Let's solve the problem step by step. ### **Step 1: Calculate the total charge (Q) passed** The total charge (Q) passed in coulombs is given by: \[ Q = I \times t \] where \(I = 2500 \, \text{A}\) (current) \(t = 15.0 \, \text{s}\) (time) \[ Q = 2500 \times 15.0 = 37,500 \, \text{C} \] **Explanation:** We multiply current by time to get the total electric charge that passed through the cell. --- ### **Step 2: Calculate the moles of electrons transferred** The number of moles of electrons (\(n\)) is given by: \[ n = \frac{Q}{F} \] where \(F = 96,485 \, \text{C/mol}\) (Faraday’s constant) \[ n = \frac{37,500}{96,485} = 0.3885 \, \text{mol e}^- \] **Explanation:** We divide the total charge by Faraday’s constant to find out how many moles of electrons were involved in the process. --- ### **Step 3: Calculate the mass of aluminum produced** The reduction half-reaction for aluminum is: \[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \] So, 3 moles of electrons produce 1 mole of Al. Moles of Al produced: \[ \text{mol Al} = \frac{0.3885}{3} = 0.1295 \, \text{mol} \] Now, find the mass: \[ \text{mass} = \text{mol Al} \times \text{molar mass of Al} \] Molar mass of Al = 26.98 g/mol \[ \text{mass} = 0.1295 \times 26.98 = 3.49 \, \text{g} \] **Explanation:** Each Al atom needs 3 electrons, so we divide the moles of electrons by 3 to get the moles of Al. Then, we multiply by the molar mass to get the mass of Al. --- ## **Final Answer** **The mass of pure aluminum produced is \(\boxed{3.49 \text{ g}}\) (to 3 significant figures).**

The concept used in this calculation is based on electrolysis principles, specifically the relationship between electric charge, moles of electrons transferred, and the amount of substance reduced or oxidized. First, the total charge passed through the cell is determined by multiplying current and time. Then, using Faraday’s constant, the number of moles of electrons involved in the process is calculated. Since the reduction of aluminum ions (Al³⁺) to aluminum metal requires three electrons per atom, the moles of aluminum produced are found by dividing the moles of electrons by three. Finally, multiplying the moles of aluminum by its molar mass yields the mass of aluminum deposited. This approach combines fundamental electrochemical concepts with stoichiometry to quantify the amount of substance formed during electrolysis.