Electrochemistry Result Question 12
12. Given below are the half-cell reactions
(2014 Main)
$ \begin{gathered} \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{gathered} $
The $E^{\circ}$ for $3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}$ 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
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Answer:
Correct Answer: 12. ( a )
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{aligned} & \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{aligned} $
$\therefore$ 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.
BETA
