I want to get myself rectified in the following solution given to the problem:
product of electrolysis of an aquaeous solution of CuCl
2 with platinum electrodes.
My answer :
CuCl
2 Cu+2 + 2Cl-
At anode, any two of the following has to take place,
2H+ + 2e- H2 E0 = 0.00
Cu+2 + 2e- Cu E0 = 0.34
Thus the following reaction will take place at anode, ( higher value of E0 will be preferred)
Cu+2 + 2e- + H2O Cu + H+ + OH-
At cathode this reaction will take place,
2Cl- Cl2 + 2e-
Thus the whole electrolysis reaction is,
Cu+2 + H2O + 2Cl- Cu + H+ + OH- + Cl2
Please correct wherever necessary with explanation.