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
Answer
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
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.
Hmm, that seems like a lot of calculations. Are you sure you want to know the answer? I hope you're not just trying to weigh down the electrode with all that aluminum! But if you really want to know, the answer is b. 3.3 x 101 g. Now, how about we lighten the mood with a joke? Why don't scientists trust atoms? Because they make up everything!
The mass of aluminum deposited at the electrode is 33.2 g.
Therefore, the correct answer is b. 3.3 x 101 g.
To answer this question, we need to calculate the amount of charge transferred during the electrolysis process, and then convert it to the amount of aluminum deposited.
First, we need to find the amount of charge transferred. We are given the current (2.75 amperes) and the time (36 hours). To calculate the charge, we multiply the current by the time in seconds:
Charge (Q) = Current (I) * Time (t) = 2.75 A * 36 hours * 3600 seconds/hour = 3.56 * 10^5 coulombs
Next, we need to calculate the number of electrons involved in the reduction of Al3+ to Al(s). For this, we can use Faraday's law, which states that one mole of electrons is equivalent to 1 Faraday (F) of charge, which is approximately 96,485 coulombs:
Number of electrons = Charge (Q) / 1.60 * 10^-19 C = 3.56 * 10^5 C / 1.60 * 10^-19 C = 2.23 * 10^24 electrons
Since it takes 3 electrons to reduce one Al3+ ion to Al(s), we can calculate the number of Al atoms deposited:
Number of Al atoms deposited = Number of electrons / 3 = 2.23 * 10^24 electrons / 3 = 7.43 * 10^23 atoms
To convert this to moles of Al, we need to divide by Avogadro's number, which is 6.022 * 10^23 atoms/mole:
Number of moles of Al deposited = Number of Al atoms deposited / Avogadro's number = 7.43 * 10^23 atoms / 6.022 * 10^23 atoms/mole = 1.23 moles
Finally, we can calculate the mass of aluminum deposited by multiplying the number of moles by the molar mass of aluminum, which is approximately 27.0 g/mol:
Mass of aluminum deposited = Number of moles of Al * Molar mass of Al = 1.23 moles * 27.0 g/mol = 33.2 g
Therefore, the correct answer is b. 3.3 x 10^1 g (or 33.2 g) of aluminum is deposited at the electrode.