Question

What is the total charge required for the reduction of 2/3 moles of Al\(_2\)O\(_3\) to form Al?

• A. \( 2 \times 96500 \, \text{C} \)
• B. \( 4 \times 96500 \, \text{C} \)
• C. \( 3 \times 96500 \, \text{C} \)
• D. \( 6 \times 96500 \, \text{C} \)

Hint:

To calculate the total charge required in a redox reaction, use the number of electrons involved in the reaction and the Faraday constant (\( 96500 \, \text{C/mol} \)).

Answer 1

The Correct Option is B

Step 1: Understanding the Reaction.
In the given reaction, \( \text{Al}_2\text{O}_3 \) is reduced to aluminum (Al). The reduction of aluminum requires a charge based on the number of moles of aluminum being formed.
Step 2: Calculating the Total Charge.
The number of moles of electrons required to reduce aluminum is 3 moles of electrons per mole of Al (as per the half-reaction: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \)). For 2/3 moles of \( \text{Al}_2\text{O}_3 \), the total charge required is: \[ \text{Charge} = \frac{2}{3} \times 3 \times 96500 = 4 \times 96500 \, \text{C} \]
Step 3: Conclusion.
The correct answer is (B) \( 4 \times 96500 \, \text{C} \).