If $\mathrm{E}^{\circ}$ cell for $\mathrm{Cd}_{(\mathrm{s})}\left|\mathrm{Cd}_{(\mathrm{IM})}^{2+} \square…
If $\mathrm{E}^{\circ}$ cell for $\mathrm{Cd}_{(\mathrm{s})}\left|\mathrm{Cd}_{(\mathrm{IM})}^{2+} \square \mathrm{Ag}_{(\mathrm{IM})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}$ is 1.2 V . What is the emf of the cell at $25^{\circ} \mathrm{C}$ ?
$\quad-1.2 \mathrm{~V}$
2.4 V
-2.4 V
1.2 V
Solution
The electrode reaction is:
$\mathrm{Cd}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cd}^{2+}+2 \mathrm{Ag}$
For this cell reaction, $\mathrm{n}=2$.
Using Nernst equation at 298 K :
$\begin{aligned}
\mathrm{E}_{\text {cell }} & =\mathrm{E}_{\text {cell }}^o-\frac{0.0592}{2} \log _{10} \frac{\left[\mathrm{Cd}^{2+}\right]}{\left.\mathrm{Ag}^{+}\right]^2} \\
& =1.2-\frac{0.0592}{2} \log _{10} \frac{1}{1} \\
\therefore \quad \mathrm{E}_{\text {cell }} & =1.2-0=1.2 \mathrm{~V} \quad\left(\because \log _{10} 1=0\right)
\end{aligned}$