Predict the products of electrolysis of an aqueous solution of `CuBr_(2)` .

Explanation

The probable reactions at cathode are :
`(i) Cu^(2+) (aq) + 2e^(-) to Cu (s) , " " E^(Theta) = 0.34 V`
(ii) `2 H_(2) O (l) + 2e^(-) to H_(2) (g) + 2 OH^(-) (aq) , E^(Theta) = -0.83` V
Since the reduction potential of `Cu^(2+)` ions is higher than that of water , copper will be reduced preferably at cathode ,
The probable reactions at anode are :
`2 Br^(-) (aq) to Br_(2) (g) + 2e^(-) " " E^(Theta) = 1.08 V`
`H_(2) O(l) to (1)/(2) O_(2) (g) + 2 H^(+) + 2e^(-) E^(Theta) = 1.23` V
Since the reduction potential of `Br^(-)` is less than water , it will be readily oxidised to `Br_2` at anode . Therefore , the reactions are :
At cathode : `Cu^(2+) (aq) + 2 e^(-) to Cu (s)`
At anode : `2 Br^(-) (aq) to Br_(2) (g) + 2 e^(-)`