Let's balance this step-by-step using the half-reaction method: Oxidation: Ag → Ag+ Reduction: 1/2O2 + H+ → H2O Balanced half-reactions: Ag → Ag+ 1/2O2 + H+ → H2O Overall balanced reaction: 1Ag + 1/2O2 + 1H+ → 1Ag+ + 1H2O The correct coefficients are: 1, 1/2, 1, 1, 1 The answer is 1.

1. Electrochemistry
• Spontaneous reactions that involve
electron transfer can be used to
generate electricity (Ex. Battery).
• Non-spontaneous reactions that involve
electron transfer can be forced to
proceed by the addition of an electric
current (Ex. electrolysis).

2. Chapter 20- Electrochemistry
• Section 20.1- Oxidation numbers

3. Oxidation Numbers
In order to keep
track of what
loses electrons
and what gains
them, we
assign oxidation
numbers.

4. Oxidation and Reduction

5. Oxidation and Reduction
• A species is oxidized when it loses electrons.

6. Oxidation and Reduction
• A species is oxidized when it loses electrons.
 Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.

7. Oxidation and Reduction

8. Oxidation and Reduction
• A species is reduced when it gains electrons.

9. Oxidation and Reduction
• A species is reduced when it gains electrons.
 Here, each of the H+ gains an electron and they
combine to form H2.

10. Oxidation and Reduction
• What is reduced is the oxidizing agent.
 H+ oxidizes Zn by taking electrons from it.

11. Oxidation and Reduction
• What is reduced is the oxidizing agent.
 H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.
 Zn reduces H+ by giving it electrons.

12. Assigning Oxidation Numbers
1. Elements in their elemental form have
an oxidation number of 0.

13. Assigning Oxidation Numbers
1. Elements in their elemental form have
an oxidation number of 0.
2. The oxidation number of a monatomic
ion is the same as its charge.

14. Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.

15. Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
 Oxygen has an oxidation number of −2,
except in the peroxide ion in which it has
an oxidation number of −1.

16. Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
 Oxygen has an oxidation number of −2,
except in the peroxide ion in which it has
an oxidation number of −1.
 Hydrogen is −1 when bonded to a metal,
+1 when bonded to a nonmetal.

17. Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
 Oxygen has an oxidation number of −2,
except in the peroxide ion in which it has
an oxidation number of −1.
 Hydrogen is −1 when bonded to a metal,
+1 when bonded to a nonmetal.
 Fluorine always has an oxidation number
of −1.

18. Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
 The other halogens have an oxidation
number of −1 when they are negative;
they can have positive oxidation
numbers, however, most notably in
oxyanions.

19. Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.

20. Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.
5. The sum of the oxidation numbers in a
polyatomic ion is the charge on the
ion.

21. For the reaction given below,
what substance is oxidized and
what is reduced?
3 NO2- + Cr2O72- + 8 H+ 2 Cr3+ + 3 NO3- + 4 H2O
1. N of NO2- is reduced, Cr of Cr2O72- is oxidized
2. N of NO2- is oxidized, Cr of Cr2O72- is reduced
3. O of NO2- is oxidized, Cr of Cr2O72- is reduced
4. Cr3+ is reduced, N of NO2- is oxidized
5. N of NO3- is oxidized, Cr3+ is reduced

22. For the reaction given below,
what substance is oxidized and
what is reduced?
3 NO2- + Cr2O72- + 8 H+ 2 Cr3+ + 3 NO3- + 4 H2O
1. N of NO2- is reduced, Cr of Cr2O72- is oxidized
2. N of NO2- is oxidized, Cr of Cr2O72- is reduced
3. O of NO2- is oxidized, Cr of Cr2O72- is reduced
4. Cr3+ is reduced, N of NO2- is oxidized
5. N of NO3- is oxidized, Cr3+ is reduced

23. Which species is oxidized and which is
reduced in the following reaction:
Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
1. Zn, oxidized; H+, reduced
2. H+, reduced; Zn, oxidized
3. Zn2+, oxidized; H2, reduced
4. H2, oxidized; Zn2+, reduced

24. Correct Answer:
1. Zn, oxidized; H+, reduced
2. H+, reduced; Zn, oxidized
3. Zn2+, oxidized; H2, reduced
4. H2, oxidized; Zn2+, reduced
The oxidation state of Zn goes
from 0 to +2 while the oxidation
state of H goes from +1 to 0.

25. PRACTICE EXERCISE
Identify the oxidizing and reducing agents in
the oxidation-reduction reaction

26. PRACTICE EXERCISE
Identify the oxidizing and reducing agents in
the oxidation-reduction reaction
Answer: Al(s) is the reducing agent; MnO4–
(aq) is the oxidizing agent.

27. Chapter 20- Electrochemistry
• Section 20.2- Balancing redox reaction
by the half-reaction method.

28. Balancing Oxidation-Reduction
Equations
Perhaps the easiest way to balance the
equation of an oxidation-reduction
reaction is via the half-reaction method.

29. Balancing Oxidation-Reduction
Equations
This involves treating (on paper only) the
oxidation and reduction as two separate
processes, balancing these half reactions,
and then combining them to attain the
balanced equation for the overall reaction.

30. Half-Reaction Method
1. Assign oxidation numbers to
determine what is oxidized and what is
reduced.

31. Half-Reaction Method
1. Assign oxidation numbers to
determine what is oxidized and what is
reduced.
2. Write the oxidation and reduction half-
reactions.

