Asked by Anonymous

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


I think it is either B or E
Answered by DrBob222
What does either B or E mean? To me it means you don't know and you're guessing because they aren't close to each other.
How many coulombs did you use. That's amperes x seconds = coulombs.
Then you know that 27/3 = 9 g Al will be deposited for every 96,485 coulombs. That makes the answer ......?
Answered by Anonymous
The first time i tried this question I got
3.3 x 101 g and I recalculated it I got 3.0 x 102 g. That's why I was confused which one to pick.
Answered by DrBob222
OK. I told you how to do it. Do you need more assistance or do you have the right answer now?
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