Chemistry - Electrochemistry Question with Solution | TestHub

For concentration cell ${\mathrm{E}}_{\mathrm{cell}}^{\underset{¯}{\mathrm{o}}}=0$

Anode : $\mathrm{Cu}\left(\mathrm{s}\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\mathrm{A}$

Cathode : ${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\mathrm{c}\to \mathrm{Cu}\left(\mathrm{s}\right)$

Overall : $\overline{{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\mathrm{c}\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)}\mathrm{A}$

As $\mathrm{\Delta G}=-\mathrm{nF}{\mathrm{E}}_{\mathrm{cell}}$

If $\mathrm{\Delta G}=-\mathrm{ve}$ then ${\mathrm{E}}_{\mathrm{cell}}$ is positive.

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\underset{¯}{\mathrm{o}}}-\frac{0.059}{2}\log \frac{{\mathrm{C}}_1}{\mathrm{C}2}$

${\mathrm{E}}_{\mathrm{cell}}=\frac{-0.059}{2}\log \frac{{\mathrm{C}}_1}{\mathrm{C}2}$

${\mathrm{E}}_{\mathrm{cell}}>0$

$\Rightarrow$ ${\mathrm{C}}_2>{\mathrm{C}}_1$