Q6 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
What is the standard electrode potential at 298 K for the reaction: $\mathrm{Cu}^{2+}+1 \mathrm{e}-\rightarrow \mathrm{Cu}^{+1}$ ?
Given: $\mathrm{E}_0 \mathrm{Cu}^{+1} / \mathrm{Cu}=0.5 \mathrm{~V} \quad \& \quad \mathrm{E}_0 \mathrm{Cu}^{+2} / \mathrm{Cu}=0.335 \mathrm{~V}$
- A0.34 V
- Correct answerB0.17 V
- C0.492 V
- D0.410 V
Solution
The standard electrode potential for the reaction $\mathrm{Cu}^{2+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+}$ can be determined using the Gibbs free energy change $\Delta G^{\circ} = -nFE^{\circ}$.
For the reaction $\mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}$, $\Delta G^{\circ}_{1} = -2 \times F \times 0.335 = -0.67F$.
For the reaction $\mathrm{Cu}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}$, $\Delta G^{\circ}_{2} = -1 \times F \times 0.5 = -0.5F$.
The target reaction is $\mathrm{Cu}^{2+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+}$, which is obtained by subtracting the second reaction from the first:
$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2} = -0.67F - (-0.5F) = -0.17F$.
Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{3}$ with $n=1$, we have $-1 \times F \times E^{\circ}_{3} = -0.17F$.
Therefore, $E^{\circ}_{3} = 0.17 \text{ V}$.
Answer: 0.17 V
For the reaction $\mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}$, $\Delta G^{\circ}_{1} = -2 \times F \times 0.335 = -0.67F$.
For the reaction $\mathrm{Cu}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}$, $\Delta G^{\circ}_{2} = -1 \times F \times 0.5 = -0.5F$.
The target reaction is $\mathrm{Cu}^{2+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+}$, which is obtained by subtracting the second reaction from the first:
$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2} = -0.67F - (-0.5F) = -0.17F$.
Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{3}$ with $n=1$, we have $-1 \times F \times E^{\circ}_{3} = -0.17F$.
Therefore, $E^{\circ}_{3} = 0.17 \text{ V}$.
Answer: 0.17 V
Hence, the correct answer is option: 'B'.