In a solution of `CuSO_(4)` how much time will be required to preciitate 2 g copper by 0.5 ampere current?

Options

A. 12157.48 sec
B. 102 sec
C. 510 sec
D. 642 sec

Answer

A

Explanation

`Cu^(2+)+underset(2xx96500 C)(2e^(-)) to underset(1 "mol" =63.5 g)(Cu)`
To deposite 63.5 g of Cu electricity required `=2xx96500 C`
To deposite 2 g of Cu electricity required `=(2xx96500)/(63.5)xx2`
`Q=Ixxt` or `t=Q//I`
`t=(2xx96500xx2)/(63.5xx0.5)=12157.48`s