If $△ G^{\circ}$ for the half cell $M n O_{4}^{-} | M n O_{2}$ in an acid solution is $x F$ then find the value of $x$ .
(Given: $E_{M n O_{4}^{-} | M n^{2 +}}^{\circ} = 1.5 V; E_{M n O_{2} | M n^{2 +}}^{\circ} = 1.25 V$ )

- 0.75

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- 0.95

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- 0.85

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None of these

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Solution

The correct option is A $- 0.75$
The expression for the standard free energy change is $Δ G^{0} = - n F E^{0}$ But it is equal to xF.
Hence, $x = \frac{- n F E^{0}}{F} = - n E^{0}$
But for the given half reaction, $n = 3$ and $E^{0} = E_{M n O_{4}^{-} | M n^{2 +}}^{0} - E_{M n O_{2} | M n^{2 +}}^{0} 1.5 - 1.25 = 0.25 V$
Hence, $x = - n E^{0} = - 3 \times 0.25 = - 0.75$

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Similar questions

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

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

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

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

M n O_{4}^{-} + M n^{2 +} \to M n O_{2}

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

E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 v o l t, E_{M n O_{2} / M n^{2 +}}^{o} = 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

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