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Balancing Redox Reactions and Standard Reduction Potentials, Slides of Electrochemistry

A series of redox reactions, balanced reactions, and standard reduction potentials for various chemical species. It includes examples of disproportionation reactions, electrolysis reactions, and thermodynamic considerations for electrochemical cells.

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Homework #6
Chapter 11
Electrochemistry
Chapter 4
83. a) Oxidation ½ Reaction
Fe + HCl HFeCl4
Fe + 4HCl HFeCl4
Fe + 4HCl HFeCl4 + 3H+
Fe + 4HCl HFeCl4 + 3H+ + 3e-
Reduction ½ Reaction
H2
2H+ H2
2H+ + 2e- H2
Balanced Reaction
2(Fe + 4HCl HFeCl4 + 3H+ + 3e-)
3(2H+ + 2e- H2)
2Fe(s) + 8HCl(aq) 2HFeCl4(aq) + 3H2(g)
b) Oxidization ½ Reaction
I- I3-
3I- I3-
3I- I3- + 2e-
Reduction ½ Reaction
IO3- I3-
3IO3- I3-
3IO3- I3- + 9H2O
3IO3- + 18H+ I3- + 9H2O
3IO3- + 18H++ 16e- I3- + 9H2O
Balanced Reaction
3IO3- + 18H++ 16e- I3- + 9H2O
8(3I- I3- + 2e-)
3IO3-(aq) + 18H+(aq) + 24I-(aq) 9I3-(aq) + 9H2O(l)
Divide trough by 3
IO3-(aq) + 6H+(aq) + 8I-(aq) 3I3-(aq) + 3H2O(l)
c) Oxidation ½ Reaction
Cr(NCS)64- Cr3+ + NO3- + CO2 + SO42-
Cr(NCS)64- Cr3+ + 6NO3- + 6CO2 + 6SO42-
Cr(NCS)64- + 54H2O Cr3+ + 6NO3- + 6CO2 + 6SO42-
Cr(NCS)64- + 54H2O Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+
Cr(NCS)64- + 54H2O Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e-
Reduction ½ Reaction
Ce4+ Ce3+
Ce4+ + e- Ce3+
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13

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Homework #

Chapter 11

Electrochemistry

Chapter 4

8 3. a) Oxidation ½ Reaction

Fe + HCl  HFeCl 4

Fe + 4HCl  HFeCl 4

Fe + 4HCl  HFeCl 4

+ 3H

Fe + 4HCl  HFeCl 4

+ 3H

  • 3e

Reduction ½ Reaction

 H

2

2H

 H

2

2 H

  • 2e-  H 2

Balanced Reaction

2(Fe + 4HCl  HFeCl 4

+ 3H

  • 3e

3( 2 H

  • 2e-  H 2

2Fe(s) + 8HCl(aq)  2HFeCl 4

(aq) + 3H 2

(g)

b) Oxidization ½ Reaction

I

 I

3

3I

 I

3

3I

 I

3

  • 2e

Reduction ½ Reaction

IO

3

 I

3

3 IO

3

 I

3

3IO

3

 I

3

+ 9H

2

O

3 IO

3

+ 18H

 I

3

+ 9H

2

O

3 IO

3

+ 18H

  • 16e

 I

3

+ 9H

2

O

Balanced Reaction

3IO

3

+ 18H

  • 16e

 I

3

+ 9H

2

O

8 (3I

 I

3

  • 2e

3 IO

3

(aq) + 18H

(aq) + 24I

(aq)  9 I 3

(aq) + 9H 2

O(l)

