The magnitude of the change in oxidising powe | vedclass

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Question

(A) The reduction half-reaction is:$MnO_4^- + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$Using the Nernst equation for the electrode potential:$E = E^0

Vedclass · Accepted Answer

$3776$

Vedclass · Answer

(A) The reduction half-reaction is:$MnO_4^- + 8H^{+} + 5e^- ightarrow Mn^{2+} + 4H_2O$Using the Nernst equation for the electrode potential:$E = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^{+}]^8}$Case $I$: When $[H^{+}] = 1 \, M$$E_1 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]}$Case $II$: When $[H^{+}] = 10^{-4} \, M$$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-] 	imes (10^{-4})^8}$$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]} + \frac{0.059}{5} \log (10^{-4})^8$$E_2 = E^0 - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^-]} + \frac{0.059}{5} 	imes (-32)$Calculating the magnitude of change $|E_1 - E_2|$:$|E_1 - E_2| = \frac{0.059}{5} 	imes 32$$|E_1 - E_2| = 0.0118 	imes 32 = 0.3776 \, V$Given the change is $x 	imes 10^{-4} \, V$:$0.3776 \, V = 3776 	imes 10^{-4} \, V$Therefore,$x = 3776$.

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