Chemistry

tester light for each solution. After testing, rinse CLEANUP AND DISPOSAL
the tester probes with distilled water. Remove
any excess moisture with a paper towel. 9. Clean your lab station. Clean all equip-
ment, and return it to its proper place.
4. Place 18 drops of distilled water in each of six Dispose of chemicals and solutions in
wells in your 24-well plate. Add 2 drops of containers designated by your teacher. Do not
1.0 M HCl to the first well to make a total of pour any chemicals down the drain or throw any-
20 drops of solution. Mix the contents of this thing in the trash unless your teacher directs you
well thoroughly by picking the contents up in a to do so. Wash your hands thoroughly after all
pipet and returning them to the well. work is finished and before you leave the lab.

5. Repeat this procedure by taking 2 drops of the ANALYSIS AND INTERPRETATION
previous dilution and placing it in the next well
containing 18 drops of water. Return any unused 1. Resolving Discrepancies: How did the conduc-
solution in the pipet to the well from which it tivity of the 1.0 M HCl solution compare with
was taken. Mix the new solution with a new that of the 1.0 M CH3COOH solution? Why do
pipet. (You now have 1.0 M HCl in the well from you think this was so?
Procedure step 2, 0.10 M HCl in the first dilution
well, and 0.010 M HCl in the second dilution.) 2. Organizing Data: What is the H3O+ concentra-
tion of the HCl solution that most closely
6. Continue diluting in this manner until you have matched the conductivity of the acetic acid?
six successive dilutions. The [H3O+] should now
range from 1.0 M to 1.0 × 10−6 M. Write the con- 3. Drawing Conclusions: What was the H3O+
centrations in the first column of your data table. concentration of the 1.0 M CH3COOH
solution? Why?
7. Using the conductivity tester, test the cells con-
taining HCl in order from most concentrated to CONCLUSIONS
least concentrated. Note the brightness of the
tester bulb, and compare it with the brightness of 1. Applying Models: The acid ionization expression
the bulb when it was placed in the acetic acid
solution. (Retest the acetic acid well any time for for CH3COOH is the following:
comparison.) After each test, rinse the tester
probes with distilled water, and use a paper towel Ka = [H3O+][CH3COO−]
to remove any excess moisture. When the bright- [CH3COOH]
ness produced by one of the HCl solutions is
about the same as that produced by the acetic Use your answer to Analysis and Interpretation
acid, you can infer that the two solutions have
about the same hydronium ion concentration and item 3 to calculate Ka for the acetic acid solution.
that the pH of the HCl solution is equal to the
pH of the acetic acid. If the glow from the bulb is 2. Applying Models: Explain how it is possible
too faint to see, turn off the lights or build a light
shield around your conductivity tester bulb. for solutions of HCl and CH3COOH to show
the same conductivity but have different
8. Record the results of your observations by noting
which HCl concentration causes the intensity of concentrations.
the bulb to most closely match that of the bulb
when it is in acetic acid. (Hint: If the conductivity EXTENSIONS
of no single HCl concentration matches that of
the acetic acid, then estimate the value between 1. Evaluating Methods: Compare the Ka value
the two concentrations that match the best.) that you calculated with the value found on
page 606 of your text. Calculate the percent
error for this experiment.

2. Predicting Outcomes: Lactic acid
(HOOCCHOHCH3) has a Ka of 1.4 × 10−4. Predict
whether a solution of lactic acid would cause the
conductivity tester to glow brighter or dimmer than
a solution of acetic acid with the same concentra-
tion. How noticeable would the difference be?

C H E M I C A L E Q U I L I B R I U M 629

CHAPTER 19

Oxidation-Reduction
Reactions

Copper oxidizes in air to form the
green patina you see on this building.

Columbus Tower, San Francisco

Oxidation and SECTION 1
Reduction
OBJECTIVES
O xidation-reduction reactions involve a transfer of electrons. Oxida-
Assign oxidation numbers to
tion involves the loss of electrons, whereas reduction involves the gain reactant and product species.
of electrons. Reduction and oxidation half-reactions must occur simul-
taneously. These processes can be identified through the understanding Define oxidation and
and use of oxidation numbers (oxidation states). reduction.

Explain what an oxidation-
reduction reaction (redox
reaction) is.

Oxidation States

Oxidation states were defined in Chapter 7. The oxidation number
assigned to an element in a molecule is based on the distribution of elec-
trons in that molecule.The rules by which oxidation numbers are assigned
were given in Chapter 7. These rules are summarized in Table 1.

TABLE 1 Rules for Assigning Oxidation Numbers Example

Rule The oxidation number of Na(s) is 0.

1. The oxidation number of any pure element is 0. The oxidation number of
Cl− is −1.
2. The oxidation number of a monatomic ion equals the
charge on the ion. The oxidation number
of O in NO is −2.
3. The more electronegative element in a binary compound is assigned
the number equal to the charge it would have if it were an ion. The oxidation number of
F in LiF is −1.
4. The oxidation number of fluorine in a compound is always −1.
The oxidation number
5. Oxygen has an oxidation number of −2 unless it is combined with F, of O in NO2 is −2.
in which it is +1 or +2, or it is in a peroxide, in which it is −1. The oxidation number
of H in LiH is −1.
6. Hydrogen’s oxidation state in most of its compounds is +1
unless it is combined with a metal, in which case it is −1. The oxidation number
of Ca in CaCO3 is +2.
7. In compounds, Group 1 and 2 elements and aluminum have The oxidation number of
oxidation numbers of +1, +2, and +3, respectively. C in CaCO3 is +4.
The oxidation number of
8. The sum of the oxidation numbers of all atoms P in H2PO−4 is +5.
in a neutral compound is 0.

9. The sum of the oxidation numbers of all atoms in a
polyatomic ion equals the charge of the ion.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 631

FIGURE 1 The color of solutions
containing chromium compounds
changes with the oxidation state of
chromium.

+2 +3 +6 +6
Oxidation state

Chromium provides a very visual example of oxidation numbers.
Different oxidation states of chromium have dramatically different col-
ors, as can be seen in Figure 1. Solutions with the same oxidation state
show less dramatic differences. The chromium(II) chloride solution is
blue, chromium(III) chloride solution is green, potassium chromate
solution is yellow, and potassium dichromate solution is orange.

FIGURE 2 Sodium and chlorine Oxidation
react violently to form NaCl. The
synthesis of NaCl from its elements Processes in which the atoms or ions of an element experience an
illustrates the oxidation-reduction increase in oxidation state are oxidation processes. The combustion of
process. metallic sodium in an atmosphere of chlorine gas is shown in Figure 2.
The sodium ions and chloride ions produced during this strongly
632 C H A P T E R 1 9 exothermic reaction form a cubic crystal lattice in which sodium cations
are ionically bonded to chloride anions. The chemical equation for this
reaction is written as follows.

2Na(s) + Cl2(g) ⎯→ 2NaCl(s)

The formation of sodium ions illustrates an oxidation process because
each sodium atom loses an electron to become a sodium ion. The oxi-
dation state is represented by placing an oxidation number above the
symbol of the atom and the ion.

0 +1

Na ⎯→ Na+ + e−

The oxidation state of sodium has changed from 0, its elemental
state, to the +1 state of the ion (Rules 1 and 7, Table 1). A species whose
oxidation number increases is oxidized. The sodium atom is oxidized to
a sodium ion.

Reduction

Processes in which the oxidation state of an element decreases are reduc-
tion processes. Consider the behavior of chlorine in its reaction with
sodium. Each chlorine atom accepts an electron and becomes a chloride
ion. The oxidation state of chlorine decreases from 0 to −1 for the chlo-
ride ion (Rules 1 and 2, Table 1).

0 −1

Cl2 + 2e− ⎯→ 2Cl−

A species that undergoes a decrease in oxidation state is reduced. The
chlorine atom is reduced to the chloride ion.

Oxidation and Reduction as a Process FIGURE 3 Copper is oxidized and
nitrogen dioxide is produced when
Electrons are released in oxidation and acquired in reduction. There- this penny is placed in a concentrat-
fore, for oxidation to occur during a chemical reaction, reduction must ed nitric acid solution.
also occur. Furthermore, the number of electrons produced in oxidation
must equal the number of electrons acquired in reduction. This makes www.scilinks.org
sense when you recall that electrons are negatively charged and that for Topic: Redox Reactions
charge to be conserved, the number of electrons lost must equal the Code: HC61281
number of electrons gained. You learned in Chapter 8 that mass is con-
served in any chemical reaction. Therefore, the masses of the elements
that undergo oxidation and reduction and the electrons that are
exchanged are conserved.

A transfer of electrons causes changes in the oxidation states of one or
more elements. Any chemical process in which elements undergo changes
in oxidation number is an oxidation-reduction reaction. This name is
often shortened to redox reaction. An example of a redox reaction can be
seen in Figure 3, in which copper is being oxidized and NO3− from nitric
acid is being reduced. The part of the reaction involving oxidation or
reduction alone can be written as a half-reaction. The overall equation for
a redox reaction is the sum of two half-reactions. Because the number of
electrons involved is the same for oxidation and reduction, they cancel
each other out and do not appear in the overall chemical equation.
Equations for the reaction between nitric acid and copper illustrate the
relationship between half-reactions and the overall redox reaction.

0 +2 (oxidation half-reaction)

3Cu ⎯→ Cu2+ + 2e−

+5 −2 +1 +4 −2 +1 −2
2NO3− + 2e− + 4H+ ⎯→ 2NO2 + 2H2O (reduction half-reaction)

0 +5 +2 +4
3Cu + 2NO3− + 4H+ ⎯→ Cu2+ + 2NO2 + 2H2O
(redox reaction)

Notice that electrons lost in oxidation appear on the product side of
the oxidation half-reaction. Electrons are gained in reduction and

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 633

appear as reactants in the reduction half-reaction. When metallic copper
reacts in nitric acid, one copper atom is oxidized to Cu2+ as two nitro-

Photochromic Lenses gen atoms are reduced from a +5 oxidation state to a +4 oxidation state.

Photochromic eyeglasses darken when Atoms are conserved. This is illustrated by the balanced chemical equa-
exposed to ultraviolet light and become
transparent again in the absence of tion for the reaction between copper and nitric acid.
ultraviolet light. This process is the result
of oxidation-reduction reactions. Silver If none of the atoms in a reaction change oxidation state, the reaction
chloride and copper(I) chloride are
embedded in the lenses. The chloride is not a redox reaction. For example, sulfur dioxide gas, SO2, dissolves in
ions absorb photons, and the silver chlo- water to form an acidic solution containing a low concentration of sul-
ride dissociates and forms chlorine
atoms and silver atoms. The elemental furous acid, H2SO3.
silver darkens the lenses. Note that the
chlorine ions are oxidized and the silver +4 −2 +1 −2 +1 +4 −2
atoms are reduced. Then, the copper(I)
ions reduce the chlorine atoms and SO2 + H2O ⎯→ H2SO3
form copper(II) ions. In the reverse
process, the copper(II) ions oxidize The oxidation states of all elemental species remain unchanged in this
the silver atoms back to the transpar- composition reaction. Therefore, it is not a redox reaction.
ent silver ions.
When a solution of sodium chloride is added to a solution of silver
634 C H A P T E R 1 9 nitrate, an ion-exchange reaction occurs and white silver chloride
precipitates.

+1 −1 +1 +5 −2 +1 +5 −2 +1 −1
Na+ + Cl − + Ag+ + NO3− ⎯→ Na+ + NO3− + AgCl

The oxidation state of each monatomic ion remains unchanged. Again,
this reaction is not an oxidation-reduction reaction.

Redox Reactions and Covalent Bonds

Both the synthesis of NaCl from its elements and the reaction between
copper and nitric acid involve ionic bonding. Substances with covalent
bonds also undergo redox reactions. An oxidation number, unlike an
ionic charge, has no physical meaning. That is, the oxidation number
assigned to a particular atom is based on its electronegativity relative to
the other atoms to which it is bonded in a given molecule; it is not based
on any real charge on the atom. For example, an ionic charge of 1−
results from the complete gain of one electron by an atom or other neu-
tral species, whereas an oxidation state of −1 means an increased attrac-
tion for a bonding electron. A change in oxidation number does not
require a change in actual charge.

When hydrogen burns in chlorine, a covalent bond forms from the
sharing of two electrons. The two bonding electrons in the hydrogen
chloride molecule are not shared equally. Rather, the pair of electrons
is more strongly attracted to the chlorine atom because of its higher
electronegativity.

00 +1 −1

H2 + Cl2 ⎯→ 2HCl

As specified by Rule 3 in Table 1, chlorine in HCl is assigned an oxida-
tion number of −1. Thus, the oxidation number for the chlorine atoms
changes from 0, its oxidation number in the elemental state, to −1; chlo-
rine atoms are reduced. As specified by Rule 1, the oxidation number of
each hydrogen atom in the hydrogen molecule is 0. As specified by Rule
6, the oxidation state of the hydrogen atom in the HCl molecule is +1; the

hydrogen atom is oxidized. Neither atom has totally lost or totally gained
any electrons. Hydrogen has donated a share of its bonding electron to
the chlorine but has not completely transferred that electron. The assign-
ment of oxidation numbers allows an approximation of the electron dis-
tribution of a molecule.An element can have different oxidation numbers
in different compounds. This difference in oxidation numbers can reveal
the difference in electron distribution of the compounds.

Reactants and products in redox reactions are not limited to monatom-
ic ions and uncombined elements. Elements in molecular compounds or
polyatomic ions can also be oxidized and reduced if they have more than
one nonzero oxidation state.An example of this is provided in the reaction
between the copper penny and nitric acid in which the nitrate ion, NO3−, is
converted to nitrogen dioxide, NO2. Nitrogen is reduced in this reaction.
Usually, we refer to the oxidation or reduction of the entire molecule or
ion. Instead of saying that the nitrogen atom is reduced, we say the nitrate
ion is reduced to nitrogen dioxide.

+5 +4

. . . + NO3− ⎯→ NO2 + . . .

