[mrreluctant]

I want to get myself rectified in the following solution given to the problem:
product of electrolysis of an aquaeous solution of CuCl₂ with platinum electrodes.

My answer :

CuCl<sub>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.</sub>

[Hunter2]

looks good