Question

In electrolysis of very dilute NaOH using platinum electodes:

• A. $H_2$ is evolved at cathode and $O_2$ at anode
• B. $N{H}_3$ is produced at anode
• C. $C{l}_2$ is obtained at cathode
• D. $O_2$ is produced at cathode and $H_2$ at anode

Answer 1

Answer: (A)

$H_2$ is evolved at cathode and $O_2$ at anode

The following half equations happens at the two electrodes:

At the cathode: $2H^{+}+2e^{-}\to H_2$

At the anode: $4O{H}^{-}-4e^{-}\to O_2+2H_{2}O$

So, option A is correct.