Question

If 0.224 litre of $H_2$ gas is formed at the cathode, in the electrolysis of water the volume of $O_2$ gas formed at the anode under identical conditions is:

• A. $0.20$ L
• B. $0.015$ L
• C. $0.112$ L
• D. None of these

Answer 1

The correct option is C $0.112$ L

Electrolysis of water:

$H_{2}O\to H_2+\frac{1}{2}{O}_2$

Under identical conditions of T, P, and V

Moles are directly proportional to the volume

Therefore the volume of $O_2$ will be half of $H_2$

Given: $V_{H_2}=0.224L$

$\therefore V_{O_2}=\frac{0.224}{2}L$

C is correct