The standard cell potential of the following cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|…
The standard cell potential of the following cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right| \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \mathrm{Fe}$ is 0.32 V . Calculate the standard Gibbs energy change for the reaction :
$\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$
(Given : $1 \mathrm{~F}=96487 \mathrm{C}$ )
$-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$+5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$+61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\Delta_r G^{\ominus}=-n F E_{\text {cell }}^{\ominus}$
For the given reaction, $n=2$
$\therefore \quad \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-2 \times 96487 \times 0.32$
$=-61751.68 \mathrm{~J} \mathrm{~mol}^{-1}$
$=-61.751 \mathrm{~kJ} \mathrm{~mol}^{-1}$