32. Half-Reaction Method
3. Balance each half-reaction.

33. Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.

34. Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O.

35. Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O.
c. Balance H by adding H+.

36. Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding electrons.

37. Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding electrons.
4. Multiply the half-reactions by integers
so that the electrons gained and lost
are the same.

38. Half-Reaction Method
5. Add the half-reactions, subtracting
things that appear on both sides.

39. Half-Reaction Method
5. Add the half-reactions, subtracting
things that appear on both sides.
6. Make sure the equation is balanced
according to mass.

40. Half-Reaction Method
5. Add the half-reactions, subtracting
things that appear on both sides.
6. Make sure the equation is balanced
according to mass.
7. Make sure the equation is balanced
according to charge.

41. 1. Yes
2. No

42. 1. Yes
2. No

43. Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq) → Mn2+(aq) + CO2(aq)

44. Half-Reaction Method
First, we assign oxidation numbers.
+7 +3 +2 +4
MnO4− + C2O42- → Mn2+ + CO2

45. Half-Reaction Method
First, we assign oxidation numbers.
+7 +3 +2 +4
MnO4− + C2O42- → Mn2+ + CO2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.

46. Oxidation Half-Reaction
C2O42− → CO2

47. Oxidation Half-Reaction
C2O42− → CO2
To balance the carbon, we add a
coefficient of 2:
C2O42− → 2 CO2

48. Oxidation Half-Reaction
C2O42− → 2 CO2

49. Oxidation Half-Reaction
C2O42− → 2 CO2
The oxygen is now balanced as well.
To balance the charge, we must add 2
electrons to the right side.
C2O42− → 2 CO2 + 2 e−

50. Reduction Half-Reaction
MnO4− → Mn2+

51. Reduction Half-Reaction
MnO4− → Mn2+
The manganese is balanced; to balance
the oxygen, we must add 4 waters to
the right side.
MnO4− → Mn2+ + 4 H2O

52. Reduction Half-Reaction
MnO4− → Mn2+ + 4 H2O

53. Reduction Half-Reaction
MnO4− → Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+
to the left side.
8 H+ + MnO4− → Mn2+ + 4 H2O

54. Reduction Half-Reaction
8 H+ + MnO4− → Mn2+ + 4 H2O

55. Reduction Half-Reaction
8 H+ + MnO4− → Mn2+ + 4 H2O
To balance the charge, we add 5 e− to
the left side.
5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O

56. Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42− → 2 CO2 + 2 e−
5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O

57. Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42− → 2 CO2 + 2 e−
5 e− + 8 H+ + MnO4− → Mn2+ + 4 H2O
To attain the same number of electrons
on each side, we will multiply the first
reaction by 5 and the second by 2.

58. Combining the Half-Reactions
5 C2O42− → 10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4− → 2 Mn2+ + 8 H2O

59. Combining the Half-Reactions
5 C2O42− → 10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4− → 2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− →
2 Mn2+ + 8 H2O + 10 CO2 +10 e−

60. Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− →
2 Mn2+ + 8 H2O + 10 CO2 +10 e−

61. Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− →
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− →
2 Mn2+ + 8 H2O + 10 CO2

62. Balance the following oxidation-reduction
reaction that occurs in acidic solution:
C2O42− + MnO4− → Mn2+ + CO2
1. 8 H+ + 5 C2O42− + MnO4− → Mn2+ + 4 H2O + 10 CO2
2. 16 H+ + 2 C2O42− + 2 MnO4− → 2 Mn2+ + 8 H2O + 4 CO2
3. 16 H+ + 5 C2O42− + 2 MnO4− → 2 Mn2+ + 8 H2O + 10
CO2
4. C2O42−+ MnO4− → Mn2+ + 2 CO2 + 2O2

63. Correct Answer:
1. 8 H+ + 5 C2O42− + MnO4− → Mn2+ + 4 H2O + 10 CO2
2. 16 H+ + 2 C2O42− + 2 MnO4− → 2 Mn2+ + 8 H2O + 4 CO2
3. 16 H+ + 5 C2O42− + 2 MnO4− → 2 Mn2+ + 8 H2O + 10
CO2
4. C2O42−+ MnO4− → Mn2+ + 2 CO2 + 2O2
Conservation of mass and charge must
be maintained on both reactants’ and
products’ side; practice using the
method of half-reactions.

64. When the following reaction is
balanced, what are the coefficients
for each substance?
__Ag + __O2 + __H+  __Ag+ + __H2O
1. 1, 1, 2, 1, 1
2. 1, 1, 4, 1, 2
3. 1, 1, 2, 1, 2
4. 4, 1, 2, 1, 2
5. 4, 1, 4, 4, 2

65. When the following reaction is
balanced, what are the coefficients
for each substance?
__Ag + __O2 + __H+  __Ag+ + __H2O
1. 1, 1, 2, 1, 1
2. 1, 1, 4, 1, 2
3. 1, 1, 2, 1, 2
4. 4, 1, 2, 1, 2
5. 4, 1, 4, 4, 2

66. PRACTICE EXERCISE
Complete and balance the following
equations using the method of half-
reactions. Both reactions occur in acidic
solution.

67. PRACTICE EXERCISE
Complete and balance the following
equations using the method of half-
reactions. Both reactions occur in acidic
solution.

