Q.
Find the emf of the cell in which the following reaction takes place at $298 \,K$
$Ni ( s )+2 Ag ^{+}(0.001 \,M ) \rightarrow Ni ^{2+}(0.001 \,M )+2 Ag ( s )$
(Given that $E_{\text {cell }}^{\circ}=10.5\, V , \frac{2.303 \text { RT }}{ F }=0.059$ at $298\, K )$
Solution: