Ncert Exercises & Solutions

The half-cell reaction at the anode during the electrolysis of aqueous sodium chloride solution is represented by :

1. Na⁺(aq) + e^- ⟶ Na(s) ; $E_{cell}^{o} \ = \ -2.71 \ V$

2. 2H₂O(l) ⟶ O₂(g) + 4H⁺(aq) + 4e^-; $E_{cell}^{o}$ = 1.23 V

3. H⁺(aq) + e^- ⟶ $\frac{1}{2}$ H₂(g) ; $E_{cell}^{o}$ = 0.00 V

4. Cl^-(aq) ⟶ $\frac{1}{2}$ Cl₂(g) + e- ; $E_{cell}^{o}$ $= 1.36 V$ $= 1.36 V$

HINT: Higher the value of oxidation potential, the higher will be the tendency to oxidize.

Explanation:

In the case of electrolysis of aqueous NaCl, an oxidation reaction occurs at the anode as follows

$\begin{array}{l} {Cl}^{-} (a q) ⟶ \frac{1}{2} {Cl}_{2} (g) + e^{-} \end{array}$

$E ° = 1 \cdot 36 V 2 H_{2} O (l) ⟶ O_{2} (g) + 4 H^{+} (aq) + 4 e^{-}$

${E^{°}}_{cell} = 1.23 V$

But due to lower

$E_{Cell}^{°}$ value water should get oxidized in preference of Cl^- (aq).

However, the actual reaction taking place in the concentrated solution of NaCl is (d) and not (b) i.e., Cl₂is produced and not O₂.

This unexpected result is explained on the basis of the concept of ‘overvoltage’, i.e., water needs greater voltage for oxidation to O₂ (as it is a kinetically slow process) than that needed for oxidation of Cl^- ions to Cl₂. Thus, the correct option is (4) not (2).

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