Question

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten cryolite Na3AlF6, resulting in the reduction of the Al2O3 to pure aluminum. Suppose a current of 620.A is passed through a Hall-Heroult cell for 90.0 seconds. Calculate the mass of pure aluminum produced. Be sure your answer has a unit symbol and the correct number of significant digits.

Answer 1

Answer:

1.13 × 10⁶ g

Explanation:

Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.

Al³⁺ + 3 e⁻ → Al

We can establish the following relations:

• 1 Ampere = 1 Coulomb / second
• The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
• 1 mole of Al is produced when 3 moles of electrons circulate
• The molar mass of Al is 26.98 g/mol.

The mass of aluminum produced under these conditions is:

$90.0 s \times \frac{1s}{620c} \times \frac{96,468c}{1mole^{-} } \times \frac{3mole^{-}}{1molAl} \times \frac{26.98gAl}{1molAl} =1.13 \times 10^{6} g Al$