What is the standard electrode potential for the electrode $M n O_{4}^{-} / M n O_{2}$ in solution?
(Given: $E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = 1.51 v o l t, E_{M n O_{2} / M n^{2 +}}^{\circ} = 1.23 v o l t$ ).

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Solution

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{\circ} = 1.51 v o l t$
$M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{\circ} = 1.23 v o l t$
Subtracting
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O; E^{\circ} = ?$
$E^{\circ} = \frac{5 \times 1.51 - 2 \times 1.23}{3} = \frac{7.55 - 2.46}{3} = \frac{5.09}{3} = 1.70 v o l t$ .

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