Copper sulphate solution `(250 ML)` was electrolyzed using a platinum anode and a copper cathode. A constant current of `2mA` was passed for `16 mi n`. It was found that after electrolysis the absorbance of the solution was reducted to `50%` of its original value . Calculate the concentration of copper sulphate in the solution to begin with.

Answer

0.792xx10^(-4)M`

Explanation

From the first law of Faraday `:`
`W=ZxxIxxt`
where `Z=` electrochemical equivalent
`((63.5//2)/(96500)fo r Cu)`
`I=` Electric current `=2mA=2xx10^(-3)A`
`t=` Time of flow of current `=16mi n =16xx60s`
`W=ZxxIxxt`
`=(63.5)/(2xx96500)xx2xx10^(-3)xx16xx60g`
Weight of `Cu` at `50%` electrolysis of `CuSO_(4)`
`=(63.5xx16xx60xx10^(-3))/(96500)g`
Weight of `Cu` at `100%` electrolysis of `CuSO_(4)`
`=(2xx63.5xx16xx60xx10^(-3))/(96500)g`
`=0.198xx63.5xx10^(-4)g`
`CuSO_(4)-=Cu`
`63.5+32+64-=63.5`
or `159g CuSO_(4)-=63.5g Cu`
Weight of `CuSO_(4)` in `250mL` of solution
`=(159.5)/(63.5)xx63.5xx0.188xx10^(-4)g`
Moles of `CuSO_(4)` in `250mL=0.198xx10^(-4)mol`
Conc of `CuSO_(4)=0.198xx10^(-4)xx(1000)/(250)M`
`=0.792xx1^(-4)M`