Divide trough by 3

IO

3

(aq) + 6H

(aq) + 8I

(aq)  3 I 3

(aq) + 3H 2

O(l)

c) Oxidation ½ Reaction

Cr(NCS) 6

4 -

 Cr

3+

+ NO

3

+ CO

2

+ SO

4

2 -

Cr(NCS) 6

4 -

 Cr

3+

+ 6NO

3

+ 6CO

2

+ 6SO

4

2 -

Cr(NCS) 6

4 -

+ 54H

2

O  Cr

3+

+ 6NO

3

+ 6CO

2

+ 6SO

4

2 -

Cr(NCS) 6

4 -

+ 54H

2

O  Cr

3+

+ 6NO

3

+ 6CO

2

+ 6SO

4

2 -

+ 108H

Cr(NCS) 6

4 -

+ 54H

2

O  Cr

3+

+ 6NO

3

+ 6CO

2

+ 6SO

4

2 -

+ 108H

  • 97e

Reduction ½ Reaction

Ce

4+

 Ce

3+

Ce

4+

  • e

 Ce

3+

Balanced Reaction

Cr(NCS) 6

4 -

+ 54H

2

O  Cr

3+

+ 6NO

3

+ 6CO

2

+ 6SO

4

2 -

+ 108H

  • 97e

97 (Ce

4+

  • e

 Ce

3+

Cr(NCS) 6

4 -

(aq)+ 54H 2

O(l) + 97Ce

4+

(aq)

 Cr

3+

(aq) + 6NO 3

(aq) + 6CO 2

(g) + 6SO 4

2

(aq) + 108H

(aq) + 97Ce

3+

(aq)

d) Oxidation ½ Reaction

CrI 3

 CrO 4

2 -

+ IO

4

CrI 3

 CrO 4

2 -

+ 3IO

4

CrI 3

+ 16H

2

O  CrO 4

2 -

+ 3IO

4

CrI 3

+ 16H

2

O  CrO 4

2 -

+ 3IO

4

+ 32H

CrI 3

+ 16H

2

O  CrO 4

2 -

+ 3IO

4

+ 32H

  • 27e

Reduction ½ Reaction

Cl 2

 Cl

Cl 2

 2Cl

Cl

2

  • 2e

 2 Cl

Balanced Equation

2(CrI 3

+ 16H

2

O  CrO 4

2 -

+ 3IO

4

+ 32H

  • 27e

27(Cl 2

  • 2e

 2Cl

2CrI 3

(s) + 32H 2

O(l) + 27Cl 2

(g)  2CrO 4

2 -

(aq) + 6IO 4

(aq) + 64H

(aq) + 54Cl

(aq)

The solution is basic not acidic, add 64 OH

to both sides

2CrI 3

(s) + 27Cl 2

(g) + 64OH

(aq)  2CrO 4

2 -

(aq) + 6IO 4

(aq) + 32H 2

O(l) + 54Cl

(aq)

e) Oxidation ½ Reaction

Fe(CN) 6

4 -

 Fe(OH) 3

+ CO

3

2 -

+ NO

3

Fe(CN) 6

4 -

 Fe(OH) 3

+ 6CO

3

2 -

+ 6NO

3

Fe(CN) 6

4 -

+ 39H

2

O  Fe(OH) 3

+ 6CO

3

2 -

+ 6NO

3

Fe(CN) 6

4 -

+ 39H

2

O  Fe(OH) 3

+ 6CO

3

2 -

+ 6NO

3

+ 75H

Fe(CN) 6

4 -

+ 39H

2

O  Fe(OH) 3

+ 6CO

3

2 -

+ 6NO

3

+ 75H

  • 61e

Reduction ½ Reaction

Ce

4+

 Ce(OH)

3

Ce

4+

+ 3H

2

O  Ce(OH) 3

Ce

4+

+ 3H

2

O  Ce(OH) 3

+ 3H

Ce

4+

+ 3H

2

O + e

 Ce(OH) 3

+ 3H

Balance Reaction

Fe(CN) 6

4 -

+ 39H

2

O  Fe(OH) 3

+ 6CO

3

2 -

+ 6NO

3

+ 75H

  • 61e

61 (Ce

4+

+ 3H

2

O + e

 Ce(OH) 3

+ 3H

Fe(CN) 6

4 -

(aq) + 222H 2

O(l) + 61Ce

4+

(aq)

 Fe(OH) 3

(s) + 6CO 3

2 -

(aq) + 6NO 3

(aq)+ 258H

(aq) + 61Ce(OH) 3

(s)

