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

Given Time t= 10 minutes `=10 xx 60=600` sec, I= 1.5A, Q= It `=1.5 xx 600=900C`
`CuSO_(4) hArr Cu^(2+) + SO_(4)^(2-)`
In reaction at Cathode is `underset(1"mole")(Cu^(2+)) + underset(2F)(2e^(-)) rarr Cu`
`2 xx 96500 C` deposit Cu= 63.5g
900C deposite `Cu= (63.5)/(2 xx 96500) xx 900= 0.2961g`