Question

Given below are the half-cell reactions:

$M{n}^{2+}+2e^{-}\to Mn;E^0=-1.18V$

$2(M{n}^{3+}+e^{-}\to M{n}^{2+});E^0=+1.51V$.

The $E^0$ for $3M{n}^{2+}\to Mn+2M{n}^{3+}$ will be:

• A. $-0.33V;$ the reaction will not occur
• B. $-0.33V;$ the reaction will occur
• C. $-2.69V;$ the reaction will not occur
• D. $-2.69V;$ the reaction will occur

Answer 1

Answer: (C)

$-2.69V;$ the reaction will not occur

Adding equation $2$ from $1$ gives $3{Mn}^{2+}\to Mn+2{Mn}^{3+}$

Hence, $E^0$ for this reaction $=(-1.18)-(+1.51)V$

$=-2.69V$.

The reaction will not occur as $E_{cell}^0$ is negative value as for spontaneous reaction value should be positive.

Option C is correct.