The solution is basic not acidic, add 258OH

to both sides

Fe(CN) 6

4 -

(aq) + 258 OH

(aq) + 61Ce

4+

(aq)

 Fe(OH) 3

(s) + 6 CO 3

2 -

(aq) + 6 NO 3

(aq)+ 61 Ce(OH) 3

(s) + 36H 2

O(l)

b) Reactions of Interest

Cu

2+

  • 2e

 Cu E°= 0.34 V

Mg

2+

  • 2e

 Mg E°=-2.37 V

Cu

2+

  • 2e

 Cu E°= 0.34 V

Mg Mg

2+

  • 2e

E° =2.37 V

Mg(s) + Cu

2+

(aq)  Cu(s) + Mg

2+

(aq) E° = 0.34 V + 2.37 V = 2.71 V

c) Reactions of Interest

IO

3

+ 6H

  • 5e

 ½I

2

+3H

2

O E°=1.20 V

Fe

3 +

  • e

 Fe

2+

E°= 0.77 V

IO

3

+ 6H

  • 5e

 ½I

2

+3H

2

O E° = 1.20 V

5(Fe

2+

 Fe

3+

  • e

) E°=-0.77 V

IO

3

(aq) + 6H

(aq) + 5Fe

2+

(aq)  ½I 2

(s) +3H 2

O(l) + 5Fe

3+

(aq) E°=1.20 V- 0 .77V= 0.43 V

Note: I 2

does not conduct electricity; therefore, the Pt electrode is used

d) Reactions of Interest

Ag

  • e

 Ag E°= 0.80 V

Zn

2+

  • 2e

 Zn E°= - 0.76 V

2(Ag

  • e

Ag) E° = 0.80 V

Zn  Zn

2+

  • 2e

E° = 0.76 V

2 Ag

(aq) + Zn(s)  2Ag(s) + Zn

2+

(aq) E° = 0.80 V + 0.76 V = 1.56 V

  1. a) Cl 2
  • 2e

 2Cl

E°= 1.36 V

2Br

 Br 2

  • 2e

E°=-1.09 V

Cl 2

(g) + 2Br

(aq)  2Cl

(aq) + Br 2

(aq) E°=1.36 V – 1.09 V= 0.27 V

Note: Br 2

is a liquid at room temperature

b)

5(IO

3

+ 2H

  • 2e

 IO

3

+H

2

O) E°= 1.60 V

2(Mn

2+

+ 4H

2

O  MnO 4

+ 8H

  • 5e

) E°= - 1.51 V

5 IO

3

(aq)+ 2 Mn

2+

(aq)+ 3 H 2

O(l)5IO 3

(aq)+2MnO 4

(aq)+6H

(aq) E°= 1.60V–1.51V=0.09V

b) For a galvanic cell E° must be positive therefore Cd must be oxidized

given:

VO

2

+ 2H

  • e

 VO

2+

+ H

2

O E° = 1.00V

Cd

2+

  • 2 e

 Cd E° = - 0.40 V

2(VO

2

+ 2H

  • e

 VO

2+

+ H

2

O) E° = 1.00V

Cd  Cd

2+

  • 2 e

E° = 0.40 V

2VO

2

(aq) +4H

(aq) + Cd(s)  2VO

2 +

(aq) +2H 2

O(l) + Cd

2+

(aq) E° = 1.40 V

  1. a) Cu

2+

  • 2e

 Cu E° = 0.34 V

In order for the cell to be a galvanic cell Cu

2+

must be reduced, therefore, SCE is

oxidized and at the anode.

cell

= 0.34 V + - 0.242 V = 0.10 V

b) Fe

3 +

  • e

 Fe

2+

E° = 0.77 V

In order for the cell to be a galvanic cell, Fe

3+

must be reduced, therefore, SCE is oxidized

and at the anode.

cell

= 0.77 V + - 0.242 V = 0.53 V

c) AgCl + e

 Ag + Cl

E° = 0.22 V

In order for the cell to be a galvanic cell, Ag must be oxidized, therefore, SCE is reduced

and at the cathode.