68. Balancing in Basic Solution
• If a reaction occurs in basic solution, one
can balance it as if it occurred in acid.

69. Balancing in Basic Solution
• If a reaction occurs in basic solution, one
can balance it as if it occurred in acid.
• Once the equation is balanced, add OH−
to each side to “neutralize” the H+ in the
equation and create water in its place.

70. Balancing in Basic Solution
• If a reaction occurs in basic solution, one
can balance it as if it occurred in acid.
• Once the equation is balanced, add OH−
to each side to “neutralize” the H+ in the
equation and create water in its place.
• If this produces water on both sides, you
might have to subtract water from each
side.

71. Balance the following oxidation-reduction
reaction that occurs in basic solution:
CN− + MnO4− → CNO− + MnO2
1. CN− + MnO4− + 2 OH− → CNO− + MnO2 + H2O
2. 2 CN− + 2 MnO4− + 2 OH− →
2 CNO− + 2 MnO2 + 4 OH−
3. 2 CN− + MnO4− → 2 CNO− + MnO2 + O2
4. 3 CN− + 2 MnO4− → 3 CNO− + 2 MnO2 + 2 OH−

72. Correct Answer:
1. CN− + MnO4− + 2 OH− → CNO− + MnO2 + H2O
2. 2 CN− + 2 MnO4− + 2 OH− →
2 CNO− + 2 MnO2 + 4 OH−
3. 2 CN− + MnO4− → 2 CNO− + MnO2 + O2
4. 3 CN− + 2 MnO4− → 3 CNO− + 2 MnO2 + 2 OH−
Conservation of mass and charge must
be maintained on both reactants’ and
products’ side; practice using the
method of half-reactions.

73. PRACTICE EXERCISE
Complete and balance the following
equations for oxidation-reduction reactions
that occur in basic solution:

74. PRACTICE EXERCISE
Complete and balance the following
equations for oxidation-reduction reactions
that occur in basic solution:

75. Chapter 20- Electrochemistry
• Section 20.3- Voltaic cells

76. Voltaic Cells
In spontaneous
oxidation-reduction
(redox) reactions,
electrons are
transferred and
energy is released.

77. Voltaic Cells
• We can use that
energy to do work if
we make the
electrons flow
through an external
device.
• We call such a setup
a voltaic cell.

78. Voltaic Cells
• A typical cell looks
like this.
• The oxidation occurs
at the anode.
• The reduction
occurs at the
cathode.

79. Voltaic Cells
Once even one
electron flows from
the anode to the
cathode, the
charges in each
beaker would not be
balanced and the
flow of electrons
would stop.

80. Voltaic Cells
• Therefore, we use a
salt bridge, usually a
U-shaped tube that
contains a salt
solution, to keep the
charges balanced.
 Cations move toward
the cathode.
 Anions move toward
the anode.

81. Voltaic Cells
• In the cell, then,
electrons leave the
anode and flow
through the wire to
the cathode.
• As the electrons
leave the anode, the
cations formed
dissolve into the
solution in the anode
compartment.

82. Voltaic Cells
• As the electrons
reach the cathode,
cations in the
cathode are
attracted to the now
negative cathode.
• The electrons are
taken by the cation,
and the neutral
metal is deposited
on the cathode.

84. because positive charge builds up in the anode and
must be neutralized as oxidation takes place there.

86. the surface Zn atoms

87. SAMPLE EXERCISE 20.4 Reactions in a Voltaic Cell
The oxidation-reduction reaction
is spontaneous. The voltaic cell utilizing this redox
reaction generates an electric current.
Indicate the reaction occurring at the
anode, the reaction at the cathode, the
direction of electron migration, the
direction of ion migration, and the signs of
the electrodes.

88. SAMPLE EXERCISE 20.4 Reactions in a Voltaic Cell
The oxidation-reduction reaction
is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of
KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not
react with either solution (such as platinum foil) is suspended in each solution, and the two conductors
are connected with wires through a voltmeter or some other device to detect an electric current. The
resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the
reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs
of the electrodes.
Solve: In one half-reaction, Cr2O72–(aq) is converted into Cr3+(aq). Starting with these ions and then completing
and balancing the half-reaction, we have
In the other half-reaction, I–(aq) is converted to I2(s):
Now we can use the summary in Figure 20.6 to help us describe the voltaic cell. The first half-reaction
is the reduction process (electrons shown on the reactant side of the equation), and by definition, this
process occurs at the cathode. The second half-reaction is the oxidation (electrons shown on the product
side of the equation), which occurs at the anode. The I– ions are the source of electrons, and the Cr2O72–
ions accept the electrons. Hence, the electrons flow through the external circuit from the electrode
immersed in the KI solution (the anode) to the electrode immersed in the K2Cr2O7 – H2SO4 solution (the
cathode). The electrodes themselves do not react in any way; they merely provide a means of transferring
electrons from or to the solutions. The cations move through the solutions toward the cathode, and the
anions move toward the anode. The anode (from which the electrons move) is the negative electrode, and
the cathode (toward which the electrons move) is the positive electrode.

89. PRACTICE EXERCISE
The two half-reactions in a voltaic cell are
(a) Indicate which reaction occurs at the
anode and which at the cathode. (b) Which
electrode is consumed in the cell reaction?
(c) Which electrode is positive?

