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
For 3 what?
3erered
To determine the mass of copper and volume of oxygen produced, we need to first calculate the amount of charge passed through the solution.
1. Calculate the total charge:
Total charge = current (A) × time (s)
Total charge = 0.5 A × 3 s
Total charge = 1.5 C
Next, we use Faraday's laws of electrolysis to determine the moles of copper and oxygen produced.
2. Calculate moles of copper:
1 mole of electrons = 96,485 C (Faraday's constant)
moles of copper = Total charge (C) / (96,485 C/mol)
moles of copper = 1.5 C / 96,485 C/mol
Note: In this reaction, the same number of moles of copper ions (Cu2+) is reduced to copper metal (Cu).
3. Calculate the mass of copper:
To calculate the mass of copper, we need the molar mass of copper.
Molar mass of copper (Cu) = 63.55 g/mol (atomic mass of copper)
mass of copper = moles of copper × molar mass of copper
4. Calculate volume of oxygen:
According to the balanced chemical equation, when copper is deposited, oxygen gas is liberated in a 1:1 ratio by volume.
Thus, the volume of oxygen is equal to the volume of copper.
Now, we can proceed with the calculations.
5. Calculate the mass of copper:
mass of copper = moles of copper × molar mass of copper
6. Calculate the volume of oxygen:
Volume of oxygen = volume of copper
It is important to note that the given conditions (25°C and 760 mmHg) correspond to standard temperature and pressure (STP), where 1 mole of any gas occupies 22.4 liters.
Let's input the values and perform the calculations:
moles of copper = 1.5 C / 96,485 C/mol
moles of copper = 0.0000155 mol (rounded to 5 decimal places)
mass of copper = 0.0000155 mol × 63.55 g/mol
mass of copper = 0.000983525 g (rounded to 6 decimal places)
Volume of oxygen = 0.000983525 L (rounded to 6 decimal places)
Therefore, the mass of copper produced is approximately 0.001 g, and the volume of oxygen produced is approximately 0.000984 L at the given conditions.
To calculate the mass of copper (Cu) and the volume of oxygen (O2) produced by passing a current through a CuSO4 solution, you need to consider the stoichiometry of the reaction and Faraday's law.
1. Write the balanced chemical equation for the electrolysis of copper sulfate (CuSO4):
2CuSO4(aq) + 2H2O(l) -> 2H2SO4(aq) + 2Cu(s) + O2(g)
From the balanced equation, you can see that for every 2 moles of electrons (2F) passed through the solution, 1 mole of oxygen gas is produced.
2. Convert the current (I) and time (t) to the quantity of electric charge (Q):
Q = I * t
In this case, Q = 0.5 A * 3 s = 1.5 C.
3. Apply Faraday's law to calculate the number of moles of electrons (n) passed through the solution:
n = Q / F
where F is Faraday's constant, which is equal to 96,485 C/mol e-.
n = 1.5 C / 96,485 C/mol e- = 1.553 x 10^-5 mol e-
4. Use stoichiometry to determine the moles of oxygen gas (O2) produced:
According to the balanced equation, 2 moles of electrons (2F) produce 1 mole of O2 gas. Therefore,
n(O2) = (1.553 x 10^-5 mol e-) / (2 mol e-/1 mol O2) = 7.765 x 10^-6 mol O2
5. Convert moles of oxygen gas (O2) to volume at standard temperature and pressure (STP):
At STP (0°C or 273.15 K and 1 atm), 1 mole of any gas occupies 22.4 L.
Volume of O2(g) = n(O2) * 22.4 L/mol = 7.765 x 10^-6 mol O2 * 22.4 L/mol = 0.0001738 L or 0.1738 mL
6. Calculate the mass of copper (Cu):
The molar mass of copper (Cu) is approximately 63.55 g/mol.
Mass of Cu = n(Cu) * molar mass of Cu
According to the balanced equation, 2 moles of electrons (2F) produce 2 moles of copper. Therefore,
n(Cu) = (1.553 x 10^-5 mol e-) / (2 mol e-/ 2 mol Cu) = 1.553 x 10^-5 mol Cu
Mass of Cu = 1.553 x 10^-5 mol Cu * 63.55 g/mol = 9.881 x 10^-4 g or 0.0009881 g
So, the mass of copper produced is approximately 0.0009881 g, and the volume of oxygen produced is approximately 0.1738 mL at STP.