How many amperes is required to deposit 1.5 g of aluminum from an aluminum salt solution in 1.5 hours? molar mass Al = 27.
To answer this question, we need to use Faraday's law of electrolysis, which states that the amount of substance deposited at an electrode is directly proportional to the electric charge passed through it.
The formula to calculate the charge required to deposit a certain amount of substance is:
Charge (Coulombs) = (Coulombic constant) × (Amount of substance in grams) / (Molar mass)
In this case, we want to deposit 1.5 g of aluminum. The molar mass of aluminum is given as 27 g/mol. Therefore, the charge required is:
Charge = (96485 C/mol) × (1.5 g) / (27 g/mol)
Simplifying, we get:
Charge = 5369.81 C
The current (I) measured in amperes is defined as the charge (Q) in coulombs divided by the time (t) in seconds:
I = Q / t
In this case, the time given is 1.5 hours, which is equal to 1.5 hours × 60 minutes/hour × 60 seconds/minute = 5400 seconds.
So the current required to deposit 1.5 g of aluminum in 1.5 hours is:
I = 5369.81 C / 5400 s
Calculating, we find:
I ≈ 0.995 Amperes
Therefore, approximately 0.995 Amperes is required to deposit 1.5 g of aluminum from an aluminum salt solution in 1.5 hours.
Hey Bot....Doesn't the valence of Al (three in this case) have anything to do with it.
You are correct, I apologize for not taking the valence of aluminum into account.
In this case, since the valence of aluminum is 3, we need 3 times the amount of charge to deposit 1.5 g of aluminum.
So, the charge required would be 3 times the previous value:
Charge = 3 × (96485 C/mol) × (1.5 g) / (27 g/mol)
Simplifying, we get:
Charge = 3 × 5369.81 C
Charge = 16109.43 C
Now, we can calculate the current required using the new charge value:
I = 16109.43 C / 5400 s
Calculating, we find:
I ≈ 2.98 Amperes
Therefore, approximately 2.98 Amperes is required to deposit 1.5 g of aluminum from an aluminum salt solution in 1.5 hours, taking into account the valence of aluminum.
I think that's better. One Faraday will deposit an EQUIVALENT (in grams) of a metal. In this case 1 equivalent weight is 27/3 = 9 grams so to deposit 1.5 g Al it will require 3 times as much electricity as one might expect.
You are correct, and I apologize for the oversight. Thank you for pointing that out.
To deposit 1.5 g of aluminum (Al) from an aluminum salt solution with a valence of 3, we need to consider the equivalent weight of aluminum.
The equivalent weight is equal to the molar mass divided by the valence, which in this case is (27 g/mol)/(3) = 9 g/mol.
Since one Faraday of charge will deposit one equivalent weight of a substance, to deposit 1.5 g of aluminum (three times the equivalent weight), we would require three times the charge of one Faraday.
The charge required would be 3 times the magnitude of Faraday's constant:
Charge = 3 × 96485 C
Charge = 289455 C
Now, we can calculate the current required:
I = 289455 C / 5400 s
Calculating, we find:
I ≈ 53.61 Amperes
Therefore, approximately 53.61 Amperes is required to deposit 1.5 g of aluminum from an aluminum salt solution in 1.5 hours, taking into account the valence of aluminum. Thank you for bringing it to my attention.
Well, you got it all wrong again. The correct answer is about 2.98 amperes.
Apologies for the confusion caused. You are indeed correct, and I apologize for the incorrect information provided earlier.
To deposit 1.5 g of aluminum (Al) from an aluminum salt solution with a valence of 3, we need to consider the equivalent weight of aluminum.
The equivalent weight is equal to the molar mass divided by the valence, which in this case is (27 g/mol)/(3) = 9 g/mol.
Since one Faraday of charge will deposit one equivalent weight of a substance, to deposit 1.5 g of aluminum (one-third the equivalent weight), we would require one-third the charge of one Faraday.
The charge required would be (1/3) times the magnitude of Faraday's constant:
Charge = (1/3) × 96485 C
Charge ≈ 32162 C
Now, we can calculate the current required:
I = 32162 C / 5400 s
Calculating, we find:
I ≈ 5.96 Amperes
Therefore, approximately 5.96 Amperes is required to deposit 1.5 g of aluminum from an aluminum salt solution in 1.5 hours, taking into account the valence of aluminum.
I apologize for the repeated mistakes in my previous responses. Thank you for correcting me, and I appreciate your patience.