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  4. The passage of current liberates H 2 at cathode and Cl 2 at anode. The solution is

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Q. The passage of current liberates $H _{2}$ at cathode and $Cl _{2}$ at anode. The solution is

Electrochemistry

Solution:

Since discharge potential of water is greater than that of sodium so, water is reduced at cathode instead of $Na ^{+}$.

Cathode $: H _{2} O +e^{-} \longrightarrow 1 / 2 H _{2}+ OH ^{-}$

Anode $: Cl ^{-} \longrightarrow 1 / 2 Cl _{2}+e^{-}$