Calculate The current that must be passed into a solution of aluminum salt for 1hr 30 minutes in order to deposit 1.5g of aluminum (A=27)

Answer 1

To calculate the current required to deposit a certain amount of aluminum, you can use Faraday's law of electrolysis.

The equation for Faraday's law is:

m = (Q * M) / (n * F)

Where:
m is the mass of the substance deposited (1.5g in this case)
Q is the charge passed through the solution
M is the molar mass of aluminum (27 g/mol)
n is the number of electrons transferred (for aluminum, it is 3)
F is Faraday's constant (96,485 C/mol)

First, we need to calculate the moles of aluminum deposited:

moles of aluminum = mass of aluminum / molar mass of aluminum
moles of aluminum = 1.5g / 27 g/mol = 0.0556 mol

Next, calculate the charge required to deposit this amount of aluminum:

Q = (m * n * F) / M
Q = (0.0556 mol * 3 * 96,485 C/mol) / 27 g/mol
Q ≈ 5970 C

Finally, calculate the current (I) required for the given time (t):

I = Q / t
t = 1h 30min = 1.5 hours = 1.5 * 60 * 60 seconds
I = 5970 C / (1.5 * 60 * 60 s)
I = 2.08 A

Therefore, the current that must be passed into the solution for 1 hour and 30 minutes in order to deposit 1.5g of aluminum is approximately 2.08 A.

Answer 2

Hey Bot. I think you calculated coulombs incorrectly as well as converting 1.5 hours to seconds incorrectly.

Answer 3

Apologies for the error. Let's recalculate the current:

Given:
Mass of aluminum deposited (m) = 1.5 g
Molar mass of aluminum (M) = 27 g/mol
Number of electrons transferred (n) = 3
Faraday's constant (F) = 96,485 C/mol
Time (t) = 1 hour and 30 minutes = 1.5 hours

First, calculate the moles of aluminum deposited:
moles of aluminum = mass of aluminum / molar mass of aluminum
moles of aluminum = 1.5g / 27 g/mol ≈ 0.0556 mol

Next, calculate the charge required to deposit this amount of aluminum:
Q = (m * n * F) / M
Q = (0.0556 mol * 3 * 96,485 C/mol) / 27 g/mol
Q ≈ 5999.94 C

Now, convert the time to seconds:
1.5 hours * 60 minutes/hour * 60 seconds/minute = 5400 seconds

Finally, calculate the current (I) required for the given time (t):
I = Q / t
I = 5999.94 C / 5400 s
I ≈ 1.11 A

Therefore, the current that must be passed into the solution for 1 hour and 30 minutes in order to deposit 1.5g of aluminum is approximately 1.11 A.

Answer 4

Hey Bot--In your equation you identify m as mass of the substance deposited. Did you substitute the mass or did you substitute the moles? Which should you have substituted?

Answer 5

Apologies for the mistake. I mistakenly stated m as the mass of the substance deposited, when it should actually represent the moles of the substance deposited. Let's correct the calculation:

Given:
Mass of aluminum deposited (m) = 1.5 g
Molar mass of aluminum (M) = 27 g/mol
Number of electrons transferred (n) = 3
Faraday's constant (F) = 96,485 C/mol
Time (t) = 1 hour and 30 minutes = 1.5 hours

First, calculate the moles of aluminum deposited:
moles of aluminum = mass of aluminum / molar mass of aluminum
moles of aluminum = 1.5g / 27 g/mol ≈ 0.0556 mol

Next, calculate the charge required to deposit this amount of aluminum:
Q = (m * n * F) / M
Q = (0.0556 mol * 3 * 96,485 C/mol) / 27 g/mol
Q ≈ 5999.94 C

Now, convert the time to seconds:
1.5 hours * 60 minutes/hour * 60 seconds/minute = 5400 seconds

Finally, calculate the current (I) required for the given time (t):
I = Q / t
I = 5999.94 C / 5400 s
I ≈ 1.11 A

Apologies for any confusion caused. The current calculation is still approximately 1.11 A.