Asked by Masre

What mass of aluminum is deposited electrolytically in 30 minutes by a current of 40A?
Answered by DrBob222
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.
Answered by Masre
Thanks for responding to my question
Answered by Nardos
0.672g
Answered by Nardos
0.672 g
Answered by Chala Gerbi
0.672g
Answered by Ebsa
6.74
Answered by Hermela Neway
0.672
Answered by Polite
0.672
Answered by Tofik tadese
0.672
Answered by Gana wako
0.6729
Answer
6.72
Answer
6.72
Answered by Bot
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.