Find the magnitude of the standard electrode potential of $M n O_{4}^{-} / M n O_{2}$ . The standard electrode potentials of $M n O_{4}^{-} / M n^{2 +} = 1.51 V$ and $M n O_{2} / M n^{2 +} = 1.23 V$ .

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Solution

Given:
(i) $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O . . . . △ G_{1}^{o}$

(ii) $M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O . . . . △ G_{2}^{o}$

substracting equation (ii) from (i) give the required equation, i.e.,
$M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O . . . . △ G_{3}^{o}$

$∴ △ G_{3}^{o} = △ G_{1}^{o} - △ G_{2}^{o}$
$- 3 F E^{o} = - 5 F E_{1}^{o} - (- 2 F E_{2}^{o})$
$E_{3}^{o} = \frac{(5 \times 1.51) V - (2 \times 1.23) V}{3}$
$E_{3}^{o} = 1.7 V$

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Similar questions

M n O {_{4}}^{-} / M n O_{2}

E {^{o}}_{M n O {_{4}}^{-} / M n^{2 +}} = 1.51 V

E {^{o}}_{M n O_{2} / M n^{2 +}} = 1.23 V

M n O_{4}^{-} / M n O_{2}

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

M n O_{4}^{-} | M n O_{2}

E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V E_{M n O_{2} / M n^{2 +}}^{0} = 1.23 V

M n O_{4}^{-} / M n O_{2}

E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V

E_{M n O_{2} / M n^{2 +}}^{o} = 1.23 V

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

E^{\circ} = 1.51 V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O

E^{\circ} = 1.23 V

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