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)
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.
Hey Bot. I think you calculated coulombs incorrectly as well as converting 1.5 hours to seconds incorrectly.
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.
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?
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.
To calculate the current required to deposit a certain amount of aluminum, we need to use Faraday's law of electrolysis. The equation is:
m = (M * I * t) / (z * F)
Where:
m is the mass of the substance (in grams) to be deposited,
M is the molar mass of the substance (in grams per mole),
I is the current (in amperes),
t is the time (in seconds),
z is the number of moles of electrons transferred per mole of substance, and
F is Faraday's constant (96485 C/mol).
In this case, we want to deposit 1.5g of aluminum, which has a molar mass (M) of 27 g/mol.
First, we need to calculate the number of moles (n) of aluminum using the formula:
n = m / M
n = 1.5g / 27 g/mol
n = 0.05556 mol
Since each aluminum atom gains 3 electrons during electrolysis, z = 3. Therefore, we can rewrite the equation as:
m = (M * I * t) / (3 * F)
To isolate I (current), we rearrange the equation:
I = (m * 3 * F) / (M * t)
Plugging in the values:
I = (1.5g * 3 * 96485 C/mol) / (27 g/mol * (1hr 30min * 60s/min))
Converting 1 hour 30 minutes to seconds:
1hr 30min = 1.5 * 60s = 90s
I = (1.5g * 3 * 96485 C/mol) / (27 g/mol * 90s)
Calculating the current:
I = 1066.39 C/s
Therefore, the current that must be passed into the aluminum salt solution for 1 hour 30 minutes in order to deposit 1.5g of aluminum is approximately 1066.39 amperes.
To calculate the current that must be passed into a solution of aluminum salt to deposit a certain amount of aluminum, we can use Faraday's law of electrolysis. The equation is:
Q = I * t * M / (n * F)
Where:
Q is the amount of substance in mol (in this case, aluminum)
I is the current in amperes (A)
t is the time in seconds (s)
M is the molar mass of the substance (in this case, aluminum)
n is the number of electrons transferred in the balanced half-reaction
F is the Faraday's constant (approximately 96485 C/mol)
First, let's calculate the number of moles of aluminum using its molar mass (M = 27 g/mol):
n = 1.5 g / 27 g/mol = 0.0556 mol
Next, let's convert the time of 1 hour 30 minutes to seconds:
t = 1 hour * 60 minutes/hour * 60 seconds/minute + 30 minutes * 60 seconds/minute
t = 1 * 60 * 60 + 30 * 60 = 5400 seconds
Now, we have all the values needed to calculate the current:
I = (Q * n * F) / (t * M)
I = (0.0556 mol * 1 * 96485 C/mol) / (5400 s * 27 g/mol)
I ≈ 0.0374 A (amperes)
Therefore, a current of approximately 0.0374 amperes should be passed into the solution of aluminum salt for 1 hour 30 minutes to deposit 1.5 grams of aluminum.