During the electrolysis of aqueous nitric acid solution using `Pt` electrodes

Options

A. `O_(2)(g)` is liberated at the anode
B. `N_(2)(g)` is liberated at the anode
C. `O_(2)(g)` is liberated at the cathode
D. `H_(2)(g)` is liberated at the anode

Answer

A

Explanation

The species we should consider for half reactions are `H^(+)(aq.), NO_(3)^(-)(aq.)` and `H_(2)O(l)`. At the cathode, the possible reduction half reactions are
`2H^(+)(aq.)+2E^(-) rarr H_(2)(g) E^(@) = 0.0 V`
`2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-) E^(@) = -0.83 V`
On account of higher reduction potential, `H^(+)` ions are reduced at the cathode of liberate `H_(2)(g)`.
Since the nitrate ion (having highest oxidation state of `N`) is not oxidized during the electrolysis, the only possible oxidation reaction at the anode is
`2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq.)+4e^(-)`
Thus, `O_(2)(g)` is liberated at the anode.