How many grams of silver will be obtained when an aqueous silver nitrate solution is electrolyzed for 22.0min with a constant current of 2.28A ?
Coulombs = amperes x seconds = approx 3000.
107 g Ag will be deposited for every 96,485 coulombs. So Ag deposited is
107g Ag x (approx 3000/96,485) = ? g Ag.
To determine the number of grams of silver obtained during electrolysis, we can use the formula:
Mass = Current × Time × Atomic Mass of Silver / Faraday's Constant
First, let's calculate Faraday's constant. Faraday's constant is the charge carried by one mole of electrons and is equal to 96,485 C/mol.
Next, we need to find the atomic mass of silver. The atomic mass of silver (Ag) is 107.87 g/mol.
Now, we can plug the given values into the formula:
Current = 2.28 A
Time = 22.0 min = 22.0 min × (60 s/min) = 1320 s
Atomic Mass of Silver = 107.87 g/mol
Faraday's Constant = 96,485 C/mol
Mass = 2.28 A × 1320 s × (107.87 g/mol) / 96,485 C/mol
Simplifying the equation:
Mass = 3.201 g
Therefore, approximately 3.201 grams of silver will be obtained when an aqueous silver nitrate solution is electrolyzed for 22.0 minutes with a constant current of 2.28 A.