A current of 2.34 A is delivered to an electrolytic cell for 85 minutes. How many grams of Au will be deposited from an aqueous solution of AuCl3?
To calculate the mass of gold (Au) deposited from an aqueous solution of AuCl3, we need to use Faraday's law of electrolysis. Faraday's law states that the mass of a substance deposited or liberated during electrolysis is directly proportional to the quantity of electric charge passed through the cell.
The formula to calculate the mass (m) of a substance deposited or liberated during electrolysis is:
m = (Q * M) / (F * z)
Where:
m = mass of the substance deposited or liberated (in grams)
Q = electric charge passed through the cell (in coulombs)
M = molar mass of the substance (in grams per mole)
F = Faraday's constant (96485 C/mol)
z = number of electrons involved in the reaction
To solve this problem, we need to find the value of Q.
Q can be calculated using the formula:
Q = I * t
Where:
I = current (in amperes)
t = time (in seconds)
Given:
Current (I) = 2.34 A
Time (t) = 85 minutes = 85 * 60 seconds
Calculating Q:
Q = 2.34 A * (85 * 60 s)
Q = 2.34 A * 5100 s
Q = 11934 C
Now that we have the value of Q, we can calculate the mass of Au deposited.
To find M, we need to know the molar mass of Au. The molar mass of Au is 196.97 g/mol.
Plugging in the values into the formula:
m = (11934 C * 196.97 g/mol) / (96485 C/mol * z)
The value of z depends on the balanced chemical equation. Since the balanced equation for the deposition of gold from AuCl3 is not provided, we cannot determine the exact value of z. It is usually the same as the number of electrons involved in the oxidation or reduction half-reaction.
Assuming the balanced equation is:
2 AuCl3 + 6 e- → 2 Au + 6 Cl-
In this case, z would be 6 since 6 electrons are involved in the reaction.
Finally, substituting the values:
m = (11934 C * 196.97 g/mol) / (96485 C/mol * 6)
m = 0.385 g
Therefore, approximately 0.385 grams of gold (Au) will be deposited from the solution.