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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? (Molar mass of Cu=63.5 g mol^(-1))

Answer»

Solution :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`


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