Molten Al2O3 is reduced by electrolysis at low potentials and high currents. If 1.5 x 10^4 amperes of current is passed through molten Al2O3 for 8.0 hours, what mass of Al is produced? Assume 100% current efficiency?
Mdep = ([1.5 x 10^ x 28800s ]/ 96500) x (26.98/6e)
= 20130g
gives me the wrong answer tho?
That's 2 Al for 6e or 1 Al for 3e so that factor should be 2*26.98/6e or 26.98/3e. That doubles the answer. Some books use 96,485 instead of 96,500. Thanks for showing your work. That always makes it easier to find the error.
To calculate the mass of Al produced during the electrolysis of molten Al2O3, we can use Faraday's laws of electrolysis and the molar masses of Al2O3 and Al.
First, let's determine the number of moles of electrons transferred during the electrolysis. According to Faraday's laws, 1 Faraday (F) of electricity is equal to the charge carried by 1 mole of electrons, which is approximately 96,500 Coulombs (C).
Given that 1.5 x 10^4 amperes of current are passed through the molten Al2O3 for 8.0 hours (28800 seconds), we can calculate the total charge passed using the formula:
Total charge (Q) = Current (I) x Time (t)
Q = (1.5 x 10^4 A) x (28800 s)
Q = 4.32 x 10^8 C
Now, we can calculate the number of moles of electrons transferred by dividing the total charge by the Faraday constant:
moles of electrons (n) = Total charge / Faraday constant
n = (4.32 x 10^8 C) / (96,500 C/mol)
n ≈ 4480 mol
Since each mole of Al2O3 produces 4 moles of Al during electrolysis, we need to divide the number of moles of electrons by 4:
moles of Al = moles of electrons / 4
moles of Al = 4480 mol / 4
moles of Al = 1120 mol
Finally, we can calculate the mass of Al produced by multiplying the moles of Al by its molar mass:
mass of Al = moles of Al x molar mass of Al
mass of Al = (1120 mol) x (26.98 g/mol)
mass of Al ≈ 30,211.6 g
Therefore, the mass of Al produced during the electrolysis of molten Al2O3 under the given conditions is approximately 30,211.6 grams.