Answer: To calculate the mass of silver deposited during electrolysis, we can use Faraday's laws of electrolysis. Faraday's first law states that the mass of a substance deposited or liberated during electrolysis is directly proportional to the amount of charge passed through the electrolyte. The equation to calculate the mass (m) of the substance deposited is:
m = (Q * M) / (n * F)
where:
m = mass of substance deposited (in grams)
Q = total charge passed through the electrolyte (in Coulombs)
M = molar mass of the substance (in grams per mole)
n = number of moles of electrons transferred in the reaction (the stoichiometry of the reaction)
F = Faraday's constant, which is the charge of 1 mole of electrons, approximately 96485 Coulombs per mole of electrons
Given:
Current (I) = 4.6 A
Time (t) = 90 minutes = 90 * 60 seconds = 5400 seconds
First, calculate the total charge (Q) passed through the electrolyte:
Q = I * t
Q = 4.6 A * 5400 seconds
Q = 24840 Coulombs
Next, determine the number of moles of electrons transferred in the reaction. In the case of the electrolysis of silver ions (Ag⁺) to form solid silver (Ag), one mole of silver ion (Ag⁺) gains one mole of electrons (n = 1).
Now, find the molar mass of silver (M). The molar mass of silver (Ag) is approximately 107.87 g/mol.
Now, use the formula to calculate the mass of silver deposited (m):
m = (Q * M) / (n * F)
m = (24840 Coulombs * 107.87 g/mol) / (1 * 96485 Coulombs/mol)
m = 2673.45 g
So, the mass of silver deposited when a current of 4.6 A is passed through the silver salt solution for 90 minutes is approximately 2673.45 grams.