Question

On electrolysis of very dilute aqueous solution of NaCl using platinum electrodes :

• A. $\mathrm{H_2}$ gas is evolved at anode.
• B. Na is evolved at cathode.
• C. $\mathrm{O_2}$ gas is evolved at anode.
• D. $\mathrm{Cl_2}$ gas is evolved at cathode.

Hint:

Dilute solution → water reacts at both electrodes: H₂ at cathode, O₂ at anode.

Answer 1

The Correct Option is C

Concept: When we pass current through a solution, water can also take part. At each electrode, whichever species is easier to discharge is the one that reacts. In a very dilute NaCl solution there is plenty of water but very little chloride, so water tends to win.

Step 1: What happens at the cathode
At the cathode reduction takes place. Sodium ions $\mathrm{Na^+}$ are very hard to reduce, so water is reduced instead and hydrogen gas comes off: \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]

Step 2: What happens at the anode
At the anode oxidation takes place. Chloride could be oxidised, but the solution is very dilute, so there is hardly any $\mathrm{Cl^-}$ around. So water is oxidised instead and oxygen gas comes off: \[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

Step 3: Read off the answer
So we get hydrogen at the cathode and oxygen at the anode. The option that says oxygen at the anode is the correct one.

Answer: Option (C), $\mathrm{O_2}$ gas is evolved at the anode.