The magnitude of the change in oxidising power of

Question:

The magnitude of the change in oxidising power of the $\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}$ couple is $\mathrm{x} \times 10^{-4} \mathrm{~V}$, if the $\mathrm{H}^{+}$concentration is decreased from $1 \mathrm{M}$ to $10^{-4} \mathrm{M}$ at $25^{\circ} \mathrm{C}$. (Assume concentration of $\mathrm{MnO}_{4}^{-}$and $\mathrm{Mn}^{2+}$ to be same on change in $\mathrm{H}^{+}$concentration). The value of $\mathrm{x}$ is______________ . (Rounded off to the nearest integer) [Given : $\frac{2303 \mathrm{R} T}{\mathrm{~F}}=0.059$ ]

Solution:

(3776)

$\mathrm{Se}^{-}+\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_{2} \mathrm{O}$

$Q=\frac{\left[\mathrm{Mn}^{+2}\right]}{\left[\mathrm{H}^{+}\right]^{8}\left[\mathrm{MnO}_{4}^{-}\right]} \quad \Rightarrow \quad \mathrm{E}_{1}=\mathrm{E}^{\circ}-\frac{0.059}{5} \log \left(\mathrm{Q}_{1}\right)$

$\mathrm{E}_{2}=\mathrm{E}^{\circ}-\frac{0.059}{5} \log \left(\mathrm{Q}_{2}\right) \quad \Rightarrow \quad \mathrm{E}_{2}-\mathrm{E}_{1}=\frac{0.059}{5} \log \left(\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}\right)$

$=\frac{0.059}{5} \log \left\{\frac{\left[\mathrm{H}^{+}\right]_{\Pi}}{\left[\mathrm{H}^{+}\right]_{1}}\right\} \Rightarrow=\frac{0.059}{5} \log \left(\frac{10^{-4}}{1}\right)^{8}$

$\left(\mathrm{E}_{2}-\mathrm{E}_{1}\right)=\frac{0.059}{5} \times(-32) \quad \Rightarrow \quad\left|\left(\mathrm{E}_{2}-\mathrm{E}_{1}\right)\right|=32 \times \frac{0.059}{5}=\mathrm{x} \times 10^{-4}$

$=\frac{32 \times 590}{5} \times 10^{-4}=\mathrm{x} \times 10^{-4} \Rightarrow=3776 \times 10^{-4} \quad \mathrm{x}=3776$