Calculate the mass of copper and volume of oxygen (at 25°C and 760 mmHg) that would be produced by passing a current of 0.5 A through a CuSO4 solution between Pt electrodes for 3 hours
The balanced chemical equation for the reaction that occurs is:
2CuSO4 + 2H2O → 2Cu + O2 + 2H2SO4
From the equation, we can see that 2 moles of CuSO4 produces 1 mole of O2.
First, let's calculate the amount of Cu that is produced:
1. Calculate the number of electrons transferred from the anode to the cathode:
- 0.5 A = 0.5 C/s (Coulombs per second)
- 1 hour = 60 × 60 = 3600 seconds
- 3 hours = 3 × 3600 = 10800 seconds
- Total charge passed = 0.5 C/s × 10800 s = 5400 C
- Each mole of electrons corresponds to a Faraday constant of 96485 C/mol
- Moles of electrons = 5400 C ÷ 96485 C/mol = 0.056 mol
2. Calculate the amount of Cu produced from the moles of electrons:
- From the balanced equation, 2 moles of electrons produce 2 moles of Cu
- Moles of Cu produced = 0.056 mol ÷ 2 = 0.028 mol
- Mass of Cu produced = 0.028 mol × 63.55 g/mol = 1.7786 g
Next, let's calculate the volume of O2 produced:
1. Calculate the moles of O2 produced from the moles of CuSO4:
- From the balanced equation, 2 moles of CuSO4 produce 1 mole of O2
- Moles of CuSO4 used = 0.028 mol
- Moles of O2 produced = 0.028 mol ÷ 2 = 0.014 mol
2. Calculate the volume of O2 produced using the ideal gas law:
- PV = nRT
- P = 760 mmHg (atm) = 0.987 atm (converted using the conversion factor: 1 atm = 760 mmHg)
- V = ?
- n = 0.014 mol
- R = 0.0821 L·atm/mol·K (gas constant)
- T = 25°C = 298 K
- V = nRT/P = (0.014 mol)(0.0821 L·atm/mol·K)(298 K)/(0.987 atm) = 0.404 L or 404 mL
Therefore, the mass of Cu produced is 1.7786 g and the volume of O2 produced is 404 mL at 25°C and 760 mmHg.