In acidic medium $\mathrm{MnO}_{2}$ is an oxidant as $\mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{H}_{2} \mathrm{O}$
If the $\mathrm{pH}$ of solution is decreased by one unit, the electrode potential of the half cell $\mathrm{Pt}: \mathrm{MnO}_{2}, \mathrm{Mn}^{2+}$ will change by

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{E} &=\mathrm{E}^{\circ}-\frac{0.0592}{2} \log \frac{\left[\mathrm{Mn}^{2+}ight]}{\left[\mathrm{H}^{+}ight]^{4}} \ &=\mathrm{E}^{\circ}-\frac{0.0592 	imes 4}{2} \log \frac{\left[\mathrm{Mn}^{2+}ight]^{1 / 4}}{\left[\mathrm{H}^{+}ight]} \ &=\mathrm{E}^{\circ}-0.0592 	imes 2\left(\log \left[\mathrm{Mn}^{2+}ight]^{1 / 4}+\mathrm{pH}ight) \ & \Delta \mathrm{E}=\mathrm{E}_{2}-\mathrm{E}_{1}=0.0592 	imes 2\left(\mathrm{pH}_{1}-\mathrm{pH}_{2}ight) \ &=0.118 	imes 1=0.118 \mathrm{~V} \end{aligned}$

In acidic medium $\mathrm{MnO}_{2}$ is an oxidant as $\mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{H}^{+}+2…

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