Chemistry, The Central Science, 10th edition

Theodore L. Brown; H. Eugene LeMay, Jr.;and Bruce E. Bursten

Chapter 20Electrochemistry

Troy WoodUniversity of BuffaloBuffalo, NY 2006, Prentice Hall

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+, reduced2. H+, reduced; Zn, oxidized3. Zn2+, oxidized; H2, reduced4. H2, oxidized; Zn2+, reduced

Correct Answer:

The oxidation state of Zn goes from 0 to +2 while the oxidation state of H goes from +1 to 0.

1. Zn, oxidized; H+, reduced2. H+, reduced; Zn, oxidized3. Zn2+, oxidized; H2, reduced4. H2, oxidized; Zn2+, reduced

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

Correct Answer:

Conservation of mass and charge must be maintained on both reactants’ and products’ side; practice using the method of half-reactions.

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

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

Balance the following oxidation-reduction reaction that occurs in basic solution:

CN + MnO4 CNO + MnO2

Correct Answer:

Conservation of mass and charge must be maintained on both reactants’ and products’ side; practice using the method of half-reactions.

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

1. +0.76 V2. +1.52 V3. 0.76 V4. 1.52 V

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.

Correct Answer:

Zn is the anode, hydrogen at the Pt wire is the cathode.

1. +0.76 V2. +1.52 V3. 0.76 V4. 1.52 V

E°cell = E°cathode E°anode

E°cell = E°cathode E°anode = 0.00 V (0.76 V)

E°cell = +0.76 V

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 V2. +0.78 V

3. 0.78 V4. +1.53 V

Correct Answer:

Zn is being oxidized at the anode and Fe3+ is being reduced at the cathode. Thus,

E°cell = E°cathode E°anode

E°cell = E°cathode E°anode = 0.77 V (0.76 V)

E°cell = +1.53 V

1. +0.01 V2. +0.78 V3. 0.78 V4. +1.53 V

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. Spontaneous2. Nonspontaneous

Correct Answer:In this case, the reduction

process is Zn2+ Zn while the

oxidation process is Fe2+ Fe3+. Thus:

1. Spontaneous2. Nonspontaneous

E° = E°red (reduction) E°red (oxidation)

E° = 0.76 V − (0.77 V) = 1.53 V

A negative E° indicates a nonspontaneous process.

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 V1. +1.47 V2. +1.53 V3. +1.59 V

Correct Answer:

Qn

EE log(0.0592)

°

2 3

22 2

Fe

ZnFe log

2

(0.0592)1.53

E

2

2

1.0

1.00.1 log

2

(0.0592)1.53 E

1.59 0.0592 1.53 (0.01) log 2

(0.0592)1.53 E

1. +1.47 V2. +1.53 V3. +1.59 V

A primary battery cannot be recharged. Which of the following batteries fits this category?

1. Lead-acid battery 2. Nickel-cadmium3. Alkaline battery4. Lithium ion

Correct Answer:

In this list, only the alkaline battery is a primary battery and is thus nonrechargeable.

1. Lead-acid battery 2. Nickel-cadmium3. Alkaline battery4. Lithium ion

Based on the standard reduction potentials, which metal would not provide cathodic protection to iron?

1. Magnesium2. Nickel3. Sodium4. Aluminum

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. Magnesium2. Nickel3. Sodium4. Aluminum

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 g2. 0.00297 g3. 0.444 g4. 0.888 g

Correct Answer:

Fn

FWti

mass

C/mol) 96,500(2

g/mol) (58.7s) (600. A2.43mass

g 0.444mass

1. 0.00148 g2. 0.00297 g3. 0.444 g4. 0.888 g