Question

A dilute aqueous solution of sulphuric acid is electrolysed using platinum electrodes $0.48$ g of oxygen is liberated at the anode and in the same time, the volume of gas liberated at the cathode (assuming one mole of gas at room temperature and pressure occupies $24$ $d{m}^3$ and both gases are insoluble in dil. $H_{2}S{O}_4$) is:

• A. $0.48$ $d{m}^3$
• B. $12.00$ $d{m}^3$
• C. $0.72$ $d{m}^3$
• D. $1.20$ $d{m}^3$

Answer 1

The correct option is C $0.72$ $d{m}^3$

Dilute aqueous solution of $H_{2}S{O}_4$ liberates $O_2$ gas at anode and $H_2$ gas at cathode.

$2H_{2}O\to 2H_2+O_2$

2 mol 2 mol 1 mol

$O_2$ produced =0.48g $=\frac{0.48}{32}=0.015mol$

$H_2$ produced=$2\times 0.015mol=0.03mol$

Volume of $H_2$ produced=$24\times 0.3d{m}^3=0.72d{m}^3$