The amount of electricity required to deposit one mole of Al from a solution of `AlCl_(3)` will be

Options

A. 3.0 Faraday
B. 1.0 Faraday
C. 1.33 Faraday
D. 0.33 Faraday

Answer

A

Explanation

`AlCl_(3) to Al^(3+)+ 3Cl^(-)`
`Al^(3+) + 3e^(-) to Al`
1 mol `3"mol"^(-)` 1 mol
1 mol 3F 1 mol
Thus, 3.0 Faraday electricity is required to deposit one mole of Al from a solution of `AlCl_(3)`