Question

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

• A. \(H_2\) gas is evolved at anode.
• B. Na is produced at cathode.
• C. \(O_2\) gas is evolved at anode.
• D. \(H_2\) gas is evolved at cathode.

Hint:

In very dilute NaCl solution, water gives \(O_2\) at anode and \(H_2\) at cathode. In concentrated brine, \(Cl_2\) is evolved at anode.

Answer 1

The Correct Option is C, D

Concept:
In electrolysis of aqueous sodium chloride solution, the discharge depends on concentration. In concentrated NaCl solution, chloride ions are discharged at anode to give chlorine gas. But in very dilute NaCl solution, water is oxidised at anode and oxygen gas is evolved. At the cathode, water is reduced to hydrogen gas instead of sodium ion being reduced to sodium metal.

Step 1: Reaction at cathode.
At cathode, reduction takes place. Sodium ion is not discharged easily because sodium has very high tendency to remain as \(Na^+\) in aqueous solution. Instead, water is reduced. \[ 2H_2O+2e^- \rightarrow H_2+2OH^- \] So hydrogen gas is evolved at cathode.

Step 2: Reaction at anode.
At anode, oxidation takes place. Since the solution is very dilute, water is oxidised more preferably than chloride ion. \[ 2H_2O \rightarrow O_2+4H^+ +4e^- \] So oxygen gas is evolved at anode.

Step 3: Important exam note.
Strictly, both of the following statements are correct: Oxygen gas is evolved at anode and Hydrogen gas is evolved at cathode Therefore, both (C) and (D) are chemically correct. If the question expects one answer due to the phrase ``very dilute'', then the distinguishing answer is usually: \[ \boxed{\text{(C) }O_2\text{ gas is evolved at anode}} \] But do not ignore that cathodic hydrogen evolution is also correct. Hence: \[ \boxed{\text{(C) and (D) are chemically correct.}} \]