Q18 / 90 · JEE Main 2024 (01 Feb Shift 2) · Chemistry

The amount of electricity in Coulomb required for the oxidation of $1 mol$ of $H_{2} O$ to $O_{2}$ is ______ $\times 10^{5} C .$

Correct answer: 2

Solution

$H_{2} O \to \frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-}$

Electricity required to oxidise $1$ mol $H_{2} O to O_{2} = 2 F$

$\frac{W}{E} = \frac{Q}{96500}$

mole $\times n$ -factor $= \frac{Q}{96500}$

$1 \times 2 = \frac{Q}{96500}$

$Q = 2 \times 96500 C$

$= 1.93 \times 10^{5} C$

Hence, the correct answer is 2.