SECTION REVIEW c. NaOH(aq ) + HCl(aq ) ⎯→ NaCl(aq ) + H2O(l )

1. How are oxidation numbers assigned? d. H2(g ) + Cl2(g ) ⎯→ 2HCl(g )
2. Label each of the following half-reactions as either
e. SO3(g ) + H2O(l ) ⎯→ H2SO4(aq )
an oxidation or a reduction half-reaction:
4. For each redox equation identified in the previous
0 −1 question, determine which element is oxidized and
which is reduced.
a. Br2 + 2e − ⎯→ 2 Br −
Critical Thinking
0 +1
5. ANALYZING INFORMATION Use the following
b. Na ⎯→ Na+ + e − equations for the redox reaction between Al3+ and

−1 0 Na to answer the questions below.

c. 2 Cl − ⎯→ Cl2 + 2e − 0 +1 (oxidation)

0 −1 3 Na ⎯→ 3 Na+ + 3e −

d. Cl2 + 2e − ⎯→ 2 Cl − +3 0 (reduction)

+1 0 Al3+ + 3e − ⎯→ Al

e. Na+ + e − ⎯→ Na 0 +3 +1 0
3 Na + Al3+ ⎯→ 3 Na+ + Al
0 +2 (redox reaction)

f. Fe ⎯→ Fe2+ + 2e − a. Explain how this reaction illustrates that charge
is conserved in a redox reaction.
+2 0
b. Explain how this reaction illustrates that mass
g. Cu2+ + 2e − ⎯→ Cu is conserved in a redox reaction.

+3 +2 c. Explain why electrons do not appear as react-
ants or products in the combined equation.
h. Fe3+ + e − ⎯→ Fe2+

3. Which of the following equations represent redox
reactions?
a. 2KNO3(s ) ⎯→ 2KNO2(s ) + O2(g )
b. H2(g ) + CuO(s ) ⎯→ Cu(s ) + H2O(l )

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 635

Skunk-Spray Remedy

Have you ever given your pet a Skunk spray gets its odor from chemi- sulfur-containing compound that is
tomato juice bath to get rid of the cals called mercaptans. in skunk spray.
smell of skunk spray on its coat?
Chemistry has a much better way of Because the odor in a skunk’s spray RSH + 3H2O2 + NaHCO3 ⎯→
conquering skunk spray. also comes from compounds con- RSO3Na + 4H2O + CO2
taining sulfur in a low oxidation
Paul Krebaum, the inventor of a state, Mr. Krebaum thought his The pet should be thoroughly
new “deskunking” formula, says solution might also work on this washed with the mixture, and care
that while working as a materials age-old problem. He mixed up a should be taken to avoid the eyes. If
engineer, he constantly had to deal milder version to try out on the the mixture is left on for a few min-
with the less-than-pleasant smell of pet: 1 qt of a 3% hydrogen peroxide utes—long enough for the reaction
the hydrogen sulfide gas that was solution, 1/4 cup of baking soda, and to occur—and then rinsed away with
released from one of his experi- 1 tsp of liquid soap. His friend tried it tap water, the smell will disappear.
ments. Venting off the gas only par- out, and the result was one wet and
tially solved the problem. A better unhappy—but much less smelly— The formula does not bleach or
solution would be to eliminate the pet. cause any other negative side
smell entirely. effects. Mr. Krebaum does have one
Mr. Krebaum says that the hydro- warning: Mix the formula just before
Mr. Krebaum rifled through his old gen peroxide in the remedy actually using it, because the mixture breaks
chemistry books and found that oxidizes the compounds, while the down quickly. The reaction releases
hydrogen peroxide could oxidize baking soda reduces the acidity of oxygen, so the formula should not
these sulfur-containing compounds the mixture and the soap helps to be kept in a sealed container.
to much less smelly components. He wash out the greasy skunk spray. Pressure will build up, and the lid
immediately whipped up a hydrogen This reaction can be seen in the fol- could eventually blow off. For this
peroxide mixture, and it worked like lowing equation. The symbol R repre- reason, bottles of “Krebaum’s
a charm. sents all the other elements in the Skunkinator” will not be appearing
on drugstore shelves any time soon.
The equation below shows that
hydrogen sulfide reacts with H2O2 to Questions
form sulfate compounds that do not
have a bad odor. 1. How did Paul Krebaum’s
research into the properties of
2NaOH + 4H2O2 + H2S ⎯→ H2S result in a benefit to dog
Na2SO4 + 6H2O owners?

“The receptors that are in your 2. What are some possible packag-
nose are sensitive to sulfur in its low ing designs that Paul Krebaum
oxidation state,” says Mr. Krebaum. could have used if he had
“However, they are not sensitive to wanted to sell his formula?
sulfur in its high oxidation state.”

Some time later, a friend of
Mr. Krebaum’s complained to him
that a skunk had sprayed his pet.

636 C H A P T E R 1 9

Balancing Redox SECTION 2
Equations
OBJECTIVES
E quations for simple redox reactions can be balanced by inspection,
Explain what must be con-
which you learned to do in Chapter 8. Most redox equations, however, served in redox equations.
require more systematic methods. The equation-balancing process
requires the use of oxidation numbers. In a balanced equation, both Balance redox equations
charge and mass are conserved. Although oxidation and reduction half- by using the half-reaction
reactions occur together, their reaction equations are balanced sepa- method.
rately and then combined to give the balanced redox-reaction equation.

Half-Reaction Method

The half-reaction method, or ion-electron method, for balancing redox CROSS-DISCIPLINARY
equations consists of seven steps. Oxidation numbers are assigned to all
atoms and polyatomic ions to determine which species are part of the Go to go.hrw.com for a full-length
redox process. The oxidation and reduction equations are balanced sep- article on redox reactions in photo-
arately for mass and charge. They are then added together to produce a synthesis.
complete balanced equation. These seven steps are applied to balance
the reaction of hydrogen sulfide and nitric acid. Sulfuric acid, nitrogen Keyword: HC6OXRX
dioxide, and water are the products of the reaction.

1. Write the formula equation if it is not given in the problem. Then
write the ionic equation.

Formula equation: H2S + HNO3 ⎯→ H2SO4 + NO2 + H2O
Ionic equation: H2S + H+ + NO3− ⎯→ 2H+ + SO42− + NO2 + H2O

2. Assign oxidation numbers. Delete substances containing only el-
ements that do not change oxidation state.

+1 −2 +1 +5 −2 +1 +6 −2 +4 −2 +1 −2
H2S + H+ + NO3− ⎯→ 2H+ + SO42− + NO2 + H2O

The sulfur changes oxidation state from −2 to +6. The nitrogen
changes oxidation state from +5 to +4. The other substances are
deleted.

+1 −2 +5 −2 +6 −2 +4 −2
− SO42− + NO2
H2S + NO 3 ⎯→

The remaining species are used in step 3.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 637

3. Write the half-reaction for oxidation. In this example, the sulfur is
being oxidized.

−2 +6

H2S ⎯→ SO24−

• Balance the atoms. To balance the oxygen in this half-reaction, H2O
must be added to the left side. This gives 10 extra hydrogen atoms on

that side of the equation. Therefore, 10 hydrogen ions are added to
the right side. In basic solution, OH− ions and water may be used to

balance atoms.

−2 +6 2− 10H+
4
H2S + 4H2O ⎯→ SO +

• Balance the charge. Electrons are added to the side having the
greater positive net charge. The left side of the equation has no net
charge; the right side has a net charge of 8+. For the charges to bal-
ance, each side must have the same net charge. Therefore, 8 electrons
are added to the product side so that it has no charge and balances
with the reactant side of the equation. Notice that the oxidation of
sulfur from a state of −2 to +6 indicates a loss of 8 electrons.

−2 +6

H2S + 4H2O ⎯→ SO42− + 10H+ + 8e−

The oxidation half-reaction is now balanced.

4. Write the half-reaction for reduction. In this example, nitrogen is
being reduced from a +5 state to a +4 state.

+5 +4

NO3− ⎯→ NO2

• Balance the atoms. H2O must be added to the product side of the
reaction to balance the oxygen atoms. Therefore, two hydrogen ions
must be added to the reactant side to balance the hydrogen atoms.

+5 +4

NO3− + 2H+ ⎯→ NO2 + H2O

• Balance the charge. Electrons are added to the side having the
greater positive net charge. The left side of the equation has a net
charge of 1+. Therefore, 1 electron must be added to this side to bal-
ance the charge.

+5 +4

NO3− + 2H+ + e− ⎯→ NO2 + H2O

The reduction half-reaction is now balanced.

5. Conserve charge by adjusting the coefficients in front of the elec-
trons so that the number lost in oxidation equals the number gained
in reduction. Write the ratio of the number of electrons lost to the
number of electrons gained.

e− lost in oxidation 8
e− gained in reduction = 1

638 C H A P T E R 1 9

This ratio is already in its lowest terms. If it were not, it would need KMnO4
to be reduced. Multiply the oxidation half-reaction by 1 (it remains
unchanged) and the reduction half-reaction by 8. The number of FeSO4
electrons lost now equals the number of electrons gained. in H2SO4

΂ ΃−2 +6 FIGURE 4 As a KMnO4 solution
is titrated into an acidic solution of
1 H2S + 4H2O ⎯→ SO42− + 10H+ + 8e− FeSO4, deep purple MnO−4 ions are
reduced to colorless Mn2+ ions.
΂ ΃+5 +4 When all Fe2+ ions are oxidized,
MnO−4 ions are no longer reduced
8 NO−3 + 2H+ + e− ⎯→ NO2 + H2O to colorless Mn2+ ions. Thus, the first
faint appearance of the MnO−4 color
6. Combine the half-reactions, and cancel out anything common to indicates the end point of the
both sides of the equation. titration.

−2 +6

H2S + 4H2O ⎯→ SO42− + 10H+ + 8e−

+5 +4

8 NO3− + 16H+ + 8e− ⎯→ 8 NO2 + 8H2O

+5 − 166H+ 8e− −2
3
8NO + + + H2S + 4H2O ⎯→

+4 + 4 +6 + 10H+ + 8e−
8H2O
8NO2 + SO42−

Each side of the above equation has 10H+, 8e−, and 4H2O. These can-
cel each other out and do not appear in the balanced equation.

+5 −2 +4 +6

8NO−3 + H2S + 6H+ ⎯→ 8NO2 + 4H2O + SO42−

7. Combine ions to form the compounds shown in the original formu-
la equation. Check to ensure that all other ions balance. The NO3−
ion appeared as nitric acid in the original equation. There are only
6 hydrogen ions to pair with the 8 nitrate ions. Therefore, 2 hydrogen
ions must be added to complete this formula. If 2 hydrogen ions are
added to the left side of the equation, 2 hydrogen ions must also be
added to the right side of the equation.

8HNO3 + H2S ⎯→ 8NO2 + 4H2O + SO42− + 2H+

The sulfate ion appeared as sulfuric acid in the original equation. The
hydrogen ions added to the right side are used to complete the form-
ula for sulfuric acid.

8HNO3 + H2S ⎯→ 8NO2 + 4H2O + H2SO4

A final check must be made to ensure that all elements are correctly
balanced.

SAMPLE PROBLEM A For more help, go to the Math Tutor at the end of this chapter.

Write a balanced equation for the reaction shown in Figure 4. A deep purple solution of potassium per-
manganate is titrated with a colorless solution of iron(II) sulfate and sulfuric acid. The products are
iron(III) sulfate, manganese(II) sulfate, potassium sulfate, and water—all of which are colorless.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 639

SOLUTION 1. Write the formula equation if it is not given in the problem. Then write the ionic equation.

KMnO4 + FeSO4 + H2SO4 ⎯→ Fe2(SO4)3 + MnSO4 + K2SO4 + H2O
K+ + MnO4− + Fe2+ + SO24− + 2H+ + SO42− ⎯→

2Fe3+ + 3SO24− + Mn2+ + SO42− + 2K+ + SO42− + H2O

2. Assign oxidation numbers to each element and ion. Delete substances containing an ele-
ment that does not change oxidation state.

+1 +7 −2 +2 +6 −2 +1 +6 −2
K+ + MnO4− + Fe2+ + SO24− + 2H+ + SO42− ⎯→

+3 +6 −2 +2 +6 −2 +1 +6 −2 +1 −2
2Fe3+ + 3SO24− + Mn2+ + SO24− + 2K+ + SO42− + H2O

Only ions or molecules whose oxidation numbers change are retained.

+7 −2 +2 +3 +2
MnO−4 + Fe2+ ⎯→ Fe3+ + Mn2+

3. Write the half-reaction for oxidation. The iron shows the increase in oxidation number.
Therefore, it is oxidized.

+2 +3

Fe2+ ⎯→ Fe3+

• Balance the mass. The mass is already balanced.

• Balance the charge.

+2 +3

Fe2+ ⎯→ Fe3+ + e−

4. Write the half-reaction for reduction. Manganese shows a change in oxidation number
from +7 to +2. It is reduced.

+7 +2

MnO−4 ⎯→ Mn2+

• Balance the mass. Water and hydrogen ions must be added to balance the oxygen atoms
in the permanganate ion.

+7 +2

MnO4− + 8H+ ⎯→ Mn2+ + 4H2O

• Balance the charge.

+7 +2

MnO4− + 8H+ + 5e− ⎯→ Mn2+ + 4H2O

5. Adjust the coefficients to conserve charge.
e− lost in oxidation 1

e− gained in reduction = 5

5(Fe2+ ⎯→ Fe3+ + e−)
1(MnO4− + 8H+ + 5e− ⎯→ Mn2+ + 4H2O)

640 C H A P T E R 1 9

6. Combine the half-reactions and cancel.

5Fe2+ ⎯→ 5Fe3+ + 5e−
MnO−4 + 8H+ + 5e− ⎯→ Mn2+ + 4H2O
MnO4− + 5Fe2+ + 8H+ + 5e− ⎯→ Mn2+ + 5Fe3+ + 4H2O + 5e−

7. Combine ions to form compounds from the original equation. The iron(III) product
appears in the original equation as Fe2(SO4)3. Every iron(III) sulfate molecule requires
two iron ions. Therefore, the entire equation must be multiplied by 2 to provide an even
number of iron ions.

2(5Fe2+ + MnO−4 + 8H+ ⎯→ 5Fe3+ + Mn2+ + 4H2O)
10Fe2+ + 2MnO−4 + 16H+ ⎯→ 10Fe3+ + 2Mn2+ + 8H2O

The iron(II), iron(III), manganese(II), and 2 hydrogen ions in the original equation are
paired with sulfate ions. Iron(II) sulfate requires 10 sulfate ions, and sulfuric acid
requires 8 sulfate ions. To balance the equation, 18 sulfate ions must be added to each
side. On the product side, 15 of these ions form iron(III) sulfate, and 2 of them form man-
ganese(II) sulfate. That leaves 1 sulfate ion unaccounted for. The permanganate ion
requires the addition of 2 potassium ions to each side. These 2 potassium ions form
potassium sulfate on the product side of the reaction.

