Question

During electrolysis of aqueous $NaOH,4g$ of $O_2$ gas is liberated at NTP at anode, $H_2$ gas liberated at cathode is:-

• A. $2.5$ litres
• B. $5.6$ litres
• C. $11.2$ litres
• D. $22.4$ litres

Answer 1

The correct option is B $5.6$ litres

Reaction for electrolysis of aq. NaOH is given as -

$2NaOH\to 2Na+O_2+H_2$

That means for every mole of $O_2$ liberated at anode same no of moles are liberated at cathode.

$n\left(O_2\right)=n\left(H_2\right)$

$W\left(O_2\right)/M\left(O_2\right)=W\left(H_2\right)/M\left(H_2\right)$

$4/32=W\left(H_2\right)/2$

$W\left(H_2\right)=2\times 4/32$

$W\left(H_2\right)=0.25g$

$Hence$, 0.25 g of $H_2$ is liberated at cathode.

$H_2:O_2$

Ratio:- 2:1

∴ The volume of $H_2$

​ liberated will be twice of $O_2$.

∴ Volume of $O_2\to$ 2.8litre

Volume of $H_2\to$ 5.6litre.

Hence, the correct option is B