Question

On electrolysis using platinum electrodes dilute nitric acid yields:

• A. hydrogen at cathode
• B. oxygen at anode
• C. both $A$ and $B$
• D. oxygen at cathode and hydrogen at anode

Answer 1

The correct option is C both $A$ and $B$

In dilute $HN{O}_3$,
Oxidation potential of $O{H}^{-}>$ oxidation potential of $N{O}_3^{-}$.

$\therefore O{H}^{-}$ oxidises to $O_2$ at anode in the following way:

$4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$

$H^{+}$ reduces to $H_2$ at cathode in the following way:

$2H^{+}+2e^{-}\to H_2\uparrow$

Option C is correct.