`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 electrollysis 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

`1250 S, 0.064M`

Explanation

`2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)`
`n_(H^(+)) =10^(-2) n_(E) = 10^(-2)`
`n_(E)` actual `= (0.10 xx 100)/(80) rArr 0.0125`
`I = (q)/(t)`
`0.965 = (0.0125 xx 96500)/(t)`
`t =(0.0125 xx 96500)/(0.965) rArr 1250 S`
`2CuSO_(4) +4Ki rarr 2CuI +I_(2) +2K_(2)SO_(4)`
`I_(2) +2Na_(2)S_(2)O_(3) rarr 2NaI +Na_(2)S_(4)O_(6)`
Mili moles of `Na_(2)S_(2)O_(3) = 0.04 xx 35 rArr 1.4`
Mili moles of `I_(2) = (1.4)/(2)`
Mili moles of `CuSO_(4) = 2xx (1.4)/(2) rarr 1.4`
`Cu^(2+) +2e^(-) rarr Cu`
`n_(CU^(2+)) = (1)/(2) xx 10^(-2)|n_(E) = 10^(-2)`
Total moles of `Cu^(2+) = 5 xx 10^(-3) +14.2 xx 10^(-3)`
= total moles of `Cu^(2+) = 5 xx 10^(-3) +1.2 xx 10^(-3)`
`= 6.42 xx 10^(-3)`
`=(6.42 xx 10^(-3))/(100) = 0.064M`