Electrochemistry 2 Question 12
12. Given below are the half-cell reactions
$$ \begin{array}{r} \mathrm{Mn}^{2+}+2 e^{-} \longrightarrow \mathrm{Mn} ; E^{\circ}=-1.18 \mathrm{eV} \ 2\left(\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+}\right) ; E^{\circ}=+1.51 \mathrm{eV} \end{array} $$
(2014 Main)
The $E^{\circ}$ for $3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}$ will be
(a) $-2.69 \mathrm{~V}$; the reaction will not occur
(b) $-2.69 \mathrm{~V}$; the reaction will occur
(c) $-0.33 \mathrm{~V}$; the reaction will not occur
(d) $-0.33 \mathrm{~V}$; the reaction will occur
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Solution:
- Standard electrode potential of reaction $\left[E^{\circ}\right]$ can be calculated as
$$ E_{\text {cell }}^{\circ}=E_{R}-E_{P} $$
where, $E_{R}=\mathrm{SRP}$ of reactant, $E_{P}=\mathrm{SRP}$ of product If $E_{\text {cell }}^{\circ}=+\mathrm{ve}$, then reaction is spontaneous otherwise non-spontaneous.
$$ \begin{gathered} \mathrm{Mn}^{3+} \xrightarrow{E_{1}^{\circ}=1.51 \mathrm{~V}} \mathrm{Mn}^{2+} \ \mathrm{Mn}^{2+} \xrightarrow{E_{2}^{\circ}=-1.18 \mathrm{~V}} \mathrm{Mn} \end{gathered} $$
$\therefore \quad$ For $\mathrm{Mn}^{2+}$ disproportionation,
$$ E^{\circ}=-1.51 \mathrm{~V}-1.18 \mathrm{~V}=-2.69 \mathrm{~V}<0 $$
Thus, all reaction will not occur.
