Ncert Exercises & Solutions

Brine is electrolysed by using inert electrodes. The reaction at the anode is:

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

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

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

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

HINT: Cl₂ produces at anode.

Explanation:

STEP 1:

Brine is electrolysed by using inert electrodes. The possible reactions occurring at anode are :

{Cl}^{-} (aq) \to \frac{1}{2} {Cl}_{2} (g) + e^{-}

${Cl}^{-} (aq) \to \frac{1}{2} {Cl}_{2} (g) + e^{-}$ ;

$E_{Cell}^{⊝}$ =1.36V

$2 H_{2} O (l) \to O_{2} (g) + 4 H^{+} + 4 e^{-}$ ;

$E_{Cell}^{⊝}$ = 1.23 V

STEP 2:

The reaction at anode with lower value of E° is preferred and therefore water should get oxidised in preference to

${Cl}^{-}$ (aq). However,

${Cl}_{2}$ is produced instead of

$O_{2}$ .

This unexpected result is explained on the basis of the fact that water needs greater voltage for oxidation to

$O_{2}$ (as it is kinetically slow process) than that needed for oxidation of

${Cl}^{-}$ ions to

${Cl}_{2}$