90. PRACTICE EXERCISE
The two half-reactions in a voltaic cell are
(a) Indicate which reaction occurs at the
anode and which at the cathode. (b) Which
electrode is consumed in the cell reaction?
(c) Which electrode is positive?
Answer: (a) The first reaction occurs at the anode, the
second reaction at the cathode. (b) The anode (Zn) is
consumed in the cell reaction. (c) The cathode is
positive.

91. Chapter 20- Electrochemistry
• Section 20.4- Standard CELL EMF

92. Electromotive Force (emf)
• Water only
spontaneously flows
one way in a
waterfall.
• Likewise, electrons
only spontaneously
flow one way in a
redox reaction—from
higher to lower
potential energy.

93. Electromotive Force (emf)
• The potential difference between the
anode and cathode in a cell is called the
electromotive force (emf).
• It is also called the cell potential, and is
designated Ecell.

94. Cell Potential
Cell potential is measured in volts (V).
J
1V=1
C

95. Positive cell potentials =
spontaneous processes

97. YES, positive voltages = spontaneous

98. Standard Reduction Potentials
Reduction
potentials for
many
electrodes
have been
measured and
tabulated.

99. Given the following reaction, which is true?
Cu(s)+ 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° =+0.46 V
1. Plating Ag onto Cu is a spontaneous process.
2. Plating Cu onto Ag is a spontaneous process.
3. Plating Ag onto Cu is a nonspontaneous process.
4. Plating Cu onto Ag is a nonspontaneous process.
5. Energy will have to be put in for the reaction to
proceed.

100. Given the following reaction, which is true?
Cu(s)+ 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° =+0.46 V
1. Plating Ag onto Cu is a spontaneous process.
2. Plating Cu onto Ag is a spontaneous process.
3. Plating Ag onto Cu is a nonspontaneous process.
4. Plating Cu onto Ag is a nonspontaneous process.
5. Energy will have to be put in for the reaction to
proceed.

102. –
1 atm pressure for Cl2(g) and 1 M solution for Cl (aq).

103. Standard Hydrogen Electrode
• Their values are referenced to a standard
hydrogen electrode (SHE).
• By definition, the reduction potential for
hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e− → H2 (g, 1 atm)

104. Standard Cell Potentials
The cell potential at standard conditions
can be found through this equation:
Ecell = Ered (cathode) − Ered (anode)
° ° °
Because cell potential is based on
the potential energy per unit of
charge, it is an intensive property.

105. Cell Potentials
• For the oxidation in this cell,
Ered = −0.76 V
°
• For the reduction,
Ered = +0.34 V
°

106. Cell Potentials
Ecell = Ered (cathode) − Ered (anode)
° ° °
= +0.34 V − (−0.76 V)
= +1.10 V
Positive values = spontaneous

107. SAMPLE EXERCISE 20.5 Calculating Eºred from Ecell
º
For the Zn-Cu2+ voltaic cell shown in Figure
20.5, we have
Given that the standard reduction potential
of Zn2+ to Zn(s) is –0.76 V, calculate the
for the reduction of Cu2+ to Cu:

108. SAMPLE EXERCISE 20.5 Calculating Ered fromº Ecell º
For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have
Given that the standard reduction potential of Zn2+ to Zn(s) is –0.76 V, calculate the for the
reduction of Cu2+ to Cu:
Solution
Analyze: We are given and for Zn2+ and asked to calculate for Cu2+.
Plan: In the voltaic cell, Zn is oxidized and is therefore the anode. Thus, the given for Zn2+ is
(anode). Because Cu2+ is reduced, it is in the cathode half-cell. Thus, the unknown reduction potential
for Cu2+ is (cathode). Knowing and (anode), we can use Equation 20.8 to solve for
(cathode).
Solve:
Check: This standard reduction potential agrees with the one listed in Table 20.1.
Comment: The standard reduction potential for Cu2+ can be represented as and that
for Zn2+ as The subscript identifies the ion that is reduced in the reduction half-
reaction.

109. PRACTICE EXERCISE
A voltaic cell is based on the half-reactions
The standard emf for this cell is 1.46 V.
Using the data in Table 20.1, calculate
for the reduction of In3+ to In+.

110. PRACTICE EXERCISE
A voltaic cell is based on the half-reactions
The standard emf for this cell is 1.46 V.
Using the data in Table 20.1, calculate
for the reduction of In3+ to In+.
Answer: –0.40 V

111. SAMPLE EXERCISE 20.6 Calculating Eºcell from Ered
º
Using the standard reduction potentials listed
in Table 20.1, calculate the standard emf for
the voltaic cell described in Sample Exercise
20.4, which is based on the reaction

112. SAMPLE EXERCISE 20.6 Calculating Ecell from Eºred
º
Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic
cell described in Sample Exercise 20.4, which is based on the reaction
Solution
Analyze: We are given the equation for a redox reaction and asked to use data in Table 20.1 to
calculate the standard emf (standard potential) for the associated voltaic cell.
Plan: Our first step is to identify the half-reactions that occur at the cathode and the anode, which we
did in Sample Exercise 20.4. Then we can use data from Table 20.1 and Equation 20.8 to calculate
the standard emf.
Solve: The half-reactions are
According to Table 20.1, the standard reduction potential for the reduction of Cr2O72– to Cr3+ is +1.33
V, and the standard reduction potential for the reduction of I2 to I– (the reverse of the oxidation half-
reaction) is +0.54 V. We then use these values in Equation 20.8.
Although the iodide half-reaction at the anode must be multiplied by 3 in order to obtain a balanced equation
for the reaction, the value of is not multiplied by 3. As we have noted, the standard reduction potential is
an intensive property, so it is independent of the specific stoichiometric coefficients.
Check: The cell potential, 0.79 V, is a positive number. As noted earlier, a voltaic cell must have a
positive emf in order to operate.

