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Redox Reactions

What is the s...

Question

What is the standard electrode potential for the electrode $M n O {_{4}}^{-} / M n O_{2}$ in the solution?
Given: $E {^{o}}_{M n O {_{4}}^{-} / M n^{2 +}} = 1.51 V$ and $E {^{o}}_{M n O_{2} / M n^{2 +}} = 1.23 V$ :

1.70 V

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- 1.70 V

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5.10 V

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- 5.10 V

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Solution

The correct option is A $1.70 V$
The relationship between standard Gibbs energy change for a reaction and its cell potential is ${Δ G}^{o} = - n F E_{c e l l}^{o}$ .
For manganese electrodes, ${Δ G}_{M n O_{4}^{-} | M n^{2 +}}^{o} = {Δ G}_{M n O_{4 -} | M n O_{2}}^{o} + {Δ G}_{M n O_{2} | M n^{2 +}}^{o} - n F E_{M n O_{4}^{-} | M n^{2 +}}^{o} = - n F E_{M n O_{4}^{-} | M n O_{2}}^{o} - n F E_{M n O_{2} | M n^{2 +}}^{o} 5 E_{M n O_{4}^{-} | M n^{2 +}}^{o} = 3 \times E_{M n O_{4}^{-} | M n O_{2}}^{o} + 2 \times E_{M n O_{2} | M n^{2 +}}^{o}$
Substitute values in the above equation,
$5 \times 1.51 V = 3 \times E_{M n O_{4}^{-} | M n O_{2}}^{o} + 2 \times 1.23 V E_{M n O_{4}^{-} | M n O_{2}}^{o} = 1.70 V$

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