10FeSO4 + 2KMnO4 + 8H2SO4 ⎯→ 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O
Final inspection shows that atoms and charges are balanced.

PRACTICE Answers in Appendix E

1. Copper reacts with hot, concentrated sulfuric acid to form cop- Go to go.hrw.com for
per(II) sulfate, sulfur dioxide, and water. Write and balance the more practice problems
equation for this reaction. that ask you to balance
redox equations.
2. Write and balance the equation for the reaction between nitric
acid and potassium iodide. The products are potassium nitrate, Keyword: HC6OXRX
iodine, nitrogen monoxide, and water.

SECTION REVIEW Critical Thinking

1. What two quantities are conserved in redox 4. RELATING IDEAS When heated, elemental phos-
equations? phorus, P4, produces phosphine, PH3, and phos-
phoric acid, H3PO4. How many grams of phosphine
2. Why do we add H+ and H2O to some half- are produced if 56 g P4 have reacted?
reactions and OH− and H2O to others?

3. Balance the following redox reaction:
Na2SnO2 + Bi(OH)3 ⎯→ Bi + Na2SnO3 + H2O

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 641

SECTION 3 Oxidizing and
Reducing Agents
OBJECTIVES
A reducing agent is a substance that has the potential to cause another
Relate chemical activity to
oxidizing and reducing substance to be reduced. Reducing agents lose electrons; they attain a
strength. more positive oxidation state during an oxidation-reduction reaction.
Explain the concept of Therefore, the reducing agent is the oxidized substance.
disproportionation.
An oxidizing agent is a substance that has the potential to cause an-
642 C H A P T E R 1 9 other substance to be oxidized. Oxidizing agents gain electrons and
attain a more negative oxidation state during an oxidation-reduction
reaction. The oxidizing agent is the reduced substance. Table 2 helps
clarify the terms describing the oxidation-reduction process.

Strengths of Oxidizing
and Reducing Agents

Different substances can be compared and rated by their relative
potential as reducing and oxidizing agents. For example, the order of the
elements in the activity series (see Table 3 on page 286) is related to each
element’s tendency to lose electrons. Elements in this series lose elec-
trons to the positively charged ions of any element below them in the
series. The more active an element is, the greater its tendency to lose
electrons and the better a reducing agent it is. The greater the distance
is between two elements in the list, the more likely it is that a redox
reaction will take place between them.

These elements and some other familiar substances are arranged in
Table 3 according to their activity as oxidizing and reducing agents. The
fluorine atom is the most highly electronegative atom. It is also the most

TABLE 2 Oxidation-Reduction Terminology

Term Change in Change in
oxidation number electron population

Oxidation in a positive direction loss of electrons

Reduction in a negative direction gain of electrons

Oxidizing agent in a negative direction gains electrons

Reducing agent in a positive direction loses electrons

active oxidizing agent. Because of its strong attraction for its own elec- TABLE 3 Relative Strength
trons, the fluoride ion is the weakest reducing agent. The negative ion of of Oxidizing and Reducing
a strong oxidizing agent is a weak reducing agent. Agents

The positive ion of a strong reducing agent is a weak oxidizing agent. Reducing Oxidizing
As shown in Table 3, Li atoms are strong reducing agents because Li is agents agents
a very active metal. When Li atoms oxidize, they produce Li+ ions, which
are unlikely to reacquire electrons, so Li+ ions are weak oxidizing ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯In⎯cr⎯ea⎯sin⎯g s⎯tr⎯en⎯gt⎯h⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Li Li+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯I⎯nc⎯re⎯as⎯ing⎯st⎯re⎯ng⎯th⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
agents.
K K+
The left column of each pair also shows the relative abilities of met-
als listed in the table to displace other metals from their compounds. Ca Ca2+
Zinc, for example, appears above copper. Thus, zinc is the more active
reducing agent, and it displaces copper ions from solutions of copper Na Na+
compounds, as illustrated in Figure 5. The copper(II) ion, on the other
hand, is a more active oxidizing agent than the zinc ion. Mg Mg2+

Nonmetals and some important ions also are included in the series in Al Al3+
Table 3. Any reducing agent is oxidized by the oxidizing agents below
it. Observe that F2 displaces Cl−, Br−, and I− ions from their solutions. Zn Zn2+
Cl2 displaces Br− and I− ions, and Br2 displaces I− ions. The equation for
the displacement of Br− by Cl2 is as follows. Cr Cr3+

Fe Fe2+

Ni Ni2+

Sn Sn2+

Cl2 + 2Br−(aq) ⎯→ 2Cl−(aq) + Br2 Pb Pb2+

−1 0 H2 H3O+

2 Br− ⎯→ Br2 + 2e− (oxidation) H2S S
(reduction) Cu Cu2+
0 −1
I− I2
Cl2 + 2e− ⎯→ 2 Cl − MnO−4
MnO24− Fe3+
In every redox reaction, there is one reducing agent and one oxi- Fe2+
dizing agent. In the preceding example, Br− is the reducing agent and
Hg Hg22+
Cl2 is the oxidizing agent. Ag Ag+

NO2− NO −
Br− 3

Mn2+ Br2

MnO2

SO2 H2SO4 (conc.)
Cr3+ Cr2O27−
Cl −
Mn2+ Cl2
F− MnO4−

F2

FIGURE 5 Zinc displaces copper ions from a copper(II) sulfate solution.
Metallic copper precipitates.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 643

Wear safety goggles and an apron.

Redox Reactions Discussion Materials
1. Write balanced equations
Procedure • aluminum foil
showing what happened in • beaker, 250 mL
Record all of your results in a each of the reactions. • 1 M copper(II) chloride solu-
data table.
2. Write a conclusion for the two tion, CuCl2
1. Put 10 mL of hydrogen perox- • 3% hydrogen peroxide
experiments. • manganese dioxide
ide in a test tube, and add a Wooden • metric ruler
small amount of manganese splint • scissors
dioxide (equal to the size of • test-tube clamp
about half a pea). What is the Clamp • test tube, 16 × 150 mm
result? • wooden splint
H2O2
2. Insert a glowing wooden splint

into the test tube (see dia-
gram). What is the result? If
oxygen is produced, a glowing
wooden splint inserted into the
test tube will glow brighter.

3. Fill the 250 mL beaker halfway

with the copper(II) chloride
solution.

4. Cut foil into 2 cm × 12 cm

strips.

5. Add the aluminum strips to the

copper(II) chloride solution.
Use a glass rod to stir the mix-
ture, and observe for 12 to
15 minutes. What is the result?

Disproportionation

Some substances can be both reduced and oxidized easily. For example,
peroxide ions, O22−, have a relatively unstable covalent bond between
the two oxygen atoms. The electron-dot formula is written as follows.

O O 2−

Each oxygen atom has an oxidation number of −1. The peroxide ion
structure represents an intermediate oxidation state between O2 and
O2−. Therefore, the peroxide ion is highly reactive.

Hydrogen peroxide, H2O2, is a covalent compound. It decomposes
into water and molecular oxygen, as shown in the equation below.

−1 −2 0

2H2O2 ⎯→ 2H2O + O2

644 C H A P T E R 1 9

FIGURE 6 A bombardier beetle
can repel large predators such as
frogs with a chemical defense mech-
anism that uses the disproportiona-
tion of hydrogen peroxide.

Notice that in this reaction, hydrogen peroxide is both oxidized and
reduced. Oxygen atoms that become part of gaseous oxygen molecules
are oxidized. The oxidation number of these oxygen atoms increases
from −1 to 0. Oxygen atoms that become part of water are reduced. The
oxidation number of these oxygen atoms decreases from −1 to −2. A
process in which a substance acts as both an oxidizing agent and a reduc-
ing agent is called disproportionation. A substance that undergoes dis-
proportionation is both self-oxidizing and self-reducing.

The bombardier beetle defends itself by spraying its enemies with an
unpleasant hot chemical mixture, as shown in Figure 6. The catalyzed
disproportionation of hydrogen peroxide produces hot oxygen gas. This
gas gives the insect an ability to eject irritating chemicals from its
abdomen with explosive force.

SECTION REVIEW 2. The photo on the left depicts two redox reactions.
Both nails are in a sulfuric acid solution. Answer
1. Describe the chemical activity of the alkali metals the following questions:
and of the halogens on the basis of oxidizing and a. When zinc is wrapped around an iron nail,
reducing strength. is the iron or zinc oxidized?
b. When copper is wrapped around an iron
nail, is the iron or copper oxidized?

3. Would Cl2 be reduced by I−? Explain.

4. Which is the stronger oxidizing agent in each of
the following pairs: Cu2+ or Al3+, I2 or S, F2 or Li+?

5. What is meant by disproportionation?

Critical Thinking

6. ORGANIZING IDEAS In general, where in the
periodic table are the elements found that in
elemental form are the strongest oxidizing
agents? Explain.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 645

CHAPTER HIGHLIGHTS

Oxidation and Reduction • Oxidation numbers are assigned by the set of rules listed in

Vocabulary Table 1. Oxidation numbers are based on the distribution of
oxidation
oxidized electrons in a molecule.
reduction
reduced • Oxidation-reduction reactions consist of two half-reactions
oxidation-reduction reaction
redox reaction that must occur simultaneously.
half-reaction
• Oxidation-reduction reactions are identified by examining the

changes in the oxidation numbers of atoms in the reactants

and products.

• Oxidation involves the loss of electrons, and reduction involves

the gain of electrons.

• A species whose oxidation number increases is oxidized. A

species whose oxidation number decreases is reduced.

Balancing Redox Equations

• Charge and mass are conserved in a balanced redox equation.
• In the half-reaction method for balancing equations, the atoms

and charge of oxidation and reduction equations are balanced

separately. Then, they are combined to give a complete balanced

equation.

• In a half-reaction, the charge on the reactant side must equal

the charge on the product side, but these charges do not need

to be zero.

• For the half-reaction method, the atoms in each half-reaction

are balanced by adding H+ ions and H2O molecules in acidic
solutions. If the solution is basic, OH− ions and H2O molecules
are added to balance the atoms in each half-reaction.

• The number of electrons lost in the oxidation half-reaction

must equal the number of electrons gained in the reduction

half-reaction. The two half-reactions must be multiplied by

appropriate factors to ensure that the same number of elec-

trons are transferred.

Oxidizing and Reducing Agents

Vocabulary • The substance that is reduced in redox reactions is the oxidiz-
reducing agent
oxidizing agent ing agent because it acquires electrons from the substance that
disproportionation
is oxidized.

• The substance that is oxidized in a redox reaction is the reducing

agent because it supplies the electrons to the substance that is

reduced.

• Strong reducing agents are substances that easily give up elec-

trons.

• Disproportionation is a process in which a substance is both an

oxidizing agent and a reducing agent.

646 C H A P T E R 1 9

CHAPTER REVIEW

For more practice, go to the Problem Bank in Appendix D. 7. Determine the oxidation number of each atom

Oxidation and Reduction indicated in the following:

SECTION 1 REVIEW a. H2 f. HNO3
b. H2O g. H2SO4
1. a. Distinguish between the processes of oxida- c. Al h. Ca(OH)2
i. Fe(NO3)2
tion and reduction. d. MgO j. O2

b. Write an equation to illustrate each process. e. Al2S3

2. Which of the following are redox reactions?

a. 2Na + Cl2 ⎯→ 2NaCl

b. C + O2 ⎯→ CO2 Balancing Redox Equations

c. 2H2O ⎯→ 2H2 + O2 SECTION 2 REVIEW

d. NaCl + AgNO3 ⎯→ AgCl + NaNO3 8. Label the following half-reactions as either
+ − reduction or oxidation half-reactions.
e. NH3 + HCl ⎯→ NH 4 + Cl a. H2S ⎯→ S + 2e− + 2H+
b. SO2 + 4e− + 2H2O ⎯→ S + 4OH−
f. 2KClO3 ⎯→ 2KCl + 3O2 c. ClO3− + 6H+ + 6e− ⎯→ Cl− + 3H2O
d. Mn(CN)46− ⎯→ Mn(CN)63− + e−
g. H2 + Cl2 ⎯→ 2HCl
9. What are the oxidation states of the elements
h. H2SO4 + 2KOH ⎯→ K2SO4 + 2H2O that changed oxidation states in the half-
reactions in the above question?
i. Zn + CuSO4 ⎯→ ZnSO4 + Cu
10. Balance the equation for the following reaction
3. For each oxidation-reduction reaction in the in a basic solution. Give balanced equations for
both half-reactions and the balanced equation
previous question, identify what is oxidized and for the overall reaction.
KMnO4 + NaIO3 ⎯→ MnO2 + NaIO4
what is reduced.

PRACTICE PROBLEMS

4. Each of the following atom/ion pairs undergoes

the oxidation number change indicated below.

For each pair, determine whether oxidation or

reduction has occurred, and then write the elec-

tronic equation indicating the corresponding

number of electrons lost or gained. PRACTICE PROBLEMS

a. K ⎯→ K+ e. H2 ⎯→ H+ 11. For each requested step, use the half-reaction
b. S ⎯→ S2− f. O2 ⎯→ O2− method to balance the oxidation-reduction
c. Mg ⎯→ Mg2+ g. Fe3+ ⎯→ Fe2+ equation below. (Hint: See Sample Problem A.)
K + H2O ⎯→ KOH + H2
d. F − ⎯→ F2 h. Mn2+ ⎯→ MnO4– a. Write the ionic equation, and assign oxida-
tion numbers to all atoms to determine what
5. Identify the following reactions as redox or is oxidized and what is reduced.
b. Write the equation for the reduction, and
nonredox: balance it for both atoms and charge.
c. Write the equation for the oxidation, and
a. 2NH4Cl(aq) + Ca(OH)2(aq) ⎯→ balance it for both atoms and charge.
2NH3(aq) + 2H2O(l) + CaCl2(aq) d. Multiply the coefficients of the oxidation and
reduction equations so that the number of
b. 2HNO3(aq) + 3H2S(g) ⎯→ electrons lost equals the number of electrons
2NO(g) + 4H2O(l) + 3S(s) gained. Add the two equations.
e. Add species as necessary to balance the
c. [Be(H2O)4]2+(aq) + H2O(l) ⎯→ overall formula equation.
H3O+(aq) + [Be(H2O)3OH]+(aq)

6. Arrange the following in order of increasing

oxidation number of the xenon atom: CsXeF8,
Xe, XeF2, XeOF2, XeO3, and XeF.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 647

CHAPTER REVIEW

12. Use the method in the previous problem to bal- 18. Arrange the following in order of decreasing
ance each of the reactions below.
a. HI + HNO2 ⎯→ NO + I2 + H2O oxidation number of the nitrogen atom: N2,
b. FeCl3 + H2S ⎯→ FeCl2 + HCl + S NH3, N2O4, N2O, N2H4, and NO3−.
19. Balance the following redox equations:
13. Balance the equation for the reaction in
which hot, concentrated sulfuric acid reacts with a. SbCl5 + KI ⎯→ KCl + I2 + SbCl3
zinc to form zinc sulfate, hydrogen sulfide, b. Ca(OH)2 + NaOH + ClO2 + C ⎯→
and water.
NaClO2 + CaCO3 + H2O
Oxidizing and Reducing 20. Balance the following equations in basic
Agents
solution:
SECTION 3 REVIEW
a. PbO2 + KCl ⎯→ KClO + KPb(OH)3
14. a. Identify the most active reducing agent b. KMnO4 + KIO3 ⎯→ MnO2 + KIO4
among all common elements. c. K2MnO4 ⎯→ MnO2 + KMnO4
21. Balance the following equations in acidic
b. Why are all of the elements in its group in the
periodic table very active reducing agents? solution:
a. MnO−4 + Cl− ⎯→ Mn2+ + HClO
c. Identify the most active oxidizing agent b. NO3− + I2 ⎯→ IO3− + NO2
among the common elements. c. NO−2 ⎯→ NO + NO−3

15. Use Table 3 to identify the strongest and weak- CRITICAL THINKING
est reducing agents among the substances listed
within each of the following groupings: 22. Interpreting Graphics Given the activity table
a. Ca, Ag, Sn, Cl− below, determine whether a reaction will occur or
b. Fe, Hg, Al, Br− not. If the reaction will occur, give the products.
c. F−, Pb, Mn2+, Na
Reducing L+
16. Use Table 3 to respond to each of the following: Agents M+
a. Would Al be oxidized by Ni2+? P+
b. Would Cu be oxidized by Ag+? L T+
c. Would Pb be oxidized by Na+? M
d. Would F2 be reduced by Cl−? P
e. Would Br2 be reduced by Cl−? T

MIXED REVIEW Oxidizing
Agents
17. Identify the following reactions as redox or
nonredox: a. L and M+
a. Mg(s) + ZnCl2(aq) ⎯→ Zn(s) + MgCl2(aq) b. P and M+
b. 2H2(g) + OF2(g) ⎯→ H2O(g) + 2HF(g) c. P and T+
c. 2KI(aq) + Pb(NO3)2(aq) ⎯→ 23. Drawing Conclusions A substance has an ele-
PbI2(s) + 2KNO3(aq) ment in one of its highest possible oxidation
d. CaO(s) + H2O(l) ⎯→ Ca(OH)2(aq) states. Is this substance more likely to be an oxi-
e. 3CuCl2(aq) + 2(NH4)3PO4(aq) ⎯→ dizing agent or a reducing agent? Explain your
6NH4Cl(aq) + Cu3(PO4)2(s) reasoning.
f. CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(g)

648 C H A P T E R 1 9

CHAPTER REVIEW

24. Drawing Conclusions Use Table 3 to decide if RESEARCH & WRITING
a redox reaction would occur between the two
species, and if so, write the balanced equation. 28. Oxidizing agents are used in the cleaning indus-
Explain your reasoning. try. Research three different oxidizing agents
a. Cl2 and Br2 used in this area, and write a report on the advan-
b. Sn2+ and Zn tages and disadvantages of these compounds.

25. Drawing Conclusions An element that dispro- 29. Oxidizing and reducing agents play important
portionates must have at least how many differ- roles in biological systems. Research the role of
ent oxidation states? Explain your reasoning. one of these agents in a biological process.
Write a report describing the process and the
USING THE HANDBOOK role of oxidation and reduction.

26. Several reactions of aluminum are shown in the ALTERNATIVE ASSESSMENT
common reactions section for Group 13 of the
Elements Handbook. Use these reactions to 30. Boilers are used to convert water to steam in
answer the following: electric power plants. Dissolved oxygen in the
a. Which of the five reactions shown are water promotes corrosion of the steel used in
oxidation-reduction reactions? How do boiler parts. Explain how dissolved oxygen is
you know? removed from the water in boilers.
b. For each redox reaction you listed in item a,
identify what is oxidized and what is reduced. 31. Performance For one day, record situations
c. Write half-reactions for each equation you that show evidence of oxidation-reduction reac-
listed in item a. tions. Identify the reactants and the products,
and determine whether there is proof that a
27. Aluminum is described in Group 13 of the chemical reaction has taken place.
Elements Handbook as a self-protecting metal.
This property of aluminum results from a redox
reaction.
a. Write the redox equation for the oxidation of
aluminum.
b. Write the half-reactions for this reaction, and
show the number of electrons transferred.
c. What problems are associated with the
buildup of aluminum oxide on electrical
wiring made of aluminum?

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 649

Math Tutor BALANCING REDOX EQUATIONS

A redox equation must conserve both mass and charge. So, to balance a redox equation,
you must balance both atoms and charge (electrons). The problem-solving tips and sam-
ple below show how to balance an equation for a redox reaction in basic solution.

Problem-Solving S

To balance redox equations for reactions in basic solution:

• Add OH− and H2O to balance oxygen and hydrogen in the redox half-reactions.
• Add OH− ions to the side of the equation that needs oxygen atoms. Make sure

you add enough OH− ions so that the number of oxygen atoms added is twice the
number needed.
• Then, add enough H2O molecules to the other side of the equation to balance the
hydrogen atoms.

SAMPLE

The following unbalanced equation represents a redox reaction that takes place in a basic

solution containing KOH. Balance the redox equation.

Br2(l) + KOH(aq) ⎯→ KBr(aq) + KBrO3(aq)
Write the full ionic equation, assign oxidation numbers, and eliminate species whose oxi-

dation numbers do not change. The result is the following equation:

0 ⎯→ B−r1− + B+5rO−3

Br2

Divide this equation into half-reactions. Note that Br2 is the reactant in both half-reactions.
Reduction: Br2 ⎯→ Br− Oxidation: Br2 ⎯→ BrO−3
Add H2O and OH− to balance atoms in basic solution. Then, add electrons to balance charge.
Reduction: Br2 + 2e− ⎯→ 2Br− (no need to add H2O or OH−)
Oxidation: 12OH− + Br2 ⎯→ 2BrO−3 + 6H2O + 10e−

To balance transferred electrons, you must multiply the reduction half-reaction by 5 so that

both reactions have 10e−.

5 × (Br2 + 2e− ⎯→ 2Br−) = 5Br2 + 10e− ⎯→ 10Br−

Combining the two half-reactions gives

5Br2 + 12OH− + Br2 + 10e− ⎯→ 10Br− + 2BrO−3 + 6H2O + 10e−

Canceling common species gives

6Br2 + 12OH− ⎯→ 10Br− + 2BrO−3 + 6H2O

Returning the potassium ions to the equation gives

6Br2 + 12KOH ⎯→ 10KBr + 2KBrO3 + 6H2O, or 3Br2 + 6KOH ⎯→
5KBr + KBrO3 + 3H2O

PRACTICE PROBLEMS

1. Balance the following equation for a redox 2. Balance the following equation for a redox
reaction that takes place in basic solution: reaction that takes place in basic solution:
MnO2(s) + NaClO3(aq) + NaOH(aq) ⎯→ N2O(g) + KClO(aq) + KOH(aq) ⎯→
NaMnO4(aq) + NaCl(aq) + H2O(l) KCl(aq) + KNO2(aq) + H2O(l)

650 C H A P T E R 1 9

Standardized Test Prep

Answer the following items on a separate piece of paper. 7.Arrange the following in order of increasing
S2O32−,
MULTIPLE CHOICE oxidation number of the sulfur atom:
S4O62−, HSO−4, and H2S.
1.In the following reaction, which species is
reduced? A. H2S, S2O32−, S4O62−, HSO4−
2K + Br2 ⎯→ 2K+ + 2Br−
A. K B. S2O32−, H2S, S4O26−, HSO−4
B. Br2
C. All of the above C. H2S, S2O23−, HSO−4, S4O62−
D. None of the above
D. HSO4−, S2O23−, S4O26−, H2S
2.The oxidation number of the sulfur atom in the
SO42− ion is 8.Which answer contains the correct information
A. +2.
B. −2. about the following reaction:
C. +6. 2Pb(NO3)2 ⎯→ 2PbO + 4NO2 + O2
D. +4.
A. This reaction is a decomposition reaction and
3.A half-reaction
A. involves a change in the oxidation state of an not a redox reaction.
element.
B. always contains H2O molecules. B. This reaction is a redox reaction in which the
C. always contains H+ ions.
D. All of the above lead is reduced and the oxygen is oxidized.

4.In the following reaction, which is the oxidizing C. This reaction is a disproportionation reaction.
agent?
AgNO2 + Cl2 + 2KOH ⎯→ D. This reaction is a redox reaction in which the
AgNO3 + 2KCl + H2O
A. AgNO2 nitrogen is reduced and the oxygen is oxidized.
B. Cl2
C. KOH SHORT ANSWER
D. KCl
9.Determine the oxidation numbers for Cu in the
5.What are the oxidation states (in increasing superconductor YBa2Cu3O7. Yttrium (Y) has an
order) of the element that undergoes dispropor- oxidation number of +3. (Cu does not have oxi-
tionation in the following reaction: dation numbers greater than +3.) Give only
Cl2 + H2O ⎯→ HCl + HOCl integer oxidation numbers.
A. −1, 0, +2
B. −1, 0, +1 10.What is an oxidizing agent?
C. −2, −1, 0
D. None of the above EXTENDED RESPONSE

6.Which reaction is a redox reaction? 11.B, F, K, and L are four unknown reducing
A. Al2O3 + 6HCl ⎯→ 2AlCl3 + 3H2O agents that oxidize to singly charged cations.
B. 2HCO3− ⎯→ CO2 + CO23− + H2O Using the following information, construct a
C. SiBr4 + 3H2O ⎯→ H2SiO3 + 4HBr table showing the relative strengths of the oxi-
D. H2O + PbO2 + NaOH + KCl ⎯→ dizing and reducing agents. Data: F reduces K+,
KClO + NaPb(OH)3 B+, and L+. B+ oxidizes K and F, but not L.

12.Balance the equation for the following reaction
in basic solution:

ClO2 ⎯→ KClO3 + KClO2

Give the balanced equation for each half-
reaction and for the overall reaction. Give the
oxidizing agent and the reducing agent.

If you are short on time, quickly
scan the unanswered questions to see which might
be easiest to answer.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 651

CHAPTER LAB SEE PRE-LAB MICRO

VOLUMETRIC ANALYSIS LAB

Reduction of Manganese
in Permanganate Ion

OBJECTIVES BACKGROUND
• Demonstrate proficiency in performing In Chapter 15, you studied acid-base titrations in
which an unknown amount of acid is titrated with a
redox titrations and recognizing end points carefully measured amount of base. In this procedure,
of a redox reaction. a similar approach called a redox titration is used. In a
• Write a balanced oxidation-reduction redox titration, the reducing agent, Fe2+, is oxidized to
equation for a redox reaction. Fe3+ by the oxidizing agent, MnO4−. When this process
• Determine the concentration of a solution occurs, the Mn in MnO4− changes from a +7 to a +2
by using stoichiometry and volume data oxidation state and has a noticeably different color.
from a titration. You can use this color change to signify a redox reac-
tion “end point.” When the reaction is complete, any
MATERIALS excess MnO−4 added to the reaction mixture will give
• 0.0200 M KMnO4 the solution a pink or purple color. The volume data
• 1.0 M H2SO4 from the titration, the known molarity of the KMnO4
• 100 mL graduated cylinder solution, and the mole ratio from the balanced redox
• 125 mL Erlenmeyer flasks, 4 equation will give you the information you need to
• 250 mL beakers, 2 calculate the molarity of the FeSO4 solution.
• 400 mL beaker
• burets, 2 SAFETY
• distilled water
• double buret clamp For review of safety, please see Safety in the
• FeSO4 solution Chemistry Laboratory in the front of your book.
• ring stand
• wash bottle PREPARATION
1. In your lab notebook, prepare a data table like
652 C H A P T E R 1 9 the one shown on the next page.

2. Clean two 50 mL burets with a buret brush and
distilled water. Rinse each buret at least three
times with distilled water to remove contaminants.

3. Label one 250 mL beaker “0.0200 M KMnO4”
and the other “FeSO4.” Label three of the flasks
“1,” “2,” and “3.” Label the 400 mL beaker
“Waste.” Label one buret “KMnO4” and the other
“FeSO4.”

Trial Initial KMnO4 DATA TABLE Initial FeSO4 Final FeSO4
volume (mL) Final KMnO4 volume (mL) volume (mL)
volume (mL)
1

2
3

4. Measure approximately 75 mL of 0.0200 M 2. Slowly add KMnO4 from the buret to the FeSO4
KMnO4, and pour it into the appropriately in the flask while swirling the flask. When the
labeled beaker. Obtain approximately 75 mL color of the solution matches the color standard
of FeSO4 solution, and pour it into the appro- you prepared in Preparation step 8, record in
priately labeled beaker. your data table the final readings of the burets.

5. Rinse one buret three times with a few milliliters 3. Empty the titration flask into the waste beaker.
of 0.0200 M KMnO4 from the appropriately Repeat the titration procedure in steps 1 and 2
labeled beaker. Collect these rinses in the waste with the flasks labeled “2” and “3.”
beaker. Rinse the other buret three times with
small amounts of FeSO4 solution from the CLEANUP AND DISPOSAL
appropriately labeled beaker. Collect these 4. Dispose of the contents of the waste
rinses in the waste beaker. beaker in the container designated by
your teacher. Also, pour the color-
6. Set up the burets as instructed by your teacher. standard flask into this container. Wash your
Fill one buret with approximately 50 mL of hands thoroughly after cleaning up the area
0.0200 M KMnO4 from the beaker, and fill the and equipment.
other buret with approximately 50 mL of the
FeSO4 solution from the other beaker. ANALYSIS AND INTERPRETATION
1. Organizing Ideas: Write the balanced equation
7. With the waste beaker underneath its tip, open for the redox reaction of FeSO4 and KMnO4.
the KMnO4 buret long enough to be sure the
buret tip is filled. Repeat the process for the 2. Evaluating Data: Calculate the number of moles
FeSO4 buret. of MnO4− reduced in each trial.

