a voltaic cell.
(a) Why do the ions move? Take your answer and convert it into another “why”
question. Now answer this question.
(b) In what direction do the cations and anions move?
8. Draw and label a diagram for a galvanic cell constructed from some (not all) of the
following materials:
strip of cadmium metal
voltmeter
strip of nickel metal
connecting wires
solid cadmium sulfate
glass U-tube
solid nickel(II) sulfate
cotton
solid potassium sulfate
various beakers
distilled water
porous porcelain cup
9. Redesign the galvanic cell in question 8 by changing at least one electrode and one
electrolyte. The net reaction should remain the same for the redesigned cell.
700 Chapter 9
NEL
Section 9.5
Standard Cells and Cell Potentials
The investigations and activities you have completed show that the design and composition
of a cell affect its operation. To make comparisons and scientific study easier, chemists specify
the composition of a cell and the conditions under which the cell operates. A standard cell
is a galvanic cell in which each half-cell contains all entities shown in the half-reaction equation at SATP conditions, with a concentration of 1.0 mol/L for the aqueous solutions. If a metal
is not part of a half-cell, then an inert electrode is used to construct the standard cell. For
example, for a standard dichromate-zinc cell, the cell description is
+
3
2
C(s) | Cr2O72
(aq) , H(aq) , Cr (aq) || Zn (aq) | Zn(s)
1.0 mol/L
at SATP
1.0 mol/L
The standard cell potential E° is the maximum electric potential difference (voltage)
of the cell operating under standard conditions; E ° represents the energy difference
(per unit of charge) between the cathode and the anode. The degree sign (°) indicates
that standard 1.0 mol/L and SATP conditions apply. Based on the idea of competition
for electrons, a standard reduction potential Er° represents the ability of a standard halfcell to attract electrons, thus undergoing a reduction. The half-cell with the greater attraction for electrons—that is, the one with the more positive reduction potential—gains
electrons from the half-cell with the lower reduction potential. The standard cell potential is the difference between the reduction potentials of the two standard half-cells.
E°
cell
E r°
cathode
E r°
anode
It is impossible to determine experimentally the reduction potential of a single halfcell because electron transfer requires both an oxidizing agent and a reducing agent.
Note that a voltmeter can only measure a potential difference, E °. In order to assign
values for standard reduction potentials, we measure the “reducing” strength of all possible half-cells relative to an accepted, standard half-cell. The half-cell used for this purpose is the standard hydrogen half-cell. A half-cell such as this, that is chosen as a reference
and arbitrarily assigned an electrode potential of exactly zero volts, is called a reference
half-cell.
Standard Hydrogen Half-Cell
standard cell a galvanic cell in
which each half-cell contains all
entities shown in the half-reaction
equation at SATP conditions, with a
concentration of 1.0 mol/L for the
aqueous entities
standard cell potential E° is the
maximum electric potential difference (voltage) of a cell operating
under standard conditions
standard reduction potential Er°
represents the ability of a standard
half-cell to attract electrons in a
reduction half-reaction
reference half-cell a half-cell
arbitrarily assigned an electrode
potential of exactly zero volts; the
standard hydrogen half-cell
LEARNING
TIP
Think of the standard cell potential
E° as representing the difference
in ability of two half-cells to gain
electrons. This potential difference
can only be measured accurately if
no current is allowed to flow. A
good-quality voltmeter has a large
internal resistance to prevent
current flow.
IUPAC now recommends the use
of SATP as standard conditions for
reduction potentials. The change
from 101.325 kPa to 100 kPa will
not noticeably affect measuring
and reporting values previously
determined at 101.325 kPa.
The standard hydrogen half-cell (Figure 6) consists of an inert platinum electrode
immersed in a 1.00 mol/L solution of hydrogen ions, with hydrogen gas at a pressure of
100 kPa bubbling over the electrode. The pressure and temperature of the cell are kept
at SATP conditions. Standard reduction potentials for all other half-cells are measured
relative to that of the standard hydrogen half-cell. The reduction potential of the hydrogen
ion reduction half-reaction is defined to be exactly zero volts.
2 H+(aq) + 2 e–
H2(g)
E °r = 0.00 V
As a result, a numerical value can be assigned to the reduction potential associated with
every other other reaction. When a half-reaction, written as a reduction, has a positive
reduction potential, we conclude that the oxidizing agent in that half-reaction is a stronger
oxidizing agent than hydrogen ions. Therefore, if a half-cell based on that half
reaction were connected to a standard hydrogen half-cell, electrons would be drawn away
from the standard hydrogen electrode. A negative reduction potential means that the oxidizing agent in the half-cell connected to the hydrogen half-cell attracts electrons less strongly
NEL
Electric Cells 701
LEARNING
TIP
connecting wire
Although the reference half-cell
potential is exactly zero volts,
the value is often stated as
0.00 V to correspond to the
measured values of the other
half-cell potentials.
H2(s) at SATP
+
1.00 mol/L H(aq) at 25°C
platinum
Figure 6
The standard hydrogen half-cell is
used internationally as the reference
half-cell in electrochemical research.
LEARNING
TIP
Determining the Cathode
and Anode of a Cell
A voltmeter has two terminals,
positive (red) and negative
(black). Connect these to the
electrodes of any cell so that the
voltmeter gives a positive
reading. Whatever electrode is
connected to the positive terminal will be the cathode, and
the other electrode will be the
anode.
Pt(s)
l
H2(g), H+
(aq)
Er° = 0.00 V
than hydrogen ions do. The choice of the standard hydrogen half-cell as a reference is the
accepted convention. If a different half-cell had been chosen as the reference, individual
reduction potentials would be different, but their relative values would remain the same.
Measuring Standard Reduction Potentials
The standard reduction potential of a half-cell can be measured by constructing a standard cell using a hydrogen reference half-cell and the half-cell whose reduction potential you want to measure. There are two things you need to know—the voltage and the
direction of the current. The magnitude of the voltage determines the numerical value
of the half-cell potential and the direction of the current determines the sign of the halfcell potential. The cell potential is measured with a voltmeter, which will also show the
direction that the electrons tend to flow from the sign of the voltage. If E° is positive,
then the positive terminal on the voltmeter is connected to the cathode and the oxidizing agent at the cathode is stronger than hydrogen ions.
The cell shown in Figure 7 can be represented as follows:
Cu (s) | Cu 2
(aq) || H2(g), H (aq) | Pt (s)
cathode
E ° = +0.34 V
anode
The voltmeter shows that the copper electrode is the cathode and is 0.34 V higher in
potential than the platinum anode (Figure 7). If the voltmeter is replaced by a connecting wire so that the current is allowed to flow, the blue colour of the copper(II) ion
disappears and the pH of the hydrogen half-cell decreases as more hydrogen ions are
produced and the solution becomes more acidic. Based on this evidence, copper(II)
ions are being reduced to copper metal and hydrogen molecules are being oxidized to
hydrogen ions. Since this redox reaction is spontaneous, copper(II) ions are stronger
oxidizing agents than hydrogen ions.
702 Chapter 9
NEL
Section 9.5
0.34 V
H2(g)
Cu(s)
Pt(s)
1.00 mol/L
+
H(aq) at 25°C
1.00 mol/L
2+
Cu(aq) at 25°C
Figure 7
A copper-hydrogen standard cell.
The standard cell potential, ∆E ° = 0.34 V, is the difference between the reduction
potentials of these two half-cells;
cathode
Cu2
(aq) 2 e → Cu(s)
H2(g) → 2 H
(aq) 2 e
anode
net
Cu2
(aq) H2(g) → Cu(s) 2 H(aq)
E° 0.34 V 0.00 V
E° (V)
E° 0.34 V
+0.34
Suppose a standard aluminum half-cell is set up with a standard hydrogen half-cell
(Figure 8).
3
Pt(s) | H
(aq) H2(g) || Al (aq) | Al(s)
cathode
NEL
anode
E° = +1.66 V
0.00
2+
Cu(aq)
+ 2e-
Cu(s)
+
2H(aq)
+ 2e-
H2(g)
0.34 V
Figure 8
As you already know from the redox
table (Appendix C11), copper(II) ions
are stronger oxidizing agents than
hydrogen ions. The cell potential
provides a quantitative measurement of how much stronger
Electric Cells 703
LEARNING
TIP
Altering the coefficients in a
half-reaction equation does not
affect the reduction potential.
1.66 V
H2(g)
Al(s)
Pt(s)
Figure 9
An aluminum-hydrogen standard
cell
1.00 mol/L
3+
Al(aq)
at 25°C
1.00 mol/L
+
H(aq) at 25°C
According to the voltmeter, the platinum electrode is the cathode and the aluminum
electrode is the anode. This indicates that hydrogen ions are stronger oxidizing agents
than aluminum ions, by 1.66 V. Since the reduction potential of hydrogen ions is defined
as 0.00 V, the reduction potential of the aluminum ions must be 1.66 V below that of
hydrogen, or –1.66 V (Figure 10).
E° (V)
0.00
+
2 H(aq)
+ 2 e-
2 H+(aq) + 2 e–
H2(g)
3+
(aq)
Al
1.66 V
–
+ 3e
H2(g)
E °r = 0.00 V
Al(s)
E °r = –1.66 V
The standard cell potential, E ° 1.66 V, is the difference between the reduction
potentials of these two half-cells. To obtain the net or overall cell reaction, add the reduction and oxidation half-reactions, but remember to balance and cancel the electrons.
cathode
3 [2 H
(aq) 2 e → H2(g)]
2 [Al(s) → Al 3
(aq) 3 e ]
anode
-1.66
3+
Al(aq)
+
3 e-
Al(s)
Figure 10
On a redox table, hydrogen ions are
stronger oxidizing agents than aluminum ions. The cell potential tells
us the hydrogen ions are 1.66 V
above aluminum ions.
