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 cathode.
• B. $Ag$ is liberated at cathode and $O_2$ is liberated at j anode.
• C. $Ag$ at cathode and nitric acid at anode is liberated.
• D. No reaction takes place.

Answer 1

Answer: (A)

$AgN{O}_3\to A{g}^{+}+N{O}_3^{-}$ (aqueous solution)
At cathode :
$A{g}_{\left(aq\right)}+e^{-}\to Ag;E^{\circ }=+0.80V$
$2H_2{O}_{\left(l\right)}+2e^{-}\to H_2+2O{H}^{-};E^{\circ }=-0.83V$
Since reduction potential of $A{g}^{+}$ is higher than $H_{2}O,A{g}^{+}$ are reduced to $Ag$.
At anode :
$Ag\to A{g}_{\left(aq\right)}^{+}+e^{-};E^{\circ }=-0.80V$
$2H_{2}O\to O_2+4H^{+}+4e^{-};E^{\circ }=-1.23V$
Since oxidation potential of $Ag$ is higher than $H_{2}O$, $Ag$ of silver anode is oxidised. The oxidation potential of $N{O}_3^{-}$ is lower than $H_{2}O$.