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Question
Predict the product of electrolysis in the following:
An aqueous solution of CuCl2 with platinum electrodes.
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Solution 1
\[\ce{CuCl2_{(aq)} -> Cu^2+_{( aq)} + 2Cl^-_{( aq)}}\]
\[\ce{H2O <=> H+ + OH-}\]
At cathode: Cu2+ ions are preferentially converted to H+ ions.
\[\ce{Cu^2+ + 2e- -> Cu}\]
At anode: Cl− ions are preferentially oxidised over OH− ions.
\[\ce{Cl- -> Cl + e-}\],
\[\ce{Cl + Cl -> Cl2_{(g)}}\]
The cathode deposits Cu, whereas the anode releases Cl2.
Solution 2
\[\ce{CuCl2_{(aq)} -> Cu^{2+}_{( aq)} + 2Cl^-_{( aq)}}\]
\[\ce{H2O <=> H+ + OH-}\]
At cathode: Cu2+ ions undergo preferential reduction and copper metal is deposited at the cathode because Cu2+ (+0.34 V) has a higher reduction potential than water (−0.83 V).
\[\ce{Cu^{2+}_{( aq)} + 2e- -> Cu_{(s)}}\]
At anode: There is a possibility of the following reactions occurring.
\[\ce{H2O_{(l)} -> \frac{1}{2} O2_{(g)} + 2H^+_{( aq)} + 2e-}\]; E° = +1.23 V
\[\ce{2Cl^-_{( aq)} -> Cl_{(g)} + 2e-}\]; E° = +1.36 V
Water will preferentially oxidise at the anode and produce O2 gas because its reduction potential is lower than that of \[\ce{Cl^-_{( aq)}}\].
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