Chemistry - Electrochemistry Question with Solution | TestHub

$2{\mathrm{Cu}}^{+}⟶{\mathrm{Cu}}^{+2}+\mathrm{Cu}$ $2{\mathrm{e}}^{-}+{\mathrm{Cu}}^{+2}⟶\mathrm{Cu};{\mathrm{E}}_1^{\circ }=0.34\mathrm{V};\dots$ (i) ${\mathrm{e}}^{-}+{\mathrm{Cu}}^{+2}⟶{\mathrm{Cu}}^{+};{\mathrm{E}}_2^{\circ }=0.15\mathrm{V};\dots$ (ii) ${\mathrm{Cu}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Cu};{\mathrm{E}}_3^{\circ }=?$ ...(iii) Now, $\Delta {\mathrm{G}}_1^{\circ }=-{\mathrm{nFE}}_1^{\circ }=-2\times 0.34\mathrm{F}$ $\Delta {\mathrm{G}}_2^{\circ }=-1\times 0.15\mathrm{F},\Delta {\mathrm{G}}_3^{\circ }=-1\times {\mathrm{E}}_3^{\circ }\mathrm{F}$ Again, $\Delta {\mathrm{G}}_1^{\circ }=\Delta {\mathrm{G}}_2^{\circ }+\Delta {\mathrm{G}}_3^{\circ }$ $\Rightarrow -0.68\mathrm{F}=-0.15\mathrm{F}-{\mathrm{E}}_3\mathrm{F}$ $0\Rightarrow {\mathrm{E}}_3^{\circ }=0.68-0.15=0.53\mathrm{V}$ ${\mathrm{E}}_{\text{cell }}^{\circ }={\mathrm{E}}_{\text{cathode }}^{\circ }\left({\mathrm{Cu}}^{+}/\mathrm{Cu}\right)$ $=0.53-0.15=0.38\mathrm{V}$