Question

Which among the following is correct for electrolysis of brine solution ?

• A. $O_2$ gas is liberated at cathode
• B. Sodium metal is collected at anode
• C. $H_2$ gas is liberated at cathode
• D. $C{l}_2$ gas is liberated at cathode

Answer 1

Answer: (C)

In electrolysis of brine solution, $H_2$ gas is liberated at cathode. The solution contains four ions $N{a}^{+},C{l}^{\oplus },H^{+}$and $\stackrel{\ominus }{O}H$ ions. There occurs a race amongst them for their discharge at their respective electrodes.
Following electrode reactions are possible:
Cathode $2H_{2}O(l)+2e^{-}⟶H_2(g)+2\bar{O}H(aq)$
Anode $2Cr(aq)⟶C{l}_2(g)+2e^{-}$
In this electrolysis, $H_2$ at cathode and $C{l}_2$ at anode are given off.