Al3+ is reduced to Al(s) at an electrode. If a current of 2.75 amps is passed for 36 hours, what mass of aluminum is deposited at the electrode? Assume 100% current efficiency. a. 9.2×10–3g
b. 3.3×101g
vs. 9.9×101g
D. 1.0×102g
e. 3.0 x 102g
Answer
2.75 A for 36 hours is a charge transfer of Q = 2.75*36*3600 = 3.56*10^5 col. It is Q/1.60*10^-19 = 2.23*10^24 electrons. It takes 3 electrons to reduce an Al(3+) atom to Al, so 7.43*10^23 Al atoms are deposited; it 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 that would be 33.2 g of Al. I didn’t work that way and I don’t see anything wrong with the way you did. I agree with the answer.