Electrochemistry - Result Question 50
####52. What is the $E _{\text{cell }}^{o}$ for the reaction $Cu^{2+}(aq)+Sn^{2+}(aq)+\rightarrow Cu(s)+Sn^{4+}(aq)$ at $25^{\circ} C$ if the equilibrium constant for the reaction is $1 \times 10^{6}$ ?
[1999]
(a) $0.5328 V$
(b) $0.3552 V$
(c) $0.1773 V$
(d) $0.7104 V$
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Solution:
- (c) $E _{\text{cell }}^{\circ}=\frac{2.303 R T}{n F} \log K _{e q}=\frac{.0591}{n} \log K _{e q}$ $=\frac{.0591}{2} \log 10^{6}=.0591 \times 3=0.1773 V$
