Question

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

• A. $2H_{2}O\left(l\right)\to O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-};E_{cell}^{\Theta }=1.23V$
• B. $C{l}^{-}\left(aq\right)\to \frac{1}{2}C{l}_2\left(g\right)+e^{-};E_{cell}^{\Theta }=1.36V$
• C. $H^{+}\left(aq\right)+e^{-}\to \frac{1}{2}{H}_2\left(g\right);E_{cell}^{\Theta }=0.00V$
• D. $N{a}^{+}\left(aq\right)+e^{-}\to Na\left(s\right);E_{cell}^{\Theta }=-2.71V$

Answer 1

The correct option is B $C{l}^{-}\left(aq\right)\to \frac{1}{2}C{l}_2\left(g\right)+e^{-};E_{cell}^{\Theta }=1.36V$

Half cell reactions at cathode

At the cathode, there is competition between the following reduction occurs reactions:

$N{a}^{+}\left(aq\right)+e^{-}\to Na\left(s\right);E_{cell}^{\Theta }=-2.71V$

$H^{+}\left(aq\right)+e^{-}\to \frac{1}{2}{H}_2\left(g\right);E_{cell}^{\Theta }=0.00V$

The reaction with higher value of $E_{cell}^{\Theta }$ is preferred for reduction and therefore, the reaction at the cathode during electrolysis is:

$H^{+}\left(aq\right)+e^{-}\to \frac{1}{2}{H}_2\left(g\right)$

Overall net reaction at cathode:
$2H_{2}O\left(l\right)+2e^{-}\to \frac{1}{2}{H}_2\left(g\right)+O{H}^{-}\left(aq\right)$

Half cell reactions at anode

At anode, oxidation occurs. So, two reactions are possible here:

$C{l}^{-}\left(aq\right)\to \frac{1}{2}C{l}_2\left(g\right)+e^{-};E_{cell}^{\Theta }=1.36V$
$2H_{2}O\to O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-};E_{cell}^{\Theta }=1.23V$

Lower value of $E^0$ is preferred for oxidation at anode, therefore water should get oxidised in preference to $C{l}^{-}$.

However, on account of overpotential of oxygen, $C{l}^{-}\left(aq\right)\to \frac{1}{2}C{l}_2\left(g\right)+e^{-}$ reaction is preferred at anode.

Net reaction of Electrolysis of aqueous NaCl:
$Nacl\left(aq\right)+H_{2}O\left(l\right)\to N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)+\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)$

Cathode:

$2H_{2}O\left(l\right)+2e^{-}\to \frac{1}{2}{H}_2\left(g\right)+O{H}^{-}\left(aq\right)$

Anode:

$C{l}^{-}\left(aq\right)\to \frac{1}{2}C{l}_2\left(g\right)+e^{-}$

Overall reaction:

$Nacl\left(aq\right)+H_{2}O\left(l\right)\to N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)+\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)$

Hence option (D) is correct.