Question

By how much would the oxidising power of the $\left(Mn{O}_4^{-}|M{n}^{2+}\right)$ couple change if the $H^{+}$ ions concentration is decreased $100$ times at ${25}^{\circ }C$?

• A. Increases by $189mV$
• B. Decreases by $189mV$
• C. Increases by $19mV$
• D. Decreases by $19mV$

Answer 1

The correct option is B Decreases by $189mV$

$Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$

$\text{According to Nernst equation at}{25}^{\circ }C,$

$E_{\text{red}}=E_{\text{red}}^{\circ }-\frac{0.0591}{5}\text{log}\left[\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right][H^{+}{]}^8}\right]$

$\text{Let}[H^{+}{]}_{\text{initial}}=X$

$E_{\text{red(initial)}}=E_{\text{red}}^{\circ }-\frac{0.0591}{5}\text{log}\left[\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right][X{]}^8}\right]$

$[H^{+}{]}_{\text{final}}=\frac{X}{100}=\frac{X}{{10}^2}$

$E_{\text{red(final)}}=E_{\text{red}}^{\circ }-\frac{0.0591}{5}\text{log}\frac{[M{n}^{2+}\times {10}^{16}]}{\left[Mn{O}_4^{-}\right]\times [X{]}^8}$

$E_{\text{red(final)}}-E_{\text{red(initial)}}^{\circ }=-\frac{0.0591}{5}\text{log}{10}^{16}=-0.1891V$

The $E_{red}$ decreases by $0.189V$. The tendency of the half-cell to get reduced is its oxidising power. Hence, the oxidising power decreases by $0.189V$.