Question

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

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

Answer 1

Answer: (D)

$Ni(s)+2A{g}^{+}(0.001M)\to N{i}^{+2}(0.001M)+2Ag(s)$
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.059}{n}\log \frac{{\left[N{i}^{+2}\right]}^1}{{\left[A{g}^{+}\right]}^2}$
$E_{\text{cell }}=10.5-\frac{0.059}{2}\log \frac{10^{-3}}{{\left(10^{-3}\right)}^2}$
$=10.5-\frac{0.059}{2}\log 10^{+3}$
$=10.5-\frac{0.059}{2}\times 3$
$=10.4115V$