Question

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

• A. 5.6
• B. 5.60

Answer 1

Answer: (A), (B)

Electrolysis of $NaOH$solution :
At cathode : $2H^{+}+2e^{-}\to H_2$
At anode : $4H{O}^{-}\to 2H_{2}O+O_2+4e^{-}$
Number of equivalents of $O_2\left(g\right)$ liberated at anode = Number of equivalents of $H_2\left(g\right)$ liberated at cathode.
$O_2\left(g\right)$ liberated $=\frac{\text{volume of}{O}_2\left(g\right)\text{liberated at NTP}}{\text{volume occupied by 1 equivalent of}{O}_2\left(g\right)\text{at NTP}}=\frac{2.8L}{5.6L}=\frac{1}{2}$
The same number of equivalents of hydrogen liberated at the cathode $=\frac{1}{2}$
$\therefore$volume of hydrogen = number of equivalents $\times$ volume occupied by 1 equivalent of hydrogen at NTP $=\frac{1}{2}\times 11.2$
$=5.6\text{litres}$.