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Question
Predict the product of electrolysis in the following:
An aqueous solution of \[\ce{AgNO3}\] 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:
\[\ce{H2O_{(l)} -> 1/2O2_{(g)} + 2H{^{+}_{(aq)}}}\]
\[\ce{NO{^{-}_{3(aq)}} -> NO3 + e^-}\]
\[\ce{Ag_{(s)} + Ag{^{+}_{(aq)}} + e^-}\]
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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