Value of standard electrode potential for the oxidation of $C l^{-}$ ions is more positive than that of water, even then the electrolysis of aqueous sodium chloride, why is $C l^{-}$ oxidised at anode instead of water?

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Solution

In the electrolysis
At anode, oxidation reaction takes place and ions get oxidized.
At cathode, reduction reaction takes place and ions get reduced.
Thus, reaction at anode will be:

$C l^{-} \to \frac{1}{2} C l_{2} + e^{-} E^{0} = 1.36 V$
$2 H_{2} O \to O_{2} + 4 H^{-} + 4 e^{-} E^{0} = 1.23 V$

At anode, the reaction with lower value of $E^{0}$ is preferred for oxidation. Thus, at anode $O_{2}$ (g) must have been produced instead of $C l_{2}$ (g) but the formation of $O_{2}$ at anode is kinetically so slow and needs some overvoltage to be additionally provided.

Thus, In electrolysis of aqueous sodium chloride solution $C l^{-}$ is oxidised at anode.

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Oxoacids of Halogens

Standard XII Chemistry

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Value of standard electrode potential for the oxidation of Cl− ions is more positive than that of water, even then the electrolysis of aqueous sodium chloride, why is Cl− oxidised at anode instead of water?

Value of standard electrode potential for the oxidation of $C l^{-}$ ions is more positive than that of water, even then the electrolysis of aqueous sodium chloride, why is $C l^{-}$ oxidised at anode instead of water?