Question

Given:

$$\begin{gathered} \left(1\right){\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu},\mathrm{E}°=0.337\mathrm{V} \\ \left(2\right){\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{+},\mathrm{E}°=0.153\mathrm{V} \end{gathered}$$.

What will be the Electrode potential E for the reaction: ${\mathrm{Cu}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Cu}$

• A. 0.30 V
• B. 0.38 V
• C. 0.52 V
• D. 0.90 V

Answer 1

The correct option is C

0.52 V

Explanation of the correct answer:

The correct answer is option C:

From the given,

$$\begin{gathered} \left(1\right) \\ C{u}^{2+}+2e^{-}\to Cu \\ E°=0.337V \\ \Delta G=-nFE° \\ \Rightarrow \Delta G=-2*F*0.3370-\left[i\right] \\ \left(2\right) \\ C{u}^{2+}+e^{-}\to C{u}^{+} \\ E°=0.153V \\ \Delta G=-nFE° \\ \Rightarrow \Delta G=-1*F*0.153-\left[ii\right] \end{gathered}$$

Adding the equations (i) and (ii),

$$\begin{gathered} C{u}^{+}+e^{-}\to Cu \\ \Delta G°=-0.521F \\ \Delta G°=-nFE° \\ \Rightarrow -0.521F=-nFE° \\ \Rightarrow E°=0.52V \end{gathered}$$

Explanation of the correct answer:

Options A, B, and D:

1. The Electrode potential for the reaction is 0.52 V.
2. Hence, these options are incorrect.

Hence, option (C) is correct, i.e., 0.52 V.