Question

In the electrolysis of an aqueous solution of $NaOH,2.8litre$ of oxygen gas at NTP was liberated at anode. How much of hydrogen gas was liberated at cathode?

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

Answer 1

Answer: (B)

$5.6litre$

Electrolysis of $NaOH$:

Anode:

$4O{H}^{-}\to O_{2\left(g\right)}+2H_{2}O+4e^{-}$

Cathode:-

$4H_{2}O+4e^{-}\to 2H_2+4O{H}^{-}$

$H_2:O_2$

Ratio:- $2:1$

$\therefore$ The volume of $H_2$ liberated will be twice of

$O_2$.

$\therefore$ Volume of $O_2\to 2.8litre$

Volume of $H_2\to 5.6litre$.

Hence, te correct option is $\text{B}$