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

Explanation

Quantity of electricity passed `= 1.70 xx 230 = 391C`
As current efficiency = 90%
Effective charge ` = (90)/( 100) xx 391 C = 351.9 C`
`Zn^(2+) + 2e^(-) rarr Zn`
`2 xx 96500C` deposit `Zn^(2+) =1` mole
`:. ` 351.9 C will deposit ` = ( 1)/( 2xx 96500) xx 351.9 ` mol
`= 0.0018` mol
`Zn^(2+)` ions present originally in 300 mL of 0.160M
`ZnSO_(4) = ( 0.160)/( 1000) xx 300 = 0.048`mol
`:. ` Amount of `Zn^(2+)` in 300 mL of solution after deposition of Zn = 0.048-0.0018 = 0.0462mol
`:.` Molarity of `Zn^(2+)` after the deposition of Zn
`= ( 0.0462)/( 300) xx 1000= 0.154M`