Electrochemistry - Result Question 43
####45. The equilibrium constant of the reaction:
$Cu(s)+2 Ag^{+}(aq) \longrightarrow Cu^{2+}(aq)+2 Ag(s)$;
$E^{\circ}=0.46 V$ at $298 K$ is
[2007]
(a) $2.0 \times 10^{10}$
(b) $4.0 \times 10^{10}$
(c) $4.0 \times 10^{15}$
(d) $2.4 \times 10^{10}$
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Solution:
- (c) As $E _{\text{cell }}^{o}=\frac{0.0591}{n} \log K_c$
$\therefore 0.46=\frac{0.0591}{2} \log K_c$
$\therefore \log K_c=\frac{2 \times 0.46}{0.0591}=15.57$
or $K_c=$ Antilog $15.57=3.7 \times 10^{15} \approx 4 \times 10^{15}$
