Question
Explain electrolysis of concentration aqueous solution of $\text{NaCl}$

Answer

$\rightarrow$ During the electrolysis of aqueous sodium chloride solution, the products are $\ce{NaOH , Cl_2}$ and $H_2$.
$\rightarrow$ In this case besides $Na^{+}$and $Cl ^{-}$ ions we also have $H ^{+}$and $OH ^{-}$ions along with the solvent molecules, $\ce{H_2O}$.
$\rightarrow$ At the cathode there is competition between the following reduction reactions :
$\ce{Na^{+}( aq ) + e ^{-} \rightarrow Na ( s ) E _{(cell)}^{\ominus}=-2.71 V}$
$\ce{H^{+}( aq ) + e ^{-} \rightarrow \frac{1}{2} H _2(g) E _{(cell)}^{\ominus}=0.00 V}$
$\rightarrow$ The reaction with higher value of $E ^{\ominus}$ is preferred and, therefore the reaction at the cathode during electrolysis is :
$\ce{H ^{+}( aq ) + e ^{-} \rightarrow \frac{1}{2} H _2(g)...(1)}$
but $H ^{+}( aq )$ is produced by the dissociation of $H _2 O$ i.e.,
$\ce{H _2 O (l) \rightarrow H ^{+}( aq ) + OH ^{-}( aq )...(2)}$
Therefore, the net reaction at the cathode may be written as the sum of $(1)$ and $(2)$ and we have
$\ce{H_2O(l) + e^{-} \rightarrow \frac{1}{2} H_2(g) + OH ^{-}}$
At the anode the following oxidation reactions are possible :
$\ce{Cl ^{-}(aq) \rightarrow \frac{1}{2}Cl_2(g) + e^{-}E _{(cell)}^{\ominus}=1.36 V............(3)}$
$\ce{2H_2O(l) \rightarrow O_2(g) + 4H^{+}(aq) + 4e^{-}E _{(cell)}^{\ominus}=1.23 V}$
$\rightarrow$ The reaction at anode with lower value of $E ^{\ominus}$ is preferred and therefore, water should get oxidised in preference to $Cl ^{-}(aq).$ However, on account of over potential of oxygen, reaction $(3)$ is preferred. Thus, the net reactions may be summarised as:
$\ce{NaCl(aq) \rightarrow{ H_2O} Na^{+}(aq) + Cl^{-}(aq)}$
Cathode : $\ce{H_2O(l) + e^{-} \rightarrow \frac{1}{2}H_2(g) + OH^{-}(aq)}$
Anode : $\ce{Cl^{-}(aq) \rightarrow \frac{1}{2}Cl_2(g) + e ^{-}}$
Net reaction :
$\ce{NaCl (aq) + H_2O (l) \rightarrow Na ^{+}( aq ) + OH ^{-}(aq) + \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g)}$

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