Question

What will be the products of electrolysis of an aqueous solution of $AgN{O}_3$ with silver electrodes?

• A. $Ag$ from $Ag$ anode dissolves while $A{g}^{+}$ from solution gets deposited on the cathode.
• B. $Ag$ is liberated at cathode and $O_2$ is liberated at anode.
• C. $Ag$ at cathode and nitric acid at anode is liberated.
• D. No reaction takes place.

Answer 1

The correct option is A $Ag$ from $Ag$ anode dissolves while $A{g}^{+}$ from solution gets deposited on the cathode.

The chemical equation for the electrolysis of $AgN{O}_3$ is:

$AgN{O}_3\to A{g}^{+}+N{O}_3^{-}$ (aqueous solution )

Reaction at cathode :

$A{g}_{aq}^{+}+e^{-}\to A{g}^{+}$; $E^0=+0.80V$

$2H_2{O}_{(}l)+2e^{-}\to 2O{H}^{-}$; $E^0=-0.83V$

Since reduction potential of Ag is higher than $H_{2}O$, the $A{g}^{+}$ ions are reduced to Ag.

Reaction at anode :
$Ag\to A{g}_{\left(aq\right)}^{+}+e^{-};E^o=-0.80V$

$2H_{2}O\to O_2+4H^{+}+4e^{-};E^0=-1.23V$

Since the oxidation potential of Ag is higher than $H_{2}O$, Ag of silver anode is oxidized. The potential of $N{O}_{(3}^{-}$is lower than $H_{2}O.$

Therefore, option A is the correct answer.