704 Chapter 9
net
6
H
(aq)
2 Al(s) → 3 H2(g) 2 Al 3
(aq)
E ° 0.00 V (1.66 V) 1.66 V
Notice that the half-reaction equations were multiplied by appropriate factors to balance
the electrons, but the reduction potentials are not altered by the factors used to balance the electrons. Electric potential represents energy per coulomb of charge (1 V 1 J/C). Multiplying
the aluminum half-reaction by a factor of 2 doubles both the energy and the charge transferred, so that the ratio of energy (J) to charge (C), that is the voltage, is unaffected.
NEL
Section 9.5
In both of these examples, the strongest oxidizing agent reacts at the cathode and the
strongest reducing agent reacts at the anode. The measured cell potential is the difference between the reduction potentials at the cathode and at the anode.
A positive cell potential (E > 0) indicates that the net
reaction is spontaneous—a requirement for all galvanic
cells.
In Figure 11, the results from the copper-hydrogen and aluminum-hydrogen standard cells are combined. This process of measuring standard cell potentials can quickly
be extended to more and more oxidizing agents. Notice that this process, although started
with the hydrogen reference cell, does not require that it be used for all cell measurements.
For example, knowing that the reduction potential of copper(II) ions is 0.34 V, we can
now set up many cells that include a standard copper half-cell. A more extensive list of
reduction potentials is found in the table of relative strengths of oxidizing and reducing
agents in Appendix C11.
Using the table in Appendix C11, you can predict the reaction that occurs spontaneously in any galvanic cell operating under standard conditions. The standard cell
potential is predicted as follows:
E °
E °r
cathode
E °r
anode
This order of subtraction is necessary to confirm the spontaneity from the sign of
E°. If E° is positive, the reaction is spontaneous.
SUMMARY
E° (V)
+0.34
0.00
2+
Cu(aq)
+ 2 e-
Cu(s)
+
2 H(aq)
+ 2 e-
H2(g)
1.66 V
-1.66
3+
Al(aq)
+ 3 e-
Al(s)
Figure 11
Measurements of standard cell
potentials show that the reduction
potential of Cu2
(aq) is 0.34 V greater
than that of H
(aq), which is 1.66 V
3 . If you set
greater than that of Al(aq)
up a standard cell using copper and
aluminum, what would be the cell
potential, E ? (Answer: 2.00 V)
Rules for Analyzing Standard Cells
LEARNING
Given the contents of the cell, you will need to do one or more of the following steps:
• The cathode is the electrode where the strongest oxidizing agent present in the
cell reacts, i.e., the oxidizing agent in the cell that is closest to the top on the left
side of the redox table.
If required, copy the reduction half-reaction for the strongest oxidizing agent
and its reduction potential.
• The anode is the electrode where the strongest reducing agent present in the cell
reacts, i.e., the reducing agent in the cell that is closest to the bottom on the right
side of the redox table.
If required, copy the oxidation half-reaction (reverse the half-reaction by
reading from right to left) for the strongest reducing agent and the reduction
potential listed in the table.
• Balance the electrons for the two half-reaction equations (but do not change the
Ers) and add the half-reaction equations to obtain the overall or net cell reaction.
• Calculate the standard cell potential, E °.
NEL
0.34 V
TIP
To ensure a correct interpretation,
always write the cathode halfreaction of the SOA first. This will
help you to remember to subtract
the reduction potentials in the correct order.
INVESTIGATION 9.5.1
Investigating Galvanic Cells
(p. 722)
Construct galvanic cells and evaluate cell potentials.
Electric Cells 705
Example 1
A standard dichromate-lead cell is constructed. Write the cell notation, label the
electrodes, and calculate the standard cell potential.
Solution
3
2
C(s) | Cr2O72
(aq), H(aq), Cr (aq), || Pb(aq), | Pb(s)
cathode
anode
E ° 1.23 V (0.13 V) 1.36 V
Example 2
A standard copper-scandium cell is constructed and the cell potential measured. The voltmeter indicates that copper metal is the cathode.
3
Cu(s) | Cu2
(aq), || Sc (aq), | Sc(s)
E ° 2.36 V
Write and label the half-reaction and net equations and calculate the standard
reduction potential of the scandium ion.
Solution
cathode
3 [Cu 2
(aq), 2 e → Cu(s) ]
net
E °
Er° 0.34 V
2 [Sc(s) → Sc3
(aq), 3 e ]
anode
3
Cu2
(aq)
2 Sc(s) → 3 Cu(s) Er° ?
E° 2.36 V
2 Sc3
(aq),
E °Cu2 E °Sc3
2.36 V 0.34 V E °Sc3
Figure 12
(a) The water behind the gates in a
lock has a certain potential
energy, E, relative to the
bottom of the closed outlet.
(b) When the outlet is opened,
water spontaneously flows to
the lower level on the other side
of the gates. Potential energy,
E, is converted to kinetic
energy of the flowing water. The
water flowing through the outlet
is analogous to electron flow.
(c) The flow of water ceases when
the levels on both sides of the
gates become equal. The gates
open, and the ship can then
exit to the next lock.
(a)
706 Chapter 9
E °Sc 3 2.02 V
Cell Potentials Under Nonstandard Conditions
The electric potential difference or voltage of a cell decreases slowly as the cell operates.
Simultaneously, colour changes, and precipitate formation occurs. If the cell is left for a
very long time, the voltage would eventually become zero and no further changes would be
observed in the cell. When people refer to a “dead” cell or battery, this is often what is meant.
The electric potential difference of a cell is a measure of the tendency for electrons to
flow. Ideally, during a measurement of the cell potential, a voltmeter should not allow
any electrons to flow. If electrons flow, oxidation and reduction reactions occur which, in
turn, change the concentrations from the standard 1.0 mol/L value. The value that is
measured by a voltmeter represents an electric potential or stored energy just as the water
behind a lock in a canal has gravitational potential energy (Figure 12(a)). Connecting
(b)
(c)
NEL
Section 9.5
the electrodes of a cell in a circuit allows the electrons to flow from the anode to the
cathode. This is analogous to opening the valve or sluice and allowing the water to flow
from behind the gates to a lower point in front of the gates (Figure 12(b)). In both cases,
stored potential energy is converted to kinetic energy of electrons or water. If the water
available behind a lock is allowed to flow out, then eventually no more water will flow. The
level (potential energy) of the water on the two sides of the gate is equalized. An equilibrium
is reached with no potential energy difference (Figure 12(c)). A similar situation occurs
with an operating cell. If electrons are allowed to flow, eventually an equilibrium will be
reached when the flow ceases. The rate of the forward reaction, which predominates initially, decreases as the rate of the reverse reaction increases, until the two rates become equal.
This is the equilibrium condition and no net flow of electrons will occur. At this time, the
electric potential difference as measured by a voltmeter becomes zero.
Standard cell potentials can be determined readily using standard reduction potentials.
If the concentrations are not standard, the cell potential can be predicted using a relationship discovered by Walther Hermann Nernst. A simplified version of the Nernst
equation is shown below:
0.0592 V
E E° log Q
n
(at 25°C only)
where
E is the cell potential at 25°C and non-standard concentrations
E ° is the cell potential at 25°C and standard concentrations (1 mol/L)
n
is the amount, in moles, of electrons transferred according to the cell reaction
Q
is the reaction quotient
Standard Cell Potential
SUMMARY
• A standard cell is one in which all entities shown in the half-reaction equation
are present and at SATP. The concentration of aqueous entities is 1.0 mol/L.
• The standard cell potential E° is the maximum electric potential difference
between the cathode and anode of a galvanic cell at standard conditions.
E°
cell
E °r
cathode
E °r
anode
• A positive standard cell potential (E° > 0) indicates that the overall cell reaction
is spontaneous.
• The standard reduction potential E°r represents the ability of a standard half-cell
to attract electrons, relative to the reference half-cell.
• The reference half-cell is Pt(s) H2(g), H
(aq), which has, by definition, a standard
reduction potential of exactly zero volts.
NEL
Electric Cells 707
Practice
Understanding Concepts
Answers
10. (a) 0.77 V
(b) 0.45 V
(c) 1.23 V
11. (a) 0.47 V
(b) 0.50 V
(c) 0.77 V
12. 0.34 V
13. 3.38 V; 2.28 V
10. For each of the following cells, write the equations for the reactions occurring at the
cathode and at the anode, and an equation for the overall or net cell reaction.
Calculate the standard cell potential. (Use the redox table in Appendix C11.)
2
(a) Sn (s) | Sn2
(aq) || Cr (aq) | Cr(s)
2
(b) C(s) | SO4(aq), H(aq), H2SO3(aq) || Co2
(aq) | Co(s)
| Pt
(c) Pt (s) | OH
,
O
||
H
,
OH
(aq)
(aq)
2(g)
2(g)
(s)
11. For each of the following standard cells, refer to the redox table of relative strengths
of oxidizing and reducing agents in Appendix C11, to represent the cell using the
standard cell notation. Identify the cathode and anode and calculate the standard cell
potential without writing half-reaction equations.
(a) copper-lead standard cell
(b) nickel-zinc standard cell
(c) iron(III)-hydrogen standard cell
12. One experimental design for determining the position of a half-cell reaction that is
not included in a table of oxidizing and reducing agents is shown below. Use the following standard cell, refer to the standard reduction potential of gold in Appendix
C11, and calculate the reduction potential for the indium(III) ion.
3
Au (s) | Au 3
(aq) || In (aq) | In(s)
cathode
E ° 1.84 V
anode
13. Any standard half-cell could have been chosen as the reference half-cell—the zero
point of the reduction potential scale. What would be the standard reduction potentials for copper and zinc half-cells, assuming that the standard lithium cell were
chosen as the reference half-cell, with its reduction potential defined as 0.00 V?
14. A zinc-iron cell is constructed and allowed to operate until the measured potential
difference becomes zero. What interpretation can be made about the chemical
system at this point?
