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

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 ......?

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.

OK. I told you how to do it. Do you need more assistance or do you have the right answer now?

To calculate the mass of aluminum deposited at the electrode, we need to use Faraday's law of electrolysis. Faraday's law states that the mass of a substance deposited at an electrode is directly proportional to the charge passed through the electrode.

The formula we can use is:

Mass = (I * t * M) / (n * F)

Where:
- Mass is the mass of the material deposited (in grams).
- I is the current (in amperes).
- t is the time (in seconds).
- M is the molar mass of the substance deposited (in grams/mol).
- n is the number of moles of electrons transferred (for a given reaction).
- F is the Faraday's constant (approximately 96,485 C/mol).

In this case, we are depositing aluminum (Al), and the reaction is:

2 Al3+ + 6 e- -> 2 Al

From the reaction, we can see that for every 6 moles of electrons transferred, we get 2 moles of aluminum deposited.

Now, let's calculate the number of moles of electrons transferred:

n = (I * t) / (F * 6)

Given:
I = 2.75 A
t = 36 hours = 36 * 60 * 60 seconds
F = 96,485 C/mol

Plugging these values into the equation, we get:

n = (2.75 * 36 * 60 * 60) / (96,485 * 6)

Next, let's calculate the mass using the formula:

Mass = (I * t * M) / (n * F)

Given:
I = 2.75 A
t = 36 hours = 36 * 60 * 60 seconds
M = molar mass of aluminum = 26.98 g/mol
n = calculated from above
F = 96,485 C/mol

Plugging these values into the equation, we get:

Mass = (2.75 * 36 * 60 * 60 * 26.98) / (n * 96,485 * 6)

Now, let's substitute the calculated value of "n" into the equation and solve for the mass:

Mass = (2.75 * 36 * 60 * 60 * 26.98) / ((2.75 * 36 * 60 * 60) / (96,485 * 6) * 96,485 * 6)

After simplifying the equation, we find that the mass of aluminum deposited is:

Mass = 0.033 g

So, the correct answer is option B: 3.3 x 10^1 g.