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)`]

Explanation

`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.