How many grams of Al would be deposited from an aqueous solution of Al3+ with a current of 1.430A after 5hrs of deposition.
How many coulombs do you have? That will be 1.430 A x (60m/hr) x (60s/min) = ??
96,485 C will deposit 27/3 or 9 g Al.
So (1.430 x 3600/96,485) x 9 = ??
To calculate the amount of aluminum (Al) deposited from an aqueous solution of Al3+ with a certain current over a given time, you need to know the molar mass of aluminum and use Faraday's law of electrolysis.
The first step is to determine the charge (Q) passed in coulombs (C) over the given time period. This can be done by multiplying the current (I) in amperes (A) by the time (t) in seconds (s):
Q = I * t
In this case, the current is given as 1.430 A, and the time is 5 hours. However, we need to convert the time to seconds:
5 hours * 60 minutes/hour * 60 seconds/minute = 18,000 seconds
So, the charge (Q) passed is:
Q = 1.430 A * 18,000 s = 25,740 C
Next, we need to determine the number of moles of electrons (n) involved in the electrodeposition of Al3+. The balanced half-reaction for the reduction of Al3+ to Al is:
2 Al3+ + 6 e- → 2 Al
From the stoichiometry of this reaction, we can see that for every 6 moles of electrons involved, 2 moles of Al are deposited. Therefore, the number of moles of electrons is:
n = Q / (F * 6)
Where F is Faraday's constant, approximately equal to 96,485 C/mol e-. Plugging in the values:
n = 25,740 C / (96,485 C/mol e- * 6) = 0.07036 mol e-
Finally, we can calculate the mass of aluminum deposited using the molar mass of aluminum (26.98 g/mol):
Mass of Al = n * molar mass of Al
Mass of Al = 0.07036 mol * 26.98 g/mol = 1.894 g
Therefore, approximately 1.894 grams of aluminum would be deposited from the aqueous solution of Al3+ with a current of 1.430 A after 5 hours of deposition.