The product of electrolysis of aqueous NaCl solution on anode and...

In case of electrolysis of aqueous NaCl , oxidation and reduction reaction occurs at anode and cathode respectively as shown below: At cathode $\mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{OH}^{-}(\mathrm{aq})$ At anode $\mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})+\mathrm{e}^{-}$ Net cell reaction is as follows $\mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})$ $+\mathrm{OH}^{-}(\mathrm{aq})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})$ Thus, the product of electrolysis of NaCl solution on cathode and anode is $\mathrm{H}_2$ and $\mathrm{Cl}_2$.