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

Explanation

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`