`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)`.

Explanation

`a. Ph=1,[H^(o+)]=0.1M` after `t` seconds
`1000mL` contains `0.1mol `of `H^(o+)`.
`100 mL` contains `=(0.1xx100)/(1000)=0.01 mol of H^(o+)`
`1F =1 mol `of `H^(o+)=1 Eq` of `H^(o+)`
`0.01F=0.01mol of H^(o+)=0.01Eq of H^(o+)`
`0.01xx96500C=(0.965xxtxx(60)/(100))`
`(` since `60% ` efficiency `)`
`:. t=1666.67s=0.463h`
`b.` `Cu^(2+)+KI rarr Cu_(2)I_(2)" "(Cu^(2+)rarrCu^(1+)) `
`mEq` of `Na_(2)S_(2)O_(3)=mEq` of `Cu^(2+)`
`implies0.041xx1xx35`
`(2S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+2e^(-))" "(n=(2)/(2)=1)`
`:. mEq` of `Cu^(2+)` deposited as `Cu`
`(Cu^(2+)+2e^(-)rarr Cu)`
`2Fimplies1 mol `of `Cu^(2+)` deposited as `Cu`
`-0.01Fimplies(1)/(2)xx0.01=0.005mol`
So `Cu^(2+)` ions present initially `=0.005+1.4xx10^(-3)`
`=6.4xx10^(-3)mols`
Molarity `=(mols)/(Volume i n L)=(4xx10^(-3))/(100//1000)=0.064M`