what must be the current in amperes to plate out 7.437g of Al from am aqueous solution of Al3+ in 5.4x10^2 minutes?
A x seconds = coulombs.
It will take 96,485 coulombs to plate (26.98/3) g Al.
Coulombs = 96,485 x (7.437 x 3/26.98) = ??
Substitute into the first equation and solve for A. Time is 540 min x 60 s/min.
To determine the current required to plate out a certain mass of aluminum (Al) from an aqueous solution of Al3+ ions, we can use Faraday's laws of electrolysis. The key equation we will use is:
Mass plated = (Current × Time × Molar mass of Al) / (n × Faraday's constant)
Where:
- Mass plated is the mass of Al that needs to be plated out (7.437g in this case)
- Current is the electric current we're trying to find (in amperes)
- Time is the duration of the electrolysis process (5.4x10^2 minutes)
- Molar mass of Al is the molar mass of aluminum (26.98 g/mol)
- n is the number of electrons involved in the redox reaction (in this case, 3 since Al3+ is reduced to Al)
- Faraday's constant is the Faraday constant, which is approximately 96,485 C/mol
Let's plug in the values and solve for the current:
7.437g = (Current × 5.4x10^2 minutes × 26.98 g/mol) / (3 × 96,485 C/mol)
Rearranging the equation to solve for the current:
Current = (7.437g × 3 × 96,485 C/mol) / (5.4x10^2 minutes × 26.98 g/mol)
Now, let's calculate the current:
Current = (7.437g × 3 × 96,485 C/mol) / (5.4x10^2 minutes × 26.98 g/mol)
Current = 0.00215 A
Therefore, the current required to plate out 7.437g of Al from the aqueous solution in 5.4x10^2 minutes is approximately 0.00215 Amperes.