First slide

Faraday's Laws

Question

When 0.1 mole of ${MnO}_{4}^{2 -}$ is oxidised, the quantity of electricity required to completely oxidise ${MnO}_{4}^{2 -}$ to ${MnO}_{4}^{-}$

Moderate

A

96500 C
B

2 x 96500 C
C

9650 C
D

96.50 C

Solution

$\underset{+ 6}{{MnO}_{4}^{2 -}} ⇌ \underset{(+ 7)}{{MnO}_{4}^{-}}$
As per the equation, for 1 mole of ${MnO}_{4}^{2 -}, 1$ F of electricity is required. Thus, for 0.1 mole of ${MnO}_{4}^{2 -} to {MnO}_{4}^{-}$ F of electricity is required.
Since; 1 F = 96500 C
$∴ 0.1 F = 0.1 \times 96500 C = 9650 C$
Hence, 9650 C of electricity is required to completely oxidise ${MnO}_{4}^{2 -} to {MnO}_{4}^{-}$ .

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