The standard cell potential of the following cell $\mathrm{Zn}\left|\mathrm{Zn}_{(a q)}^{2+}\right|\

Question

The standard cell potential of the following cell $\mathrm{Zn}\left|\mathrm{Zn}_{(a q)}^{2+}\right|\left|\mathrm{Fe}_{(a q)}^{2+}\right| \mathrm{Fe}$ is 0.32 V . Calculate the standard Gibbs energy change for the reaction $\mathrm{Zn}_{(s)}+\mathrm{Fe}_{(a q)}^{2+} \longrightarrow \mathrm{Zn}_{(a q)}^{2+}+\mathrm{Fe}_{(s)}$
(Given : $1 \mathrm{~F}=96487 \mathrm{C}$ )

Professor PC Thomas Classes · Answer

(a) : $\Delta G^{\circ}=-n F E_{	ext {cell }}^{\circ}$ Given, $E_{	ext {cell }}^{\circ}=0.32 \mathrm{~V}$ and $1 \mathrm{~F}=96487 \mathrm{C}$ \ For the given reaction, $n=2$ $ \Delta G^{\circ}=-2 	imes 96487 	imes 0.32=-61751.68 \mathrm{~J} \mathrm{~mol}^{-1}=-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1} $