8. Add 50 mL of distilled water to one of the 3. Analyzing Information: Calculate the number
125 mL Erlenmeyer flasks, and add one drop of moles of Fe2+ oxidized in each trial.
of 0.0200 M KMnO4 to the flask. Set this mix-
ture aside to use as a color standard. It can be 4. Applying Conclusions: Calculate the average
compared with the titration mixture to deter- concentration (molarity) of the iron(II)
mine the end point. sulfate solution.

PROCEDURE EXTENSIONS
1. Designing Experiments: What possible sources
1. Record in your data table the initial buret read- of error can you identify with this procedure?
ings for both solutions. Add 10 mL of the If you can think of ways to eliminate them, ask
hydrated iron(II) sulfate solution, FeSO4•7H2O, your teacher to approve your plan, and run the
to the flask labeled “1.” Add 5 mL of procedure again.
1 M H2SO4 to the FeSO4 solution in this flask.
The acid will help keep the Fe2+ ions in the
reduced state, which will allow you time to
titrate.

O X I D A T I O N - R E D U C T I O N R E A C T I O N S 653

CHAPTER 20

Electrochemistry

Electrochemical reactions provide
energy in all kinds of applications.

Artist’s rendering of the Mars Rover

Introduction to SECTION 1
Electrochemistry
OBJECTIVES
O xidation-reduction reactions involve energy changes. Because these
Identify parts of an electro-
reactions involve electron transfer, the net release or net absorption of chemical cell and their
energy can occur in the form of electrical energy rather than as heat. functions.
This property allows for a great many practical applications of redox
reactions. The branch of chemistry that deals with electricity-related Write electrode half reactions
applications of oxidation-reduction reactions is called electrochemistry. for cathodes and anodes.

Electrochemical Cells

Oxidation-reduction reactions involve a transfer of electrons. If the two www.scilinks.org
substances are in contact with one another, a transfer of energy as heat Topic: Electrochemical Cells
accompanies the electron transfer. In Figure 1 a zinc strip is in contact Code: HC60478
with a copper(II) sulfate solution. The zinc strip loses electrons to the
copper(II) ions in solution. Copper(II) ions accept the electrons and fall
out of solution as copper atoms. As electrons are transferred between
zinc atoms and copper(II) ions, energy is released as heat, as indicated
by the rise in temperature.

29 Before 29
28
28 After
27
27
26 26
25
25 24
23
24
22
23 21

22

21

Zinc strip

CuSO4 FIGURE 1 Energy as heat given
off when electrons are transferred
Cu directly from Zn atoms to Cu2+ ions
causes the temperature of the aque-
ous CuSO4 solution to rise.

E L E C T R O C H E M I S T R Y 655

FIGURE 2 An electrochemical Zinc electrode Conducting wire
cell consists of two electrodes. Each Copper electrode
electrode is in contact with an elec-
trolyte; the electrode and the elec-
trolyte make up a half-cell. The two
electrodes are connected by a wire,
and a porous barrier separates the
two electrolytes.

Copper(II) sulfate
Electrolyte

Zinc sulfate
Electrolyte

Anode Porous barrier
Cathode

If, however, we separate the substance that is oxidized from the sub-
stance that is reduced, the electron transfer is accompanied by a trans-
fer of electrical energy instead of energy as heat. One means of
separating oxidation and reduction half-reactions is with a porous bar-
rier, or salt bridge. This barrier prevents the metal atoms of one half-
reaction from mixing with the ions of the other half-reaction. Ions in the
two solutions can move through the porous barrier, which keeps a
charge from building up on the electrodes. Electrons can be transferred
from one side to the other through an external connecting wire. Electric
current moves in a closed loop path, or circuit, so this movement of elec-
trons through the wire is balanced by the movement of ions in solution.

Altering the system in Figure 1 as just described would simply involve
separating the copper and zinc, as shown in Figure 2. The Zn strip is in an
aqueous solution of ZnSO4. The Cu strip is in an aqueous solution of
CuSO4. Both solutions conduct electricity, so, as you learned in
Chapter 12, they are classified as electrolytes. An electrode is a conductor
used to establish electrical contact with a nonmetallic part of a circuit, such
as an electrolyte. In Figure 2, the Zn and Cu strips are electrodes. A sin-
gle electrode immersed in a solution of its ions is a half-cell.

The Half-Cells

In the half-cell that contains the Zn electrode in aqueous ZnSO4 solution,
the half-reaction is Zn(s) ⎯→ Zn2+(aq) + 2e−. The Zn metal loses two
electrons to form Zn2+ ions in solution, and therefore oxidation is taking
place in this half-cell. The electrode where oxidation occurs is called the
anode. In the half-cell that contains the Cu electrode in aqueous CuSO4
solution, the half-reaction is Cu2+(aq) + 2e− ⎯→ Cu(s). In this half-
reaction, the Cu2+ ions in solution gain electrons to become Cu solid; that
is, reduction is taking place. The electrode where reduction occurs is called
the cathode.

656 C H A P T E R 2 0

Recall from Chapter 19 that these two half-reactions cannot occur www.scilinks.org
separately. Both oxidation and reduction must occur in an electro- Topic: Electrical Energy
chemical reaction. The two half-cells taken together make an electro- Code: HC60475
chemical cell. In the Zn/Cu electrochemical cell, the electrons move
from the Zn electrode through the wire and down the Cu electrode to Electrodes Porous barrier
the Cu2+ ions at the electrode-solution interface. The Cu2+ ions are
reduced to solid Cu, and the resulting Cu atoms attach themselves to FIGURE 3 The light bulb is
the surface of the Cu electrode. For this reaction, charge is carried powered by the reaction in this cell.
through the barrier by a combination of Zn2+(aq) ions moving from the
anode to the cathode and the SO24−(aq) ions moving from the cathode
to the anode.

The Complete Cell

An electrochemical cell may be represented by the following notation:

anode electrode⏐anode solution⏐⏐cathode solution⏐cathode electrode

The double line represents the salt bridge, or porous barrier, between
the two half-cells. For the present cell, the cell notation is

Zn(s)⏐Zn2+(aq)⏐⏐Cu2+(aq)⏐Cu(s).

The electrochemical reaction can be found by adding the anode half-
reaction to the cathode half-reaction. This overall (or net) reaction is
the following redox reaction:

Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s)

Although the two half-reactions occur at the same time, they occur
at different places in the cell. Thus, for the reaction to proceed, electrons
must pass through the wire that connects the two half-cells.

An electrochemical cell that consists of the Zn and Cu reaction
described above is called the Daniell Cell, named for the English
chemist John Frederick Daniell. The Daniell Cell can generate enough
electricity to light up the light bulb shown in Figure 3. In electrochemi-
cal cells, either a chemical reaction produces electrical energy or an
electric current produces a chemical change.

SECTION REVIEW 3. Write the half-reaction in which I−(aq) changes to
I2(s). Would this reaction occur at the anode or
1. Why is the use of a salt bridge or porous barrier cathode?
necessary in an electrochemical cell?
Critical Thinking
2. Given the Cu2+(aq)⏐Cu(s) and Mg2+(aq)⏐Mg(s)
half-reactions, where Cu2+(aq)⏐Cu(s) is the cath- 4. RELATING IDEAS Is the net chemical result of an
ode reaction, electrochemical cell a redox reaction? Explain your
answer.
a. write the overall reaction.

b. write the cell notation.

E L E C T R O C H E M I S T R Y 657

SECTION 2 Voltaic Cells

OBJECTIVES V oltaic cells use spontaneous oxidation-reduction reactions to convert

Describe the operation of chemical energy into electrical energy. Voltaic cells are also called galvan-
voltaic cells, including dry ic cells. The most common application of voltaic cells is in batteries.
cells, lead-acid batteries, and
fuel cells. How Voltaic Cells Work

Identify conditions that lead Figure 4 shows an example of a voltaic cell: the Zn⏐⏐Cu electrochemi-
to corrosion and ways to cal cell discussed in the previous section.
prevent it.
Electrons given up at the anode pass along the external connecting
Describe the relationship wire to the cathode. The movement of electrons through the wire must
between voltage and the be balanced by the movement of ions in the solution. Thus, in Figure 4,
movement of electrons. sulfate ions in the CuSO4 solution can move through the barrier into
the ZnSO4 solution.
Calculate cell voltage/poten-
tials from a table of standard
electrode potentials.

Zinc atoms losing two e– e– Porous barrier
electrons to become ions
Copper
Zinc atom, Zinc ion, Zinc strip
Zn Zn2+ strip

e–

Water Copper(II) sulfate,
CuSO4, solution

molecule,

H2O Sulfate
ion, SO42–

Zinc sulfate, e–
ZnSO4, solution

Anode Cathode

FIGURE 4 In a voltaic cell, electrons Copper Copper
spontaneously flow from anode to cathode. ion, Cu2+ atom, Cu
The copper strip gains mass as copper ions
become copper atoms. The zinc strip loses Copper(II) ions gaining two
mass as the zinc atoms become zinc ions. electrons to become atoms

658 C H A P T E R 2 0

The dry cells pictured in Figure 5 are common sources of electrical Alkaline Zinc carbon
energy. Like the wet cell previously described, dry cells are voltaic cells. batteries batteries
The three most common types of dry cells are the zinc-carbon battery,
the alkaline battery, and the mercury battery. They differ in the sub- Mercury
stances being oxidized and reduced. battery

Zinc-Carbon Dry Cells

Batteries such as those used in flashlights are zinc-carbon dry cells.
These cells consist of a zinc container, which serves as the anode, filled
with a moist paste of MnO2, carbon black, NH4Cl, and ZnCl2, as illus-
trated in Figure 6a. When the external circuit is closed, zinc atoms are
oxidized at the negative electrode, or anode.

0 ⎯→ +2 + 2e− FIGURE 5 Many common batter-
ies are simple voltaic dry cells.
Zn(s) Zn2+(aq)

Electrons move across the circuit and reenter the cell through the carbon

rod. The carbon rod is the cathode or positive electrode. Here MnO2 is
reduced in the presence of H2O according to the following half-reaction.

+4 + H2O(l) + 2e− ⎯→ +3 + 2OH−(aq)

2MnO2(s) Mn2O3(s)

Alkaline Batteries www.scilinks.org
Topic: Batteries
The batteries found in a portable compact disc player or other small elec- Code: HC60139
tronic device are frequently alkaline dry cells. These cells do not have a
carbon rod cathode, as in the zinc-carbon cell. The absence of the carbon
rod allows them to be smaller. Figure 6b shows a model of an alkaline bat-
tery. This cell uses a paste of Zn metal and potassium hydroxide instead of
a solid metal anode. The half-reaction at the anode is as follows.

0 + 2OH −(aq) ⎯→ +2 + 2e−

Zn(s) Zn(OH)2(s)

The reduction half-reaction, the reaction at the cathode, is exactly the
same as that for the zinc-carbon dry cell.

(a) zinc dry cell (b) alkaline dry cell FIGURE 6 (a) In a zinc dry cell,
Carbon rod zinc is oxidized to Zn2+ at the anode,
(cathode) Zn-KOH and manganese(IV) is reduced to
anode paste manganese(III) at the cathode.
Spacer (b) KOH makes the electrolyte
Brass current paste in this battery basic. Thus, it is
Zinc shell collector called an alkaline dry cell.
(anode)
KOH
electrolyte

Spacer
MnO2,
cathode mix

Steel jacket

Moist electrolytic paste of ZnCl2,
MnO2, NH4Cl, and carbon black

E L E C T R O C H E M I S T R Y 659

FIGURE 7 It is important that Zn in KOH (anode)
mercury batteries be recycled and
not just discarded because mercury Steel jacket
is a poisonous substance. Separator
HgO, carbon (cathode)

Mercury Batteries

The tiny batteries found in hearing aids, calculators, and camera flashes
are mercury batteries, as shown in Figure 7. The anode half-reaction is
identical to that found in the alkaline dry cell. However, the cathode, or
reduction, half-reaction is different. The cathode half-reaction is
described by the following equation.

+2 + H2O(l ) + 2e− ⎯→ 0 ) + 2OH−(aq)

HgO(s) Hg(l

www.scilinks.org Fuel Cells
Topic: Fuel Cells
Code: HC60625 A fuel cell is a voltaic cell in which the reactants are being continuous-
ly supplied and the products are being continuously removed.
FIGURE 8 The reactions in Therefore, unlike a battery, a fuel cell could, in principle, work forever,
this fuel cell take place at carbon changing chemical energy into electrical energy.
electrodes that contain metal
catalysts. The water formed is Fuel cells based on the reactions listed below and shown in Figure 8
removed as a gas. are used in the United States space program.

660 C H A P T E R 2 0 Cathode: O2(g) + 2H2O(l) + 4e− ⎯→ 4OH−(aq)
Anode: 2H2(g) + 4OH−(aq) ⎯→ 4e− + 4H2O(l)

Net reaction: 2H2 + O2 ⎯→ 2H2O

Fuel cells are very efficient and have very low emissions.

Excess H2(g) + H2O(g) Excess O2(g) + H2O(g)
H2(g)
e– O2(g)
e–

K+
OH_

Porous graphite Electrolyte solution Porous graphite
anode cathode

Corrosion and Its Prevention

Corrosion is an electrochemical process that has a large economic www.scilinks.org
impact. Approximately 20% of all the iron and steel produced is used Topic: Corrosion
to repair or replace corroded structures. One of the metals most com- Code: HC60360
monly affected by corrosion is iron. Rust, hydrated iron(III) oxide,
forms by the following overall reaction. FIGURE 9 The cathodic reaction
happens where the O2 concentration
4Fe(s) + 3O2(g) + xH2O(l) ⎯→ 2Fe2O3•xH2O(s) is high. The anodic reaction happens
in a region where the O2 concentra-
The amount of hydration of the iron oxide, reflected by the coefficient tion is low, such as a pit in the metal.
x in the equation above, varies, and affects the color of the rust formed.

