Question

the following equation refer to the electrolysis of copper sulphate solution using copper electrodes, during the process of electrolysis what happens to the colour of the copper sulphate solution. give reason for your answer

Answer 1

think this should help

Anode: 4OH(-) -----> 2H2O + O2 + 4e(-)

Cathode: Cu(2+) + 2e(-) -----> Cu

During electrolysis:the cathode gets coated with copper. bubbles of oxygen are given off at the anode As the copper ions are discharged as copper atoms at the cathode, the blue colour of the solution gradually fades and an oxidation reaction occurs which is the 4e- (electron loss).

The negative sulphate ions (SO42-) or the traces of hydroxide ions (OH–) are attracted to the positive electrode. But the sulfate ion is too stable and nothing happens. Instead either hydroxide ions or water molecules are discharged and oxidised to form oxygen.