The electrochemical cell shown below is a concentration cell. $\mathrm{M} \mid \mathrm{M}^{2+}$ (saturated…
The electrochemical cell shown below is a concentration cell. $\mathrm{M} \mid \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt, $\left.\mathrm{MX}_{2}\right) \| \mathrm{M}^{2 \perp}\left(0.001 \mathrm{~mol} \mathrm{dm}^{-3}\right) \mid \mathrm{M}$.
The emf of the cell depends on the difference in concentrations of $\mathrm{M}^{2+}$ ions at the two electrodes. The emf of the cell at $298 \mathrm{~K}$ is $0.059 \mathrm{~V}$.
Question: The value of $\Delta \mathrm{G}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ for the given cell is (take 1 $\mathrm{F}=96500 \mathrm{Cmol}^{-1}$ )
$-5.7$
$5.7$
$11.4$
$-11.4$
Solution
At anode : $M(\mathrm{~s})+2 X^{-}(\mathrm{aq}) \longrightarrow M X_{2}(\mathrm{aq})+2 \mathrm{e}^{-}$
At cathode $: M^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow M(\mathrm{~s})$
Thus, here $n=2$
$\Delta G=-n F E_{\text {cell }}$
$=-2 \times 96500 \times 0.059 \times 10^{-3} \mathrm{~kJ} / \mathrm{mole}=-11.4 \mathrm{~kJ} / \mathrm{mole}$