Question

Predict the product of electrolysis in the following:

An aqueous solution of AgNO₃ with silver electrodes.

Answer 1

AgNO₃ ionizes in aqueous solutions to form Ag⁺ and \[\ce{NO^-_3}\] ions.

On electrolysis, either Ag⁺ ions or H₂O molecules can be reduced at the cathode. But the reduction potential of Ag⁺ ions is higher than that of H₂O.

\[\ce{Ag+_{ (aq)} + e- -> Ag_{(s)}};\] E° = + 0.80 V

\[\ce{2H_2O_{(l)} + 2e- -> H_{2(g)} + 2OH^-_{ (aq)}}\]; E° = −0.83 V

Hence, Ag⁺ ions are reduced at the cathode. Similarly, Ag metal or H₂O molecules can be oxidised at the anode. But the oxidation potential of Ag is higher than that of H₂O molecules.

\[\ce{Ag_{(s)} -> Ag^+_{ (aq)} + e-}\];  E° = −0.80 V

\[\ce{2H_2O_{(l)} -> O_{2(g)} + 4H+_{ (aq)} + 4e-}\]; E° = −1.23 V

Therefore, Ag metal gets oxidised at the anode.

Answer 2

\[\ce{AgNO3_{(aq)} -> \frac{1}{2} O2_{(g)} + NO^-_3_{(aq)}}\]

\[\ce{H2O <=> H+ + OH-}\]

At cathode: Ag⁺ will be reduced preferentially, and silver metal will be deposited at the cathode because its reduction potential (+0.80 V) is higher than that of water (−0.83 V).

\[\ce{Ag{^+_{(aq)}} + e- -> Ag_{(s)}}\]

At anode: The Following reactions may take place:

\[\ce{H2O_{(l)} -> \frac{1}{2} O2_{(g)} + 2H+_{(aq)} + 2e-}\]

\[\ce{NO^-_3_{(aq)} -> NO3 + e-}\]

\[\ce{Ag_{(s)} -> Ag{^+_{(aq)}} + e-}\]

Silver has the lowest reduction potential among these reactions. As a result, the silver anode will oxidise to produce Ag⁺ ions, which will then enter the solution.

\[\ce{Ag_{(s)} -> Ag{^+_{(ag)}}}\]