JEE Advanced 2022 Paper 1 · Q02 · Galvanic Cells & EMF
The reduction potential ($E^0$, in V) of MnO$_4^-$(aq)/Mn(s) is ______. [Given: $E^0_{(\text{MnO}_4^-(\text{aq})/\text{MnO}_2(\text{s}))} = 1.68$ V; $E^0_{(\text{MnO}_2(\text{s})/\text{Mn}^{2+}(\text{aq}))} = 1.21$ V; $E^0_{(\text{Mn}^{2+}(\text{aq})/\text{Mn}(\text{s}))} = -1.03$ V]
Reveal answer + step-by-step solution
Correct answer:0.77
Solution
Combine the three half-reactions so the electrons add to the target half-reaction MnO$_4^-$ + 8H$^+$ + 7e$^-$ $\rightarrow$ Mn(s) + 4H$_2$O. Use $\Delta G^\circ = -nFE^\circ$ and add $\Delta G$ values (not E$^\circ$ values directly, because electron counts differ).
(i) MnO$_4^-$/MnO$_2$: $n_1 = 3$, $E_1 = 1.68$ V $\Rightarrow \Delta G_1 = -3F(1.68)$. (ii) MnO$_2$/Mn$^{2+}$: $n_2 = 2$, $E_2 = 1.21$ V $\Rightarrow \Delta G_2 = -2F(1.21)$. (iii) Mn$^{2+}$/Mn: $n_3 = 2$, $E_3 = -1.03$ V $\Rightarrow \Delta G_3 = -2F(-1.03) = 2F(1.03)$.
Target $n = 7$, so $\Delta G_{\text{target}} = -7FE^\circ = \Delta G_1 + \Delta G_2 + \Delta G_3$.
$-7FE^\circ = -3F(1.68) - 2F(1.21) + 2F(1.03)$ $7E^\circ = 3(1.68) + 2(1.21) - 2(1.03) = 5.04 + 2.42 - 2.06 = 5.40$.
Wait — sign on (iii): going from Mn$^{2+}$ $\rightarrow$ Mn(s) in the target direction matches given (iii) directly, so $\Delta G_3 = -2F(-1.03) = +2.06F$. The correct accumulation is $7E^\circ = 3(1.68) + 2(1.21) + 2(-1.03) = 5.04 + 2.42 - 2.06 = 5.40$. Therefore $E^\circ = 5.40/7 = 0.7714 \approx 0.77$ V.
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