How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten Al2O3?

Question

How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten Al2O3? (A) 1 F (B) 3 F (C) 2 F (D) 4 F

Tardigrade Admin · Accepted Answer

Al2O3 → 2Al3+ + 3O2- or, Al3+ underset3 mole+ 3e- → underset1 mole(27 g)Al ∵ 27 g of aluminium needs = 3 moles of electrons = 3 × 96500 coulombs ∴ 40.0 g of aluminium will need = (3 × 96500 × 40.0/27) coulombs = 4.28888 × 105 coulombs (4.44 F)

Q. How much electricity in terms of Faraday is required to produce $40.0 g$ of $A l$ from molten $A l_{2} O_{3}$ ?

$1 F$

$3 F$

$2 F$

$4 F$

Q. How much electricity in terms of Faraday is required to produce 40.0g of Al from molten Al2​O3​?

1F

3F

2F

4F

Al2​O3​→2Al3++3O2− or, Al3+3mole+3e−​→1mole(27g)Al​ ∵27g of aluminium needs =3 moles of electrons =3×96500 coulombs ∴40.0g of aluminium will need =273×96500×40.0​ coulombs =4.28888×105 coulombs (4.44F)

Q. How much electricity in terms of Faraday is required to produce $40.0 g$ of $A l$ from molten $A l_{2} O_{3}$ ?

$1 F$

$3 F$

$2 F$

$4 F$