Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{Cu}}^{2+}+1{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{+};\Delta {\mathrm{G}}_1^0=-{\mathrm{n}}_1{\mathrm{E}}_1^0\mathrm{F}]$ $\frac{{\mathrm{Cu}}^{+}+1{\mathrm{e}}^{-}\to \mathrm{Cu};\Delta {\mathrm{G}}_2^0=-{\mathrm{n}}_2{\mathrm{E}}_2^0\mathrm{F}}{{\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}\Delta {\mathrm{G}}^{\circ }=\Delta {\mathrm{G}}_1^0+\Delta {\mathrm{G}}_2^0}$ $-\mathrm{n}{\mathrm{E}}^{\circ }\mathrm{F}=-1{\mathrm{n}}_1{\mathrm{E}}_1^0\mathrm{F}+(-1){\mathrm{n}}_2{\mathrm{E}}_2^0\mathrm{F}$ $-\mathrm{n}{\mathrm{E}}^{\circ }\mathrm{F}=-\mathrm{F}\left({\mathrm{n}}_1{\mathrm{E}}_1^0+{\mathrm{n}}_2{\mathrm{E}}_2^0\mathrm{F}\right)$ ${\mathrm{E}}^{\circ }=\frac{{\mathrm{n}}_1{\mathrm{E}}_1^0+{\mathrm{n}}_2{\mathrm{E}}_2^0}{\mathrm{n}}=\frac{0.15\times 1+0.50\times 1}{2}=0.325$