Question

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (A12O 3) dissolved in molten cryolite (Na3AIF6) ,resulting in the reduction of the AI�3 to pure aluminum. Suppose a current of 6800.A is passed through a Hall-Heroult cell for 54.0 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:

34.2 g

Explanation:

In the Hall-Heroult process, Al³⁺ (from Al₂O₃) is reduced to Al. The reduction half-reaction is:

Al³⁺ + 3 e⁻ ⇒ Al

We can establish the following relations:

• 1 A = 1 c/s
• 1 mole of e⁻ has a charge of 96468 c (Faraday's constant)
• 1 mol of Al is produced when 3 moles of e⁻ circulate
• The molar mass of Al is 26.98 g/mol.

Suppose a current of 6800 A is passed through a Hall-Heroult cell for 54.0 seconds. The mass of Al produced is:

$54.0s.\frac{6800c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molAl}{3mole^{.} } .\frac{26.98gAl}{1molAl} =34.2gAl$