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Calculate the time to deposit 1.27 g of ...

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

Text Solution

Verified by Experts

The correct Answer is:
1930 s
Reaction at cathode is :
`underset(63.5" g")(Cu^(2+))(aq)+2e^(-) to underset(2xx96500" C")(Cu(s))`
63.5 g of copper is deposited by passing charge `=2xx96500" C"`
1.27 g of copper of deposited by passing charge `=((2xx96500C)xx(1.27 g))/((63.5 g))=3860" C"=3860" As"`
Time to deposit 1.27 g of copper `=((3860" As"))/((2A))=1930" S "`
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