113. PRACTICE EXERCISE
Using data in Table 20.1, calculate the
standard emf for a cell that employs the
following overall cell reaction:

114. PRACTICE EXERCISE
Using data in Table 20.1, calculate the
standard emf for a cell that employs the
following overall cell reaction:
Answer: +2.20 V

115. SAMPLE EXERCISE 20.7 From Half-Reactions to Cell EMF
A voltaic cell is based on the following two
standard half-reactions:
By using the data in Appendix E, determine
(a) the half-reactions that occur at the
cathode and the anode, and
(b) the standard cell potential.

116. SAMPLE EXERCISE 20.7 From Half-Reactions to Cell EMF
A voltaic cell is based on the following two standard half-reactions:
By using the data in Appendix E, determine (a) the half-reactions that occur at the cathode and the
anode, and (b) the standard cell potential.
Solve: (a) According to Appendix E,
The
standard reduction potential for Sn2+ is more positive (less negative) than that for Cd2+; hence, the
reduction of
Sn2+ is the reaction that occurs at the cathode.
The anode reaction therefore is the loss of electrons by Cd.
(b) The cell potential is given by Equation 20.8.
Notice that it is unimportant that the values of both half-reactions are negative; the negative values
merely indicate how these reductions compare to the reference reaction, the reduction of H+(aq).
Check: The cell potential is positive, as it must be for a voltaic cell.

117. PRACTICE EXERCISE
A voltaic cell is based on a Co2+/Co half-cell
and an AgCl/Ag half-cell.
(a) What reaction occurs at the anode? (b)
What is the standard cell potential?

118. PRACTICE EXERCISE
A voltaic cell is based on a Co2+/Co half-cell
and an AgCl/Ag half-cell.
(a) What reaction occurs at the anode? (b)
What is the standard cell potential?

120. True

121. Oxidizing and Reducing Agents
• The strongest
oxidizers have the
most positive
reduction potentials.
• The strongest
reducers have the
most negative
reduction potentials.

122. SAMPLE EXERCISE 20.8 Determining the Relative Strengths of
Oxidizing Agents
Using Table 20.1, rank the following
ions in order of increasing strength as
oxidizing agents:
NO3–(aq), Ag+(aq), Cr2O72–(aq).
Solution
Plan: The more readily an ion is reduced (the more positive its value), the stronger it is as an
oxidizing agent.
Solve: From Table 20.1, we have
Because the standard reduction potential of Cr2O72– is the most positive, Cr2O72– is the strongest
oxidizing agent of the three. The rank order is Ag+ < NO3– < Cr2O72–.

123. PRACTICE EXERCISE
Using Table 20.1, rank the following
species from the strongest to the weakest
reducing agent: I–(aq), Fe(s), Al(s).

124. PRACTICE EXERCISE
Using Table 20.1, rank the following
species from the strongest to the weakest
reducing agent: I–(aq), Fe(s), Al(s).
Answer: Al(s) > Fe(s) > I–(aq)

126. Pb(s)

127. Oxidizing and Reducing Agents
The greater the
difference between
the two, the greater
the voltage of the
cell.

128. Which substance is the stronger oxidizing
agent?
• Br2
• O2
• NO3-
• H+
• Cl2

129. Which substance is the stronger oxidizing
agent?
• Br2
• O2
• NO3-
• H+
• Cl2

130. Which substance is the stronger reducing
agent?
• H2O2
• Mn2+
• NO
• I-
• Ag

131. Which substance is the stronger reducing
agent?
• H2O2
• Mn2+
• NO
• I-
• Ag

132. Calculate the emf of the following cell:
Zn(s)|Zn2+(aq, 1 M)|| H+(aq, 1 M)|H2(g, 1 atm)|Pt
E° (Zn/Zn2+)= −0.76 V.
1. +0.76 V
2. +1.52 V
3. −0.76 V
4. −1.52 V

133. Correct Answer:
1. +0.76 V E°cell = E°cathode − E°anode
2. +1.52 V
3. −0.76 V Zn is the anode, hydrogen at the
4. −1.52 V Pt wire is the cathode.
E°cell = E°cathode − E°anode = 0.00 V − (−0.76 V)
E°cell = +0.76 V

134. Calculate the emf produced by the
following voltaic cell reaction:
Zn + 2 Fe3+ → Zn2+ + 2 Fe2+
Zn2+ + 2 e− → Zn E° = −0.76 V
Fe3+ + e− → Fe2+ E° = 0.77 V
1. +0.01 V 3. −0.78 V
2. +0.78 V 4. +1.53 V

135. Correct Answer:
1. +0.01 V E°cell = E°cathode − E°anode
2. +0.78 V
3. −0.78 V
Zn is being oxidized at the
4. +1.53 V anode and Fe3+ is being reduced
at the cathode. Thus,
E°cell = E°cathode − E°anode = 0.77 V − (−0.76 V)
E°cell = +1.53 V

136. As written, is the following oxidation-
reduction equation spontaneous or non-
spontaneous?
Zn2+ + 2 Fe2+ → Zn + 2 Fe3+
Zn2+ + 2 e− → Zn E° = −0.76 V
Fe3+ + e− → Fe2+ E° = 0.77 V
1. Spontaneous
2. Nonspontaneous

137. Correct Answer:
In this case, the reduction
1. Spontaneous process is Zn2+ → Zn while the
2. Nonspontaneous oxidation process is Fe2+ →
Fe3+. Thus:
E° = E°red (reduction) − E°red (oxidation)
E° = −0.76 V − (0.77 V) = −1.53 V
A negative E° indicates a nonspontaneous process.