cell

= 0.242 V – 0.22 V = 0.02 V

d) Al

3 +

  • 3e

 Al E° = - 1.66 V

In order for the cell to be a galvanic cell, Al must be oxidized, therefore, SCE is reduced

and at the cathode.

cell

= 0.242 V + 1.66 V = 1.90 V

e) Ni

2 +

  • 2e

 Ni E° = - 0.23 V

In order for the cell to be a galvanic, cell Ni must be oxidized, therefore, SCE is reduced

and at the cathode.

cell

= 0.242 V + 0.23 V = 0.47 V

  1. a) Cu  Cu

2+

+2e

E°= - 0.34 V

2H

  • 2e

 H

2

E°= 0.0 V

No H

cannot oxidize Cu

b) 2I

I

2

+2e

E°=-0.54 V

Fe

3+

  • e

 Fe

2+

E°= 0.77 V

Yes Fe

3+

is capable of oxidizing I

if it is going to Fe

2+

2I

I

2

+2e

E°=-0.54 V

Fe

3+

  • 3 e

 Fe E°= - 0.04 V

No Fe

3+

cannot oxidize I

if it is going to Fe

c) Ag

  • e

 Ag E°=0.80 V

H

2

 2H

+2e

E°= 0.0 V

Yes H 2

is capable of reducing Ag

d) Cr

3+

  • e

 Cr

2+

E° = - 0.50 V

Fe

2+

 Fe

3+

  • e

E°=-0.77 V

No Fe

2+

is not capable of reducing Cr

3 +

to Cr

2+

  1. The oxidizing agent is the species that is reduced. Therefore, the best oxidizing reagent is the

species that has the largest E° value for the reduction ½ reaction.

K

< H

2

O < Cd

2+

< I

2

< AuCl 4

< IO

3

  1. The reducing agent is the species that is oxidized. Therefore, the best reducing agent is the

species that has the smallest E° value for the reduction ½ reaction.

F

< H

2

O < I

2

< Cu

< H

< K

  1. Choices

Cl 2

  • 2e

 2 Cl

E° = 1.36 V

Ag

  • e

 Ag E° = 0.80 V

Pb

2+

  • 2e-  Pb E° = - 0.13 V

Zn

2+

  • 2e

 Zn E° = - 0.76 V

Na

  • e

 Na E° = - 2.71 V

Underlined are possible answers

a) The oxidizing agent is the species that is being reduced or species in the reduction

reaction. Therefore Cl

, Ag, Pb, and Zn can be eliminated because if they were on the

reactants side of the reaction the reaction would be an oxidation reaction, therefore,

they can only be reducing agents. Out of the remaining species the best oxidizing agent

is the one with the largest E° value in the reduction ½ reactions.

Ag

b) The reducing agent is a species that is being oxidized or the species in the oxidizing

reaction. Therefore Ag

, Zn

2 +

, and Na

can be eliminated because if you flip the reaction

to make it an oxidation reaction they are on the product side of the reaction and not the

reactants, therefore, they can only be oxidizing agents. Out of the remaining species the

best oxidizing agent is the one with the most negative E° value in the reduction ½

reactions (the E° value will become positive when the reaction is flipped).

Zn

c) SO 4

2 -

+ 4H

  • 2e

 H

2

SO

3

+ H

2

O E°= 0.20 V

In order for SO 4

2 -

to oxidize a species the species has to be in an oxidation reaction;

therefore, Ag

, Zn

2+

, and Na

can be eliminated because they are on the reactant side of

reduction reactions. Of the remaining species the E° value of the reduction ½ reaction

must be smaller than 0. 20 V.

Pb and Zn

  1. a) Cu

2+

  • 2e

 Cu E° = 0.34 V

Cu

2+

  • e

 Cu

E° = 0.16 V

In order to reduce Cu

2+

to Cu but not to Cu

the E˚ values of the reduction ½ reaction

must be greater than 0.16 V and less than 0.34 V. The species also must be on the

product side of the reduction reaction.

