Question

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 $298\, K )$

• A. $1.05 \,V$
• B. $1.0385\, V$
• C. $1.385\,V$
• D. None

Answer 1

Answer: (D)

$Ni ( s )+2 Ag ^{+}(0.001 M ) \rightarrow Ni ^{+2}(0.001 M )+2 Ag ( s )$
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{ n } \log \frac{\left[ Ni ^{+2}\right]^{1}}{\left[ Ag ^{+}\right]^{2}}$
$E _{\text {cell }}=10.5-\frac{0.059}{2} \log \frac{10^{-3}}{\left(10^{-3}\right)^{2}}$
$=10.5-\frac{0.059}{2} \log 10^{+3}$
$=10.5-\frac{0.059}{2} \times 3$
$=10.4115 \,V$