138. Chapter 20- Electrochemistry
• Section 20.5- Free Energy & EMF

139. Free Energy
Positive Ecell values indicate a
spontaneous process, so they must
have negative ΔG values.

140. SAMPLE EXERCISE 20.9 Spontaneous or Not?
Using standard reduction potentials
(Table 20.1), determine whether the
following reactions are spontaneous
under standard conditions.

141. SAMPLE EXERCISE 20.9 Spontaneous or Not?
Using standard reduction potentials (Table 20.1), determine whether the following reactions are
spontaneous under standard conditions.
Solution
Analyze: We are given two equations and must determine whether or not each is spontaneous.
Plan: To determine whether a redox reaction is spontaneous under standard conditions, we first need
to write its reduction and oxidation half-reactions. We can then use the standard reduction potentials
and Equation 20.10 to calculate the standard emf, E°, for the reaction. If a reaction is spontaneous, its
standard emf must be a positive number.
Solve: (a) In this reaction Cu is oxidized to Cu2+ and H+ is reduced to H2. The corresponding half-
reactions and associated standard reduction potentials are
Notice that for the oxidation, we use the standard reduction potential from Table 20.1 for the reduction
of Cu2+ to Cu. We now calculate E° by using Equation 20.10:

142. SAMPLE EXERCISE 20.9 continued
Because E° is negative, the reaction is not spontaneous in the direction written. Copper metal does
not react with acids in this fashion. The reverse reaction, however, is spontaneous and would have an
E° of +0.34 V:
Cu2+ can be reduced by H2.
(b) We follow a procedure analogous to that in (a):
In this case
Because the value of E° is positive, this reaction is spontaneous and could be used to build a voltaic
cell.

143. PRACTICE EXERCISE
Using the standard reduction potentials
listed in Appendix E, determine which of the
following reactions are spontaneous under
standard conditions:

144. PRACTICE EXERCISE
Using the standard reduction potentials
listed in Appendix E, determine which of the
following reactions are spontaneous under
standard conditions:
Answer: Reactions (b) and (c) are spontaneous.

145. Free Energy
ΔG for a redox reaction can be found by
using the equation
ΔG = −nFE
where n is the number of moles of
electrons transferred, and F is a
constant, the Faraday.
1 F = 96,485 C/mol = 96,485 J/V-mol

146. Free Energy
Under standard conditions,
ΔG° = −nFE°

147. Given the following reaction, which is true?
Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° =+0.46V

148. Given the following reaction, which is true?
Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° =+0.46V

149. SAMPLE EXERCISE 20.10 Determining ΔG° and K
(a) Use the standard reduction potentials
listed in Table 20.1 to calculate the standard
free-energy change, ΔG°, and the equilibrium
constant, K, at room temperature (T = 298 K)
for the reaction
(b) Suppose the reaction in part (a) was written
What are the values of E°, ΔG°, and K when
the reaction is written in this way?

150. SAMPLE EXERCISE 20.10 Determining ΔG° and K
(a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy
change, ΔG°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction
(b) Suppose the reaction in part (a) was written
What are the values of E°, ΔG°, and K when the reaction is written in this way?
Solution
Analyze: We are asked to determine ΔG° and K for a redox reaction, using standard reduction potentials.
Plan: We use the data in Table 20.1 and Equation 20.10 to determine E° for the reaction and then use E° in
Equation 20.12 to calculate ΔG°. We will then use Equation 19.22, ΔG° = –RT in K, to calculate K.
Solve: (a) We first calculate E° by breaking the equation into two half-reactions, as we did in Sample
Exercise 20.9, and then obtain values from Table 20.1 (or Appendix E):
Even though the second half-reaction has 4 Ag, we use the value directly from Table 20.1
because emf is an intensive property.
Using Equation 20.10, we have

151. SAMPLE EXERCISE 20.10 continued
The half-reactions show the transfer of four electrons. Thus, for this reaction n = 4. We now use
Equation 20.12 to calculate ΔG°:
The positive value of E° leads to a negative value of ΔG°.
Now we need to calculate the equilibrium constant, K, using ΔG° = –RT ln K. Because ΔG° is a large
negative number, which means the reaction is thermodynamically very favorable, we expect K to be
large.
K is indeed very large! This means that we expect silver metal to oxidize in acidic environments, in air,
to Ag+. Notice that the voltage calculated for the reaction was 0.43 V, which is easy to measure.
Directly measuring such a large equilibrium constant by measuring reactant and product
concentrations at equilibrium, on the other hand, would be very difficult.

