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

Options

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

Answer

A

Explanation

`Cu^(3+)+underset(2"mole 2F")(2e^(-))rarrunderset("1mole" 63.4gm)(Cu)`
To deposit 63.5gm of copper, electricity needed
`=2xx96500C`
To d eposit 2 gm of copper, electricity needed
`=2xx(96500)/(63.5)xx2=6078.74C`
`Q=6078.74C`
`1=0.5` ampere
Q=It
t(in seconds)`=(Q)/(1)=(6078.74)/(0.5)=12157.48s`