The mechanism for the corrosion of iron contains the following elec-
trochemical reactions.

Anode: Fe(s) ⎯→ Fe2+(aq) + 2e−

Cathode: O2(g) + 2H2O(l) + 4e− ⎯→ 4OH−(aq)

The anode and cathode reactions occur at different regions of the metal
surface. The electric circuit is completed by electron flow through the
metal itself, which acts like the wire in an electrochemical cell. The
water on the surface of the metal serves as the salt bridge. Thus, for cor-
rosion to occur, water and oxygen must be present with the iron.

As shown in Figure 9, when the iron is exposed to water and oxygen, the
iron metal at the anodic site is oxidized to Fe2+ ions. The electrons released
at this site travel along the metal (like the wire of a cell) to the cathodic
region, where oxygen is reduced. The Fe2+ ions travel along the moisture
toward the cathodic regions. At the cathode, the Fe2+ ions are further oxi-
dized to Fe3+ ions and form rust, which is hydrated iron oxide, Fe2O3•xH2O.

2Fe2+(aq) + (3 + x)H2O(l) ⎯→ Fe2O3•xH2O(s) + 6H+(aq) + 2e−

Water molecule, H2O
Oxygen molecule, O2
Iron(II) ion, Fe2+
Hydroxide ion, OH−

Anodic site: Ion conduction Cathodic site:
Fe(s) ⎯→ 2e− + Fe2+(aq) O2(aq) + 2H2O(l) + 4e− ⎯→ 4OH−(aq)

Iron, Fe Paint

Water layer Rust

e−
Electron conduction

E L E C T R O C H E M I S T R Y 661

The presence of salt or high acidity speed the corrosion process,
because the movement of electrons is facilitated by the extra ions present.

How can corrosion be prevented? One way, which makes use of elec-
trochemical properties of metals, is to coat steel with zinc in a process
called galvanizing. Zinc is more easily oxidized than iron; therefore, zinc
will react before the iron is oxidized. This is called cathodic protection,
and the more easily oxidized metal used is called a sacrificial anode.

The Alaskan oil pipeline, shown in Figure 10, is an example of steel that
is cathodically protected. However, instead of coating, zinc is connected to
the pipe by a wire. The zinc will oxidize before the iron in the steel does.
As the zinc anode corrodes, it gives electrons to the cathode, the steel, and
as the zinc anode does so, it prevents the steel from corroding. As the zinc
dissolves, it needs to be replaced. As long as zinc metal that is able to cor-
rode is present, the steel will be protected from corrosion.

FIGURE 10 The Alaskan pipeline Electrical Potential
is cathodically protected by a
parallel zinc cable. In a voltaic cell, the oxidizing agent at the cathode pulls the electrons
through the wire away from the reducing agent at the anode. The “pull,”
Careers in Chemistry or driving force on the electrons, is called the electric potential. Electric
potential, or voltage, is expressed in units of volts (V), which is the
Go to go.hrw.com for a full-length potential energy per unit charge. Current is the movement of the elec-
article on careers in chemical trons and is expressed in units of amperes, or amps (A).
engineering.
Electrical potential is analogous to gravitational potential. Just as water
Keyword: HC6ELEX flows from a position of higher gravitational potential to a position of lower
gravitational potential, electrons flow from higher electric potential to
662 C H A P T E R 2 0 lower electrical potential.

Electrode Potentials

Reconsider the voltaic cell shown in Figure 4. There are two electrodes,
Zn and Cu. These two metals each have different tendencies for accept-
ing electrons. This tendency for the half-reaction of either copper or zinc
to occur as a reduction half-reaction in an electrochemical cell can be
quantified as a reduction potential. There are two half-cells in Figure 4: a
strip of zinc placed in a solution of ZnSO4 and a strip of copper placed in
a solution of CuSO4. The difference in potential between an electrode and
its solution is known as electrode potential. When these two half-cells are
connected and the reaction begins, a difference in potential is observed
between the electrodes. This potential difference, or voltage, is propor-
tional to the energy required to move a certain electric charge between
the electrodes. A voltmeter connected across the Zn⏐⏐Cu voltaic cell
measures a potential difference of about 1.10 V when the solution con-
centrations of Zn2+ and Cu2+ ions are each 1 M.

The potential difference measured across the complete voltaic cell is
easily measured and equals the sum of the electrode potentials for the
two half-reactions. But an individual electrode potential cannot be meas-

ured directly. This is because there can be no transfer of electrons unless H2(g)
both the anode and the cathode are connected to form a complete cir-
cuit. A relative value for the potential of a half-reaction can be deter- Pt wire
mined by connecting it to a standard half-cell as a reference. This H2(g)
standard half-cell, shown in Figure 11, is called a standard hydrogen
electrode, or SHE. It consists of a platinum electrode dipped into a Pt black
1.00 M acid solution surrounded by hydrogen gas at 1 atm pressure and electrode
25°C. Other electrodes are ranked according to their ability to reduce
hydrogen under these conditions. FIGURE 11 A hydrogen
electrode is the standard reference
The anodic reaction for the standard hydrogen electrode is described electrode for measuring electrode
by the forward half-reaction in the following equilibrium equation. potentials. The electrode surface in
contact with the solution is actually
0 +1 a layer of hydrogen adsorbed onto
the surface of the platinum.
H2(g) ←⎯⎯→ 2H+(aq) + 2e−

The cathodic half-reaction is the reverse. An arbitrary potential of

0.00 V is assigned to both of these half-reactions. The potential of a half-

cell under standard conditions measured relative to the standard hydro-
gen electrode is a standard electrode potential, E0. Electrode potentials

are expressed as potentials for reduction. These reduction potentials

provide a reliable indication of the tendency of a substance to be

reduced. Figure 12 shows how the SHE is used to find the electrode

potentials of the zinc and copper half-cells. Half-reactions for some

common electrodes and their standard electrode potentials are listed in

Table 1 on the next page.
Effective oxidizing agents, such as Cu2+ and F2, have positive E0 val-

ues. Half-reactions with negative reduction potentials prefer oxidation
over reduction. Negative E0 values indicate that the metal or other elec-

trode is more willing to give up electrons than hydrogen. Effective
reducing agents, such as Li and Zn, have negative E0 values.

When a half-reaction is written as an oxidation reaction, the sign of its

electrode potential is reversed, as shown for the oxidation and reduction

half-reactions for zinc.

Zn2+ + 2e− ⎯→ Zn E0 = −0.76 V
Zn ⎯→ Zn2+ + 2e− E0 = +0.76 V

Zinc strip, Copper
Zn strip, Cu

FIGURE 12 The electrode poten-
tials of zinc and copper half-cells are
measured by coupling them with a
standard hydrogen electrode.

E L E C T R O C H E M I S T R Y 663

TABLE 1 Standard Reduction Potentials

Half-cell reaction Standard Half-cell reaction Standard
electrode electrode
F2 + 2e− ←⎯⎯→ 2F − potential, E0 Fe3+ + 3e− ←⎯⎯→ Fe potential, E0
MnO−4 + 8H+ + 5e− ←⎯⎯→ Mn2+ + 4H2O (in volts) Pb2+ + 2e− ←⎯⎯→ Pb (in volts)
Au3+ + 3e− ←⎯⎯→ Au Sn2+ + 2e− ←⎯⎯→ Sn
Cl2 + 2e− ←⎯⎯→ 2Cl− +2.87 Ni2+ + 2e− ←⎯⎯→ Ni − 0.04
Cr2O72− + 14H+ + 6e− ←⎯⎯→ 2Cr3+ + 7H2O +1.49 Co2+ + 2e− ←⎯⎯→ Co − 0.13
MnO2 + 4H+ + 2e− ←⎯⎯→ Mn2+ + 2H2O +1.42 Cd2+ + 2e− ←⎯⎯→ Cd − 0.14
Br2 + 2e− ←⎯⎯→ 2Br− +1.36 Fe2+ + 2e− ←⎯⎯→ Fe − 0.23
Hg2+ + 2e− ←⎯⎯→ Hg +1.33 S + 2e− ←⎯⎯→ S2− − 0.28
Ag+ + e− ←⎯⎯→ Ag +1.21 Cr3+ + 3e− ←⎯⎯→ Cr − 0.40
Hg22+ + 2e− ←⎯⎯→ 2Hg +1.07 Zn2+ + 2e− ←⎯⎯→ Zn − 0.41
Fe3+ + e− ←⎯⎯→ Fe2+ +0.85 Al3+ + 3e− ←⎯⎯→ Al − 0.51
MnO4− + e− ←⎯⎯→ MnO42− +0.80 Mg2+ + 2e− ←⎯⎯→ Mg − 0.74
I2 + 2e− ←⎯⎯→ 2I− +0.80 Na+ + e− ←⎯⎯→ Na − 0.76
Cu2+ + 2e− ←⎯⎯→ Cu +0.77 Ca2+ + 2e− ←⎯⎯→ Ca − 1.66
Cu2+ + e− ←⎯⎯→ Cu+ +0.56 Ba2+ + 2e− ←⎯⎯→ Ba − 2.37
S + 2H+(aq) + 2e− ←⎯⎯→ H2S(aq) +0.54 K+ + e− ←⎯⎯→ K − 2.71
2H+(aq) + 2e− ←⎯⎯→ H2 +0.34 Li+ + e− ←⎯⎯→ Li − 2.76
+0.16 − 2.90
+0.14 − 2.93
− 3.04
0.00

The potential difference across the zinc/hydrogen cell is −0.76 V, so
zinc is considered to have an electrode potential of −0.76. The negative
number indicates that electrons flow from the zinc electrode, where zinc

is oxidized, to the hydrogen electrode, where aqueous hydrogen ions

are reduced.

A copper half-cell coupled with the standard hydrogen electrode gives
a potential difference measurement of +0.34 V. This positive number
indicates that Cu2+(aq) ions are more readily reduced than H+(aq) ions.

Standard electrode potentials can be used to predict if a redox reac-

tion will occur spontaneously. A spontaneous reaction will have a posi-
tive value for E0cell, which is calculated using the following equation.

E0cell = E0cathode − E 0
anode

The half-reaction that has the more negative standard reduction potential
will be the anode. Oxidation occurs at the anode, so the anode half-cell
reaction will be the reverse of the reduction reaction found in Table 1. For
this reason, the total potential of a cell is calculated by subtracting the stan-
dard reduction potential for the reaction at the anode (E0anode) from the
standard reduction potential for the reaction at the cathode (E0cathode).

664 C H A P T E R 2 0

SAMPLE PROBLEM A For more help, go to the Math Tutor at the end of this chapter.

Write the overall cell reaction, and calculate the cell potential for a voltaic cell consisting of the following half-
cells: an iron (Fe) electrode in a solution of Fe(NO3)3 and a silver (Ag) electrode in a solution of AgNO3.

SOLUTION

1 ANALYZE Given: A half-cell consists of Fe(s) with Fe(NO3)3(aq) and a second half-cell consists of

Ag(s) with AgNO3(aq).
0
Unknown: E cell

2 PLAN 1. Look up E0 for each half-reaction (written as reductions) in Table 1.
Fe3+(aq) + 3e− ⎯→ Fe(s) E0 = −0.04 V Ag+(aq) + e− ⎯→ Ag(s) E0 = +0.80 V

2. Determine the cathode and anode.

Fe in Fe(NO3)3 is the anode because it has a lower reduction potential than Ag. Ag in
Ag(NO3) is therefore the cathode.

3 COMPUTE 1. Determine the overall cell reaction. Multiply the Ag half-reaction by 3 so that the num-
4 EVALUATE ber of electrons lost in that half-reaction equals the number of electrons gained in the
oxidation of iron. Reverse the iron half-reaction to be an oxidation half-reaction.

3Ag+(aq) + Fe(s) ⎯→ 3Ag(s) + Fe3+(aq)
2. Calculate the cell potential by E0cell = E0cathode − E0anode. Note that when a half-reaction is

multiplied by a constant, the E0 value is not multiplied by that constant but remains the same.

E0cell = E0cathode − E0anode = +0.80 V − (−0.04 V) = +0.84 V

The calculated value for E0cell is positive, which confirms that it is a voltaic cell, as the prob-
lem states.

PRACTICE Answers in Appendix E

1. For each pair of half-cells, determine the overall electrochemical Go to go.hrw.com for
reaction that proceeds spontaneously and the E0 value. more practice problems
that ask you to calculate
a. Cr2O72−/Cr3+ and Ni2+/Ni cell potential.
b. SHE and Fe2+/Fe3+
Keyword: HC6ELEX

SECTION REVIEW 4. Given the MnO2/Mn2+ and Cr3+/Cr half-cells,
determine the overall electrochemical reaction
1. What is a voltaic cell? that occurs spontaneously and the E0 value.

2. What is electrode potential, and how is it used to Critical Thinking
calculate information about an electrochemical cell?
5. EVALUATING IDEAS A sacrificial anode is allowed
3. Given the Na+/Na and K+/K half-cells, determine to corrode. Why is use of a sacrificial anode con-
the overall electrochemical reaction that proceeds sidered to be a way to prevent corrosion?
spontaneously and the E0 value.

