Question

Given below are the half-cell reactions
$M{n}^{2+}+2e^{-}\to Mn;E^{\circ }=-1.18eV$
$2\left(M{n}^{3+}+e^{-}\to M{n}^{2+}\right);E^{\circ }=+1.51eV$
The $E^{\circ }$ for $3M{n}^{2+}\to Mn+2M{n}^{3+}$ will be

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

Answer 1

Answer: (A)

Standard electrode potential of reaction $\left[E^{\circ }\right]$ can be calculated as
$E_{\text{cell }}^{\circ }=E_R-E_P$
where, $E_R=$ SRP of reactant,$E_P=$ SRP of product
If $E_{\text{cell }}^{\circ }=+ve$, then reaction is spontaneous otherwise non-spontaneous.
$M{n}^{3+}\stackrel{E_1^0=1.51V}{\to }M{n}^{2+}$
$M{n}^{2+}\stackrel{E_2^0=-1.18V}{\to }Mn$
$\therefore$ For $M{n}^{2+}$ disproportionation,
$E^{\circ }=-1.51V-1.18V=-2.69V<0$
Thus, all reaction will not occur.