Given below are the half-cell reactions: \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn} ; \mathrm{E}^{0}=-1.18 \mathrm{~V}\) \(2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\right) ; \mathrm{E}^{0}=+1.51 \mathrm{~V}\) The \(\mathrm{E}^{0}\) for \(3 \mathrm{Mn}^{2+} \longrightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\) will be:

(C) -2.69 V ; the reaction will not occur

(A) -0.33 V ; the reaction will not occur

(B) -0.33 V ; the reaction will occur

(D) -2.69 V ; the reaction will occur

JEE Main · 2014 · 06 Apr 2014 · Offline · Redox Reactions and Electrochemistry

Given below are the half-cell reactions:

Mn^{2 +} + 2 e^{-} ⟶ Mn; E^{0} = - 1.18 V

2 (Mn^{3 +} + e^{-} ⟶ Mn^{2 +}); E^{0} = + 1.51 V

The

E^{0}

for

3 Mn^{2 +} ⟶ Mn + 2 Mn^{3 +}

will be:

Answer: (C)

Build the target change

3 Mn X^{2 +} Mn + 2 Mn X^{3 +}

from the two given half-reactions. One is the reduction

Mn X^{2 +} + 2 e X^{-} Mn

with

E^{\circ} = - 1.18

V; the other must be run in reverse as an oxidation,

2 Mn X^{2 +} 2 Mn X^{3 +} + 2 e X^{-}

, carrying

E^{\circ} = - 1.51

V. Both involve two electrons, so the cell potential is

E^{\circ} = - 1.18 + (- 1.51) = - 2.69

V. Being negative, it tells us the change is non-spontaneous, so the reaction will not occur under standard conditions.

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option