Al3+ is reduced to Al(s) at an electrode. If a current of 2.75 ampere is passed for 36 hours, what mass of aluminum is deposited at the electrode? Assume 100% current efficiency.

a. 9.2 x 10–3 g
b. 3.3 x 101 g
c. 9.9 x 101 g
d. 1.0 x 102 g
e. 3.0 x 102 g


2.75 A for 36 hrs is a charge transfer of Q = 2.75*36*3600 = 3.56*10^5 coul. This is Q/1.60*10^-19 = 2.23*10^24 electrons. It takes 3 electrons to reduce an atom of Al(3+) to Al, so 7.43*10^23 atoms of Al are deposited; this is 7.43*10^23 / 6.022*10^23 = 1.23 moles of Al. The atomic mass of Al is 27.0 g/mol so this would be 33.2 g of Al

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  1. I didn't work it that way and I don't see anything wrong with the way you did it. I agree with the answer.

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