Question

On electrolysis of dil. $NaCl$ using Platinum $\left(Pt\right)$ electrode, the product obtained at cathode will be:

• A. $O_2\text{gas}$
• B. $Na$
• C. $C{l}_2\text{gas}$
• D. $H_2\text{gas}$

Answer 1

The correct option is D $H_2\text{gas}$

Among $H_{2}O$ and $N{a}^{+}$, discharge potential of $H_{2}O$ is lower than $N{a}^{+}$ so $H_{2}O$ will discharge at cathode and $H_2\left(g\right)$ will be obtained at cathode.
$\text{Cathode :}2H_{2}O+2e^{-}\to H_2+2O{H}^{-}$