HgCl 2

  • 2e

 2Hg + 2Cl

E° = 0.27 V

AgCl + e

 Ag + 4Cl

E° = 0.22 V

SO

4

2 -

+ 4H

  • 2e

 H

2

SO

4

+ H

2

O E°= 0.20 V

Therefore, Hg/Cl

, Ag/Cl

, and H 2

SO

3

are capable of reducing Cu

2+

to Cu but not to Cu

b) Br 2

  • 2e

2Br

E °= 1.09 V

I

2

  • 2e

 2I

E°=0.54 V

In order to reduce Br 2

to Br

but not I 2

to I

the E˚ values of the reduction ½ reaction

must be great than 0.54 V and less than 1.09 V. The species also must be on the product

side of the reduction reaction.

VO

2

+ 2H

  • e

 VO

2+

+ H

2

O E° = 1.00V

AuCl 4

  • 3e

 Au + 4Cl

E° = 0.99V

NO

3

+ 4H

  • 3e

 NO + 2H

2

O E° = 0.96V

ClO 2

  • e

 ClO 2

E°=0.954 V

2Hg

2+

  • 2e

 Hg 2

2+

E°=0.91 V

Ag

  • e

 Ag E° = 0.80 V

2Hg

2+

  • 2e

 2Hg E°=0.80 V

Fe

3 +

  • 2e

 Fe

2+

E° = 0.77 V

O

2

+ 2H

  • 2e

 H

2

O

2

E°=0.68 V

MnO 4

  • e

 MnO 4

2 -

E°=0.56 V

Therefore, VO

2+

,Au/Cl

, NO, ClO 2

, Hg 2

2+

, Ag, Hg, Fe

2+

, H

2

O

2

, and MnO 4

are capable of

reducing Br 2

to Br

but not I 2

to I

  1. a) 2(ClO 2

 ClO 2

+e

) E°=-0.95 V

Cl 2

  • 2e

 2Cl

E°= 1.36 V

2 ClO 2

(aq) + Cl 2

(g)  2 ClO 2

(g) + 2Cl

(aq) E°=1.36 V - 0.95 V = 0.41 V

Note: The Na

in the equation is just there as a spectator ion

𝐶

𝑚𝑜𝑙

𝐽

𝑚𝑜𝑙

𝑘𝐽

𝑚𝑜𝑙

𝑛𝐹𝐸°

𝑅𝑇 = 𝑒

( 2 )( 96 , 485

𝐶

𝑚𝑜𝑙

)( 0. 41 𝑉)

( 8. 3145

𝐽

𝑚𝑜𝑙∙𝐾

)( 298 𝐾)

= 7. 4 × 10

13

b) Assume acidic conditions

Reduction ½ Reaction

ClO 2

 Cl

ClO 2

 Cl

+ 2H

2

O

ClO 2

+ 4H

 Cl

+2H

2

O

ClO 2

+ 4H

  • 5e

 Cl

+2H

2

O

Oxidation ½ Reaction

ClO 2

 ClO 3

ClO 2

+ H

2

O ClO 3

ClO 2

+ H

2

O ClO 3

+ 2H

ClO 2

+ H

2

O ClO 3

+ 2H

  • e

Balanced Reaction

ClO 2

+ 4H

  • 5e

 Cl

+2H

2

O

5(ClO 2

+ H

2

O ClO 3

+ 2H

  • e

6ClO

2

(g) + 3H

2

O(l)  Cl

(aq) + 5ClO

3

(aq) + 6H

(aq)

  1. a) Reaction 1:

Unbalanced

Mn(s) + NO 3

(aq)  NO(g) + Mn

2+

(aq)

Oxidation ½ reaction

Mn  Mn

2+

Mn  Mn

2+

  • 2e

Reduction ½ reaction (The problem told you that you have nitric acid as a reactant. To

determine what species NO 3

forms, use the standard reduction potentials)

