Question

Given below are the half - cell reactions
$M{n}^{2+}+2e^{-}\to Mn;E^{\circ }=-1.18eV$
$2(M{n}^{3+}+e^{-}\to M{n}^{2+});E^0=+1.51eV$
The $E^{\circ }$for $3M{n}^{2+}\to Mn+2M{n}^{3+}$ will be

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

Answer 1

The correct option is A -2.69 V; the reaction will not occur

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