Applying Inquiry Skills
15. Develop a table of oxidizing agents and reduction potentials from experimental evidence.
Experimental Design
Several cells are investigated; each cell has at least one half-cell in common with one
of the other cells. The cell potentials are measured and the positive and negative
electrodes of each cell are identified.
Evidence
Positive electrode
C(s) | Cr2O72
(aq),
Tl(s) | Tl
(aq)
Pd(s) | Pd2
(aq)
H
(aq)
Negative electrode
|| Pd2
(aq) | Pd(s)
E ° 0.28 V
|| Ti2
(aq)| Ti(s)
E ° 1.29 V
|| Tl (aq)
E ° 1.29 V
| Tl(s)
Analysis
(a) Using the given evidence, complete a table of relative strengths of oxidizing and
reducing agents, including reduction potentials.
16. Complete the Prediction for the following investigation. Include your reasoning.
Question
What is the total electric potential difference of two cells connected in series?
Experimental Design
Copper-silver and copper-zinc standard cells are connected as shown in Figure 13.
The total electric potential difference of the two cells is measured with a voltmeter
connected to the silver and zinc electrodes.
708 Chapter 9
NEL
Section 9.5
Cu(s)
Zn(s)
Cu(s)
Ag(s)
Ag(s) | Ag +(aq) || Cu2+(aq) | Cu(s) — Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s)
Figure 13
Two standard cells
in series
Section 9.5 Questions
Understanding Concepts
1. (a) For a given cell, how is the cell potential predicted?
(b) What are the restrictions on this prediction?
2. How does the cell potential indicate spontaneity of the
reaction?
3. Why are the reactions in galvanic cells always spontaneous?
4. Define the hydrogen reference cell, including contents and
conditions.
5. Why is a reference half-cell necessary?
6. (a) What is the cell potential of a standard cobalt-zinc
cell?
(b) What is the theoretical interpretation of this cell potential?
7. For each of the following cells,
•
use the given cell notation to identify the strongest oxidizing and reducing agents;
NEL
•
write chemical equations to represent the cathode,
anode, and overall (net) cell reactions (include the
half-cell and cell potentials); and
•
draw a diagram of each cell, labelling the electrodes,
polarity (signs) of electrodes, electrolytes, direction of
electron flow, and direction of ion movement.
2
(a) Cu(s) | Cu2
(aq) || Zn(aq) | Zn(s)
2
(b) C(s) | Cr2O7(aq), H(aq) || Sn2
(aq) | Sn(s)
8. You can determine a possible identity of an unknown halfcell from the cell potential involving a known half-cell. Use
the following evidence and the table of reduction potentials
in Appendix C11 to determine the reduction potential and
possible identity of the unknown X2
(aq) | X(s) redox pair.
X
2
2 Ag(aq)
(s) → 2 Ag(s) X(aq)
E ° 1.08 V
Electric Cells 709
9.6
Figure 1
Large ships have steel hulls. The
rusting of steel involves the oxidation of iron in the steel and is a
constant headache for shipping
companies.
corrosion an electrochemical
process in which a metal reacts with
substances in the environment,
returning the metal to an ore-like
state
Corrosion
The history of civilization is often divided into different “ages”
such as the Copper, Bronze, and Steel ages. These descriptions
are based on when these metals were refined and used for tools
and weapons. The process of refining a metal is electrochemical
in nature and requires energy to recover the pure metal from its
naturally occurring compounds (ores). Corrosion is also an
electrochemical process. Because we live in an oxidizing (oxygen)
environment, spontaneous oxidation (corrosion) of a metal
occurs. In fact, we need to produce metals such as iron continually to replace the metals lost to corrosion. Preventing corrosion and dealing with the effects of corrosion are major economic
and technological problems for our society (Figure 1).
As a metal is oxidized, metal atoms lose electrons to form
positive ions. A redox table of relative strengths of oxidizing
and reducing agents provides the evidence that metals vary
greatly in their ability to be oxidized. Some metals, such as gold
and silver, are noble because they are relatively weak reducing
agents. On the other hand, Group 1 and 2 metals are very strong
reducing agents and are, therefore, easily oxidized. In general, any
metal appearing below the oxygen half-reactions in a redox
table will be oxidized in our environment. Iron (including steel)
and aluminum are such metals, and are extensively used as
structural materials. Why is the corrosion or rusting of iron
such a major problem, but the corrosion of aluminum, which
is a much stronger reducing agent, not? The answer lies primarily in the nature of the oxide
that forms on the surface of the metal. A freshly cleaned surface of aluminum rapidly oxidizes in air to form aluminum oxide.
4 Al(s) 3 O2(g) → 2 Al2O3(s)
The aluminum oxide adheres tightly to the surface of the metal. This prevents further corrosion by effectively sealing any exposed surfaces.
Unfortunately, the iron compounds that form on the surface of exposed iron do not
adhere very well. They flake off, exposing new iron to be corroded. In addition, the corrosion of iron is a complex process that is significantly affected by the presence of substances other than oxygen.
INVESTIGATION 9.6.1
The Corrosion of Iron (p. 723)
Study the factors that affect the corrosion of iron.
DID YOU
KNOW
?
Rates of Corrosion
A tin can (tin on steel) will corrode
completely in about 100 a; an aluminum can in about 400 a; and a
glass bottle in about 100 ka.
710 Chapter 9
Rusting of Iron
Studies of the corrosion of iron have shown that the presence of both oxygen and water
is required and the iron is converted into iron hydroxides and oxides. The first step of the
mechanism is thought to be the oxidation of iron at a wet exposed surface (Figure 2).
Fe(s) → Fe2
(aq) 2 e
NEL
Section 9.6
Iron(II) ions diffuse through the water on the iron surface while the electrons easily
travel through the iron metal, which is an electrical conductor. The electrons are picked
up by oxygen molecules dissolved in water on the surface at a point away from the original oxidation site (Figure 2).
1
O2(g) H2O(l) 2 e → 2 OH
(aq)
2
DID YOU
KNOW
?
Hydrated oxide
Iron(III) hydroxide can be converted to iron(III) oxide trihydrate
as shown below.
2 Fe(OH)3(s) → Fe2O3•3H2O(s)
The combination of iron(II) ions and hydroxide ions forms a low-solubility precipitate of iron(II) hydroxide, which is further oxidized by oxygen and water to form
iron(III) hydroxide, a yellow-brown solid. The familiar red-brown rust is formed by the
dehydration of iron(III) hydroxide to form a mixture of iron(III) hydroxide and hydrated
iron(III) oxide. The amount of the hydroxide and the oxide varies, so rust is referred to
as a hydrated oxide of indeterminate formula, Fe2O3• xH2O(s).
In fact, it is difficult to determine
how much of the iron(III) exists in
rust as the hydroxide or hydrated
oxide. Warming this mixture can
drive off some of the waters of
hydration.
water
air
O2(g)
rust
OH
Fe2
iron
object
e
cathode
1 O H O 2 e
2(g)
2 (l)
2
anode
2
OH
(aq)
Fe(s)
Fe2
(aq)
2
e
e
Figure 2
The corrosion of iron is a small electrochemical cell with iron oxidation
at one location (the anode) and
oxygen reduction at another location
(the cathode).
This simplified mechanism for the rusting of iron can be used to explain why certain
conditions promote rusting. If the iron is kept in a dry environment (low humidity) or
if air has been removed from the water, little or no corrosion occurs (Figure 3). Eliminating
either water or the oxygen in the water makes the reduction of aqueous oxygen impossible. Iron cannot be oxidized unless a suitable oxidizing agent is present. If oxidizing agents
other than oxygen are present, such as certain metal ions, nonmetals, or hydrogen ions,
TRY THIS activity
Home Corrosion Experiment
Soft drinks are acidic and contain electrolytes. Would different
brands of pop corrode iron at different rates?
Materials: Coca-Cola, 7-Up, 2 identical steel nails, 2 plastic
glassses
(a) Predict which drink will cause faster corrosion.
• Test your prediction, using a clean steel nail placed in a
fresh sample of each soft drink.
(b) Explain the results.
NEL
Figure 3
Rusting of exposed iron is almost
negligible when the relative humidity
is less than 50%. This iron pillar in
Delhi, India, has existed for about
1500 years because of the very dry
and unpolluted environment.
Electric Cells 711
DID YOU
KNOW
?
Fighting Corrosion
Concrete structures are reinforced
with steel bars (sometimes referred
to as rebar). These bars are made
from scrap steel that is melted,
reshaped, and then air-cooled. This
cooling step allows the carbon in
the steel to precipitate, forming
microscopic carbide “fingers” that
strengthen the steel. Unfortunately,
they also act as little batteries.
Electrons tend to flow from the iron
toward the carbide fingers. As the
iron loses electrons, it corrodes. If a
concrete structure is in an area
prone to corrosion, such as at the
water line or even near ocean spray,
this electron loss and corrosion
occur even faster, resulting in a general weakening of the steel, and
therefore, the concrete. The concrete structure may look just as
solid, but it is no longer as strong as
when it was first built. One solution
being researched is to cool the steel
rebar with water instead of air. The
steel cools more quickly, which
seems to reduce the growth of carbide fingers. The result is a less
“electrically active” steel that may
last longer.
the iron can still be corroded through spontaneous redox reactions. This helps to explain
the corrosion of iron in acidic environments, for example, why acid rain corrodes iron
more than natural rain does.
In general, electrolytes accelerate rusting. Ships rust more rapidly in seawater than
in fresh water and cars rust more rapidly in places where salt is used on roads. Chloride
ions from salt are known to inhibit the adherence of protective oxide coatings on
many metals, thus exposing more metal to be corroded. Electrolytes like sodium chloride conduct electricity and improve charge transfer, accelerating the rusting process.
