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Question
Predict the product of electrolysis in the following:
An aqueous solution of \[\ce{AgNO3}\] with silver electrodes.
Solution
\[\ce{AgNO3_{(aq)} -> Ag^+_{( aq)} + NO^-_{3( aq)}}\]
\[\ce{H2O ⇌ H^+ + OH^-}\]
At cathode: Since the reduction potential of silver (+0.80 V) is higher than that of water (−0.830 V), \[\ce{Ag+}\] will be preferentially reduced and silver metal will be deposited at the cathode.
\[\ce{Ag^+_{( aq)} + e^- -> Ag_{(s)}}\]
At anode: The following reaction will take place –
\[\ce{H2O_{(l)} -> 1/2O2_{(g)} + 2H^+_{( aq)}}\]
\[\ce{NO^-_{3( aq)} -> NO3 + e^-}\]
\[\ce{Ag_{(s)} + Ag^+_{( aq)} + e^-}\]
In these reactions, the reduction potential of copper is the lowest. Therefore, silver itself will be converted into \[\ce{Ag+}\] as a result of oxidation at the anode and the \[\ce{Ag+}\] ions will go into solution.
\[\ce{Ag_{(s)} -> Ag^+_{( aq)} + e^-}\]
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