Question

Given that the standard potential ;$\left(\mathrm{Eo}\right)$ of $\raisebox{1ex}{${\mathrm{Cu}}^{2+}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Cu}$}\right.$ and ${\mathrm{Cu}}^{+}/\mathrm{Cu}$are $0.34\mathrm{V}$and $0.522\mathrm{V}$ respectively , the ${\mathrm{E}}^0$0f ${\mathrm{Cu}}^{2+}/{\mathrm{Cu}}^{+}$is :

• A. $+0.158\mathrm{V}$
• B. $-\mathrm{O}.158\mathrm{V}$
• C. $0.182\mathrm{V}$
• D. $-0.182\mathrm{V}$

Answer 1

The correct option is A

$+0.158\mathrm{V}$

Answer(A)

Explanation for correct option:

$$\begin{gathered} {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\mathrm{Cu} \\ {\mathrm{E}}^{\circ }=0.340\mathrm{V} \\ {\mathrm{Cu}}^{-}{\mathrm{Cu}}^{+}+{\mathrm{e}}^{-} \\ {\mathrm{E}}^{\circ }={}^{-}0.522\mathrm{V} \\ {\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{+} \\ {\mathrm{E}}^{\circ }=? \end{gathered}$$

Applying $∆\mathrm{G}={\mathrm{nFE}}^{\circ }$ We get,

$$\begin{gathered} ({}^{-}1\times \mathrm{F}\times {\mathrm{E}}^{\circ })={}^{-}2\times \mathrm{F}\times 0.340+\left({}^{-}1\times \mathrm{F}\times {}^{-}0.522\right) \\ \mathrm{We}\mathrm{get}, \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{158}\mathbf{V} \end{gathered}$$

Hence (A) is correct option.