In the acidic medium, MnO₄⁻ acts as an oxidizing agent.
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
If the $[H^{+}]$ is reduced to half of its original value, calculate the electrode potential of the half-cell MnO₄⁻, Mn²⁺ | Pt. ( $lo g 2 = 0.3010$ )

Question

In the acidic medium, MnO₄⁻ acts as an oxidizing agent.

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

If the $[H^{+}]$ is reduced to half of its original value, calculate the electrode potential of the half-cell MnO₄⁻, Mn²⁺ | Pt. ( $lo g 2 = 0.3010$ )

TestHub · Accepted Answer

The electrode potential of the half-cell MnO₄⁻, Mn²⁺ | Pt is given by:

$E_{M n O_{4}^{-} / M n^{2 +}} = E_{M n O_{4}^{-} / M n_{2 +}}^{\circ} - \frac{0.059}{5} lo g \frac{[ M n ^{2 +} ]}{[ M n O _{4}^{-} ] [ H ^{+} ] ^{8}}$

When the concentration of H⁺ is reduced to half of its initial value, the electrode potential is given by:

$E_{M n O_{4}^{-} / M n^{2 +}}^{'} = E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} - \frac{0.059}{5} lo g \frac{[ M n ^{2 +} ] 2 ^{8}}{[ M n O _{4}^{-} ] [ H ^{+} ] ^{8}}$

The decrease in electrode potential is given by:

$E_{M n O_{4}^{-} / M n^{2 +}} - E_{M n O_{4}^{-} / M n^{2 +}}^{'} = \frac{0.0591}{5} lo g (2^{8}) = 0.02846 V$

Therefore, the electrode potential decreases by $28.46 mV$ .

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