When 0.1 mole of MnO42- is oxidised, the quantity of electricity required to completely oxidise MnO42- to MnO4- is

Question

When 0.1 mole of MnO42- is oxidised, the quantity of electricity required to completely oxidise MnO42- to MnO4- is (A) 96500 C (B) 2 × 96500 C  (C) 9650 C (D) 96.50 C

Tardigrade Admin · Accepted Answer

The oxidation reaction is as shown below.

M n O_{4}^{2 -} + e^{-} \to M n O_{4}^{-}

Oxidation of one mole of

M n O_{4}^{2 -}

requires

1

mole of electrons.
Hence, the oxidation of

0.1

mole of electrons will require

0.1

mole of electrons.
This corresponds to

96500 \times 0.1 = 9650 C

.

Q. When $0.1$ mole of $M n O_{4}^{2 -}$ is oxidised, the quantity of electricity required to completely oxidise $M n O_{4}^{2 -}$ to $M n O_{4}^{-}$ is

96500 C

$2 \times 96500 C$

9650 C

96.50 C

The oxidation reaction is as shown below.
$M n O_{4}^{2 -} + e^{-} \to M n O_{4}^{-}$
Oxidation of one mole of $M n O_{4}^{2 -}$ requires $1$ mole of electrons.
Hence, the oxidation of $0.1$ mole of electrons will require $0.1$ mole of electrons.
This corresponds to $96500 \times 0.1 = 9650 C$ .

Q. When 0.1 mole of MnO42−​ is oxidised, the quantity of electricity required to completely oxidise MnO42−​ to MnO4−​ is