Take E deg(Mn^3+/Mn^2+) = +1.57 V and E deg(MnO2/Mn^3+) = +0.95 V (acidic solution, F = 96500 C mol^-1). For 2Mn^3+ + 2H2O -> Mn^2+ + MnO2 + 4H^+ as written, Delta G deg and the conclusion are

-59.8 kJ; Mn^3+ disproportionates spontaneously

+59.8 kJ; Mn^3+ is stable towards disproportionation

-119.7 kJ; Mn^3+ disproportionates spontaneously

-151.5 kJ; Mn^3+ disproportionates spontaneously

JEE Advanced levelSingle correct optionDifficulty 3 of 3df075

Take $E^{\circ} (Mn X^{3 +} / Mn X^{2 +}) = + 1.57 V$ and $E^{\circ} (MnO X_{2} / Mn X^{3 +}) = + 0.95 V$ (acidic solution, $F = 96500 C mo l^{- 1}$ ). For $2 Mn X^{3 +} + 2 H X_{2} O \to Mn X^{2 +} + MnO X_{2} + 4 H X^{+}$ as written, $Δ G^{\circ}$ and the conclusion are

One

Mn X^{3 +}

is reduced (

1.57 V

cathode) and one is oxidised to

MnO X_{2}

(

0.95 V

anode), with a single electron transferred in the reaction as written (

n = 1

E_{cell}^{\circ} = 1.57 - 0.95 = 0.62 V

, and

Δ G^{\circ} = - 1 \times 96500 \times 0.62 = - 59830 J \approx - 59.8 kJ

, which is negative, so the disproportionation is spontaneous.