Chem1612 2014-N-12 Worksheet With Answers - The University Of Sydney - 2014 Page 27
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28CHEM1612
2004-N-8
November 2004
Marks
Consider the following balanced redox reaction.
2
+
3+
2+
(aq) 2In
2In(s) + 3MnO
(s) + 12H
(aq) + 3Mn
(aq) + 6H
O
2
2
o
If E
= 1.568 V, what would be the measured potential of this cell at 298 K at the
following concentrations?
+
3+
2+
[H
(aq)] = 0.25 M;
[In
(aq)] = 0.20 M;
[Mn
(aq)] = 0.42 M
The reaction quotient Q is given by:
3
2
2
3
[In (aq)] [Mn ]
Q =
12
[H (aq)]
-
The cell reaction involves a 6e
process (2In 2In(III) and 3Mn(IV)
3Mn(II)). The cell potential can be calculated using the Nernst equation,
RT
o
E =
E
ln
Q
nF
1
1
2
3
(8.314 J K mol ) (298 K)
(0.20) (0.42)
ln
= (1.568 V)
1.52 V
-1
12
(6) (96485Cmol )
(0.25)
Answer: 1.52 V
2
What is the value of the equilibrium constant for the following reaction at 298 K?
2+
2+
(aq) + Zn(s) Zn
Cu
(aq) + Cu(s)
Relevant electrode potentials can be found on the data page.
2+
2+
The standard reduction potentials for Cu
(aq) / Cu(s) and Zn
(aq) / Zn(s) are
2+
+0.34 V and -0.76 V respectively. The Zn
(aq) / Zn potential is the less positive
and is reversed. The cell potential is therefore:
o
E
= ((0.34) – (-0.76)) V = 1.10 V
cell
-
o
The equilibrium constant for this 2e
process can be calculated using
E
cell
RT
=
ln
K
:
nF
1
(2) (96485Cmol )
lnK
=
× (1.10) = 85.7
-1
-1
(8.314 JK mol ) (298K)
85.7
37
K = e
= 1.62 × 10
37
Answer: 1.62 × 10
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