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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
Reduction ½ Reaction
H
2
2H
H
2
2 H
Balanced Reaction
2(Fe + 4HCl HFeCl 4
+ 3H
3( 2 H
2Fe(s) + 8HCl(aq) 2HFeCl 4
(aq) + 3H 2
(g)
b) Oxidization ½ Reaction
I
I
3
3I
I
3
3I
I
3
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
I
3
+ 9H
2
O
Balanced Reaction
3IO
3
+ 18H
I
3
+ 9H
2
O
8 (3I
I
3
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
Reduction ½ Reaction
Ce
4+
Ce
3+
Ce
4+
Ce
3+
Balanced Reaction
Cr(NCS) 6
4 -
+ 54H
2
O Cr
3+
+ 6NO
3
+ 6CO
2
+ 6SO
4
2 -
+ 108H
97 (Ce
4+
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
Reduction ½ Reaction
Cl 2
Cl
Cl 2
2Cl
Cl
2
2 Cl
Balanced Equation
2(CrI 3
+ 16H
2
O CrO 4
2 -
+ 3IO
4
+ 32H
27(Cl 2
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
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
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+
Cu E°= 0.34 V
Mg
2+
Mg E°=-2.37 V
Cu
2+
Cu E°= 0.34 V
Mg Mg
2+
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
½I
2
+3H
2
O E°=1.20 V
Fe
3 +
Fe
2+
E°= 0.77 V
IO
3
+ 6H
½I
2
+3H
2
O E° = 1.20 V
5(Fe
2+
Fe
3+
) 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
Ag E°= 0.80 V
Zn
2+
Zn E°= - 0.76 V
2(Ag
Ag) E° = 0.80 V
Zn Zn
2+
E° = 0.76 V
2 Ag
(aq) + Zn(s) 2Ag(s) + Zn
2+
(aq) E° = 0.80 V + 0.76 V = 1.56 V
- a) Cl 2
2Cl
E°= 1.36 V
2Br
Br 2
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
IO
3
+H
2
O) E°= 1.60 V
2(Mn
2+
+ 4H
2
O MnO 4
+ 8H
) 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
VO
2+
+ H
2
O E° = 1.00V
Cd
2+
Cd E° = - 0.40 V
2(VO
2
+ 2H
VO
2+
+ H
2
O) E° = 1.00V
Cd Cd
2+
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
- a) Cu
2+
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.
E°
cell
= 0.34 V + - 0.242 V = 0.10 V
b) Fe
3 +
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.
E°
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.
E°
cell
= 0.242 V – 0.22 V = 0.02 V
d) Al
3 +
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.
E°
cell
= 0.242 V + 1.66 V = 1.90 V
e) Ni
2 +
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.
E°
cell
= 0.242 V + 0.23 V = 0.47 V
- a) Cu Cu
2+
+2e
E°= - 0.34 V
2H
H
2
E°= 0.0 V
No H
cannot oxidize Cu
b) 2I
I
2
+2e
E°=-0.54 V
Fe
3+
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+
Fe E°= - 0.04 V
No Fe
3+
cannot oxidize I
if it is going to Fe
c) Ag
Ag E°=0.80 V
H
2
2H
+2e
E°= 0.0 V
Yes H 2
is capable of reducing Ag
d) Cr
3+
Cr
2+
E° = - 0.50 V
Fe
2+
Fe
3+
E°=-0.77 V
No Fe
2+
is not capable of reducing Cr
3 +
to Cr
2+
- 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
- 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
- Choices
Cl 2
2 Cl
E° = 1.36 V
Ag
Ag E° = 0.80 V
Pb
2+
Zn
2+
Zn E° = - 0.76 V
Na
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
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
- a) Cu
2+
Cu E° = 0.34 V
Cu
2+
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
2Hg + 2Cl
E° = 0.27 V
AgCl + e
Ag + 4Cl
E° = 0.22 V
SO
4
2 -
+ 4H
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
2Br
E °= 1.09 V
I
2
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
VO
2+
+ H
2
O E° = 1.00V
AuCl 4
Au + 4Cl
E° = 0.99V
NO
3
+ 4H
NO + 2H
2
O E° = 0.96V
ClO 2
ClO 2
E°=0.954 V
2Hg
2+
Hg 2
2+
E°=0.91 V
Ag
Ag E° = 0.80 V
2Hg
2+
2Hg E°=0.80 V
Fe
3 +
Fe
2+
E° = 0.77 V
O
2
+ 2H
H
2
O
2
E°=0.68 V
MnO 4
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
- a) 2(ClO 2
ClO 2
+e
) E°=-0.95 V
Cl 2
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
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
Balanced Reaction
ClO 2
+ 4H
Cl
+2H
2
O
5(ClO 2
+ H
2
O ClO 3
+ 2H
6ClO
2
(g) + 3H
2
O(l) Cl
(aq) + 5ClO
3
(aq) + 6H
(aq)
- a) Reaction 1:
Unbalanced
Mn(s) + NO 3
(aq) NO(g) + Mn
2+
(aq)
Oxidation ½ reaction
Mn Mn
2+
Mn Mn
2+
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
NO + 2H
2
O
Balanced Reaction
3(Mn Mn
2+
2(NO
3
+ 4H
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
Reduction ½ Reaction
IO
4
IO
3
IO
4
IO
3
+ H
2
O
IO
4
+ 2H
IO
3
+ H
2
O
IO
4
+ 2H
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.