It is well known to plumbers that you cannot use steel straps or nails to hold copper
pipes in place, because corrosion of the iron will be accelerated. Any moisture that is
present sets up an electric cell similar in principle to Volta’s original discovery of electricity from dissimilar metals (section 9.4). As the cell operates, the iron corrodes to
form rust.
The rusting of iron requires the presence of oxygen and
water and is accelerated by the presence of acidic solutions, electrolytes, mechanical stresses, and contact with
less active metals.
Corrosion Prevention
Methods used for preventing or minimizing the corrosion of iron can be divided into two
categories: barrier methods that employ protective coatings and the method of cathodic
protection. In some critical situations, such as a large fuel tank, both methods may be used.
Paint and other similar coatings are a simple method of corrosion prevention. This
method works well as long as the surface is completely covered and the coating remains
intact. Unfortunately, a scratch or chip in the surface can easily expose a small surface
of iron and corrosion begins.
Both tin and zinc are used as metallic coatings. Tin, as in the familiar tin can, adheres
well to the iron and provides a strong, shiny coating. The outer surface of the tin coating
has a thin, strongly adhering layer of tin oxide that protects the tin. If a crack or break
occurs in the tin layer, moisture can collect in the crack and an electric cell with tin and
iron electrodes is established. Since iron is more easily oxidized than tin, iron becomes
the anode in this cell. The electrons released by the oxidation of iron flow to the tin and
corrosion is accelerated. Evidence of this is the typical iron rust on tin cans that have
been crushed and left outside.
A spontaneous electric cell also arises when a zinc coating on an iron object is broken.
However, in this case, the zinc is more easily oxidized than the iron. The zinc is preferentially
oxidized, preventing corrosion of the iron. Zinc plating (galvanizing) of steel or iron provides double protection—a protective layer and preferential corrosion of the zinc.
Cathodic Protection
cathodic protection a method of
corrosion prevention in which the
metal being protected is forced to
become the cathode of a cell, using
either an impressed current or a
sacrificial anode
712 Chapter 9
According to the redox theory of a cell, oxidation is the loss of electrons and occurs at
the anode of a cell. Therefore, an effective method of preventing corrosion of iron is
cathodic protection, forcing the iron to become the cathode by supplying the iron with
electrons.
For a battery or DC generator connected in a circuit, electrons flow out of the negative
terminal and into the positive terminal. If the negative terminal is connected to the iron
object and the positive terminal to an inert carbon electrode, an electric current is forced
NEL
Section 9.6
to flow to the iron, through an electrolyte such as ground water, from the carbon electrode.
The iron is forced to become the cathode and is prevented from corroding. An impressed
current is an electric current forced to flow toward an iron object by an external potential difference. This method of corrosion prevention requires a constant electric power
supply (typically 8 mV) and is used as cathodic protection for pipelines and culverts.
A less common but simpler method of cathodic protection is the use of a sacrificial
anode. A sacrificial anode is a metal more easily oxidized than iron and connected to the
iron object to be protected. The practice of zinc plating (galvanizing) iron objects is a
common example of this method. Sacrificial zinc anodes are also connected to the exposed
underwater metal surfaces of ships and boats to prevent the corrosion of the iron in the
steel. Blocks of magnesium can also be used as sacrificial anodes (Figure 4). In all cases,
the more active metal (appearing below iron in a half-reaction table) is slowly consumed
or sacrificed at the anode, forcing the iron object to be the cathode of the cell.
underground
steel tank
electron
flow
copper wire
magnesium
NEL
Figure 4
Corrosion of iron involves the oxidation of iron at the anode of a cell. If
the iron is attached directly or connected electrically to a metal that is
more easily oxidized (a sacrificial
anode), then a spontaneous cell
develops in which iron is the
cathode. The electrolyte of the cell
is the moisture in the ground.
Electric Cells 713
Section 9.6 Questions
Understanding Concepts
1. What are the minimum requirements for the corrosion of
iron?
2. List some factors that accelerate or promote the corrosion
of iron.
3. Write the balanced net ionic equation for the corrosion of
iron to iron(II) ions in the presence of oxygen and water.
4. Although the corrosion of iron is a serious problem, other
metals are also corroded in air or other environments. For
each of the following situations, use your knowledge of
writing and balancing redox equations to write and label
the half-reaction and net ionic equations:
(a) Zinc is an active metal that oxidizes quickly when
exposed to air and water.
(b) A lead pipe corrodes if it is used to transport acidic
solutions that also contain dissolved oxygen.
(c) In dry air, minute quantities of hydrogen sulfide gas
can slowly react with silver objects to produce
hydrogen gas and silver sulfide, recognized by the dark
tarnish on the surface of the silver.
5. You may have noticed that when rusting appears on a car
body, the rust appears around the break or chip in the paint
but the damage may extend under the painted surface for
some distance.
(a) What is the evidence for damage extending well
beyond the break in the paint?
(b) Suggest an explanation why the damage may extend
far from the break in the paint.
Figure 5
When this pipeline was being constructed, a zinc wire was
attached to and buried with the pipe.
6. Would a basic solution prevent or slow down the corrosion
of iron? Provide your reasoning.
7. Why is a zinc coating on iron better than a tin coating?
Making Connections
8. What are the two methods of cathodic protection and how
are they similar?
10. A zinc wire is connected to and buried with a pipeline
when it is built (Figure 5).
(a) Why is this done? Include a brief description of the
principles involved.
(b) Is this the only type of corrosion protection used with
major pipelines?
(c) Discuss the environmental and safety issues associated with protecting and also not protecting pipelines.
Applying Inquiry Skills
9. The following investigation looks at the reactivity of oxygen
and various acids with a metal. Evidence of a spontaneous
reaction would be the corrosion of the metal.
Question
What effect do oxygen and various acids have on the corrosion of copper metal?
Experimental Design
Several test tubes are set up with a clean piece of copper
metal in each. Various possible oxidizing agents will be
tested for reaction wth copper: oxygen gas only, oxgyen
bubbled into water, oxygen bubbled into each of dilute
hydrochloric acid, nitric acid, sulfuric acid, and phosphoric
acid. (If oxygen is not available, use air.)
(a) Identify the independent, dependent, and controlled
variables.
(b) Prepare a list of materials and write a procedure
including safety and disposal instructions. When your
teacher has approved your work, conduct and report
on this experiment.
714 Chapter 9
GO
www.science.nelson.com
11. State several examples of metal corrosion of manufactured
materials. Which examples involve environmental, health, or
safety issues? Are there examples of corrosion that are
desirable? Discuss briefly.
12. Search the Internet, using the key words “iron corrosion.”
How many different titles did the search find? What is the
implication of this number? Find a general site and list the
different classes of iron corrosion.
NEL
Chapter 9
LAB ACTIVITIES
INVESTIGATION 9.1.1
Single Displacement Reactions
In this investigation you will observe some common single
displacement reactions and then interpret the changes in
terms of electron transfer.
Purpose
The purpose of this investigation is to gather some evidence
about single displacement reactions.
What are the products of the single displacement reactions for
the following sets of reactants?
copper and aqueous silver nitrate
aqueous chlorine and aqueous sodium bromide
magnesium and hydrochloric acid
zinc and aqueous copper(II) sulfate
aqueous chlorine and aqueous potassium iodide
Prediction
(a) According to the single displacement generalization,
predict the balanced chemical equations for each set
of reactants listed above.
Small quantities of reactants are mixed and diagnostic tests
(Appendix A6) are used to determine the products of each
reaction.
(b) For each chemical equation listed in your prediction,
record diagnostic tests that you will use. (Some diagnostic tests will be very specific and some will be a
general observation you expect.)
Materials
NEL
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
The substances used in this experiment are toxic
and corrosive. Avoid skin contact. Wash any
splashes on the skin or clothing with plenty of water.
If any chemical is splashed in your eyes, rinse for at
least 15 min and inform your teacher.
Keep the cyclohexane sealed to avoid evaporation
and inhalation of the vapours. Dispose of the hydrocarbon mixtures as directed by your teacher.
Make sure matches are extinguished by dipping in
water.
Procedure
1. Set up five test tubes, each filled to a depth of 2–3 cm
with one of the five aqueous solutions.
2. Add the appropriate element to each solution (see
Question) in each test tube.
3. Perform diagnostic tests on each of the five mixtures.
Record your observations.
Experimental Design
lab apron
eye protection
five small test tubes
two test-tube stoppers
test-tube rack
emery paper or steel wool
wash bottle
matches
copper wire or strip
magnesium metal ribbon
zinc strip
Inquiry Skills
Magnesium and cyclohexane are highly flammable.
Keep away from open flame.
Question
•
•
•
•
•
Unit 5
dropper bottles of 0.1 mol/L
—aqueous silver nitrate
—aqueous sodium bromide
—aqueous copper(II) sulfate
—aqueous potassium iodide
—hydrochloric acid
chlorine water (or bleach)
cyclohexane
4. Dispose of the solutions as directed by your teacher.
Evidence
(c) Design a table to record your observations.
Analysis
(d) Interpret your evidence and record the products that
you can reasonably conclude you obtained in each
reaction.
Evaluation
(e) Evaluate the experimental design, materials, and procedure by considering any possible flaws and
improvements.
(f) Use your answer to (a). What is your judgment of the
quality and quantity of evidence obtained?
(g) How confident are you in the answer obtained?
Provide your reasons.
(h) Evaluate the Prediction and provide your reasons.
Electric Cells 715
LAB EXERCISE 9.1.1
Inquiry Skills
Oxidation States of Vanadium
Vanadium is a transition metal that forms many different
ions with different oxidation states (Table 1). Vanadium and
its compounds have many different uses, including colouring
for glass, ceramics, and plastics.
