During the electrolysis of brine, by using inert electrodes,
- O2 liberates at anode
- H2 liberates at anode
- Na deposits on cathode
- Cl2 liberates at anode
The Correct Option is D
Approach Solution - 1
During the electrolysis of brine (aqueous NaCl solution) using inert electrodes, the ions present are:
- At Cathode (Reduction): Na+, H2O molecules
- At Anode (Oxidation): Cl−, H2O molecules
At Cathode (negative electrode):
Reduction potentials:
- Na+(aq) + e− → Na(s) (Highly negative potential)
- 2H2O(l) + 2e− → H2(g) + 2OH−(aq) (Lower negative potential)
H₂ gas is liberated at cathode since reduction of water is more favorable than Na⁺ reduction.
At Anode (positive electrode):
Oxidation potentials:
- 2Cl−(aq) → Cl2(g) + 2e− (Favorable oxidation)
- 2H2O(l) → O2(g) + 4H+(aq) + 4e− (Less favorable than chloride oxidation in concentrated solutions)
Cl₂ gas is liberated at anode, as oxidation of chloride ions is preferred due to the higher concentration and easier oxidation potential.
Overall reaction for brine electrolysis:
2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq)
Conclusion:
The correct option is: (D) Cl₂ liberates at anode
Correct Answer: Option (D)
Approach Solution -2
During the electrolysis of brine (aqueous sodium chloride, \( \text{NaCl} \)), the reactions that occur at the anode and cathode depend on the ions present in the solution:
At the anode (positive electrode), oxidation occurs. The chloride ions \( \text{Cl}^- \) from sodium chloride undergo oxidation to form chlorine gas \( \text{Cl}_2 \). \[ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \] So, chlorine gas (\( \text{Cl}_2 \)) is liberated at the anode.
At the cathode (negative electrode), reduction occurs. The hydrogen ions \( \text{H}^+ \) (from water) are reduced to form hydrogen gas \( \text{H}_2 \): \[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]
Sodium ions (\( \text{Na}^+ \)) do not get deposited on the cathode because the reduction potential for sodium is very high compared to that of hydrogen, so sodium remains in the solution.
Thus, the correct answer is \( \text{Cl}_2 \) liberates at the anode.
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