Question

How much electricity in terms of Faraday is required to produce 40 g of Al from molten Al$_2$O$_3$? (Given: Atomic mass of Al = 27 u)

Hint:

Use: \(\text{Charge} = \text{mol} \times \text{n} \, \text{(electrons)}\)
Always check the charge on the ion to find number of electrons transferred.

Answer 1

Step 1: Reaction at cathode:
\[\text{Al}^{3+} + 3e^- \rightarrow \text{Al (s)}\]
Step 2: Molar mass of Al = 27 g/mol
Moles of Al in 40 g = \(\frac{40}{27} \approx 1.481 \, \text{mol}\)
Step 3: Since 3 Faradays are needed to produce 1 mol of Al:
Total charge = \(1.481 \times 3 = 4.44 \, \text{F}\)
Conclusion: 4.44 Faraday of electricity is required.