In the electrolysis of aqueous sodium chloride with inert electrodes, the products obtained at anode and cathode respectively are $C{l}_2$ and $H_2.$
The solution contains four ions $N{a}^{+},C{l}^{-},H^{+},OH.$
Following reaction takes place:
At cathode
$\begin{array}{c}N{a}^{+}\left(aq\right)+e^{-}\to Na\left(s\right);\\ E_{N{a}^{+}/Na}^{-}=-2.71V\\ 2H_{2}O\left(l\right)+2e^{-}\to H_2\left(g\right)+2\stackrel{-}{O}H\left(aq\right);\\ E_{H_{2}O/H_2}^o=-0.83V\end{array}$
At anode
$\begin{array}{c}2C{l}^{-}\left(aq\right)\to C{l}_2\left(g\right)+2e^{-};E_{C{l}^{-}/C{l}_2}^o=-1.36V\\ 2H_2{O}_2\left(l\right)\to O\left(g\right)+4H^{+}\left(aq\right)+4e^{-};\\ E_{H_{2}O/O_2}^o=-1.23V\end{array}$
$E_{H_{2}O/H_2}^o>E_{N{a}^{+}/Na}^o.$ is formed at cathode.
Whereas $E_{H_{2}O/O_2}^o>E_{C{l}^{-}/C{l}_2}^o.$ So, formed at anode.