Question

Given below are the half-cell reactions:
${Mn}^{2+}+2e^{-}\to Mn;E^o=-1.18V$
$2({Mn}^{+3}+e^{-}\to {Mn}^{2+});E^o=+1.51V$
The $E^o$ for $3{Mn}^{2+}\to Mn+2{Mn}^{3+}$ will be:

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

Answer 1

Answer: (C)

Standard potential of reaction $\left[E^0\right]$ is given by,

$E_0^{cell}=E_R-E_P$

$M{n}^{2+}+2e^{-}⟶Mn{E}^0=-1.18V2M{n}^{2+}⟶2M{n}^{3+}+2e^{-}{E}^0=-1.51V\underset{–}{}$

$3M{n}^{2+}⟶Mn+2M{n}^{3+}$ $E^0=-1.18+(-1.51)=-2.69V$

The reaction is non spontaneous, $E^0=-2.69V$