Table 1 Colours of Vanadium Ions
Ion name
Ion formula
vanadate(V)
VO3 (aq)
vanadate (IV)
2
VO(aq)
vanadium(III)
vanadium(II)
Colour
3
V(aq)
2
V(aq)
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Evidence
Table 2 Reactions of Vanadium Ions
Procedure
Final solution colours
(1) ammonium vanadate(V)
dissolved in sulfuric acid
yellow
(2) yellow solution with
three subsequent additions
of small quantities of
zinc dust
yellow turned blue,
then green, then violet
(3) violet solution left
sitting in an open container
slowly turned green
(4) yellow solution mixed with
potassium iodide solution
very dark, almost black
(5) blue solution mixed with
potassium iodide solution
stayed blue; no change
(6) violet solution slowly mixed
with acidic potassium
permanganate
violet to green to blue to
yellow
Analysis
(a) Using Table 1, identify the vanadium ions in the
sequence of reactions in Table 2.
Purpose
(b) In each case, is the vanadium being oxidized or
reduced? Justify your answer, using oxidation numbers.
The purpose of this lab exercise is to investigate some redox
chemistry of vanadium compounds.
(c) Explain the observations made in (3) to (6) in Table 2
above. Suggest what is causing these changes.
Question
What are the oxidation states and changes in oxidation
number for vanadium ions?
INVESTIGATION 9.3.1
Spontaneity of Redox Reactions
In the past, we have usually assumed that all chemical reactions are spontaneous; that is, they occur of their own accord
once reactants are placed in contact, without a continuous
addition of energy to the system. Spontaneous redox reactions in solution generally provide visible evidence of a reaction within a few minutes.
716 Chapter 9
(d) Vanadium ion chemistry can be quite complicated. For
example, the initial yellow solution is likely an equilib and VO . Does this alter your
rium between VO3(aq)
2(aq)
analysis in the previous questions? Explain briefly.
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Purpose
The scientific purpose of this investigation is to test the
assumption that all single displacement reactions are
spontaneous.
NEL
LAB EXERCISE 9.1.1
Inquiry Skills
Oxidation States of Vanadium
Vanadium is a transition metal that forms many different
ions with different oxidation states (Table 1). Vanadium and
its compounds have many different uses, including colouring
for glass, ceramics, and plastics.
Table 1 Colours of Vanadium Ions
Ion name
Ion formula
vanadate(V)
VO3 (aq)
vanadate (IV)
2
VO(aq)
vanadium(III)
vanadium(II)
Colour
3
V(aq)
2
V(aq)
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Evidence
Table 2 Reactions of Vanadium Ions
Procedure
Final solution colours
(1) ammonium vanadate(V)
dissolved in sulfuric acid
yellow
(2) yellow solution with
three subsequent additions
of small quantities of
zinc dust
yellow turned blue,
then green, then violet
(3) violet solution left
sitting in an open container
slowly turned green
(4) yellow solution mixed with
potassium iodide solution
very dark, almost black
(5) blue solution mixed with
potassium iodide solution
stayed blue; no change
(6) violet solution slowly mixed
with acidic potassium
permanganate
violet to green to blue to
yellow
Analysis
(a) Using Table 1, identify the vanadium ions in the
sequence of reactions in Table 2.
Purpose
(b) In each case, is the vanadium being oxidized or
reduced? Justify your answer, using oxidation numbers.
The purpose of this lab exercise is to investigate some redox
chemistry of vanadium compounds.
(c) Explain the observations made in (3) to (6) in Table 2
above. Suggest what is causing these changes.
Question
What are the oxidation states and changes in oxidation
number for vanadium ions?
INVESTIGATION 9.3.1
Spontaneity of Redox Reactions
In the past, we have usually assumed that all chemical reactions are spontaneous; that is, they occur of their own accord
once reactants are placed in contact, without a continuous
addition of energy to the system. Spontaneous redox reactions in solution generally provide visible evidence of a reaction within a few minutes.
716 Chapter 9
(d) Vanadium ion chemistry can be quite complicated. For
example, the initial yellow solution is likely an equilib and VO . Does this alter your
rium between VO3(aq)
2(aq)
analysis in the previous questions? Explain briefly.
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Purpose
The scientific purpose of this investigation is to test the
assumption that all single displacement reactions are
spontaneous.
NEL
Unit 5
INVESTIGATION 9.3.1 continued
Question
Which combinations of copper, lead, silver, and zinc metals
and their aqueous metal ion solutions produce spontaneous
reactions?
Prediction
(a) State and justify your answer to the question.
Experimental Design
A drop of each metal ion solution is placed in separate locations on a clean area of each of the four metal strips.
The solutions used are toxic—especially the lead
solution—and irritants. Avoid skin contact. Remember
to wash your hands before leaving the laboratory.
Dispose of reaction products according to your
teacher’s instructions.
Rinse all of the metal strips thoroughly and return
them so they can be used again.
Procedure
(b) Write a brief procedure for this investigation. Have
your teacher approve your procedure before you start.
Evidence
(c) Design a table to record your observations.
Materials
lab apron
eye protection
reusable strips of copper, lead, silver, and zinc metals
0.10 mol/L solutions of copper(II) nitrate, lead(II) nitrate,
silver nitrate, and zinc nitrate in dropper bottles
steel wool or emery paper
LEARNING
TIP
Distinguishing Lead and Zinc
When cleaned, lead and zinc metals look very similar.
However, lead is much softer. You can distinguish between
them because lead strips bend and scratch much more
easily than zinc strips.
LAB EXERCISE 9.3.1
Building a Redox Table
Suppose that a research team is developing a table of relative
strengths of oxidizing and reducing agents. One team member
had completed an investigation summarized in Table 3, Section
9.3, and another had completed the investigation reported
in Practice question 9, Section 9.3. A third member used the
combination of metals, nonmetals, and solutions shown
below. By completing this exercise, you will see how scientists have developed more extensive tables of relative strengths
of oxidizing and reducing agents.
NEL
Analysis
(d) Based on your evidence, which combinations of
reactants produced spontaneous reactions?
Evaluation
(e) Were the experimental design, materials, and procedure
sufficient to answer the question? Justify your answer.
(f) Suggest an improvement to increase the certainty of
the evidence.
(g) Evaluate the prediction, including your reasons.
(h) What does your answer to (g) tell you about the
assumption on which the prediction was based? What
should be done with this assumption?
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Purpose
The purpose of this lab exercise is to construct a table of
relative strengths of oxidizing and reducing agents.
Question
What is the table of relative strengths of oxidizing and reducing
agents for copper, silver, bromine, and iodine?
Electric Cells 717
Unit 5
INVESTIGATION 9.3.1 continued
Question
Which combinations of copper, lead, silver, and zinc metals
and their aqueous metal ion solutions produce spontaneous
reactions?
Prediction
(a) State and justify your answer to the question.
Experimental Design
A drop of each metal ion solution is placed in separate locations on a clean area of each of the four metal strips.
The solutions used are toxic—especially the lead
solution—and irritants. Avoid skin contact. Remember
to wash your hands before leaving the laboratory.
Dispose of reaction products according to your
teacher’s instructions.
Rinse all of the metal strips thoroughly and return
them so they can be used again.
Procedure
(b) Write a brief procedure for this investigation. Have
your teacher approve your procedure before you start.
Evidence
(c) Design a table to record your observations.
Materials
lab apron
eye protection
reusable strips of copper, lead, silver, and zinc metals
0.10 mol/L solutions of copper(II) nitrate, lead(II) nitrate,
silver nitrate, and zinc nitrate in dropper bottles
steel wool or emery paper
LEARNING
TIP
Distinguishing Lead and Zinc
When cleaned, lead and zinc metals look very similar.
However, lead is much softer. You can distinguish between
them because lead strips bend and scratch much more
easily than zinc strips.
LAB EXERCISE 9.3.1
Building a Redox Table
Suppose that a research team is developing a table of relative
strengths of oxidizing and reducing agents. One team member
had completed an investigation summarized in Table 3, Section
9.3, and another had completed the investigation reported
in Practice question 9, Section 9.3. A third member used the
combination of metals, nonmetals, and solutions shown
below. By completing this exercise, you will see how scientists have developed more extensive tables of relative strengths
of oxidizing and reducing agents.
NEL
Analysis
(d) Based on your evidence, which combinations of
reactants produced spontaneous reactions?
Evaluation
(e) Were the experimental design, materials, and procedure
sufficient to answer the question? Justify your answer.
(f) Suggest an improvement to increase the certainty of
the evidence.
(g) Evaluate the prediction, including your reasons.
(h) What does your answer to (g) tell you about the
assumption on which the prediction was based? What
should be done with this assumption?
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Purpose
The purpose of this lab exercise is to construct a table of
relative strengths of oxidizing and reducing agents.
Question
What is the table of relative strengths of oxidizing and reducing
agents for copper, silver, bromine, and iodine?
Electric Cells 717
Synthesis
LAB EXERCISE 9.3.1 continued
Evidence
Table 3 Reactions of Metals and Nonmetals with
Solutions of Ions
I2(aq)
2+
Cu(aq)
Ag+
(aq)
Br2(aq)
I
(aq)
X
X
√
√
Cu(s)
√
X
√
√
Ag(s)
X
X
X
√
Br (aq)
X
X
X
X
X no evidence of a redox reaction
√ evidence redox reaction occurred
Analysis
(b) Compare Table 3 in Section 9.3, your analysis table
from Practice question 9 in section 9.3, and your
table from (a) above. Note that there are several substances that appear in two of these tables. Combine
all three tables in one larger table showing the order
of oxidizing and reducing agents.
LEARNING
TIP
Tables of Oxidizing and
Reducing Agents
All tables of oxidizing and
reducing agents follow the
same general format.
OA n e e RA
(a) Using the results from Table 3, prepare a table of
relative strengths of oxidizing and reducing agents.
INVESTIGATION 9.3.2
The Reaction of Sodium with Water
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
This demonstration will provide practice in both predicting
and testing a chemical reaction.
