Question

How much charge in coulombs is required for the reduction of one mole of Al\(^{3+}\) to Al?

• A. \( 1.930 \times 10^4 \) C
• B. \( 2.895 \times 10^5 \) C
• C. \( 2.895 \times 10^4 \) C
• D. \( 1.930 \times 10^5 \) C

Hint:

To calculate the charge for the reduction or oxidation of a species, multiply the number of electrons involved by the Faraday constant.

Answer 1

The Correct Option is B

Step 1: Understanding the concept.
For the reduction of Al\(^{3+}\) to Al, 3 moles of electrons are required for each mole of Al\(^{3+}\). The charge of one mole of electrons is 96,500 C (Faraday’s constant).

Step 2: Applying the values.
For 1 mole of Al\(^{3+}\), the total charge required is: \[ \text{Charge} = 3 \times 96,500 \, \text{C} = 2.895 \times 10^5 \, \text{C} \]
Step 3: Conclusion.
The correct answer is (B) \( 2.895 \times 10^5 \) C, as this is the required charge for the reduction of one mole of Al\(^{3+}\).