Question

$Al _{2} O _{3}$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^{4}$ amperes of current is passed through molten $Al _{2} O _{3}$ for 6 hours, what mass of aluminium is produced ? (Assume $100 \%$ current efficiency, at. mass of $Al =27\, g\, mol ^{-1}$ ) -

• A. $9.0 \times 10^3 g $
• B. $8.1 \times 10^4 g $
• C. $2.4 \times 10^5 g $
• D. $1.3 \times 10^4 g $

Answer 1

The Correct Option is B

To solve this problem, we need to calculate the mass of aluminium produced by electrolysis of molten alumina (Al₂O₃) using Faraday's laws of electrolysis.

The reaction taking place during the electrolysis of Al₂O₃ can be represented as:

2Al_2O_3 \rightarrow 4Al + 3O_2

This indicates that 2 moles of Al₂O₃ produce 4 moles of Al. Therefore, 1 mole of Al is produced from 1/2 mole of Al₂O₃.

Given:

• Current (I) = 4.0 \times 10^4 amperes
• Time (t) = 6 hours = 6 × 3600 seconds = 21600 seconds
• Molar mass of Al = 27 g/mol
• Charge of 1 mole of electrons = Faraday constant, F \approx 96500 C/mol
• Assuming 100% current efficiency

First, calculate the total charge (Q) passed through the Al₂O₃ solution:

Q = I \times t = 4.0 \times 10^4 \, A \times 21600 \, s = 8.64 \times 10^8 \, C

The electrochemical equivalent for aluminium can be worked out using the fact that 3 moles of electrons are needed to produce 2 moles of Al, which means 1 mole of Al requires 1.5 moles of electrons. Therefore, the charge required to produce 1 mole of Al is:

1.5 \times 96500 \, C/mol = 144750 \, C/mol

Now, calculate the number of moles of Al produced:

\text{moles of Al} = \frac{Q}{144750} = \frac{8.64 \times 10^8}{144750} \approx 5971.19

Finally, calculate the mass of aluminium produced:

\text{mass of Al} = \text{moles of Al} \times \text{molar mass of Al} = 5971.19 \times 27 \approx 8.1 \times 10^4 \, \text{g}

Thus, the mass of aluminium produced is 8.1 \times 10^4 \, \text{g}.

Hence, the correct answer is:

1. 8.1 \times 10^4 \, \text{g}