A solution of `CuSO_(4)` is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

Explanation

`1=10`min `=10xx60=600s`
Charge `=` Current`xx` Time
`=1.5Axx600s=600C`
According to the reaction `:(CuSO_(4) rarr Cu^(2+)+SO_(4)^(2-))`
`Cu^(2+)(aq)+2e^(-) rarr Cu(s)`
`2e^(-)=2F` or `2xx 96487 C = 1 mol of Cu=63 g `of `Cu`
For `900C` the mass of `Cu` deposited
`=(63g mol ^(-1)xx 900 C)/(2xx96487 C mol ^(-1))=0.2938g`