Question

Electrolysis of $AgN{O}_3\left(aq\right)$ using platinum (inert) electrodes is carried out. Which of the following is true?

• A. $O_2\left(g\right)$ is liberated at anode.
• B. $H_2\left(g\right)$ is liberated at anode
• C. $H_2\left(g\right)$ is liberated at cathode
• D. Silver is deposited at anode.

Answer 1

The correct option is A $O_2\left(g\right)$ is liberated at anode.

Electrolysis of aqueous $AgN{O}_3$:
Platinum electrode is an inert electrode.
This metal electrode doesn’t involve in the redox reaction of the cell.
Hence, this electrode can conduct electrons into or out of the cell but cannot take part in the half-reactions.

At cathode:
$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right);E_{A{g}^{+}/Ag}^0=0.80V$

$2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$
$E_{H_{2}O/H_2}^0=-0.83V$

The reaction with a higher reduction value of $E^0$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:
Since, $Pt$ electrode are inert, it does not involve in redox reaction of cell.
Therefore, $O{H}^{-}$ or $N{O}_3^{-}$ ions can be oxidized at anode.
But $O{H}^{-}$ ions having higher oxidation potential, therefore they get oxidised to $O_2$
$N{O}_3^{-}$ can't be further oxidised because already nitrogen attains its highest oxidation state (+5).
Anode reaction:
$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$

Hence, option (a) is correct.