Question

Predict the product of electrolysis in the following:

An aqueous solution of CuCl₂ with platinum electrodes.

Answer 1

In aqueous solutions, CuCl₂ ionizes to give Cu²⁺ and Cl^– ions as:

\[\ce{CuCl_{2(aq)} -> Cu^{2+}_{ (aq)} + 2Cl-_{ (aq)}}\]

On electrolysis, either of Cu²⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu²⁺ is more than that of H₂O molecules.

\[\ce{Cu^{2+}_{ (aq)} + 2e- -> Cu_{(aq)}}\]; E° = + 0.34V

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

Hence, Cu²⁺ ions are reduced at the cathode and get deposited.

Similarly, at the anode, either of Cl^– or H₂O is oxidized. The oxidation potential of H₂O is higher than that of Cl^–

\[\ce{2Cl-_{ (aq)} -> Cl_{2(g)} + 2e-}\]; E° = -1.36 V

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

But oxidation of H₂O molecules occurs at a lower electrode potential than that of Cl^–ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl^– ions are oxidized at the anode to liberate Cl₂ gas.