Chem1612 2014-N-12 Worksheet With Answers - The University Of Sydney - 2014 Page 13
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28CHEM1612
2009-N-16
November 2009
Marks
What is the value of the equilibrium constant for the following reaction at 298 K?
•
3
3+
2+
2Fe
(aq) + 3Sn(s) → 2Fe(s) + 3Sn
(aq)
Relevant electrode potentials can be found on the data page.
The relevant reduction potentials are:
3+
-
Fe
(aq) + 3e
à Fe(s)
E° = -0.04 V
2+
-
Sn
(aq) + 2e
à Sn(s)
E° = -0.14 V
2+
As the Sn
/ Sn couple is the more negative, it is reversed giving:
E° = (-0.04 + 0.14) V = 0.10 V
The equilibrium constant, K, is related to the standard reduction potential using:
E° = (RT/nF) × lnK
lnK = nFE° / RT = (6 × 96485 × 0.10) / (8.314 × 298) = 23.37
23.37
10
K = e
= 1.4 × 10
10
Answer: 1.4 × 10
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