Question

What will be the products of electrolysis of an aqueous solution of $AgNO_3$ with silver electrodes?

• A. $Ag$ from $Ag$ anode dissolves while $Ag^+$ 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)

$AgNO_3 \to Ag^+ + NO_3^-$ (aqueous solution)
At cathode :
$Ag_{\left(aq\right)} + e^{-} \to Ag ; E^{\circ} = + 0.80\,V$
$2H_{2}O_{\left(l\right)}+ 2e^{-} \to H_{2} + 2OH^{-} ; E^{\circ} = -0.83 \,V$
Since reduction potential of $Ag^{+}$ is higher than $H_{2}O, Ag^{+}$ are reduced to $Ag$.
At anode :
$Ag\to Ag^{+}_{\left(aq\right)} + e^{-} ; E^{\circ} = - 0.80\,V$
$2H_{2}O \to O_{2} + 4H^{+} +4e^{-} ; E^{\circ} = - 1.23 \,V$
Since oxidation potential of $Ag$ is higher than $H_{2}O$, $Ag$ of silver anode is oxidised. The oxidation potential of $NO_{3}^{-}$ is lower than $H_{2}O$.