In acidic medium,MnO_4^- acts as an oxidising | vedclass

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(A) The Nernst equation for the given half-cell reaction is: $E = E^\circ - \frac{0.0591}{n} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$.Considering only

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increase by $28.46 \ mV$

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(A) The Nernst equation for the given half-cell reaction is: $E = E^\circ - \frac{0.0591}{n} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$.Considering only the concentration of $H^+$ ions,the change in potential $\Delta E$ is given by: $\Delta E = E_2 - E_1 = -\frac{0.0591}{5} \log \frac{[H^+]_2^8}{[H^+]_1^8}$.Given $[H^+]_2 = 2[H^+]_1$,we have: $\Delta E = -\frac{0.0591}{5} \log (2^8) = -\frac{0.0591 	imes 8}{5} \log 2$.$\Delta E = -0.01182 	imes 8 	imes 0.3010 = -0.02846 \ V = -28.46 \ mV$.Since the potential increases,the electrode potential will increase by $28.46 \ mV$.

In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

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