Question

In the Hall-Heroult process, a large electric current is pased through a solution of alumunium oxide (Al2O3) dissolved in molten cryolite (Na2AlF6), resulting in the reduction of the Al2O3 to pure alumunium. Suppose a current of 6800A is passed through a Hall-Heroult cell for 54.0 seconds. Calculate the mass of pure alumunium produced. Be sure your answer has a unit symbol and the correct number of significant digits.

Answer 1

The value 34.2 g is the mass of pure alumunium produced. a current of 6800A is passed through a Hall-Heroult cell for 54.0 seconds.

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 34.2 G.

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