If 2 faraday current is pass through `CuSO_(4)` solution then find out how much copper can be deposited on cathode ? [Cu=63.5 gm `"mole"^(-1)`]

Options

A. 0 gm
B. 2 gm
C. 63.5 gm
D. 127 gm

Answer

C

Explanation

`CuSO_(4)toCu_((aq))^(2+)+SO_(4(aq))^(2-)` . . . Ionization
`Cu_((aq))^(2+)+2e^(-) to Cu_((S))` . . . (Cathodic reduction)
2 mole `e^(-) to ` 1 mol copper
So 2F current produce 63.5 gm copper