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100mL CuSO(4)(aq) was electrolyzed using...

`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)`.

Text Solution

Verified by Experts

The correct Answer is:
1250 s, 0.064 M
`CuSO_(4)100ml` M molar
`("At cathode")/(Cu^(2+)+2e^(-))toCu]xx2`
`underline(("at anode")/(2H_(2)O)toO_(2)+4H^(+)+4e^(-))`
overall, ,brgt `2Cu^(2+)+2H_(2)Oto2Cu(s)+O_(2)+4H^(+)`
`pH=1implies[H^(+)]=0.1impliesH^(+)=0.1` mole
`Q=0.01` mole of `e^(-)=0.01xx96500` coulomb
`=965` coulomb.
`therefore` duration of electrolysis with `100%` efficiency `=(Q)/(i)=(965)/(0.954)=1000`sec
`therefore` duration of electrolysis with `80%` efficiency `=(1000)/(0.8)sec=1250sec`
Cu deposited `=(0.01)/(2)"mole"=5m "mole"`
`underset(1.4" m mole")(2Cu^(2+)+4I^(-))toCu_(2)I_(2)+underset(0.7" m mole")(I_(2))`
`I_(2)+underset(0.04xx35)(2Na_(2)S_(2))toNa_(2)S_(4)O_(6)+2Nal`
`0.7" m mole"=1.4" m mole"`
`therefore` initial conc of `CuSO_(4)=(5+1.4)/(100)=(6.4)/(100)=0.064M`.
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