NO

3

 NO

NO

3

 NO + 2H

2

O

NO

3

+ 4H

 NO + 2H

2

O

NO

3

+ 4H

  • 3e

 NO + 2H

2

O

Balanced Reaction

3(Mn  Mn

2+

  • 2e

2(NO

3

+ 4H

  • 3e

 NO + 2H

2

O)

3Mn(s) + 2NO 3

(aq) +8H

(aq)  3Mn

2+

(aq) + 2NO(g) + 4H 2

O(l)

Reaction 2:

Unbalanced

Mn

2+

(aq)+ IO 4

(aq) MnO 4

(aq) + IO 3

(aq)

Oxidation ½ reaction

Mn

2+

 MnO 4

Mn

2+

+ 4H

2

O MnO 4

Mn

2+

+ 4H

2

O MnO 4

+ 8H

Mn

2+

+ 4H

2

O MnO 4

+ 8H

  • 5e

Reduction ½ Reaction

IO

4

 IO

3

IO

4

 IO

3

+ H

2

O

IO

4

+ 2H

  • 2e

 IO

3

+ H

2

O

IO

4

+ 2H

  • 2e

 IO

3

+ H

2

O

c) ΔG˚ is the maximum possible work that can be done. Increasing the temperature

increases the TΔS˚ term, therefore, since this term is +, less possible work can be done.

If E° vs. T was plotted on a graph, the intercept would equal −

∆𝐻°

𝑛𝐹

and the slope of the line

would equal

∆𝑆°

𝑛𝐹

. The smaller the −

∆𝑆°

𝑛𝐹

term is the smaller the temperature dependence of E.

Therefore, cells that have small ΔS° terms are relatively temperature independent.

  1. a) Cu

 Cu

2+

  • e

E°= - 0.16 V

Cu

  • e

 Cu E°= 0.52 V

2Cu

 Cu

2+

  • Cu E°= - 0.16 V + 0.52 V = 0.36 V

This reaction is spontaneous

𝐶

𝑚𝑜𝑙

𝐽

𝑚𝑜𝑙

𝑘𝐽

𝑚𝑜𝑙

𝑛𝐹𝐸°

𝑅𝑇

= 𝑒

( 1 )( 96 , 485

𝐶

𝑚𝑜𝑙

)( 0. 36 𝑉)

( 8. 3145

𝐽

𝑚𝑜𝑙∙𝐾

)( 298 𝐾)

= 1. 2 × 10

6

b) 2(Fe

2+

 Fe

3+

+e

) E°= - 0.44 V

Fe

2+

+2e

 Fe E°= - 0.77 V

3Fe

2+

 2Fe

3+

  • Fe E°=-0.44 V+ - 0.77 V=-1.21 V

This reaction is not spontaneous

c) HClO 2

+ H

2

O  ClO 3

+ 3H

  • 2e

E°=-1.21 V

HClO 2

+ 2H

  • 2e

 HClO + H 2

O E°=1.65 V

2HClO 2

(aq)  ClO 3

(aq) + H

(aq) + HClO(aq) E°= - 1.21 V + 1.65 V =0.44 V

𝐶

𝑚𝑜𝑙

𝐽

𝑚𝑜𝑙

𝑘𝐽

𝑚𝑜𝑙

𝑛𝐹𝐸°

𝑅𝑇 = 𝑒

( 2

) ( 96 , 485

𝐶

𝑚𝑜𝑙

)

(

  1. 44 𝑉

)

( 8. 3145

𝐽

𝑚𝑜𝑙∙𝐾

)

( 298 𝐾

)

= 7. 6 × 10

14

  1. a) For galvanic cell E° must be positive

Au

3+

+3e

 Au E° = 1.50 V

3(Tl  Tl

  • e

) E° = 0.34 V

Au

3+

(aq) + 3Tl(s)  Au(s) + 3Tl

(aq) E° = 1.50 V + 0.34 V = 1.84 V

b) ∆𝐺° = −𝑛𝐹𝐸° = −( 3 )( 96 , 485

𝐶

𝑚𝑜𝑙

)( 1. 84 𝑉) = − 5. 32 × 10

5 𝐽

𝑚𝑜𝑙

𝑘𝐽

𝑚𝑜𝑙

𝑛𝐹𝐸°

𝑅𝑇 = 𝑒

( 3

) ( 96 , 485

𝐶

𝑚𝑜𝑙

)