Purpose
The purpose of this demonstration is to test the five-step
method for predicting redox reactions.
Question
What are the products of the reaction of sodium metal with
water?
Prediction
(a) Answer the question and provide your reasoning.
Experimental Design
(b) Write a general plan for this reaction.
(c) Suggest one diagnostic test for each predicted
product, using the “If [procedure], and [evidence],
then [analysis]” format for every product predicted.
(This format is described in Appendix A6.)
(d) What control(s) should be used with these tests?
718 Chapter 97
This reaction of sodium metal with water must be
demonstrated with great care, because a great deal
of heat is produced.
Evidence
(e) Record all observations, including the controls, in a
suitable table.
Analysis
(f) According to the evidence collected, what products
were obtained?
Evaluation
(g) Assuming the evidence is of suitable quality, evaluate
your prediction.
(h) Evaluate the method of writing redox reactions used
to make your prediction.
(i) This experiment is not sufficient to provide a judgment of the method of predicting redox reactions.
What should be done next?
NEL
Synthesis
LAB EXERCISE 9.3.1 continued
Evidence
Table 3 Reactions of Metals and Nonmetals with
Solutions of Ions
I2(aq)
2+
Cu(aq)
Ag+
(aq)
Br2(aq)
I
(aq)
X
X
√
√
Cu(s)
√
X
√
√
Ag(s)
X
X
X
√
Br (aq)
X
X
X
X
X no evidence of a redox reaction
√ evidence redox reaction occurred
Analysis
(b) Compare Table 3 in Section 9.3, your analysis table
from Practice question 9 in section 9.3, and your
table from (a) above. Note that there are several substances that appear in two of these tables. Combine
all three tables in one larger table showing the order
of oxidizing and reducing agents.
LEARNING
TIP
Tables of Oxidizing and
Reducing Agents
All tables of oxidizing and
reducing agents follow the
same general format.
OA n e e RA
(a) Using the results from Table 3, prepare a table of
relative strengths of oxidizing and reducing agents.
INVESTIGATION 9.3.2
The Reaction of Sodium with Water
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
This demonstration will provide practice in both predicting
and testing a chemical reaction.
Purpose
The purpose of this demonstration is to test the five-step
method for predicting redox reactions.
Question
What are the products of the reaction of sodium metal with
water?
Prediction
(a) Answer the question and provide your reasoning.
Experimental Design
(b) Write a general plan for this reaction.
(c) Suggest one diagnostic test for each predicted
product, using the “If [procedure], and [evidence],
then [analysis]” format for every product predicted.
(This format is described in Appendix A6.)
(d) What control(s) should be used with these tests?
718 Chapter 97
This reaction of sodium metal with water must be
demonstrated with great care, because a great deal
of heat is produced.
Evidence
(e) Record all observations, including the controls, in a
suitable table.
Analysis
(f) According to the evidence collected, what products
were obtained?
Evaluation
(g) Assuming the evidence is of suitable quality, evaluate
your prediction.
(h) Evaluate the method of writing redox reactions used
to make your prediction.
(i) This experiment is not sufficient to provide a judgment of the method of predicting redox reactions.
What should be done next?
NEL
Unit 5
ACTIVITY 9.4.1
Developing an Electric Cell
In this activity, an aluminum soft-drink can is both the container and one of the electrodes (Figure 1). The other electrode
is a solid conductor such as graphite from a pencil, an iron nail,
or a piece of copper wire or pipe. The electrolyte may be a
salt solution or an acidic or basic solution. Although the
overall performance of a cell depends on many factors, only
the voltage is investigated here.
The purpose of this activity is to use a technological
problem-solving (trial-and-error) approach to construct a
working electric cell with the highest possible voltage. You
will need to control variables and work in a systematic way.
Be prepared to alter your materials if your results are not
promising and to maximize your results.
voltmeter
–
+
Be careful when handling acidic and basic solutions
because they are corrosive. Wear eye protection and
work near a source of water. Electrolytes may be
toxic or irritants; follow all safety precautions. Avoid
eye and skin contact.
Dispose of solutions according to your teacher’s
instructions.
Design
• Using the same electrolyte and the aluminum can as
the control variables, test two or three different materials as the second electrode. Measure the voltage of
each cell. (Scrape the paint from the can where the
wire is attached.)
• Using the same two electrodes as the control variables,
test two or three possible electrolytes. Measure the
voltage of each cell.
• Test additional combinations, based on the analysis of
the initial trials.
Evidence
(a) Keep careful records of all observations, including
what worked and did not work.
cathode
electrolyte
aluminum
can anode
(b) Set up a table or organized list to record your observations and analysis of each trial in your problemsolving cycle.
Analysis
(c) What is the best result you obtained? Report the
details of the design and voltage.
Evaluation
Figure 1
An aluminum can cell is an efficient design, since one of the electrodes also serves as the container.
(e) If you were to continue this process, what changes or
improvements would you make?
Materials
lab apron
eye protection
various electrodes
acidic, basic, and neutral
ionic solutions
bottle of distilled water
steel wool or emery paper
NEL
(d) Evaluate the quality of your evidence, including any
sources of experimental error or uncertainty.
voltmeter
2 plug–and–clip wires
an aluminum can with the
top removed
(f) Evaluate the suitability of your final electric cell for
potential commercial development. Consider a
variety of factors.
Electric Cells 719
LAB EXERCISE 9.4.1
Inquiry Skills
Characteristics of a Hydrogen Fuel Cell
A fuel cell supplies electrical energy in the same way as any
electric cell. Ions transfer the charge within the cell, and electrons flow through an external circuit between the electrodes.
In a hydrogen fuel cell, the electrodes are not consumed, and
there is no need to reverse the cell to recharge it. The reactants, hydrogen and oxygen, are continuously supplied and
consumed to produce water and electricity.
Commercial hydrogen fuel cells, such as the Ballard cell, utilize a solid electrolyte called a proton exchange membrane
(PEM). This polymer is bonded to two porous carbon cloth
electrodes (Figure 2).
Purpose
The purpose of this lab exercise is to compare the electrical
characteristics of a hydrogen fuel cell and a dry cell.
Question
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Analysis
(a) Construct and label a graph with voltage on the vertical
axis and current on the horizontal axis. Plot and label
the line for the fuel cell and another line for the dry cell.
(b) Describe and compare the trend shown by the
voltage-current line for the fuel cell with that of the
dry cell. Does the hydrogen fuel cell behave like a
regular (dry) cell?
(c) The power output of a cell is a measure of the quantity of energy per second delivered by the cell. For
each voltage and current, calculate the power supplied by using the formula P VI. (Recall that power
is measured in watts (W) and that 1 VA = 1 W.)
Summarize your results in a table for both the fuel
cell and the dry cell.
How do the trends of the voltage-current and power-current
graphs compare for a hydrogen fuel cell and a typical dry cell?
(d) Repeat (a) and (b) to construct a power curve for
each cell. Plot power on the vertical axis and current
on the horizontal axis.
Experimental Design
(e) Use the results of your analysis to answer the Question.
Hydrogen and oxygen gases, produced from the electrolysis of
water, are passed into a hydrogen fuel cell. The voltage of the cell
is measured with an open circuit. Various resistances are added
to the circuit, and the voltage and current are measured for each
resistance (load). The fuel cell is disconnected and replaced by a
dry cell. The same procedure is followed to collect voltage and current measurements for the same set of resistances in the circuit.
Synthesis
(f) Compare the designs of fuel cells and dry cells.
(g) What makes fuel cells more practical for powering
electric cars compared to other types of cells?
individual cell
Evidence
Fuel cell
Voltage
(V)
Dry cell
Current
(mA)
Voltage
(V)
Current
(mA)
0.81
0
1.44
0
0.80
5
1.42
5
0.79
6
1.41
10
0.78
16
1.40
22
0.73
66
1.34
90
0.70
115
1.30
146
0.67
175
1.25
212
0.61
315
1.14
335
720 Chapter 9
H2
membrane electrode
assembly
O2
anode
cathode
polymer electrolyte
H2O
H+
platinum
catalyst
carbon mat
carbon mat
Figure 2
The membrane electrode assembly includes the electrodes
(carbon mats) and the proton exchange polymer. Each carbon
particle has tiny platinum particles on its surface.
NEL
Unit 5
ACTIVITY 9.5.1
Galvanic Cell Design
The purpose of this activity is to demonstrate the design and
operation of a galvanic cell used in scientific research. A cell
with only one electrolyte is compared with similar cells containing the same electrodes but two electrolytes (Figure 3).
Materials
• Use a voltmeter to determine which electrode is positive and which is negative, and measure the electric
potential difference (voltage) of each cell.
(b) According to the voltmeter test, which electrode is the
cathode and which is the anode?
(c) Why is your answer to (b) the same for all three cells?
lab apron
eye protection
silver and copper electrodes
steel wool (for cleaning electrodes)
voltmeter with leads
4 medium-sized beakers
U-tube
cotton plugs
porous cup
bottle of distilled water
solutions of sodium nitrate, silver nitrate,
copper(II) nitrate
(d) Suggest a reason why two of the voltages measured
are very similar and the third is very different.
• With the voltmeter connected, remove and then
replace the various parts of each cell.
(e) Why does the voltmeter reading go to zero when one
of the parts of the cell is removed?
(f) What common device in your home and school also
“breaks” the circuit?
Solutions used are irritants and are toxic if ingested.
Avoid contact with skin and eyes. Silver nitrate will
temporarily blacken your skin.
Dispose of solutions according to your teacher’s
instructions.
• For each cell, connect the two electrodes directly with
a wire. Record any evidence of a reaction after several
minutes, and after one or two days. Measure the
electric potential difference after several days.
(g) What is the design of a control that can be used to
compare changes with each of the three cells?
(h) State some diagnostic tests that could be done to
obtain more specific evidence for the operation of
each cell.
