Question

During electrolysis of water the volume of $O_2$ liberated is $2.24d{m}^3$. The volume of hydrogen liberated, under same conditions will be:

• A. $2.24d{m}^3$
• B. $1.12d{m}^3$
• C. $4.48d{m}^3$
• D. $0.56d{m}^3$

Answer 1

The correct option is C $4.48d{m}^3$

$\underset{2vol.}{2H_{2}O}\stackrel{Electrolysis}{\to }\underset{2vol.}{2H_2}+\underset{1vol.}{O_2}$
Thus, the volume of hydrogen liberated is twice that of the volume of oxygen liberated. When $2.24d{m}^3$ of oxygen is liberated the volume of hydrogen liberated will be $2\times 2.24d{m}^3$ or $4.48d{m}^3$