(

  1. 84 𝑉

)

( 8. 3145

𝐽

𝑚𝑜𝑙∙𝐾

)( 298 𝐾)

= 2. 26 × 10

93

c) 𝐸 = 𝐸° −

𝑅𝑇

𝑛𝐹

𝑅𝑇

𝑛𝐹

[𝑇𝑙

]

3

[𝐴𝑢

3 +

]

𝐽

𝑚𝑜𝑙∙𝐾

𝐶

𝑚𝑜𝑙

1. 0 × 10

− 4

3

1. 0 × 10

− 2

  1. 2(Cr

2+

Cr

3+

  • e

Co

2 +

  • 2e

 Co

2Cr

2+

  • Co

2 +

 2Cr

3+

𝐽

𝑚𝑜𝑙∙𝐾

𝐶

𝑚𝑜𝑙

2. 79 × 10

7

[

3 +

]

2

[

2 +

]

2

[

2 +

]

𝐽

𝑚𝑜𝑙∙𝐾

𝐶

𝑚𝑜𝑙

2

2

(

  1. 20

𝐶

𝑚𝑜𝑙

𝐽

𝑚𝑜𝑙

𝑘𝐽

𝑚𝑜𝑙

  1. Given

Al

3+

  • 3e

Al E° = - 1.66 V

Pb

2+

  • 2e

 Pb E° =- 0.13 V

2(Al Al

3+

  • 3e

) E° = 1.66 V

3(Pb

2+

  • 2e

 Pb) E° = - 0.13 V

2Al(s) + 3Pb

2+

(aq)  2Al

3+

(aq) + 2Pb(s) E° = 1.66 V + - 0.13 V= 1.53 V

Calculate the final Pb

2+

concentration:

Pb

2+

Al

3+

Initial 1.00 1.

Change - 3x +2x=0.

Final 1.00-3x 1.00+0.60=1.

x must equal 0.30 M, therefore, the Pb

2+

concentration will change by 0.90 M and the final Pb

2+

concentration equals 1.00-0.90=0.10 M.

Calculate E of cell

[𝐴𝑙

3 +

]

2

[

2 +

]

3

𝐽

𝑚𝑜𝑙∙𝐾

𝐶

𝑚𝑜𝑙

2

3

b) Hf

4+

  • 4e

 Hf

1 𝑚𝑜𝑙 𝐻𝑓

4 𝑚𝑜𝑙 𝑒

  1. 49 𝑔 𝐻𝑓

1 𝑚𝑜𝑙 𝐻𝑓

c) 2I

 I

2

+2e

1 𝑚𝑜𝑙 𝐼 2

2 𝑚𝑜𝑙 𝑒

  1. 80 𝑔 𝐼 2

1 𝑚𝑜𝑙 𝐼

2

2

d) CrO 3

+ 6H

  • 6e

 Cr + 3H 2

O

1 𝑚𝑜𝑙 𝐶𝑟

6 𝑚𝑜𝑙 𝑒

  1. 00 𝑔 𝐶𝑟

1 𝑚𝑜𝑙 𝐶𝑟

  1. The question wants you to calculate the molarity of the initial solution

We know the volume so we need to calculate the mole of Ag

The ½ reaction that we are interested in is

Ag

+e

 Ag

If we know the moles of electrons we can get the moles of Ag

. Calculate the moles of

electrons.

60 𝑠

1 𝑚𝑖𝑛

𝐶

𝑚𝑜𝑙

Calculate the moles of Ag

1 𝑚𝑜𝑙 𝐴𝑔

1 𝑚𝑜𝑙 𝑒

Calculate molarity

  1. The question asked you to calculate time to plate out 10 g Bi.

To calculate the time use

Therefore, need to determine the number of e

  • . To do this you must find a relationship between

e

and Bi. Balance the equation.

