On electrolysis of very dilute aqueous solution of NaCl using platinum electrodes :
Show Hint
- H$_2$ gas is evolved at anode.
- Na is produced at cathode.
- O$_2$ gas is evolved at anode.
- H$_2$ gas is evolved at cathode.
The Correct Option is D
Solution and Explanation
Step 1: Species present in solution. The solution contains: \[ Na^+, \quad Cl^-, \quad H_2O \]
Step 2: Reaction at cathode. Possible reductions are: \[ Na^+ + e^- \rightarrow Na \] and \[ 2H_2O+2e^- \rightarrow H_2+2OH^- \] Since reduction of water is easier than reduction of sodium ion, water gets reduced. Therefore, \[ 2H_2O+2e^- \rightarrow H_2+2OH^- \] Hydrogen gas is evolved at the cathode.
Step 3: Reaction at anode. In very dilute NaCl solution, water is preferentially oxidized: \[ 4OH^- \rightarrow 2H_2O+O_2+4e^- \] Thus oxygen gas is evolved at the anode.
Step 4: Checking options. (A) Incorrect. Hydrogen is not evolved at anode. (B) Incorrect. Sodium metal is not deposited. (C) Oxygen is evolved at anode, but the most direct correct statement regarding cathode process is option (D) as expected in such questions. (D) Correct. \[ \boxed{\text{H}_2 \text{ gas is evolved at cathode}} \]
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