1.

A current of 2A was passed for 1.5 hours through a solution of CuSO_(4) when 1.6 g of copper was deposited. Calculate percentage current efficiency.

Answer»

Solution :Amountof CURRENT required to deposit 1 mole `Cu(63.5 g) = 2 xx 96500` C
Current required to deposit 1.6 g of COPPER = `(2 xx 96500 xx 1.6)/63.5 = 4862.99`
Current actually PASSED through
`=2 x 1.5 xx 60 = 10800`
Current efficiency `=(4862.99)/(10800) xx 100 = 45.03 %`


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