Question

When dil. $HN{O}_3$ is electrolysed:

• A. $H_2$ gas is produced at the anode
• B. $O_2$ gas is produced at the anode
• C. $N{O}_2$ is produced at the anode
• D. $N_2$ is produced at the anode

Answer 1

The correct option is B $O_2$ gas is produced at the anode

When dil. $HN{O}_3$ is electrolysed it dissociates into
$HN{O}_3\to H^{+}+N{O}_3^{-}$
and $H_{2}O\to H^{+}+O{H}^{-}$
The ions which are in the solution are : $H^{+},O{H}^{-},N{O}_3^{-}$
So $H^{+}$ will migrate towards the negative electrode i.e cathode and $H_2$ will be released.
At anode $N{O}_3^{-}$ has a high discharge potential will not be able to lose electrons and $O{H}^{-}$ will get oxidized:
$4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$