• Construct the three cells shown in Figure 3.
(i) Suggest a reason why all solutions were nitrates.
(a) Which design is most similar to Volta’s invention?
Compare the three cell designs.
Figure 3
Three different cell designs
salt bridge:
Ag(s)AgNO3(aq)Cu(NO3)2(aq)Cu(s)
NEL
no porous boundary:
porous cup:
Ag(s)NaNO3(aq)Cu(s)
Ag(s)AgNO3(aq)Cu(NO3)2(aq)Cu(s)
Electric Cells 721
INVESTIGATION 9.5.1
Investigating Galvanic Cells
In this investigation, you are given the opportunity to construct galvanic cells and compare your observations with the
rules and concepts you have learned.
Purpose
The purpose of this investigation is to compare the predictions
of cell potentials and electrodes of various cells with those
measured in the lab.
Question
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Procedure
(c) Based on the equipment supplied, write a specific
procedure to collect the evidence to answer the question. Be sure to include safety and disposal instructions. Have your teacher approve your procedure
before starting.
Evidence/Analysis
In cells constructed from various combinations of copper,
lead, silver, and zinc half-cells, what are the standard cell
potentials, and which is the anode and cathode in each case?
(d) Prepare a table to record your observations and the
predicted cell potentials. Include a column for
expressing the accuracy of each result (in terms of a
percentage difference).
Prediction
(e) Note and record any unexpected observations.
(a) According to redox concepts and a redox table
(Appendix C11), prepare a table of all possible combinations of half-cells and answer the question.
Experimental Design
(b) Write a brief general plan to answer the question,
including the identification of variables.
Materials
lab apron
safety glasses
voltmeter and connecting wires
U-tube with cotton plugs, porous cups, or filter-paper
strips
four 100-mL beakers or well plate
distilled water
steel wool or emery paper
Cu(s), Pb(s), Ag(s), and Zn(s) strips
1.0 mol/L CuSO4(aq), Pb(NO3)2(aq), AgNO3(aq), NaNO3(aq),
and ZnSO4(aq)
Evaluation
(f) List all sources of experimental error or uncertainty.
Considering this list, state your judgment of the
overall quality of the evidence obtained.
(g) For each cell, compare the electrodes you predicted to
be the cathodes and the anodes with your evidence.
How well do these agree?
(h) Limitations of the equipment and materials mean
that some experimental uncertainties are unavoidable. Assuming that about 5% difference is unavoidable, is the agreement between your predicted and
measured cell potentials acceptable? Justify your
answer.
(i) Is there any pattern to the accuracy of your measured
cell potentials? Suggest some reasons to explain this.
(j) Evaluate the design, procedure, and materials. Note
any flaws or possible improvements.
Some of the solutions used are toxic and/or irritants.
Avoid skin and eye contact.
722 Chapter 9
NEL
Unit 5
INVESTIGATION 9.6.1
The Corrosion of Iron
The knowledge gained from this experiment is used to help
explain corrosion and to develop methods of corrosion prevention.
Purpose
The purpose of this investigation is to test your predictions of
factors affecting the rate of corrosion of iron.
Question
What factors, chemical and electrical, affect the rate of corrosion of iron?
Prediction
(a) Based on your experience and knowledge of electrochemistry, predict some factors that may affect the
rate of corrosion of iron. Provide your reasoning.
Experimental Design
Several iron nails or pieces of iron wire are thoroughly cleaned
with steel wool, rinsed with water, and dried with a solvent
(alcohol or acetone). In part 1, the iron is exposed to different
conditions in separate test tubes. A clean piece of iron in a
dry empty test tube is the control. All test tubes are observed
immediately and after one day. In part 2, two iron-carbon
cells are connected to 9-V batteries, with the electrodes
attached oppositely for the two cells.
Materials
(b) Prepare a list of materials. Check with your teacher to
make sure your materials are suitable and available.
Procedure
(c) Using the experimental design and your list of materials, write a procedure to answer the question.
Include safety and disposal instructions. Obtain
approval from your teacher before proceeding.
NEL
Inquiry Skills
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
The solvent used is very flammable and may be
toxic.
Evidence
(d) Prepare a table to record your observations.
Analysis
(e) Compare the evidence from the factors you tested
with the control. Which factors appear to accelerate
the rate of corrosion of iron? To the extent possible,
arrange your factors from least to greatest apparent
effect.
(f) List any factors that did not seem to affect the rate of
corrosion of iron compared to the control.
Evaluation
(g) Evaluate the design, noting any flaws or improvements that could be made. Check the variables and
control to see if these are all appropriate. Suggest
changes if necessary.
(h) Evaluate the procedure and materials. Were you able
to gather sufficient and suitable evidence? Suggest
improvements where required.
(i) What are the main sources of experimental error or
uncertainty? How serious would you judge these to
be?
(j) Using your answers to (g) to (i), state your judgment
of the quality of the evidence. How confident are you
about the evidence?
(k) Assuming your evidence is of reasonable quality, state
your judgment of the prediction.
(l) Judge the reasoning you used to make your prediction. How useful was it? Why is there a need for more
empirical and theoretical knowledge about corrosion?
Electric Cells 723
Chapter 9
SUMMARY
fuel cell
Key Expectations
• Demonstrate an understanding of oxidation and
reduction in terms of the transfer of electrons or
change in oxidation number. (9.1)
galvanic cell
• Identify and describe the functioning of the
components in electric cells. (9.1, 9.2, 9.3, 9.4, 9.5)
oxidation
• Demonstrate an understanding of oxidation–reduction reactions through experiments and analysis of
these reactions. (9.1, 9.3, 9.4, 9.5)
oxidizing agent
• Write balanced chemical equations for redox reactions
using half-reaction equations. (9.1, 9.2, 9.3)
redox spontaneity rule
• Predict the spontaneity of redox reactions and cell
potentials using a table of half-cell reduction
potentials. (9.3)
reduction
• Describe examples of common cells and evaluate their
environmental and social impact. (9.4)
standard cell
• Research and assess environmental, health, and safety
issues involving electrochemistry. (9.4, 9.6)
standard reduction potential
• Describe galvanic cells in terms of oxidation and
reduction half-cells and electric potential differences.
(9.5)
• Describe the function of the hydrogen reference halfcell in assigning reduction potential values. (9.5)
• Determine oxidation and reduction half-cell reactions,
current and ion flow, electrode polarity and cell
potentials of typical galvanic cells. (9.5)
• Explain corrosion as an electrochemical process, and
describe corrosion-inhibiting techniques. (9.6)
• Use appropriate scientific and technological
vocabulary related to electrochemistry. (all sections)
half-cell
inert electrode
oxidation number
primary cell
reducing agent
reference half-cell
secondary cell
standard cell potential
volt
Key Symbols and Equations
•
•
Er°, E°
E°
cell
Er°
cathode
E r°
anode
Problems You Can Solve
• Assign oxidation numbers and use them to identify
oxidation and reduction processes, and to balance
redox reactions.
Key Terms
• Write and balance half-reaction equations to obtain a
balanced redox equation.
ampere
• Construct a table of relative strengths of oxidizing and
reducing agents.
anode
battery
cathode
cathodic protection
corrosion
• Predict spontaneity of redox reactions using a redox
table.
• Analyze components and processes occurring in electric and galvanic cells.
coulomb
electric cell
MAKE a summary
electric current
electric potential difference (voltage)
electrode
Start with “Cell” and make a flow chart that connects as
many terms, concepts, and symbols as possible from those
listed above.
electrolyte
724 Chapter 9
NEL
Chapter 9
SELF-QUIZ
Identify each of the following statements as true, false, or
incomplete. If the statement is false or incomplete, rewrite
it as a true statement.
1. Oxidation corresponds to an increase in oxidation
number.
2. Reduction is a process in which electrons are lost or
donated by an atom or ion in a redox reaction.
3. An oxidizing agent gains electrons and one of its
atoms decreases in oxidation number.
4. The strongest oxidizing agent in a galvanic cell is
above the strongest reducing agent in the redox table,
producing a cell potential that is negative.
5. The cathode of a cell is the electrode where electrons
are lost or given up by the reducing agent.
6. Only the cell potential can be experimentally measured and a reference half-cell, assigned a zero value, is
necessary to calculate reduction potentials.
7. The cell potential of a standard lead-nickel cell is
–0.39 V.
8. Corrosion is an electrochemical process in which
electrons are transferred.
9. The development of electric cars is closely related to
advances in the hydrogen fuel cell.
Identify the letter that corresponds to the best answer to
each of the following questions.
10. A reducing agent can be described as a substance that
(a) loses electrons and causes reduction.
(b) loses electrons and becomes reduced.
(c) gains electrons and causes oxidation.
(d) gains electrons and becomes reduced.
(e) gains electrons and becomes oxidized.
11. Which of the following solutions should not be
stored in a tin-plated container?
I NaNO3(aq)
II AgNO3(aq)
(a)
(b)
(c)
(d)
(e)
NEL
III SnBr2(aq)
IV Cl2(aq)
I only
II, III, IV
II and III
II and IV
III and IV
An interactive version of the quiz is available online.
GO
www.science.nelson.com
Unit 5
12. The oxidation number of the carbon atom in the carbonate ion is
(a) 8
(b) 6
(c) 4
(d) 2
(e) 0
13. In a galvanic cell,
(a) electrons are provided by the reducing agent at
the negative electrode.
(b) electrons are gained by the oxidizing agent at the
negative electrode.
(c) electrons flow through the solution from the
anode to the cathode.
(d) electrons flow through the solution from the
cathode to the anode.
(e) electrons flow through the porous barrier from
the cathode to the anode.
14. A porous boundary, or a salt bridge, is required in a
standard cell to
(a) conduct electrons from the anode to the cathode.
(b) transfer ions between the half-cells.