152. SAMPLE EXERCISE 20.10 continued
(b) The overall equation is the same as that in part (a), multiplied by The half-reactions are
The values of are the same as they were in part (a); they are not changed by multiplying the half-
reactions by Thus, E° has the same value as in part (a):
Notice, though, that the value of n has changed to n = 2, which is the value in part (a). Thus, ΔG° is
half as large as in part (a).
Now we can calculate K as before:
Comment: E° is an intensive quantity, so multiplying a chemical equation by a certain factor will not
affect the value of E°. Multiplying an equation will change the value of n, however, and hence the
value of ΔG°. The change in free energy, in units of J/mol of reaction as written, is an extensive
quantity. The equilibrium constant is also an extensive quantity.

153. PRACTICE EXERCISE
For the reaction
(a) What is the value of n? (b) Use the
data in Appendix E to calculate ΔG°. (c)
Calculate K at T = 298 K.

154. PRACTICE EXERCISE
For the reaction
(a) What is the value of n? (b) Use the
data in Appendix E to calculate ΔG°. (c)
Calculate K at T = 298 K.
Answer:
(a) 6, (b) +87 kJ/mol, (c) K = 6 ×10–16

155. Chapter 20- Electrochemistry
• Section 20.6- Non-standard EMF

156. Nernst Equation
• Remember that
ΔG = ΔG° + RT ln Q
• This means
−nFE = −nFE° + RT ln Q

157. Nernst Equation
Dividing both sides by −nF, we get the
Nernst equation:
RT
E = E° − ln Q
nF
or, using base-10 logarithms,
2.303 RT
E = E° − log Q
nF

158. Nernst Equation
At room temperature (298 K),
2.303 RT
= 0.0592 V
F
Thus the equation becomes
0.0592
E = E° − log Q
n

159. Calculate the emf produced by the following
voltaic cell reaction.
[Zn2+] = 1.0 M, [Fe2+] = 0.1 M, [Fe3+] = 1.0 M
Zn + 2 Fe3+ → Zn2+ + 2 Fe2+
Zn2+ + 2 e− → Zn E° = −0.76 V
Fe3+ + e− → Fe2+ E° = 0.77 V
1. +1.47 V
2. +1.53 V
3. +1.59 V

160. Correct Answer:
E = E ° − (0.0592) logQ
n
1. +1.47 V (0.0592)
[Fe ] [Zn ]
2+ 2 2+
2. +1.53 V E = 1.53 − log
2 [Fe ] 3+ 2
3. +1.59 V
[0.1]2 [1.0]
E = 1.53 − (0.0592) log [ ]2
2 1.0
E = 1.53 − (0.0592) log (0.01) = 1.53 + 0.0592 = 1.59
2

161. Concentration Cells
• Notice that the Nernst equation implies that a cell
could be created that has the same substance at
both electrodes.
• For such a cell, Ecell would be 0, but Q would not.
°
• Therefore, as long as the concentrations
are different, E will not be 0.

162. SAMPLE EXERCISE 20.13 pH of a Concentration Cell
A voltaic cell is constructed with two hydrogen
electrodes. Electrode 1 has atm
and an unknown concentration of H+(aq).
Electrode 2 is a standard hydrogen electrode
([H+] = 1.00 M, atm). At 298 K the
measured cell voltage is 0.211 V, and the
electrical current is observed to flow from
electrode 1 through the external circuit to
electrode 2. Calculate[H+] for the solution at
electrode 1. What is its pH?

163. SAMPLE EXERCISE 20.13 pH of a Concentration Cell
A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has atm and an unknown
concentration of H+(aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, atm). At 298
K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1 through
the external circuit to electrode 2. Calculate[H+] for the solution at electrode 1. What is its pH?
Solution
Analyze: We are given the voltage of a concentration cell and the direction in which the current flows.
We also have the concentrations of all reactants and products except for [H+] in half-cell 1, which is
our unknown.
Plan: We can use the Nernst equation to determine Q and then use Q to calculate the unknown
concentration. Because this is a concentration cell, = 0 V.
Solve: Using the Nernst equation, we have
Because electrons flow from electrode 1 to electrode 2, electrode 1 is the anode of the cell and
electrode 2 is the cathode. The electrode reactions are therefore as follows, with the concentration of
H+(aq) in electrode 1 represented with the unknown x:

164. SAMPLE EXERCISE 20.13 continued
Thus,
At electrode 1, therefore,
and the pH of the solution is
Comment: The concentration of H+ at electrode 1 is lower than that in electrode 2, which is why
electrode 1 is the anode of the cell: The oxidation of H2 to H+(aq) increases [H+] at electrode 1.

165. PRACTICE EXERCISE
A concentration cell is constructed with two
Zn(s)-Zn2+(aq) half-cells. The first half-cell has
[Zn2+] = 1.35 M, and the second half-cell has
[Zn2+] = 3.75 × 10–4 M. (a) Which half-cell is
the anode of the cell? (b) What is the emf of
the cell?

166. PRACTICE EXERCISE
A concentration cell is constructed with two
Zn(s)-Zn2+(aq) half-cells. The first half-cell has
[Zn2+] = 1.35 M, and the second half-cell has
[Zn2+] = 3.75 × 10–4 M. (a) Which half-cell is
the anode of the cell? (b) What is the emf of
the cell?
Answer: (a) the second half-cell, (b) 0.105 V

167. Chapter 20- Electrochemistry
Applications
• Section 20.7-
of Oxidation-
Reduction Reactions

168. Batteries

169. Alkaline Batteries

170. • When the anode is gone, the battery is
“dead”.
• Rechargable batteries use electricity to
drive the reactions in the opposite
direction.
• This replaces the anode.

