Chemistry,%20The%20Central%20Science,%2010th%20edition

1
Chapter 20Electrochemistry

• Chemistry, The Central Science, 10th edition
• Theodore L. Brown H. Eugene LeMay, Jr.
• and Bruce E. Bursten

Troy Wood University of Buffalo Buffalo, NY ?
2006, Prentice Hall
2
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

3
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.
4
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

5
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.
6
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?

7
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.
8

• 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

9
Correct Answer

1. 0.76 V
2. 1.52 V
3. ?0.76 V
4. ?1.52 V

Ecell Ecathode ? Eanode
Zn is the anode, hydrogen at the Pt wire is the
cathode.
Ecell Ecathode ? Eanode 0.00 V ? (?0.76
V) Ecell 0.76 V
10

• 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
2. 0.78 V

1. ?0.78 V
2. 1.53 V

11
Correct Answer

1. 0.01 V
2. 0.78 V
3. ?0.78 V
4. 1.53 V

Ecell Ecathode ? Eanode
Zn is being oxidized at the anode and Fe3 is
being reduced at the cathode. Thus,
Ecell Ecathode ? Eanode 0.77 V ? (?0.76
V) Ecell 1.53 V
12

• 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

13
Correct Answer
In this case, the reduction process is Zn2 ? Zn
while the oxidation process is Fe2 ? Fe3. Thus

1. Spontaneous
2. Nonspontaneous

E Ered (reduction) ? Ered (oxidation)
E ?0.76 V - (0.77 V) ?1.53 V
A negative E indicates a nonspontaneous process.
14

• 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

15
Correct Answer
(0.0592)
-


Q
E
E
log
n

1. 1.47 V
2. 1.53 V
3. 1.59 V

2



2
2
Zn
Fe
(0.0592)
-


log

1.53
E


2

2

3
Fe




2

1.0
0.1
(0.0592)
-


log

1.53
E


2

2
1.0
(0.0592)



-

1.59


0.0592


1.53


(0.01)

log

1.53
E
2
16
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

17
Correct Answer

1. Lead-acid battery
2. Nickel-cadmium
3. Alkaline battery
4. Lithium ion

In this list, only the alkaline battery is a
primary battery and is thus nonrechargeable.
18
Based on the standard reduction potentials, which
metal would not provide cathodic protection to
iron?

1. Magnesium
2. Nickel
3. Sodium
4. Aluminum

19
Correct Answer
In order to provide cathodic protection, the
metal that is oxidized while protecting the
cathode must have a more negative standard
reduction potential. Here, only Ni has a more
positive reduction potential (?0.28 V) than Fe2
(?0.44 V) and cannot be used for cathodic
protection.

1. Magnesium
2. Nickel
3. Sodium
4. Aluminum

20
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

21
Correct Answer
?
?
FW
t
i

mass
?
F
n

1. 0.00148 g
2. 0.00297 g
3. 0.444 g
4. 0.888 g

(
)
?
?
g/mol)

(58.7
s)

(600.
A
2.43

mass
?
C/mol)

96,500
(2