Question

Find the number of faradays of electricity required to produce 45 g of Al from molten $Al_2O_3$.

(At. mass of Al = 27)

• A. 1 F
• B. 3 F
• C. 5 F
• D. 7 F

Answer 1

Answer: (C)

5 F

Answer 2

To determine the number of Faradays required to produce 45 g of Al from molten $Al_2O_3$, we can follow these steps:

1. Calculate moles of Al:

   The number of moles of Al can be found using the formula:

   $$\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}}$$

   Given that the mass of Al is 45 g and the molar mass of Al is 27 g/mol:

   $$\text{Moles of Al} = \frac{45~\text{g}}{27~\text{g/mol}} = \frac{5}{3} \text{ moles}$$

2. Determine electrons required per mole of Al:

   The reduction half-reaction for aluminum is:

   $$Al^{3+} + 3e^- \rightarrow Al$$

   This indicates that 3 moles of electrons are required to produce 1 mole of Al.

3. Calculate total moles of electrons:

   To find the total moles of electrons needed to produce $\frac{5}{3}$ moles of Al:

   $$\text{Total moles of electrons} = \frac{5}{3} \text{ moles Al} \times 3 \frac{\text{moles } e^-}{\text{mole Al}} = 5 \text{ moles electrons}$$

4. Apply Faraday's Law:

   1 Faraday is equivalent to 1 mole of electrons. Therefore, 5 moles of electrons correspond to 5 Faradays.

Thus, the number of Faradays of electricity required to produce 45 g of Al from molten $Al_2O_3$ is 5 F.