A solution of `CuSO_(4)` is electroysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ?
(Molar mass of `Cu=63.5 g//mol)`

Explanation

Current=1.5 amperes, time=10min=`10xx60s=600s`
`therefore`Quantity of electricity passed=Current (amp)`xx`time (s)`=1.5xx600`coulombs=900C
The reaction occuring at the cathode is: `Cu^(2+)+2e^(-)toCu`
Thus, 2F, i.e., `2xx96500C` deposit Cu=10 mol=63.5 g
900C will deposit `Cu=(63.5)/(2xx96500)xx900=0.296g`