(c) keep the electrodes from contacting.
(d) maintain standard conditions in both half cells.
(e) stop current flow when electrodes are connected.
15. If the electrodes of a standard copper-silver cell are
connected with a wire,
(a) silver is plated at the anode.
(b) a voltmeter would show a reading of 1.14 V.
(c) electrons flow from the silver to copper
electrodes.
(d) the solution at the anode becomes darker blue.
(e) the silver ion concentration increases.
16. The standard hydrogen half-cell can be represented as
(d) Pt(s) H2(g) H
(a) Pt(s) H2(g)
(aq)
(b) Pt(s) H(aq)
(e) Pt(s) H2(g) H
(aq), OH (aq),
(c) Pt(s) H2(g), H(aq)
17. The corrosion of iron is accelerated by
(a) low humidity.
(d) nonelectrolytes.
(b) lack of oxygen.
(e) low pH.
(c) low temperature.
18. Cathodic protection of iron is effective because the
iron
(a) is forced to become the anode of a cell.
(b) is forced to become the cathode of a cell.
(c) is no longer a part of any electrochemical cell.
(d) is attached to the positive terminal of a battery.
(e) is electrically prevented from gaining electons.
Electric Cells 725
Chapter 9
REVIEW
Understanding Concepts
1. Write theoretical definitions for each of the following
words in terms of both electrons and oxidation states:
(a) oxidation
(b) reduction
(c) redox reaction
(9.1)
2. Write and label balanced half-reaction equations for
each of the following redox reactions:
2
2
(a) 2 Fe3
(aq) Ni(s) → 2 Fe(aq) Ni(aq)
(b) Br2(aq) 2 I
(aq) → 2Br(aq) I2(s)
2
2
(c) Pd(aq) Sn(aq) → Pd(s) Sn4
(9.1)
(aq)
3. Assign an oxidation number to
(a) I in I2(s)
(d) H in NH3(g)
(b) I in CaI2(s)
(e) H in AlH3(s)
(c) I in HIO(aq)
9. While working on the development of a new electrochemical cell, a research chemist places selected
Period 4 transition metal strips into aqueous solutions of their ionic compounds. She observes that the
following combinations of metal and cation react
spontaneously:
2
V(s) Mn2
(aq) → V(aq) Mn(s)
(9.1)
4. Assign oxidation numbers to all atoms/ions and indicate which atom/ion is oxidized and which is
reduced.
(a) 2 Al(s) Fe2O3(s) → 2 Fe(s) Al2O3(s)
3
(b) In(s) 3 Tl
(aq) → In(aq) 3 Tl(s)
3
2
4
(c) 2 Cr(aq) Sn(aq) → 2 Cr 2
(aq) Sn(aq)
(d) Cl2(aq) 2 I
(aq) → 2Cl(aq) I2(aq)
(e) UCl4(s) 2 Ca(s) → 2CaCl2(s) U(s)
(9.1)
5. Make a list of everything that must be balanced in a
net ionic equation representing a redox reaction. (9.2)
6. The silver(II) ion, used in chemical analysis, reacts
spontaneously with water according to the following
(unbalanced) equation:
Ag 2
(aq) H2O(l) → Ag (aq) O2(g)
(a) Assign an oxidation number to each atom or ion.
(b) Balance the equation, assuming an acid solution.
(9.2)
7. Use the oxidation number method to balance the
reaction equations for the following redox reactions
in acid solutions:
(a) Cu(s) HNO3(aq) →
Cu(NO3)2(aq) NO(g) H2O(l)
(b) MnO4
(aq) H2C2O4(aq) →
Mn2
(aq) CO2(g) H2O(l)
(c) KIO3(aq) KI(aq) HCl(aq) →
KCl(aq) I2(s) H2O(l) (9.2)
8. Write equations for the reduction and oxidation halfreactions, and balanced net redox equation.
(a) O3(g) I
(aq) → IO3(aq) O2(g) (acidic)
(b) Pt(s) NO3 (aq) Cl (aq) →
2
PtCl6(aq) NO2(g) (acidic)
726 Chapter 9
(c) CN
(aq) ClO2(aq) →
CNO
(aq) Cl(aq) (basic)
(d) PH3(g) CrO42
(aq) →
P
Cr(OH)4(aq)
4(s) (basic)
2
(e) MnO4(aq) →
2 MnO (acidic) (9.2)
Mn(aq)
4(aq)
2
V2
(aq) Ti(s) → V(s) Ti(aq)
2
Co2
(aq) Mn(s) → Co(s) Mn(aq)
(a) Use this information to develop a table of oxidizing and reducing agents for these metals and
their ions.
(b) Identify the strongest oxidizing and the strongest
reducing agent in your table.
(9.3)
10. For each of the following situations, list and classify
the entities present, write the equations for the halfreactions and the overall equation, and then predict
whether a spontaneous reaction will be observed.
(a) Nitric acid is added to aqueous potassium bromide.
(b) Aqueous potassium permanganate is used to
titrate an acidic solution of iron(II) sulfate.
(c) A strip of copper is placed in a beaker of
hydrochloric acid.
(d) An iron pipe is exposed to the wind and the rain.
(e) Aqueous cobalt(II) sulfate is mixed with a basic
solution of sodium sulfite.
(f) Aqueous solutions of chromium(II) nitrate and
tin(II) nitrate are mixed together.
(9.3)
11. Calcium metal spontaneously reacts with water.
(a) Write the half-reaction and net ionic reaction
equation for this reaction.
(b) Describe diagnostic tests that could be done to
test for the predicted products.
(9.3)
12. State and describe the three main components of a
simple electric cell.
(9.4)
13. The mercury cell is a special cell for products such as
watches and hearing aids. The equations for the halfreactions are
HgO(s) H2O(l) 2 e → Hg(l) 2 OH
(aq)
Zn(s) 2 OH
(aq) → ZnO(s) H2O(l) 2 e
NEL
Unit 5
(a) Write the equation for the overall or net cell reaction.
(b) In which direction do the electrons flow—mercury to zinc or zinc to mercury? Explain briefly.
(c) Identify the anode and cathode in this cell. (9.5)
14. What is the predicted cell potential of each of the following standard cells? Include half-cell reaction equations.
(a) permanganate-silver cell
(b) tin-zinc cell
(9.5)
15. Predict the potential of the following standard cells:
2
(a) Co(s) Co2
(aq) Zn(aq) Zn(s)
2
2
(b) Cu(s) Cu(aq) Sn(aq) Sn(s)
2
2
(c) C(s)MnO4
(9.5)
(aq), H(aq), Mn(aq) Ni(aq) Ni(s)
16. A standard nickel-cadmium cell is constructed and
tested.
(a) Predict which electrode will be the cathode and
which one will be the anode.
(b) List all entities present, write the half-cell and net
cell reaction equations, and calculate the cell
potential.
(c) Sketch and label a cell diagram for a standard
nickel-cadmium cell. Specify all substances, label
important cell components, and show the direction of electron and ion movement.
(9.5)
17. Identify and describe the components of the
standard half-cell used as a reference for reduction
potential. (9.5)
18. Given the potential of the following standard cell
with a cadmium anode, predict the reduction potential of the cerium(III) ion half-cell.
2
Ce(s) Ce3
(aq) Cd(aq) Cd(s) E ˚ = 1.94 V
(9.5)
19. What is the voltage and chemical state of a “dead” galvanic cell?
(9.5)
20. One method of reducing the corrosion of an iron
object is by applying a protective coating to its surface. Describe the chemical processes involved in protecting iron with the following metals:
(a) tin
(b) zinc
(9.6)
21. Describe the following methods of cathodic protection
and use redox theory to explain how each method can
prevent the corrosion of iron pipes and tanks.
(a) impressed current
(b) sacrificial anode
(9.6)
NEL
Applying Inquiry Skills
22. A group of students are given the following materials
and told to construct an operating galvanic cell.
a strip of copper
1.0 mol/L copper(II) sulfate
a strip of lead
1.0 mol/L lead(II) nitrate
connecting wire
1.0 mol/L sodium nitrate
a U-tube and cotton
(a) Draw a diagram to show how these materials can
be used to construct an operating galvanic cell:
(b) Write the equations for the anode half-reaction;
cathode half-reaction; overall reaction.
(c) Predict the cell potential.
(9.5)
23. Your challenge is to identify three unknown solutions
using only the materials listed: 0.25 mol/L solutions
of unknowns A, B, and C; silver, zinc, and magnesium
strips; dropper bottles of 0.25 mol/L aqueous solutions of sodium sulfate, sodium carbonate, and
sodium hydroxide; steel wool; test tubes and test-tube
rack; 50-mL beakers; 400-mL waste beaker.
(a) Assuming all possible spontaneous reactions are
rapid and that the nitrate ion is a spectator ion,
write a procedure to identify which solution is
sodium nitrate, which is lead(II) nitrate, and
which is calcium nitrate.
(b) Describe the expected results.
(9.5)
Making Connections
24. For decades, the use of electric cars has been impeded
by the lack of a powerful, lightweight, inexpensive
battery. Recent advances in the hydrogen fuel cell are
facilitating the introduction of electric cars.
(a) Describe the operation of the Ballard fuel cell.
(b) Speculate on the environmental and social impacts
of the widespread use of electric cars.
(9.4)
25. In a methane fuel cell, the chemical energy of this compound is converted into electrical energy instead of the
heat that would flow during the combustion of methane.
(a) Using only the following half-reactions and
reduction potentials, write a net reaction equation and determine the approximate potential for
the methane fuel cell:
2 7 H O 8 e → CH
CO3(aq)
2 (g)
4(g) 10 OH(aq)
Er +0.17 V
O2(g) 2 H2O(g) 4 e → 4 OH
E
(l)
r +0.40 V
(b) Discuss some of the advantages and disadvantages of this technology.
(9.5)
Electric Cells 727
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