Predict the product of electrolysis of an aqueous solution of `CuCl_(2)` with an inert electrode.

Explanation

Electrolysis of aqueous solution of `CuCl_(2)` with platinum electrodes
`CuCl_(2) (s) + (aq) to Cu^(2+) (aq) + 2 Cl^(-) (aq) H_(2) O to H^(+) + OH^(-)`
At cathode : `Cu^(2+)` will be reduced in preference to `H^(+)` ions
`Cu^(2+) + 2e^(-) to Cu`
At anode : `Cl^(-)` ions will be oxidized in preference to `OH^(-)` ions
`Cl to Cl + e^(-)`
`Cl + Cl to Cl_(2)`
Thus, copper will be deposited on the cathode and `Cl_2` will be liberated at anode .