Question

In the electrolysis of acidulated $H_{2}O$, $11.2$ litres of $H_2$ gas is liberated at the cathode. The volume of $O_2$ gas liberated at the anode is:

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

Answer 1

The correct option is B $5.6$ litres

$11.2litre{H}_2=0.5molesof{H}_{2}atNTP$

$2H^{+}+2e^{-}\to H_2$

1 mole of $H_2$ requires 2 moles electrons.

$\therefore$ The number of electrons required $=0.5\times 2=1electron$

$O^{2-}\to O_2+2e^{-}$

1 mole of $O_2$ liberates 4 electrons.

$\therefore$ Moles of $O_2$ for 1 mole $e^{-}=\frac{1}{4}$

$\frac{1}{4}moleatNTP=\frac{1}{4}\times 22.4=5.6$ litres.