A current of 1.70 ampere is passed through 300mL of 0.160M solution of `ZnSO_(4)` for 230 sec with a current efficiency of 90%. Find the molarity of `Zn^(2+)` after the deposition of Zn. Assume the volume of the solution deposition remains constant during elecyrolysis

A current of 1.70 ampere is passed through 300mL of 0.160M solution of

1.	A current of 1.70 ampere is passed through 300mL of 0.160M solution of `ZnSO_(4)` for 230 sec with a current efficiency of 90%. Find the molarity of `Zn^(2+)` after the deposition of Zn. Assume the volume of the solution deposition remains constant during elecyrolysis
Answer» We know, `i=(1.70xx90)/(100)` ampere `therefore` Eq. of `Zn^(2+)` lost `=(i.t.)/(96500)` `=(1.70xx90xx230)/(100xx96500)=3.646xx10^(-3)` `therefore` Meq.of `Zn^(2+)"lost"=3.646` Initial Meq. `Zn^(2+)=300xx0.160xx2` `[because Mxx2=N" for " Zn^(2+), Meq. = NxxV_(("in.mL."))]` `=48xx2=96` `therefore` Meq. of `Zn^(2+)` left in solution `=96-3.646=92x354` `therefore [ZnSO_(4)]=(92.354)/(2xx300)=0.154M`

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A current of 1.70 ampere is passed through 300mL of 0.160M solution of `ZnSO_(4)` for 230 sec with a current efficiency of 90%. Find the molarity of `Zn^(2+)` after the deposition of Zn. Assume the volume of the solution deposition remains constant during elecyrolysis

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