What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 25 A, assuming 100% efficiency?

• A. 0.25 grams
• B. 2.52 grams
• C. 25.0 grams
• D. 9.36 grams

Answer: (D)

Verified step by step guidance

1. Identify the half-reaction for the reduction of aluminum ions to aluminum metal: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \). This indicates that 3 moles of electrons are required to produce 1 mole of aluminum.
2. Calculate the total charge passed in one hour using the formula \( Q = I \times t \), where \( I \) is the current (25 A) and \( t \) is the time in seconds (3600 seconds for one hour).
3. Use Faraday's constant (approximately 96485 C/mol) to determine the number of moles of electrons transferred. This is done by dividing the total charge by Faraday's constant.
4. Determine the moles of aluminum produced by using the stoichiometry of the half-reaction. Since 3 moles of electrons produce 1 mole of aluminum, divide the moles of electrons by 3.
5. Convert the moles of aluminum to mass using the molar mass of aluminum (approximately 26.98 g/mol). Multiply the moles of aluminum by its molar mass to find the mass produced per hour.