Final answer:
The mass of aluminum deposited when a current of 3.0 amperes is passed through an aluminum electrolyte for 2 hours is approximately 2.0155 grams. This is calculated using Faraday's laws of electrolysis and the molar mass of aluminum.
Step-by-step explanation:
The problem involves the application of Faraday's laws of electrolysis to calculate the mass of aluminum deposited during the electrolysis process. The total amount of charge Q in coulombs passed through the aluminum electrolyte can be found by the formula Q = I × t, where I is the current in amperes and t is the time in seconds.
Current (I) = 3.0 amperes
Time (t) = 2 hours = 7200 seconds (since 1 hour = 3600 seconds)
Q = I × t = 3.0 A × 7200 s = 21600 C
The amount of substance in moles, n, that will be deposited (or reacted) can then be calculated using the charge and Faraday's constant (F), as n = Q / (n × F), where n is the number of moles of electrons exchanged per mole of substance (for aluminum, Al3+, n = 3) and F is Faraday's constant (96500 C/mol).
× = moles of electrons exchanged per mole of aluminum = 3
Faraday's constant (F) = 96500 C/mol
Number of moles of aluminum (nAl) = Q / (n × F)
nAl = 21600 C / (3 × 96500 C/mol) = 21600 C / 289500 C/mol = 0.07465 mol
Finally, we calculate the mass of aluminum deposited using the molar mass of aluminum (27 g/mol) and the number of moles calculated above.
Molar mass of aluminum (MAl) = 27 g/mol
Mass of aluminum deposited (mAl) = nAl × MAl = 0.07465 mol × 27 g/mol = 2.0155 g
Therefore, the mass of aluminum deposited is approximately 2.0155 grams.