E L E C T R O C H E M I S T R Y 665

Fuel-Cell Cars Eventually, fuel-cell cars such as this The use of fuels such as methanol
one may replace gasoline-powered cars. solves the storage and fuel infrastruc-
The car of the future is quiet, has low ture problem, because these fuels can
emissions, and requires less mainte- The range of fuel-cell powered cars is be stored similar to the way gasoline
nance than cars of today do. These limited by the amount of fuel that can is stored. Therefore, these fuels can be
are the promises of fuel-cell cars. The be stored in the storage tank. Current sold at current gasoline stations.
development of fuel-cell cars has storage now limits the cars to around
been pushed by state and federal pro- 100 miles before refueling. Researchers One should also consider the emis-
grams, legislation, and funding. There in this area are looking into carbon- sions and energy use involved in
are a number of issues that need to adsorption systems, which are refriger- making and delivering the hydrogen
be resolved before the large-scale ated and pressurized tanks that can fuel in addition to the emissions from
manufacture of fuel-cell cars becomes store large amounts of hydrogen. the hydrogen fuel cell. Producing
feasible. These include the fuel- Hydrogen is very combustible, but safe hydrogen fuel itself could involve sub-
storage capacity of the car, and the hydrogen tanks have been developed. stantial emissions. Even so, fuel-cell
fueling infrastructure. So today, the experts do not have any cars are still more efficient and have
more concern about the safety of greater environmental benefits than
The technology currently favored for hydrogen storage then they do about the internal-combustion cars of today.
automobiles is the proton-exchange gasoline storage.
membrane (PEM) fuel cell. The system Currently, all major car companies
works by combining hydrogen with However, the landscape is not yet are developing and researching fuel-
oxygen taken from the outside air. The dotted with hydrogen refueling sta- cell cars. Most companies have pro-
movement of the electrons (produced tions. Current experimental users totype cars and some cars are in use
by the redox reaction) along an exter- such as governmental institutions by governmental institutions. Which
nal route generates electricity, which must have their own hydrogen refu- type will ultimately succeed in the
is used eventually to turn the car eling stations. Auto companies say mass market is yet unknown. Experts
wheels. The process is two to three there needs to be a consumer infra- agree, though, that for zero-emission
times more efficient than that of the structure for the hydrogen fuel-cell cars, hydrogen is the only major fuel
internal-combustion engine, and the cars before consumers will buy. On option that has been identified to
only products are water and a moder- the other hand, possible developers date.
ate amount of energy as heat. In most of such hydrogen stations say that
cases, a traditional car battery is used there needs to be a demand before Questions
to operate the car’s electrical system. the infrastructure can form.
1. Fuel cells that use hydrogen as
In principle, fuel cells can run on fuel produce water and moderate
any hydrogen-rich fuel. Several can- amounts of energy in the form of
didate fuels that can power fuel cells heat as the only emissions. What
are being investigated: methanol, kinds of negative effects might
ethanol, natural gas, gasoline, and cars that use hydrogen-rich fuels
liquid or compressed hydrogen. If have on the environment?
the fuel is not hydrogen gas, the
engine needs to include a reformer 2. In the PEM fuel cell, the hydro-
that can convert the other fuel (such gen ions are produced at which
as methanol) to hydrogen. electrode: the anode or cathode?

666 C H A P T E R 2 0

Electrolytic Cells SECTION 3

OBJECTIVES

S ome oxidation-reduction reactions do not occur spontaneously but Describe the nature of elec-
trolytic cells.
can be driven by electrical energy. If electrical energy is required to pro-
duce a redox reaction and bring about a chemical change in an electro- Describe the process of elec-
chemical cell, it is an electrolytic cell. Most commercial uses of redox trolysis in the decomposition
reactions make use of electrolytic cells. of water and in production of
metals.

How Electrolytic Cells Work Explain the process of
electroplating.
A comparison of electrolytic and voltaic cells can be seen in Figure 13.
The voltaic cell shown in Figure 13 has a copper cathode and a zinc Describe the chemistry of a
anode. If a battery is connected so that the positive terminal contacts the rechargeable cell.
copper electrode and the negative terminal contacts the zinc electrode,
the electrons move in the opposite direction. The battery forces the cell FIGURE 13 The direction in
to reverse its reaction; the zinc electrode becomes the cathode, and the which the electrons move reverses
copper electrode becomes the anode. The half-reaction at the anode, in if a voltaic cell is connected to a
which copper metal is oxidized, can be written as follows: direct current source to become an
electrolytic cell.
0 ⎯→ +2 + 2e− Electrolytic Cell
e– e–
Cu Cu2+

The reduction half-reaction of zinc at the cathode is written as follows:

+2 + 2e− ⎯→ 0

Zn2+ Zn

Voltaic Cell
e– e–

Zinc strip Copper strip

Zinc sulfate, Copper(II) sulfate,
ZnSO4, CuSO4, solution
solution
Cathode Cathode Anode
Anode
E L E C T R O C H E M I S T R Y 667

There are two important differences between the voltaic cell and the
electrolytic cell.
1. The anode and cathode of an electrolytic cell are connected to a bat-

tery or other direct-current source, whereas a voltaic cell serves as a
source of electrical energy.
2. Electrolytic cells are those in which electrical energy from an external
source causes nonspontaneous redox reactions to occur. Voltaic cells
are those in which spontaneous redox reactions produce electricity. In
an electrolytic cell, electrical energy is converted to chemical energy;
in a voltaic cell, chemical energy is converted to electrical energy.

www.scilinks.org Electroplating
Topic: Electroplating
Code: HC60492 An electrolytic process in which a metal ion is reduced and a solid metal is
deposited on a surface is called electroplating. An electroplating cell con-
tains a solution of a salt of the plating metal, an object to be plated (the
cathode), and a piece of the plating metal (the anode). A silver-
plating cell contains a solution of a soluble silver salt and a silver anode.
The cathode is the object to be plated. The silver anode is connected to
the positive electrode of a battery or to some other source of direct cur-
rent. The object to be plated is connected to the negative electrode.

A cell in which silver is being electroplated onto a bracelet can be
seen in Figure 14. Silver ions are reduced at the cathode according to
the following half-reaction and deposited as metallic silver when elec-
trons flow through the circuit.

A+1g+ + e− ⎯→ 0

Ag

e– Silver strip, Ag
e–
Ag(CN)2–
solution

CN–
Ag+

FIGURE 14 The bracelet in this Ag+ Ag+
cell is being coated with a thin layer
of silver. Silver ions are replaced in Power source Cathode Anode
the solution as the pure silver anode
dissolves.

668 C H A P T E R 2 0

Meanwhile, metallic silver is oxidized at the anode according to the fol-
lowing half-reaction.

0 ⎯→ A+1g+ + e−

Ag

In effect, silver is transferred from the anode to the cathode of the cell.

Rechargeable Cells

A rechargeable cell combines the oxidation-reduction chemistry of
both voltaic cells and electrolytic cells. When a rechargeable cell con-
verts chemical energy to electrical energy, it operates as a voltaic cell.
But when the cell is recharged, it operates as an electrolytic cell, con-
verting electrical energy to chemical energy.

The standard 12 V automobile battery, shown in Figure 15, is a set of
six rechargeable cells. The anode in each cell is lead submerged in a
solution of H2SO4. The anode half-reaction is described below.

Pb(s) + SO24−(aq) ⎯→ PbSO4(s) + 2e−

At the cathode, PbO2 is reduced according to the following equation.

PbO2(s) + 4H+(aq) + SO42−(aq) + 2e− ⎯→ PbSO4(s) + 2H2O(l)

The net oxidation-reduction reaction for the discharge cycle is:

Pb(s) + PbO2(s) + 2H2SO4(aq) ⎯→ 2PbSO4(s) + 2H2O(l)

A car’s battery produces the electric energy needed to start its engine.
Sulfuric acid, present as its ions, is consumed, and lead(II) sulfate accu-
mulates as a white powder on the electrodes. Once the car is running,
the half-reactions are reversed by a voltage produced by the alternator.
The Pb, PbO2, and H2SO4 are regenerated. A battery can be recharged
as long as all reactants necessary for the electrolytic reaction are pre-
sent, and all reactions are reversible.

Intercell connectors

H2SO4(aq) Pb plates FIGURE 15 The rechargeable
PbO2 plates cells of a car battery produce electric-
Intercell dividers ity from reactions between lead(IV)
oxide, lead, and sulfuric acid.

E L E C T R O C H E M I S T R Y 669

Electrolysis

Electroplating and recharging a battery are examples of electrolysis.
Electrolysis is the process of passing a current through a cell for which the
cell potential is negative and causing an oxidation-reduction reaction to
occur. That is, electrical energy is used to force a nonspontaneous chemi-
cal reaction to occur. For this cell reaction to occur, the external voltage
must be greater than the potential that would be produced by the spon-
taneous reverse cell reaction.

Electrolysis is of great industrial importance. It is used to purify
many metals from the ores in which they are found chemically com-
bined in Earth’s crust.

Electrolysis of Water

The electrolysis of water, shown in Figure 16, leads to the cell reaction
in which water is broken down into its elements, H2 and O2. Recall that
hydrogen gas and oxygen gas combine spontaneously to form water and
are used to power fuel cells, which produce electricity. Therefore, the
reverse process (electrolysis of water) is nonspontaneous and requires
electrical energy. The following half-reactions occur at the anode and
cathode.

Anode: 6H2O(l) ⎯→ 4e− + O2(g) + 4H3O+(aq)
Cathode: 4H2O(l) + 4e− ⎯→ 2H2(g) + 4OH−(aq)

FIGURE 16 Electrical energy
from the battery is used to break
down water. Hydrogen forms at
the cathode (left tube), and oxygen
forms at the anode (right tube).

670 C H A P T E R 2 0

Aluminum Production by Electrolysis Sodium Production
by Electrolysis
Aluminum is the most abundant metal in Earth’s crust. It is a relative-
ly reactive metal; therefore, in nature, it is found as its oxide in an ore Sodium is such a reactive metal that
called bauxite. Aluminum is now very useful commercially, but it was preparing it through a chemical process
not until 1886 that a process to obtain pure aluminum metal was dis- can be dangerous. Today, sodium is pro-
covered. Charles M. Hall (from the United States) and Paul Héroult duced mainly by the electrolysis of
(from France) simultaneously, but independently, determined a practi- molten sodium chloride. The melting
cal method for producing aluminum; it is an electrolytic process called point of sodium chloride is about
the Hall-Héroult process. 800°C; but mixing it with solid calcium
chloride lowers the melting point to
Bauxite ore contains not only aluminum oxide (Al2O3), but oxides of around 600°C. This mixture is intro-
iron, silicon, and titanium. Therefore, the aluminum oxide (called alu- duced into an electrolytic cell called the
mina) must first be separated from the other compounds in the ore. The Downs Cell to retrieve the sodium
pure hydrated alumina (Al2O3•nH2O) is obtained by treating bauxite metal, and the lower melting point
with sodium hydroxide, which dissolves the alumina but does not dis- allows the cell to use less electrical
solve the other compounds in the ore. The alumina solution is then sep- energy to run.
arated from the remaining solid compounds and reprecipitated to
obtain pure alumina. The purified alumina is dissolved in molten cryo-
lite, Na3AlF6, at 970°C in an electrochemical cell, and the aluminum
ions are reduced to aluminum metal. The liquid aluminum is denser
than the molten cryolite and alumina; therefore, the molten aluminum
metal settles to the bottom of the cell and is drained off periodically.

The electrolytic solution contains a large number of aluminum-
containing ions, and the chemistry of the electrochemical reaction is not
completely understood. Scientists still debate the exact species that par-
ticipate in the half-reactions; but the overall cell reaction is

2Al2O3(l) + 3C(s) ⎯→ 4Al(l) + 3CO2(g),

where carbon is the anode and steel is the cathode in the cell.
The aluminum metal produced in this process is 99.5% pure. The

Hall-Héroult process made the production of aluminum economically
feasible. However, this process is the largest single user of electrical
energy in the United States—nearly 5% of the national total. Recycling
aluminum saves almost 95% of the cost of production. Aluminum recy-
cling is one of the most economically worthwhile recycling programs
that has been developed.

SECTION REVIEW Critical Thinking

1. Describe an electrolytic cell. 5. APPLYING CONCEPTS Copper ore contains zinc
2. Explain the process of electroplating. metal, which is oxidized along with Cu during the
3. What is a rechargeable cell? electrolytic purification process. However, the Zn2+
4. Give an example of how electrolytic cells are used ions are not then reduced when the Cu2+ ions are
reduced to Cu at the cathode to obtain purified
in industry. copper metal. Explain how Zn can be oxidized
with Cu, but their ions not be reduced together.

E L E C T R O C H E M I S T R Y 671

CHAPTER HIGHLIGHTS

Introduction to Electrochemistry

Vocabulary • Electrochemistry is the branch of chemistry that deals with
electrochemistry
electrode electricity-related applications of redox reactions.
half-cell
anode • The electrode immersed in an electrolyte solution is a half-cell.
cathode • The anode is the electrode where oxidation takes place. The

cathode is the electrode where reduction occurs.

• The cell consists of electrodes connected by a wire along which

the electron travels and a salt bridge (or porous barrier)

through which ions transfer to balance the charge.

• An electrochemical cell is a system of electrodes and elec-

trolytes in which either chemical reactions produce electrical

energy or electric current produces chemical change.

Voltaic Cells • A voltaic cell, sometimes called a galvanic cell, uses a sponta-

Vocabulary neous redox reaction to produce electrical energy. Examples of
voltaic cell
reduction potential voltaic cells are batteries and fuel cells.
electrode potential
standard electrode potential • Fuel cells are voltaic cells in which the reactants are continu-

ously supplied and the products are continuously removed.

• The potential difference must be measured across a complete

cell because no transfer of electrons can occur unless both the

anode and cathode are connected to form a complete circuit.

Thus, the standard electrode potential for a half-cell is meas-

ured against the standard hydrogen electrode (SHE).

• Standard reduction potentials, E0, are stated as reduction half-

reactions. Effective oxidizing agents have positive E0 values,
while effective reducing agents have negative E0 values.

• A voltaic cell has an Ec0ell value that is positive.
• Corrosion occurs when iron is exposed to oxygen and water.

One of the best methods to prevent corrosion is by the use of

sacrificial anodes.

Electrolytic Cells • Electrolytic cells are cells in which electrical energy from an

Vocabulary external source causes a nonspontaneous reaction to occur.
electrolytic cell
electroplating • An electrolytic cell has an E 0 value that is negative.
electrolysis cell
• Electrolysis has great economic impact. Applications of

electrolytic cells are electroplating of metallic surfaces,

rechargeable batteries, aluminum production, and purification

of metals.

672 C H A P T E R 2 0

CHAPTER REVIEW

For more practice, go to the Problem Bank in Appendix D. 13. a. What information is provided by the stan-
dard reduction potential of a given half-cell?
Introduction to
Electrochemistry b. What does the relative value of the reduction
potential of a given half-reaction indicate
SECTION 1 REVIEW about its oxidation-reduction tendency?
1. In the half-cell Zn2+(aq) + 2e− ⎯→ Zn(s), what
is the electrode and is this half-reaction an 14. When the cell Ba(s)⏐Ba2+(aq)⏐⏐Sn2+(aq)⏐Sn(s)
anodic reaction or a cathodic reaction? is running, what observations can be made?
2. What role does the porous barrier play?
3. For each of the following pairs of half-cells, PRACTICE PROBLEMS
write the overall reaction and the cell notation.
Assume the first half-cell given in each pair is 15. For each of the following pairs of half-cells,
the cathodic half-cell. determine the overall electrochemical reaction
a. Ag+/Ag, Co2+/Co that proceeds spontaneously:
b. Au3+/Au, Zn2+/Zn a. Na+/Na, Ni2+/Ni
c. Hg2+/Hg, K+/K b. F2/F−, S/H2S
4. Describe the components of an electrochemical c. Br2/Br−, Cr3+/Cr
cell and how the electrical charge travels d. MnO−4/Mn2+, Co2+/Co
through these components.
16. Determine the values of E0 for the cells in the
Voltaic Cells previous problem.