- a) Cu
Cu
2+
E°= - 0.16 V
Cu
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
E°=-1.21 V
HClO 2
+ 2H
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
)
(
- 44 𝑉
)
( 8. 3145
𝐽
𝑚𝑜𝑙∙𝐾
)
( 298 𝐾
)
= 7. 6 × 10
14
- a) For galvanic cell E° must be positive
Au
3+
+3e
Au E° = 1.50 V
3(Tl Tl
) 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
)
(
- 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
- 2(Cr
2+
Cr
3+
Co
2 +
Co
2Cr
2+
2 +
2Cr
3+
𝐽
𝑚𝑜𝑙∙𝐾
𝐶
𝑚𝑜𝑙
2. 79 × 10
7
[
3 +
]
2
[
2 +
]
2
[
2 +
]
𝐽
𝑚𝑜𝑙∙𝐾
𝐶
𝑚𝑜𝑙
2
2
(
- 20
𝐶
𝑚𝑜𝑙
𝐽
𝑚𝑜𝑙
𝑘𝐽
𝑚𝑜𝑙
- Given
Al
3+
Al E° = - 1.66 V
Pb
2+
Pb E° =- 0.13 V
2(Al Al
3+
) E° = 1.66 V
3(Pb
2+
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+
Hf
−
1 𝑚𝑜𝑙 𝐻𝑓
4 𝑚𝑜𝑙 𝑒
−
- 49 𝑔 𝐻𝑓
1 𝑚𝑜𝑙 𝐻𝑓
c) 2I
I
2
+2e
−
1 𝑚𝑜𝑙 𝐼 2
2 𝑚𝑜𝑙 𝑒
−
- 80 𝑔 𝐼 2
1 𝑚𝑜𝑙 𝐼
2
2
d) CrO 3
+ 6H
Cr + 3H 2
O
−
1 𝑚𝑜𝑙 𝐶𝑟
6 𝑚𝑜𝑙 𝑒
−
- 00 𝑔 𝐶𝑟
1 𝑚𝑜𝑙 𝐶𝑟
- 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
- 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
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
- 0 𝑔
3 𝑚𝑜𝑙 𝑒
−
1 𝑚𝑜𝑙 𝐵𝑖
Determine time
𝐶
𝑚𝑜𝑙
- For electrolysis reaction E° cell
is -
a) Cathode reaction Ni
2+
Ni E°= - 0.23 V
Anode reaction 2Br
Br 2
E°= - 1.09 V
b) Cathode reaction Al
3+
Al E°= - 1.66 V
Anode reaction 2F
F
2
E°= - 2.87 V
c) Cathode reaction Mn
2+
Mn E°= - 1.18 V
Anode reaction 2I
I
2
E°= - 0.54 V
For the aqueous solution we must also consider the reactions with H 2
O
d) Cathode reaction
Ni
2+
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
E°= - 1.09 V
H
2
O O
2
+ 4H
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+
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
E°= - 2.87 V
H
2
O O
2
+ 4H
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+
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
E°=-0.54 V
H
2
O O
2
+ 4H
E°=-1.23 V
Since it is easier to oxidize I
than H 2
O the iodine reaction will occur at the anode.
- The question wants you to determine the charge on the ruthenium. The reaction that we are
interested in is: Ru
n+
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
- 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
- 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.