BiO

 Bi

BiO

 Bi + H 2

O

BiO

+ 2H

 Bi + H 2

O

BIO

+ 2H

  • 3e

 Bi + H 2

O

Note: If you balanced it in basic conditions you would also come out with 3 electrons

Determine the number of e

1 𝑚𝑜𝑙 𝐵𝑖

  1. 0 𝑔

3 𝑚𝑜𝑙 𝑒

1 𝑚𝑜𝑙 𝐵𝑖

Determine time

𝐶

𝑚𝑜𝑙

  1. For electrolysis reaction E° cell

is -

a) Cathode reaction Ni

2+

  • 2e

 Ni E°= - 0.23 V

Anode reaction 2Br

 Br 2

  • 2e

E°= - 1.09 V

b) Cathode reaction Al

3+

  • 3e

 Al E°= - 1.66 V

Anode reaction 2F

 F

2

  • 2e

E°= - 2.87 V

c) Cathode reaction Mn

2+

  • 2e

 Mn E°= - 1.18 V

Anode reaction 2I

 I

2

  • 2e

E°= - 0.54 V

For the aqueous solution we must also consider the reactions with H 2

O

d) Cathode reaction

Ni

2+

  • 2e

 Ni E°= - 0.23 V

2H

2

O + 2e

 H

2

+ 2OH

E°= - 0.83 V

Since it is easier to reduce Ni

2+

than H 2

O the nickel reaction will occur at the

cathode.

Anode reaction

2Br

 Br 2

  • 2e

E°= - 1.09 V

H

2

O  O

2

+ 4H

  • 4e

E°= - 1.23 V

Since it is easier to oxidize B

than H 2

O the bromine reaction will occur at the

anode.

e) Cathode reaction

Al

3+

  • 3e

 Al E°=-1.66 V

2H

2

O + 2e

 H

2

+ 2OH

E°= - 0.83 V

Since it is easier to reduce H 2

O than Al

3+

the water reaction will occur at the cathode.

Anode reaction

2F

 F

2

  • 2e

E°= - 2.87 V

H

2

O  O

2

+ 4H

  • 4e

E°= - 1.23 V

Since it is easier to oxidize H 2

O than F

the water reaction will occur at the anode.

f) Cathode reaction

Mn

2+

  • 2e

 Mn E°= - 1.18 V

2H

2

O + 2e

 H

2

+ 2OH

E°= - 0.83 V

Since it is easier to reduce H 2

O than Mn

2 +

the water reaction will occur at the cathode.

Anode reaction

2I

 I

2

  • 2e

E°=-0.54 V

H

2

O  O

2

+ 4H

  • 4e

E°=-1.23 V

Since it is easier to oxidize I

than H 2

O the iodine reaction will occur at the anode.

  1. The question wants you to determine the charge on the ruthenium. The reaction that we are

interested in is: Ru

n+

  • ne

 Ru. Therefore, we should be able to identify n by comparing the

moles of Ru to the moles of e-.

Calculate the moles of Ru

1 𝑚𝑜𝑙 𝑅𝑢

  1. 07 𝑔 𝑅𝑢

Calculate the moles of e

60 𝑠

1 𝑚𝑖𝑛

𝐶

𝑚𝑜𝑙

𝑚𝑎𝑥

𝐶

𝑚𝑜𝑙

= − 4. 75 × 10

5

𝐽

𝑚𝑜𝑙

𝑘𝐽

𝑚𝑜𝑙

This is the w max

for 2 moles of H 2

O. Need to find the work mass for 1.00 kg of H 2

O

1000 𝑔

1 𝑘𝑔

1 𝑚𝑜𝑙 𝐻

2

𝑂

  1. 02 𝑔 𝐻

2

𝑂

− 475 𝑘𝐽

2 𝑚𝑜𝑙 𝐻

2

𝑂

The work done can be no larger than - 13,200 kJ. Usually the max work is not done and some of

the energy is lost to heat. Fuel cells are more efficient in converting chemical energy into

electrical energy than combustion reaction. The main disadvantage of fuel cells is their cost.