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 100.A is passed through a Hall-Heroult cell for 41.0 seconds. Calculate the mass of pure aluminum produced. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

Answer 1

Answer:

0.382g

Explanation:

Step 1: Write the reduction half-reaction

Al³⁺(aq) + 3 e⁻ ⇒ Al(s)

Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s

We will use the following relationships.

• 1 A = 1 C/s

• 1 mole of electrons has a charge of 96486 C (Faraday's constant)

• 1 mole of Al is produced when 3 moles of electrons pass through the cell.

• The molar mass of Al is 26.98 g/mol.

The mass of Al produced is:

$41.0s \times \frac{100C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molAl}{3mole^{-} } \times \frac{26.98gAl}{1molAl} = 0.382gAl$