Question

Predict the products of electrolysis of an aqueous solution of CuBr₂ using inert electrodes.

Given: \[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = 0.34 V, \[\ce{E^{\circ}_{{H_{2}O}/H_{2}}}\], OH⁻ = −0.83 V, \[\ce{E^\circ_{Br/Br^-}}\] = 1.08 V and \[\ce{E^{\circ}_{\frac{1}{2} O_{2}, H{^{+}}/{H_{2}O}}}\] = 1.23 V.

Answer 1

To predict the products of electrolysis of an aqueous solution of CuBr₂ using inert electrodes, we analyse the possible reactions at the cathode and anode based on standard electrode potentials.

Electrolyte dissociation:

\[\ce{CuBr2 -> Cu}\]

At cathode (Reduction): 

Possible species:

\[\ce{Cu^{2+} + 2 e− -> Cu}\]; E° = +0.34 V

\[\ce{2H2O + 2e- -> H2 + 2OH-}\]; E° = −0.83 V

Copper ion has a higher (more positive) reduction potential, so it will be reduced:

\[\ce{Cu2+ + 2e− -> Cu_{(s)}}\]

At anode (Reduction): 

Possible species:

\[\ce{2Br− -> Br2 + 2e-}\]; E° = −1.08 V

\[\ce{2H2O -> O2 + 4H+ +4e-}\]; E° = −1.23 V

Bromide ions are more easily oxidised than water because their oxidation potential is less negative.

\[\ce{2Br− -> Br2_{(g)} + 2e-}\]

Overall reaction:

\[\ce{Cu{^{2+}_{(aq)}} + 2Br^-_{ (aq)} -> Cu{(s)} + Br2_{(g)}}\]

At Cathode: Copper metal (Cu_(s))

At Anode: Bromine gas (Br_2(g))