A current of 1.70A is passed through 300.0 mL of 0.160 M solution at `ZnSO_(4)` for 230s with a current efficiency of `90%`. Find out the molarity of `Zn^(2+)` after the deposition of Zn. Assume the volume of the solution to remain constant during the electrolysis.

Explanation

Among of charge (Q) passed through the cell
= Current (I)`xx` Current efficiency `xx` time (in s)
`=1.70xx(90)/(100)xx230=351.9C`
`351.9C=(1)/(96500)xx"351.9 mol of electrons"`
`" "[because" 1 mol of electrons = 96500 C"]`
`=3.646xx10^(-3)" mol of electrons"`
Now, the reaction is : `Zn^(2+)(aq)+2e rarr Zn(s)`
`"1 mol of "Zn^(2+)` ions is discharged by 2 mol of electrons
`therefore" "` Number of mol of `Zn^(2+)` ions discharged by
`3.646xx10^(-3)" mol electrons "=(3.646xx10^(-3))/(2)=1.823xx10^(-3)`
`therefore" Concentration of discharged "Zn^(2+)" ions "`
`=(1.823xx10^(-3))/(300)xx1000=6.076xx10^(-3)M`
`therefore" Concentration of remaining "Zn^(2+)" ions"`
`=(0.16-6.076xx10^(-3))M=0.1539M`