Generate an image of a scientific setup showcasing an electrolytic cell. There are electrodes immersed in a crystal-clear solution within it. A distinct, shimmering layer of aluminium, appearing almost magical, is present on one of the electrodes, showing an outcome of the electrolysis process.

What mass of aluminum is deposited electrolytically in 30 minutes by a current of 40A?

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0.672g

0.672 g

Coulombs = Amperes x seconds

C = 40 A x 30 min x (60 seconds/min) = 72,000
27/3 = 9 grams of Al will be deposited by 96,485 coulombs. How much will be deposited with only 72,000 C? Post your work if you get stuck.

0.672g

6.74

0.672

0.672

6.72

To calculate the mass of aluminum deposited, we need to use Faraday's constant, which relates the amount of charge (in coulombs) to the amount of substance that undergoes a reaction.

One mole of electrons carries a charge of 96,485 coulombs (i.e., Faraday's constant). The molar mass of aluminum is 27 g/mol, and the reaction involves the deposition of one mole of aluminum for every three moles of electrons transferred:

Al3+ + 3e- → Al(s)

Therefore, to deposit one mole of aluminum requires 3 x 96,485 C = 289,455 C.

In 30 minutes, the total charge transferred is:

Q = I x t = 40 A x 30 min x 60 s/min = 72,000 C

The amount of aluminum deposited can be found as follows:

moles of Al deposited = Q / (3 x 96,485) = 0.248 moles

mass of aluminum deposited = moles of Al x molar mass = 0.248 mol x 27 g/mol = 6.696 g

Therefore, about 6.7 grams of aluminum will be deposited electrolytically in 30 minutes by a current of 40A.