Question

Given below are the half-cell reactions :

${Mn}^{2+}+2 e^{-} \longrightarrow {Mn} ; E^{\circ}=-1.18 {~V}$

$2\left({Mn}^{3+}+e^{-} \longrightarrow {Mn}^{2+}\right) ; E^{\circ}=+1.51 {~V}$

The $E^{\circ}$ for $3 {Mn}^{2+} \longrightarrow {Mn}+2 {Mn}^{3+}$ will be

(a) $-0.33 {~V}$; the reaction will occur

(b) $-2.69 \mathrm{~V}$; the reaction will not occur

(c) $-2.69 \mathrm{~V}$; the reaction will occur

(d) $-0.33 \mathrm{~V}$; the reaction will not occur.

Answer 1

Correct Option : (B)

Explanation :

Overall reaction :

${Mn}^{2+}+2 e^{-} \longrightarrow {Mn} ; \quad E^{\circ}=-1.18 {V} $

$2 {Mn}^{2+} \longrightarrow 2 {Mn}^{3+}+2 e^{-} ; \quad E^{\circ}=-1.51 {V}$

$3 {Mn}^{2+} \longrightarrow {Mn}+2 {Mn}^{3+} ; \quad E^{\circ}=-1.18+(-1.51)$

$=-2.69 {V}$

As $E^{\circ}$ is negative, the reaction will not occur.