For the following sequential reaction Eo values in volts at 298 K are given- MnO4-0-564 V- MnO4-22-26 V- MnO20-95 V- Mn-31-51 V- Mn-2Calculate Eo for MnO4- Mn-2-

The correct option is C 1.51 voltGiven: Δ G^⊖ #160; = nF E _ cell where n → No. of electrons #160; F → Faraday's constant E → Reduction ...

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For the follo...

Question

For the following sequential reaction $E^{o}$ values in volts at 298 K are given:
$M n O_{4}^{-} 0.564 V \to M n O_{4}^{- 2} 2.26 V \to M n O_{2} 0.95 V \to M n^{+ 3} 1.51 V \to M n^{+ 2}$
Calculate $E^{o}$ for $M n O_{4}^{-} \to M n^{+ 2}$ .

-1.81 volt

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-1.18 volt

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1.51 volt

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1.18 volt

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M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{o} = X_{1} V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{o} = X_{2} V

E^{o}

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

M n O_{4}^{⊖} + C_{2} O_{4}^{2 -} + H^{\oplus} \to M n^{+ 2} + C O_{2} + H_{2} O

M^{x +} + M n O_{4}^{- 1} \to M O_{3}^{- 1} + M n^{+ 2}

M n O_{4}^{- 1}

M^{n +}

M O_{3}^{- 1}

M n^{2 +} + 2 e^{-} \to M n, E^{o} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}), E^{o} = + 1.51 V

E^{o}

3 M n^{2 +} \to M n + 2 M n^{3 +}

M n^{2 +} + 2 e^{-} \to M n; E^{o} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{o} = + 1.51 V

E^{o}

3 M n^{2 +} \to M n + 2 M n^{3 +}

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