Question

When an electric current of $0.5$ ampere is passed through acidulated water for two hours, then volume of hydrogen $\left(H_2\right)$ at NTP produced will be:

($1$ coulomb of electricity deposits $0.00001$ gm of hydrogen).

• A. $0.1$ litre
• B. $0.4$ litre
• C. $0.2$ litre
• D. $0.8$ litre

Answer 1

Answer: (B)

$0.4$ litre

The reaction at the cathode : ${2H}^{+}+{2e}^{-}⟶H_2$

Equivalent of $H_2=\frac{it}{96500}=\frac{0.5\times 2\times 60\times 60}{96500}=0.037$

$\therefore$ Volume of $H_2$ at $NTP=\frac{22.4\times 0.037}{2}=0.4$ litre.