`50mL` of `0.1M CuSO_(4)` solution is electrolysed with a current of `0.965A` for a period of 200sec. The reactions at electrodes are:
Cathode: `Cu^(2+) +2e^(-) rarr Cu(s)`
Anode: `2H_(2)O rarr O_(2) +4H^(+) +4e`.
Assuming no change in volume during electrolysis, calculate the molar concentration of `Cu^(2+),H^(+)` and `SO_(4)^(2-)` at the end of electrolysis.

Answer

`Cu^(2+) = 0.08M, H^(+) = 0.04M, SO_(4)^(2-) = 0.1M`

Explanation

`n_(e) = (It)/(96500) = (0.965 xx 200)/(96500) = 2xx 10^(-3)`
`n_(Cu^(2+)) = 10^(-3)`
`n_(Cu^(2+)) Initial = 5 xx 10^(-3)`
`n_(Cu^(2+))` left in solution `=4xx 10^(-3)`
`m_(Cu^(2+)) = (4 xx 10^(-3))/(50) xx 100 = 0.08M`
`2H_(2)O rarr _(2) +4H^(+) +4e^(-)`
`n_(E) = 2 xx 10^(-3) = n_(H^(+))`
`m_(H^(+)) = (2 xx 10^(-3))/(50) xx 10^(-3) = 0.04M`
Molarity of `SO_(4)^(2-)` will not be changed
`m_(SO_(4)^(2-)) = 0.1`