Cr 2 o 7 2 i 2 reducing agents i cr 3 so 4 2 2150

Cr
2
O
7
2
-
I
2
; reducing agents: I
-
Cr 3
+
SO
4
2
-
21.50 reducing agents rxn
A B A( s ) + B b
+
( aq )
B
B( s ) + A a
+
( aq )
B H
2
B( s ) + 2H
+
( aq )
B
B b
+
( aq ) + H
2
( g )
C A A( s ) + C c
+
( aq )

C( s ) + A a
+
( aq )
∴ C A B C( s ) + 2H + ( aq )
B
C
c
+ ( aq ) + H
2
( g )
(Metal C would cause formation of H
2
gas.)
21.51 a) copper b) oxidizing agent: Cu 2
+
; reducing agent: Fe
c) yes d) Cu
2
+
( aq ) + Fe( s )
B
Cu( s ) + Fe
2
+
( aq )
e) E °
cell
= E °
Cu
2
+
- E °
Fe
2
+
= 0.34V - ( - 0.44V)
= 0.78V
21.52 E
cell
= -
0.0592V
log
 
 
 
Q
n K
and ∆ G = - n FE
cell
a) When Q / K < 1, E
cell
0 and ∆ G < 0.
When Q / K = 1, E
cell
= 0 and ∆ G = 0.
When Q / K 1, E
cell
< 0 and ∆ G 0.
b) Only when Q / K < 1 will the reaction proceed spontaneously and be able to do work.

21.53 A( s ) + B
+
( aq )
B
A
+
( aq ) + B( s )
a) [A
+
] increases and [B
+
] decreases.
b) E
cell
decreases.
c) E
cell
= E °
cell
-
0.0592V
n
log
[A ]
[B ]
+
+
E
cell
will equal E °
cell
when log
[A ]
[B ]
+
+
= 0, which will occur when
[A ]
[B ]
+
+
= 1.
d) Yes. E
cell
will be less than E °
cell
when Q ( =
[A ]
[B ]
+
+
for this problem) is greater than 1.
21.54 a) E
cell
= E °
cell
-
2.303 RT
n F
log Q
E
cell
=
2.303 RT
n F
log K -
2.303 RT
n F
log Q
E
cell
=
2.303 RT
n F
log
 
 
 
K
Q
= -
2.303 RT
n F
log
 
 
 
K
Q
If Q/K < 1, E
cell
will decrease with a decrease in cell temperature. If Q / K 1, E
cell
will
increase (become less negative) with a decrease in cell temperature.
b) E
cell
= E °
cell
-
2.303 RT
n F
log
[active ion at anode]
[active ion at cathode]
E
cell
will decrease as the concentration of an active ion at the anode increases.
c) E
cell
will increase as the concentration of an active ion at the cathode increases.
d) E
cell
will increase as the pressure of a gaseous reactant in the cathode compartment increases.
21.55 In a concentration cell, the redox reaction proceeds in the direction to achieve concentration
equality in the half-cells.
cathode rxn: M + ( aq ) + e -
B
M( s )
anode rxn: M( s )
B
M
+
( aq ) + e
-
The more concentrated electrolyte is in the cathode compartment.
21.56 a) Ni( s ) + 2Ag
+
( aq )
h
2Ag( s ) + Ni 2
+
( aq )
E °
cell
= E °
Ag
+
- E °
Ni
2
+
= 0.80V - ( - 0.25V)
= 1.05V
E °
cell
=
0.0592V
n
log K
log K =
cell
0.0592V
ο
n E
=
2 x 1.05V
0.0592V
= 35.5
K = 3 x 10 35
b) 2Cr 3
+
( aq ) + 3Fe( s )
h
3Fe 2
+
( aq ) + 2Cr( s )
E °
cell
= E °
Cr
3
+
- E °
Fe
2
+
= - 0.74V - ( - 0.44V)
= - 0.30V

log K =
cell
0.0592V
ο
n E
=
6 x ( 0.30V)
0.0592V
-
= - 30.4
K = 4 x 10
-
31
21.57 a) 2Al( s ) + 3Cd 2
+
( aq )
h
2Al 3
+
( aq ) + 3Cd( s )
E °
cell
= E °
Cd
2
+
- E °
Al
3
+
= - 0.40V - ( - 1.66V)
= 1.26V
log K =
cell
0.0592V
ο
n E
=
6 x 1.26V
0.0592V
= 127.7
K = 5 x 10
127
b) I
2
( s ) + 2Br
-
( aq )
h
Br
2
( l ) + 2I
-
( aq )
E °
cell
= E °
I
2
- E °
Br
2
= 0.53V - (1.07V)
= - 0.54V
log K =
cell
0.0592V
ο
n E
=
2 x ( 0.54V)
0.0592V
-
= - 18.2
K = 6 x 10
-
19
21.58 a) 2 Ag( s ) + Mn 2
+
( aq )
h
2 Ag
+
( aq ) + Mn( s )
E °
cell
= E °
Mn
2
+
- E °
Ag
+
= - 1.18V - (0.80V)
= - 1.98V
log K =
cell
0.0592V
ο
n E
=
2 x ( 1.98V)
0.0592
-
= - 66.9
K = 1 x 10
-
67
b) Cl
2
( g ) + 2 Br
-
( aq )
h
Br
2
( l ) + 2 Cl
-
( aq )
E °
cell
= E °
Cl
2
- E °
Br
2
= 1.36V - (1.07V)
= 0.29V
log K =
cell
0.0592V
ο
n E
=
2 x (0.29V)
0.0592
= 9.8
K = 6 x 10
9
21.59 a) 2 Cr( s ) + 3 Cu 2
+
( aq )
h
2 Cr 3
+
( aq ) + 3 Cu( s )
E °
cell
= E °
Cu
2
+
- E °
Cr
3
+
= 0.34V - ( - 0.74V)
= 1.08V
log K =
cell
0.0592V
ο
n E
=
6 x (1.08V)
0.0592
= 109.5
K = 3 x 10 109

b) Sn( s ) + Pb 2
+
( aq )
h
Sn 2
+
( aq ) + Pb( s )
E ° = E °
Pb
2
+
- E °
Sn
2
+
= - 0.13V - ( - 0.14V)
= 0.01V
log K =
cell
0.0592V
ο
n E
=
2 x (0.01V)
0.0592
= 0.3
K = 2
21.60 a) ∆ G ° = - nFE °
cell
= - (2 mol e
-
) x (96.5 kJ/V ⋅ mol e
-
) x (1.05V)
= - 203 kJ
b) ∆ G ° = - nFE °
cell
= - (6 mol e - ) x (96.5 kJ/V ⋅ mol e - ) x ( - 0.30V)
= 1.7 x 10 2 kJ
21.61 a) ∆ G ° = - nFE °
cell
= - (6 mol e
-
) x (96.5 kJ/V ⋅ mol e
-
) x (1.26V)
= 7.30 x 10
2
kJ
b) ∆ G ° = - nFE °
cell
= - (2 mol e - ) x (96.5 kJ/V ⋅ mol e - ) x ( - 0.54V)
= 1.0 x 10 2 kJ
21.62 a) ∆ G ° = - nFE °
cell
= - (2 mol e
-
) x (96.5 kJ/V ⋅ mol e
-
) x