In the electrolysis of aqueous sodium chloride with inert electrodes the products obtained at anode and cathode respectively are

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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 $N a^{+}, C l^{-}, H^{+}, O H .$
Following reaction takes place:
At cathode
$\begin{matrix} N a^{+} (a q) + e^{-} \to N a (s); \\ E_{N a^{+} / N a}^{-} = - 2.71 V \\ 2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 \bar{O} H (a q); \\ E_{H_{2} O / H_{2}}^{o} = - 0.83 V \end{matrix}$
At anode
$\begin{matrix} 2 C l^{-} (a q) \to C l_{2} (g) + 2 e^{-}; E_{C l^{-} / C l_{2}}^{o} = - 1.36 V \\ 2 H_{2} O_{2} (l) \to O (g) + 4 H^{+} (a q) + 4 e^{-}; \\ E_{H_{2} O / O_{2}}^{o} = - 1.23 V \end{matrix}$
$E_{H_{2} O / H_{2}}^{o} > E_{N a^{+} / N a}^{o} .$ is formed at cathode.
Whereas $E_{H_{2} O / O_{2}}^{o} > E_{C l^{-} / C l_{2}}^{o} .$ So, formed at anode.

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