The standard electrode potential in V for the electrode $M n O_{4}^{-} / M n O_{2}$ in solution is: (round off the answer to the nearest integer)
Given: $E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V$ and $E_{M n O_{2} / M n^{2 +}}^{o} = 1.23 V$

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Solution

$M n O_{4}^{-} + 8 H^{-} + 5 e \to M n^{2 +} + 4 H_{2} O; E_{1}^{o} = 1.51 V$ ........... (i)
$∴ - Δ G_{1}^{o} = 5 \times 1.51 \times F$
$= 7.55 F$

$M n O_{2} + 4 H^{+} + 2 e \to M n^{2 +} + 2 H_{2} O; E_{2}^{o} = 1.23 V$ ........... (ii)
$∴ - Δ G_{2}^{o} = 2 \times 1.23 \times F$
$= 2.46 F$
Subtracting equation (ii) from (i)
$M n O_{4}^{-} + 4 H^{+} + 3 e \to 2 H_{2} O + M n O_{2}; E_{3}^{o} = ?$
or $- Δ G_{3}^{o} = n_{3} E_{3}^{o} F$

$∴ - Δ G_{3}^{o} = - Δ G_{1}^{o} - (- Δ G_{2}^{o})$

$3 E_{3}^{o} F = 7.55 F - 2.46 F$

$∴ E_{3}^{o} = \frac{5.09}{3} = 1.70 v o l t$

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