How many coulombs of charge are required to produce 101 g of Al from Al^{3+}(aq) by electrolysis? (Faraday's constant, F = 96,500 C/mol)

• A. 289,500 C
• B. 96,500 C
• C. 540,000 C
• D. 270,000 C

Answer: (C)

Verified step by step guidance

1. Determine the number of moles of aluminum (Al) produced using its molar mass. Use the formula: \(\text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}}\). The molar mass of Al is approximately 27 g/mol.
2. Write the half-reaction for the reduction of aluminum ions: \(\mathrm{Al^{3+} + 3e^- \rightarrow Al}\). This shows that 3 moles of electrons are required to produce 1 mole of aluminum.
3. Calculate the total moles of electrons needed by multiplying the moles of aluminum by 3 (from the half-reaction).
4. Use Faraday's constant (\(F = 96,500\) C/mol) to find the total charge in coulombs. Multiply the moles of electrons by Faraday's constant: \(\text{charge} = \text{moles of electrons} \times F\).
5. The result from step 4 gives the total coulombs of charge required to produce 101 g of aluminum by electrolysis.