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

1.13 × 10⁶ g

Step-by-step 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 * (1s)/(620c) * (96,468c)/(1mole^(-) ) * (3mole^(-))/(1molAl) * (26.98gAl)/(1molAl) =1.13 * 10^(6) g Al`