At what pH,given half cell MnO_4^{-} (0.1 M) | vedclass

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

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(A) The reduction half-reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} ightleftharpoons Mn^{2+} + 4H_2O$Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^{-}] [H^{+}]^8}$Substituting the given values: $1.282 = 1.54 - \frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} 	imes [H^{+}]^8}$$1.282 - 1.54 = -\frac{0.059}{5} \log \frac{10^{-2}}{[H^{+}]^8}$$-0.258 = -\frac{0.059}{5} \log \frac{10^{-2}}{[H^{+}]^8}$$\frac{0.258 	imes 5}{0.059} = \log (10^{-2}) - \log ([H^{+}]^8)$$21.86 = -2 + 8 \ pH$$8 \ pH = 23.86$$pH = 2.98 \approx 3$

At what $pH$,given half cell $MnO_4^{-} (0.1 \ M) \mid Mn^{2+} (0.001 \ M)$ will have electrode potential of $1.282 \ V$? (Nearest Integer) Given $E_{MnO_4^{-} / Mn^{2+}}^{o} = 1.54 \ V, \frac{2.303 RT}{F} = 0.059 \ V$

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