SECTION 2 REVIEW 17. Suppose chemists had chosen to make the I2 +
2e− ⎯→ 2I− half-cell the standard electrode and
5. Describe a voltaic cell, and give two examples had assigned it a potential of zero volts.
of a voltaic cell. a. What would be the E0 value for the Br2 +
2e− ⎯→ 2Br− half-cell?
6. What is the essential advantage of a fuel cell b. What would be the E0 value for the Al3+ +
over batteries in the generation of electrical 3e− ⎯→ Al half-cell?
energy? c. How much change would be observed in the
E0 value for the reaction involving Br2 + I− if
7. Explain why corrosion is a voltaic cell. the I2 half-cell is the standard?
8. Discuss the advantages and disadvantages of
18. If a strip of Ni were dipped into a solution of
corrosion-prevention methods. AgNO3, what would be expected to occur?
9. Which half reaction would more likely be an Explain, using E0 values and equations.

oxidation reaction: one with a standard reduc- Electrolytic Cells
tion potential of −0.42 V, or one with a standard
reduction potential of +0.42 V? SECTION 3 REVIEW
10. Why are dry-cell batteries called dry cells, even
though their chemistry involves water? 19. What reaction happens at the cathode in an
11. a. Explain what is meant by the potential electrolysis process?

difference between the two electrodes in an 20. Explain why water cannot be used in the electro-
electrochemical cell. chemical cell during the production of aluminum.
b. How is this potential difference measured?
What units are used? 21. Calculate the voltage of a cell in which the
12. The standard hydrogen electrode is assigned an overall reaction is the electrolysis of aqueous
electrode potential of 0.00 V. Explain why this cadmium chloride into its elements.
voltage is assigned.
22. According to electrochemical data, can Ni be
plated onto a zinc metal object using a nickel
nitrate solution? Explain.

E L E C T R O C H E M I S T R Y 673

CHAPTER REVIEW

23. Distinguish between a voltaic cell and an 32. How do the redox reactions for each of the fol-
electrolytic cell in terms of the nature of the lowing types of batteries differ?
reaction involved. a. zinc-carbon
b. alkaline
24. a. What is electroplating? c. mercury
b. Identify the anode and cathode in such
a process. 33. a. Why are some standard reduction potentials
positive and some negative?
MIXED REVIEW
b. Compare the E0 value for a metal with the
25. Predict whether each of the following reactions reactivity of that metal.
will occur spontaneously as written by deter-
mining the E0 value for potential reaction. CRITICAL THINKING
Write and balance the overall equation for each
reaction that does occur spontaneously. 34. Applying Models Explain how the oxidation-
a. Mg + Sn2+ ⎯→ reduction chemistry of both the voltaic cell
b. K + Al3+ ⎯→ and the electrolytic cell are combined in the
c. Li+ + Zn ⎯→ chemistry of rechargeable cells.
d. Cu + Cl2 ⎯→
35. Applying Ideas In lead-acid batteries, such as
26. Why is it possible for alkaline batteries to be your car battery, the degree of discharge of the
smaller than zinc-carbon dry cells? battery can be determined by measuring the
density of the battery fluid. Explain how this
27. Draw a diagram of a voltaic cell whose two is possible.
half-reactions consist of Ag in AgNO3 and Ni in
NiSO4. Identify the anode and cathode, and 36. Applying Ideas In lead-acid batteries, the
indicate the directions in which the electrons battery cannot be recharged indefinitely.
and ions are moving. Explain why not.

28. Can a solution of Sn(NO3)2 be stored in an 37. Interpreting Graphics A voltaic cell is
aluminum container? Explain, using E0 values. pictured below. Identify the species that is
oxidized if current is allowed to flow.
29. A voltaic cell is made up of a cadmium
electrode in a solution of CdSO4 and a zinc Digital voltmeter
electrode in a solution of ZnSO4. The two
half-cells are separated by a porous barrier. MnO4– Pt Cr2O72–
a. Which is the cathode, and which is the Mn2+ Pt Cr3+
anode?
b. In which direction are the electrons flowing? H+ H+
c. Write balanced equations for the two half-
reactions, and write a net equation for the
combined reaction.

30. Would the following pair of electrodes make a
good battery? Explain.
Cd ⎯→ Cd2+ + 2e−
Fe ⎯→ Fe2+ + 2e−

31. a. What would happen if an aluminum spoon
were used to stir a solution of Zn(NO3)2?

b. Could a strip of Zn be used to stir a solution
of Al(NO3)3? Explain, using E0 values.

674 C H A P T E R 2 0

CHAPTER REVIEW

USING THE HANDBOOK 42. In our portable society, batteries have become a
necessary power supply. As consumers, we want
38. Aluminum is described in Group 13 of the to purchase batteries that will last as long as
Elements Handbook as a self-protecting metal, possible. Advertisements tell us that some bat-
and can be used in preventing corrosion of iron teries last longer than others, but do they really?
structures. Using electrochemical data, explain Design an investigation to answer the question.
how aluminum protects iron structures. Is there a difference in longevity between the
major brands of AA batteries? Add a cost-effec-
RESEARCH & WRITING tiveness component to your design.

39. Go to the library, and find out about the 43. When someone who has a silver filling in a
electroplating industry in the United States. tooth bites down on an aluminum gum wrapper,
What are the top three metals used for plating, saliva acts as an electrolyte. The system is an
and how many metric tons of each are used for electrochemical cell that produces a small jolt of
electroplating each year in the United States? pain. Explain what occurs, using half-cell reac-
tions and E0 values.
40. Investigate the types of batteries being
considered for electric cars. Write a report on Graphing Calculator Equilibrium
the advantages and disadvantages of these Constant for an Electrochemical Cell
types of batteries. Go to go.hrw.com for a graphing calculator
exercise that asks you to calculate an equilib-
ALTERNATIVE ASSESSMENT rium constant for an electrochemical cell.

41. Performance Take an inventory of the types of Keyword: HC6ELEX
batteries used in your home. Find out the volt-
age supplied by each battery and what electro-
chemical reaction each uses. Suggest why that
electrochemical reaction is used in each case.

E L E C T R O C H E M I S T R Y 675

Math Tutor CALCULATING CELL POTENTIALS

You have learned that electrons are transferred in all oxidation-reduction reactions.
Electrons from a substance being oxidized are transferred to another substance being
reduced. A voltaic cell is a simple device that physically separates the oxidation reac-
tion from the reduction reaction, thus forcing electrons released during oxidation to
travel through a wire to reach the site of reduction. If a device, such as a light bulb or
motor, is placed in the circuit, the moving electrons can perform useful work.

The potential, or voltage, of a given voltaic cell depends on how strongly the oxida-
tion process tends to give up electrons in addition to how strongly the reduction
process tends to take them. The greater these two combined tendencies, the higher
the potential of the cell. A potential has been measured for each half-reaction, as you
can see in Table 1 on p. 664. In a voltaic cell, the process with the more-negative
reduction potential will proceed as the oxidation reaction at the anode of the cell. The
more positive reaction will proceed as reduction at the cathode. The following exam-
ple shows you how to determine the potentials of different kinds of voltaic cells.

Problem-Solving S

• In a voltaic cell, the process that has the more negative reduction potential will
proceed as the oxidation reaction at the anode of the cell.

SAMPLE

Calculate the potential of a voltaic cell in which Hg2+ ions are reduced to Hg metal while

Zn metal is oxidized to Zn2+ ions.

Reduction takes place at the cathode, so the cathode half-reaction is
Hg2+(aq) + 2e− ⎯→ Hg(l)

Oxidation takes place at the anode, so the anode half-reaction is
Zn(s) ⎯→ Zn2+(aq) + 2e−

To use the equation for cell potential, rewrite the anode half-reaction as a reduction
reaction, Zn2+(aq) + 2e− ⎯→ Zn(s).

Use Table 1 on page 664 to find the standard reduction potential for each half-reaction.

Then, calculate the cell potential.

Hg2+(aq) + 2e− ⎯→ Hg(l ) E0cathode = +0.85 V

Zn2+(aq) + 2e− ⎯→ Zn(s) E 0 = −0.76 V
anode

E0cell = E 0 − E0anode = +0.85 V − (−0.76 V) = 1.61 V
cathode

PRACTICE PROBLEMS

1. Calculate the potential of a voltaic cell in which 2. Calculate the potential of a cell in which
aluminum metal is oxidized to Al3+ ions while
Cu2+ ions are reduced to Cu+ ions. the reaction is
Pb(s) + Br2(l) ⎯→ Pb2+(aq) + 2Br−(aq).

676 C H A P T E R 2 0

Standardized Test Prep

Answer the following items on a separate piece of paper. 5.Refer to the table of reduction potentials below.
Which metal, Zn or Au, can reduce Sn2+ ions to
MULTIPLE CHOICE Sn metal when placed in an aqueous solution of
Sn2+ ions?
1.The electrode at which reduction occurs is A. Zn
A. always the anode. B. Au
B. always the cathode. C. Both Zn and Au can reduce Sn2+ ions
C. either the anode or the cathode. D. Neither Zn or Au can reduce Sn2+ ions
D. always the half-cell.
6.When silver is electroplated onto another metal,
2.Refer to the table of standard reduction poten- Ag+ is
A. oxidized at the anode.
tials below. A voltaic cell contains a strip of zinc B. reduced at the anode.
C. oxidized at the cathode.
metal in a solution containing zinc ions in one D. reduced at the cathode.

half-cell. The second is a strip of tin metal in a 7.Which metal would best provide cathodic protec-
tion from corrosion for an iron bridge?
solution containing tin ions. When this cell A. Au
B. Sn
operates, C. Cu
A. Sn is oxidized and Zn2+ is reduced. D. Mg
B. Sn is reduced and Zn2+ is oxidized.
C. Sn2+ is oxidized and Zn is reduced. SHORT ANSWER
D. Sn2+ is reduced and Zn is oxidized.
8.In the production of aluminum by the
3.When a rechargeable cell is being recharged, Hall-Héroult process, is the aluminum metal
produced at the cathode or the anode?
the cell acts as a(n)
9.The reduction potential for nitric acid is 0.96 V
A. fuel cell. C. voltaic cell. (reduction half-reaction is NO−3 + 4H+ + 3e− ⎯→
NO + 2H2O). If gold (Au) is placed in a beaker
B. electrolytic cell. D. Leclanche cell. of 1 M nitric acid, will the gold dissolve?

4.Refer to the table of standard reduction poten- EXTENDED RESPONSE

tials below. The standard potential of the cell 10.Given a voltaic cell consisting of a gold electrode
Sn⏐Sn2+⏐⏐Cr3+⏐Cr is −0.60 V. What is the stan- in a gold nitrate solution in one half-cell and a
dard reduction potential of the Cr3+/Cr zinc electrode in a zinc nitrate solution in the
other half-cell, what occurs once the wire is
electrode? connected to both electrodes?

A. +0.88 V C. −0.88 V 11.If pure water is placed in a beaker with wires
connected to a cathode and anode and electrical
B. +0.74 V D. −0.74 V voltage is applied, will electrolysis of water
occur? Explain your answer.
Half-cell reaction Standard reduction
Au3+ + 3e− ←⎯⎯→ Au potential, E0 (in volts)
Cu2+ + 2e− ←⎯⎯→ Cu
Fe2+ + 2e− ←⎯⎯→ Fe +1.50
Sn2+ + 2e− ←⎯⎯→ Sn +0.34
Zn2+ + 2e− ←⎯⎯→ Zn −0.41
Mg2+ + 2e− ←⎯⎯→ Mg −0.14
−0.76
−2.37

Remember that if you can elimi-
nate two of the four answer choices, your chances of
choosing the correct answer choice will double.

E L E C T R O C H E M I S T R Y 677

CHAPTER LAB BACKGROUND
In voltaic cells, oxidation and reduction half-reactions
Voltaic Cells take place in separate half-cells, which can consist of
a metal electrode immersed in a solution of its metal
OBJECTIVES ions. The electrical potential, or voltage, that devel-
• Construct a Cu-Zn voltaic cell. ops between the electrodes is a measure of the com-
• Design and construct two other voltaic cells. bined reducing strength of one reactant and
• Measure the potential of the voltaic cells. oxidizing strength of the other reactant.
• Evaluate cells by comparing the measured
SAFETY
cell voltages with the voltages calculated
from standard reduction potentials. For review of safety, please see Safety in the
MATERIALS Chemistry Laboratory in the front of your book.
• 0.5 M Al2(SO4)3, 75 mL
• 0.5 M CuSO4, 75 mL PREPARATION
• 0.5 M ZnSO4, 75 mL 1. Follow your teacher’s instructions to create the
• Aluminum strip, 1 cm × 8 cm data table that you will use to record your data
• Copper strip, 1 cm × 8 cm for three voltaic cells.
• Zinc strip, 1 cm × 8 cm
• Distilled water 2. Remove any oxide coating from strips of alu-
• 100 mL graduated cylinder minum, copper, and zinc by rubbing them with
• Emery cloth an emery cloth. Keep the metal strips dry until
• 150 mL beakers, 3 you are ready to use them.
• Salt bridge
• Voltmeter 3. Label three 150 mL beakers “Al2(SO4)3,”
• Wires with alligator clips, 2 “CuSO4,” and “ZnSO4.”

678 C H A P T E R 2 0 PROCEDURE
1. Pour 75 mL of 0.5 M ZnSO4 into the ZnSO4
beaker and 75 mL of 0.5 M CuSO4 into the
CuSO4 beaker.
2. Place one end of the salt bridge into the CuSO4
solution and the other end into the ZnSO4
solution.

3. Place a zinc strip into the zinc solution and a
copper strip into the copper solution.