CBSE 2025, Set 1

Question

Account for the following:

(i) On the basis of E° values, O$_2$ gas should be liberated at the anode but it is Cl$_2$ gas which is liberated in the electrolysis of aqueous NaCl.

(ii) Conductivity of CH$_3$COOH decreases on dilution.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Cl$_2$ liberated instead of O$_2$ (overpotential):** Thermodynamically, based on standard electrode potentials, O$_2$ should be preferentially discharged at the anode (oxidation of OH$^-$/H$_2$O). However, the **overpotential of O$_2$ is very high** on platinum electrodes — a significantly larger voltage than the theoretical value is required to actually oxidise water to O$_2$. The overpotential of Cl$^-$ oxidation to Cl$_2$ is much lower. Therefore, at practical electrode potentials, **Cl$_2$ is preferentially discharged** at the anode. **(ii) Conductivity of CH$_3$COOH decreases on dilution:** CH$_3$COOH is a **weak electrolyte** — only partially ionised in solution. On dilution: - The degree of ionisation ($\alpha$) **increases** (more dissociation). - However, the **total concentration of ions per unit volume decreases** (dilution effect dominates). Since electrical conductivity depends on the **number of ions per unit volume**, the overall conductivity **decreases** with dilution.

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