In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at anode?

Easy

A

${Na}^{+} (aq) + e^{-} ⟶ Na (s); E_{cell}^{\circ} = - 2 .71 V$
B

$2 H_{2} O (l) ⟶ O_{2} (g) + 4 H^{+} (aq) + 4 e^{-}; E_{cell}^{\circ} - 1 .23 V$
C

$H^{+} (aq) + e^{-} ⟶ \frac{1}{2} H_{2} (g); E_{cell}^{\circ} = 0 .00 V$
D

${Cl}^{-} (aq) ⟶ \frac{1}{2} {Cl}_{2} (g) + e^{-}; E_{cell}^{\circ} = 1 .36 V$

Solution

In case of electrolysis of aq .NaCl oxidation reaction occurs at anode as follows:
$\begin{array}{l} {Cl}^{-} (aq) ⟶ \frac{1}{2} {Cl}_{2} (g) + e^{-}; E^{\circ} = 1 \cdot 36 V \\ 2 H_{2} O (l) ⟶ O_{2} (g) + 4 H^{+} (aq) + 4 e^{- \dots}; E_{cell}^{\circ} = 1 .23 V \end{array}$
But due to lower $E_{cell}^{o}$ value, water should get oxidised in preference of Cl^- (aq)
However, the actual reaction taking place is the first one i.e. Cl₂ is produced.
This unexpected result is explained on the basis of the concept of 'over voltage', i.e. water needs greater voltage for oxidation to O₂ (as it is kinetically slow process) than that needed for oxidation of Cl^- ions to Cl₂.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME