`100mL CuSO_(4)(aq)` was electrolyzed using inert electrodes by passing `0.965A `till the `pH` of the resulting solution was 1. The solution after electrolysis was neutralized, treated with excess `KI` and titrated with `0.04M Na_(2)S_(2)O_(3)`. Volume of `Na_(2)S_(2)O_(3)` required was `35mL`. Assuming no volume change during electrolysis, calculate:
(a) duration of electrolysis if current efficiency is `80%`
(b) initial concentration `(M)` of `CuSO_(4)`.

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`