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.
First, we need to determine the amount of charge (Q) that passes through the solution:
Q = I x t
Q = 0.5 A x 3 hours x 3600 s/hour
Q = 5400 C
Next, we can use Faraday's law to calculate the moles of electrons that are transferred:
1 mole of electrons = 96,485 C
moles of electrons = Q / 96,485 C
moles of electrons = 0.056 moles
Since one mole of electrons is needed to produce one mole of copper, we know that 0.056 moles of copper will be produced. The molar mass of copper is 63.55 g/mol, so the mass of copper produced is:
mass of copper = moles of copper x molar mass of copper
mass of copper = 0.056 moles x 63.55 g/mol
mass of copper = 3.57 g
To determine the volume of oxygen produced, we need to use the ideal gas law. Since the pressure and temperature are given, we can solve for volume:
PV = nRT
where P = 760 mmHg, V = volume of oxygen, n = moles of oxygen, R = 0.08206 L atm/mol K (the ideal gas constant), and T = 298 K.
We need to determine the number of moles of oxygen that are produced. Since copper is being oxidized to copper(II) ions during the reaction, we know that two moles of electrons are being transferred for every mole of oxygen produced (from water molecules in the solution). Therefore, the moles of oxygen produced is:
moles of oxygen = moles of electrons / 2
moles of oxygen = 0.056 moles / 2
moles of oxygen = 0.028 moles
Now we can solve for volume:
V = nRT / P
V = 0.028 moles x 0.08206 L atm/mol K x 298 K / 760 mmHg
V = 0.0010 L or 1.0 mL
Therefore, 3.57 g of copper and 1.0 mL of oxygen gas would be produced by the electrolysis of the CuSO4 solution for 3 hours.