171. Hydrogen Fuel Cells

172. A primary battery cannot be recharged.
Which of the following batteries fits
this category?
1. Lead-acid battery
2. Nickel-cadmium
3. Alkaline battery
4. Lithium ion

173. Correct Answer:
1. Lead-acid battery In this list, only the
2. Nickel-cadmium alkaline battery is a
3. Alkaline battery primary battery and is
4. Lithium ion thus nonrechargeable.

174. Chapter 20- Electrochemistry
• Section 20.8- Corrosion

175. • Corrosion is the oxidation of metals to
form unwanted compounds. (Ex.
Rusting)
• Some metals form a protective oxide
coating that prevents further corrosion.
(Ex. Al forms protective Al2O3)

176. Corrosion and…

177. …Corrosion Prevention
a.k.a. cathodic
protection
Sacrificial anode

179. Al, Zn

180. Based on the standard reduction
potentials, which metal would not provide
cathodic protection to iron?
1. Magnesium
2. Nickel
3. Sodium
4. Aluminum

181. Correct Answer:
In order to provide cathodic
1. Magnesium protection, the metal that is
2. Nickel oxidized while protecting the
3. Sodium cathode must have a more negative
standard reduction potential. Here,
4. Aluminum
only Ni has a more positive
reduction potential (−0.28 V) than
Fe2+ (−0.44 V) and cannot be used
for cathodic protection.

182. Chapter 20- Electrochemistry
• Section 20.9- Electrolysis

183. Electrolysis
Voltaic cells use spontaneous redox
reactions to do work.
It is possible to use electrical energy
to force non-spontaneous redox
reactions to occur.
These electrolysis reactions occur
in electrolytic cells.

184. Cathode is still reduction
Anode is oxidation.
In electrolysis the electrons leave the
negative terminal of the voltage source
and enter the cathode where reduction
occurs

186. Electroplating
• Two electrodes immersed in a single solution
with voltage applied.
• The metal from the anode
will be oxidized into ions
that enter the solution.
• The cathode will be plated
with a thin layer of ions
from the solution.
171

188. Electroplating

189. • charge passing through circuit is measured in
coulombs.
• Coulombs = Amperes x seconds
• 96,500 C per mole of e-
YOU CAN DETERMINE # e- TRANSFERRED!!
• The number of electrons transferred is directly
proportional to the amount of substance that is
oxidized or reduced.
• Amount of solid plated is determined by the # of
electrons in the reduction 1/2 reaction.

190. Ni2+ is electrolyzed to Ni by a current of 2.43
amperes. If current flows for 600 s, how
much Ni is plated (in grams)?
(AW Ni = 58.7 g/mol)
1. 0.00148 g
2. 0.00297 g
3. 0.444 g
4. 0.888 g

191. Correct Answer:
i × t × FW
mass =
n×F
1. 0.00148 g
(2.43 A )× (600. s) × (58.7 g/mol)
2. 0.00297 g mass =
3. 0.444 g (2 × 96,500 C/mol)
4. 0.888 g

192. SAMPLE EXERCISE 20.14 Aluminum Electrolysis
Calculate the number of grams of aluminum
produced in 1.00 h by the electrolysis of
molten AlCl3 if the electrical current is 10.0 A.

193. SAMPLE EXERCISE 20.14 Aluminum Electrolysis
Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if
the electrical current is 10.0 A.
Solution
Analyze: We are told that AlCl3 is electrolyzed to form Al and asked to calculate the number of grams
of Al produced in 1.00 h with 10.0 A.
Plan: Figure 20.31 provides a road map of the problem. First, the product of the amperage and the
time in seconds gives the number of coulombs of electrical charge being used (Equation 20.18).
Second, the coulombs can be converted with the Faraday constant (F = 96,485 C/mole electrons) to
tell us the number of moles of electrons being supplied. Third, reduction of 1 mol of Al3+ to Al requires
three moles of electrons. Hence we can use the number of moles of electrons to calculate the number
of moles of Al metal it produces. Finally, we convert moles of Al into grams.
Solve: First, we calculate the coulombs of electrical charge that are passed into the electrolytic cell:
Second, we calculate the number of moles of electrons that pass into the cell:
Third, we relate the number of moles of electrons to the number of moles of aluminum being formed,
using the half-reaction for the reduction of Al3+:

194. SAMPLE EXERCISE 20.14 continued
Thus, three moles of electrons (3 F of electrical charge) are required to form 1 mol of Al:
Finally, we convert moles to grams:
Because each step involves a multiplication by a new factor, the steps can be combined into a single
sequence of factors:

195. PRACTICE EXERCISE
(a) The half-reaction for formation of
magnesium metal upon electrolysis of
molten MgCl2 is
Calculate the mass of magnesium formed
upon passage of a current of 60.0 A for a
period of 4.00 ×103s. (b) How many seconds
would be required to produce 50.0 g of Mg
from MgCl2 if the current is 100.0 A?

196. PRACTICE EXERCISE
(a) The half-reaction for formation of
magnesium metal upon electrolysis of
molten MgCl2 is
Calculate the mass of magnesium formed
upon passage of a current of 60.0 A for a
period of 4.00 ×103s. (b) How many seconds
would be required to produce 50.0 g of Mg
from MgCl2 if the current is 100.0 A?
Answer: (a) 30.2 g of Mg, (b) 3.97 × 103 s