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Question
Predict the product of electrolysis in the following:
An aqueous solution of AgNO3 with silver electrodes.
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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), 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:
\[\ce{H2O_{(l)} -> 1/2 O2_{(g)} + 2H^+_{( aq)}}\]
\[\ce{NO^-_{3(aq)} -> NO3 + e-}\]
\[\ce{Ag_{(s)} + Ag^+_{( aq)} + e-}\]
In these reactions, the reduction potential of silver is the lowest. Therefore, silver itself will be converted into Ag+ as a result of oxidation at the anode, and the Ag+ ions will go into solution.
\[\ce{Ag_{(s)} -> Ag^+_{( aq)} + e-}\]
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