Final answer:
Using a current of 3.7 A for 6 hours, the total charge passed through an electrolytic cell is 79872 C. This results in 0.8278 moles of electrons, which yields 0.2759 moles of aluminum. Thus, 7.4442 grams of aluminum would be produced.
Step-by-step explanation:
To calculate the mass of solid aluminum produced by the reduction of Al3+ ions during electrolysis at the cathode, we need to use Faraday's laws of electrolysis and the electrochemical equivalent of aluminum. First, we determine the total charge that has passed through the cell.
The total charge (Q) is the product of the current (I) and the time (t) in seconds. The given current is 3.7 amps and the time is 6 hours. Since 1 hour is 3600 seconds, we need to convert 6 hours into seconds (6 hours * 3600 seconds/hour).
Total charge Q = I * t = 3.7 A * (6 * 3600 s) = 79872 C
The next step is to use Faraday's constant to find the number of moles of electrons that have been transferred. Faraday's constant is approximately 96500 C/mol electrons. Therefore, the moles of electrons transferred can be calculated by dividing the total charge by Faraday's constant.
Moles of electrons = Q / Faraday's constant = 79872 C / 96500 C/mol = 0.8278 mol
The reduction half-reaction for aluminum at the cathode is:
Al3+ + 3e- → Al(s)
Since it takes 3 moles of electrons to reduce one mole of aluminum ions to aluminum metal, we can calculate the moles of aluminum produced by dividing the moles of electrons by 3.
Moles of aluminum = Moles of electrons / 3 = 0.8278 mol / 3 = 0.2759 mol
With the molar mass of aluminum being approximately 26.98 g/mol, the mass of Al produced can be determined as follows:
Mass of Al = Moles of Al * Molar mass of Al = 0.2759 mol * 26.98 g/mol = 7.4442 g
Therefore, after 6 hours with a current of 3.7 A, 7.4442 grams of solid aluminum would be produced at the cathode.