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Electrode Potential

KMnO 4 + Cu 2...

Question

${K M n O}_{4} + {C u}_{2} S ⟶ {M n}^{2 +} + {C u}^{2 +} + {S O}_{2}$
In above reaction, find 'n' factor of ${C u}_{2} S$ .

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Solution

$K M n O_{4} + C u_{2} S ⟶ M n^{2 +} + C u^{2 +} + S O_{2}$
Oxidation state of $C u$ in $C u_{2} S = + 1$
Oxidation state of $C u$ in $C u^{2 +} = + 2$
$∴$ Change in oxidation state of $C u = 1$
$∴$ Change in oxidation state of $C u_{2} S = 1 \times 2 = 2$
$∴$ $n -$ factor of $C u_{2} S =$ Change in oxidation state of $C u_{2} S$
$∴$ $n -$ factor of $C u_{2} S = 2$

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