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Find the emf of the cell in which the following reaction takes place at 298 K
$Ni(s)+2Ag^+(0.001\ M)\rightarrow Ni^{2+}(0.001\ M)+2Ag(s)$
(Given that $E^\circ_{cell}=10.5\ V$, $\frac{2.303RT}{F}=0.059$ at 298 K)
A
1.05 V
B
1.0385 V
C
1.385 V
D
0.9615 V
Detailed Solution
$Ni(s) + 2Ag^+(0.001\ M) \rightarrow Ni^{2+}(0.001\ M) + 2Ag(s)$, n = 2
$E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log\frac{[Ni^{2+}]}{[Ag^+]^2}$
$= 10.5 - \frac{0.059}{2}\log\frac{10^{-3}}{(10^{-3})^2} = 10.5 - \frac{0.059}{2} \times 3$
$= 10.4115$ V
The calculated answer is not given in the options (with $E^\circ_{cell} = 1.05$ V it would be 0.9615 V), so this question was declared a bonus.