During electrolysis of aqueous `CuBr_(2)` using `Pt` electrode,

Options

A. `Br_(2) (g)` is evoled at anode
B. `Cu(s)` is deposited at cathod
C. `Br_(2) (g)` is evolved at anode and `H_(2)(g)` at cathod
D. `H_2((g))` is evolved at anode .

Answer

A, B

Explanation

At cahtode : `Cu^(2+) + 2e^(-) rightarrow Cu` `E_(Cu^(2+)//Cu)^(@)=+0.34V`
`2H_(2)O + 2e^(-) rightarrow H_(2) 2OH^(-)` `d E_(H_(2)O //H_(2))^(@) =+ 0.83V`
At anode : `2Br^(-) rightarrow Br_(2) + 2e^(-)` `E_(Br)^(-) //Br_(2))^(@) =- 1.09V`
`2H_(2) O rightarrow O_(2) + 4H^(+) + 4e^(-)` `EH_(2)O //H_(2)^(@) = -1.23V`