what mass of copper is deposited from copper sulfate when a current of 0.22 A is passed through the solution for 15 minute
i thought this was physics
To determine the mass of copper deposited from copper sulfate, we need to make use of the Faraday's law of electrolysis. This law states that the amount of substance (in moles) deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (in coulombs) passed through the electrolyte.
The formula to calculate the amount of substance deposited can be derived from Faraday's law and is given as:
Mass (g) = (Current (A) × Time (s) × Molar mass (g/mol)) / (1 Faraday (C))
1. Convert the time from minutes to seconds:
Time (s) = 15 minutes × 60 seconds/minute = 900 s
2. Find the molar mass of copper (Cu):
The molar mass of Cu is approximately 63.55 g/mol.
3. Calculate the quantity of electricity (coulombs) passed through the circuit:
Quantity of electricity (C) = Current (A) × Time (s)
Quantity of electricity (C) = 0.22 A × 900 s
4. Convert the quantity of electricity from coulombs to Faradays:
1 Faraday (F) = 96,485 C
Quantity of electricity (F) = (Quantity of electricity (C)) / (1 Faraday (C))
5. Substitute the values into the formula to find the mass of copper deposited:
Mass (g) = (Current (A) × Time (s) × Molar mass (g/mol)) / (1 Faraday (C))
Following these steps, we can calculate the mass of copper deposited from copper sulfate.