Question
In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?
A.
$$Na_{\left( {aq} \right)}^ + + {e^ - } \to N{a_{\left( s \right)}};E_{{\text{cell}}}^ \circ = - 2.71\,V$$
B.
$$2{H_2}{O_{\left( l \right)}} \to {O_{2\left( g \right)}} + 4H_{\left( {aq} \right)}^ + + 4{e^ - };$$ $$E_{{\text{cell}}}^ \circ = 1.23\,V$$
C.
$$H_{\left( {aq} \right)}^ + + {e^ - } \to \frac{1}{2}{H_{2\left( g \right)}};$$ $$E_{{\text{cell}}}^ \circ = 0.00\,V$$
D.
$$Cl_{\left( {aq} \right)}^ - \to \frac{1}{2}C{l_{2\left( g \right)}} + {e^ - };$$ $$E_{{\text{cell}}}^ \circ = 1.36\,V$$
Answer :
$$Cl_{\left( {aq} \right)}^ - \to \frac{1}{2}C{l_{2\left( g \right)}} + {e^ - };$$ $$E_{{\text{cell}}}^ \circ = 1.36\,V$$
Solution :
At the anode, the following oxidation reactions are possible :
$$\left( {\text{D}} \right)Cl_{\left( {aq} \right)}^ - \to \frac{1}{2}C{l_{2\left( g \right)}} + {e^ - };$$ $$E_{{\text{cell}}}^ \circ = 1.36\,V$$
$$\left( {\text{B}} \right)2{H_2}{O_{\left( l \right)}} \to {O_{2\left( g \right)}} + 4H_{\left( {aq} \right)}^ + + 4{e^ - };$$ $$E_{{\text{cell}}}^ \circ = 1.23\,V$$
The reaction at anode with lower value of $${E^ \circ }$$ is preferred and therefore, water should get oxidised in preference to $$Cl_{\left( {aq} \right)}^ - $$
However, on account of overpotential of oxygen, reaction (D) is preferred.