1.

Calculate the time to deposit 1.27 g of copper at cathode when a current of 2 A was passed through the solution of CuSO_(4).[Molar mass of Cu = 63.5 g mol^(-1), 1 F = 96500 C mol^(-1)]

Answer»

Solution :`Cu^(2+) | 2E^(-) to Cu(s)`
No. of COULOMBS required to deposit 63.5 g Cu `=2 xx 96500`
No. of coulombs required to deposit 1.27 g Cu `=(2 xx 96500)/(63.5) xx 1.27`
`=(2 xx 96500 xx 1.27)/(63.5)xx 1.27`
`(2 xx 96500 xx 1.27)/(63.5) = 2` amperes `xx` TIME in seconds
or Time (in seconds) `=(2 xx 96500 xx 1.27)/(63.5 xx 2) = 1930` seconds.


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