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(A) The reduction half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.Using the Nernst equation: $E = E^o - \frac{0.059}{5} \log \fr

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Increase by $189 \ mV$

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(A) The reduction half-reaction is: $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$.Using the Nernst equation: $E = E^o - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$.At $25 ^oC$,taking the Nernst factor as $0.059 \ V$ (or approximately $0.06 \ V$):$E = E^o - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]} + \frac{0.059 	imes 8}{5} \log [H^+]$.Let the initial concentration be $[H^+]_1 = X$ and final concentration be $[H^+]_2 = 100X$.The change in potential $\Delta E = E_2 - E_1 = \frac{0.059 	imes 8}{5} \log \frac{[H^+]_2}{[H^+]_1}$.$\Delta E = \frac{0.472}{5} \log(100) = 0.0944 	imes 2 = 0.1888 \ V \approx 189 \ mV$.Since $[H^+]$ increases,the reduction potential $E$ increases,meaning the oxidising power increases by $189 \ mV$.

By how much would the oxidising power of the $(MnO_4^- / Mn^{2+})$ couple change if the $H^{+}$ ion concentration is increased up to $100$ times at $25 ^oC$?

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