JEE Advanced 2022 — Paper 1 — CHEM-I solution

JEE Advanced2022Paper 1CHEM-I

The reduction potential ( $E^{0}$ , in V) of $MnO_{4}^{-} (aq) / Mn(s)$ is ______.

[Given: $E_{(MnO_{4}^{-} (aq) / MnO_{2} (s))}^{0} = 1.68$ V; $E_{(MnO_{2} (s) / Mn^{2 +} (aq))}^{0} = 1.21$ V; $E_{(Mn^{2 +} (aq) / Mn(s))}^{0} = - 1.03$ V ]

Your answer

Solution

Combine three half-reactions using $Δ G^{\circ} = - n F E^{\circ}$ (Gibbs energies add; potentials do not add directly when electron counts differ).

(1) $MnO_{4}^{-} + 4 H^{+} + 3 e^{-} \to MnO_{2} + 2 H_{2} O$ , $Δ G_{1}^{\circ} = - 3 F (1.68) = - 5.04 F$ .

(2) $MnO_{2} + 4 H^{+} + 2 e^{-} \to Mn^{2 +} + 2 H_{2} O$ , $Δ G_{2}^{\circ} = - 2 F (1.21) = - 2.42 F$ .

(3) $Mn^{2 +} + 2 e^{-} \to Mn(s)$ , $Δ G_{3}^{\circ} = - 2 F (- 1.03) = + 2.06 F$ .

Adding gives $MnO_{4}^{-} + 8 H^{+} + 7 e^{-} \to Mn(s) + 4 H_{2} O$ with $n = 7$ .

$Δ G^{\circ} = (- 5.04 - 2.42 + 2.06) F = - 5.40 F = - 7 F E^{\circ}$ .

$E^{\circ} = \frac{5.40}{7} = 0.77$ V.

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