Question

If $5.6l$ of oxygen is liberated during the electrolysis of acidified water, the volume of hydrogen liberated at the cathode is :

• A. $5.6l$
• B. $2.8l$
• C. $11.2l$
• D. $22.4l$

Answer 1

Answer: (C)

$11.2l$

Reaction at anode: $4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$
Reaction at cathode: $2H^{+}+2e^{-}\to H_2$
Theoretically the gas volume ratio for $H_2:O_2$ is $2:1$. For every four electrons that flow through the circuit four hydrogen ions are reduced to two molecules of hydrogen and two molecules of water (or four hydroxide ions) are oxidized to give one molecule of oxygen. Thus volume of $H_2$ is $2\times volumeof{O}_2$ = $2\times 5.6l=11.2l$