Q77 / 180 · NEET 2026 (Re-NEET) · Chemistry
The standard electrode potential ($E^\circ$) for the half-cell reaction $\text{Fe}^{3+}+e^-\rightarrow\text{Fe}^{2+}$ at $298$ K is
(Given: $E^\circ(\text{Fe}^{3+}/\text{Fe})=-0.04$ V and $E^\circ(\text{Fe}^{2+}/\text{Fe})=-0.44$ V at $298$ K)
(Given: $E^\circ(\text{Fe}^{3+}/\text{Fe})=-0.04$ V and $E^\circ(\text{Fe}^{2+}/\text{Fe})=-0.44$ V at $298$ K)
- A$+0.92$ V
- B$+0.40$ V
- Correct answerC$+0.76$ V
- D$-0.48$ V
Solution
The given half-cell reactions are:
$\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$, $E^\circ_1 = -0.04$ V
$\Delta G^\circ_1 = -n_1FE^\circ_1 = -3 \times F \times (-0.04) = 0.12F$
$\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$, $E^\circ_2 = -0.44$ V
$\Delta G^\circ_2 = -n_2FE^\circ_2 = -2 \times F \times (-0.44) = 0.88F$
The required half-cell reaction is:
$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$
This reaction can be obtained by subtracting the second reaction from the first reaction:
$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$
$-1 \times F \times E^\circ_3 = 0.12F - 0.88F$
$-FE^\circ_3 = -0.76F$
$E^\circ_3 = +0.76$ V
Answer: $+0.76$ V
$\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$, $E^\circ_1 = -0.04$ V
$\Delta G^\circ_1 = -n_1FE^\circ_1 = -3 \times F \times (-0.04) = 0.12F$
$\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$, $E^\circ_2 = -0.44$ V
$\Delta G^\circ_2 = -n_2FE^\circ_2 = -2 \times F \times (-0.44) = 0.88F$
The required half-cell reaction is:
$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$
This reaction can be obtained by subtracting the second reaction from the first reaction:
$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$
$-1 \times F \times E^\circ_3 = 0.12F - 0.88F$
$-FE^\circ_3 = -0.76F$
$E^\circ_3 = +0.76$ V
Answer: $+0.76$ V
Hence, the correct answer is option: 'C'.