Calculate the mass of copper and volume of oxygen (at 25°c and 760mmHg) that would be produced by passing a current of 0.5A through a CuSO4 solution b/n Pt electrodes for 3 hours

Answer 1

To calculate the mass of copper produced, we need to use Faraday’s law of electrolysis.

The equation for the production of copper in the electrolysis of copper sulfate is:

Cu^2+ (aq) + 2e^- → Cu (s)

The balanced equation shows that for every 2 moles of electrons passed, 1 mole of copper is produced.

Step 1: Calculate the number of moles of electrons passed.

Given:
Current = 0.5 A
Time = 3 hours = 3 × 60 × 60 seconds = 10,800 seconds
1 Faraday (F) of electrical charge = 96,500 C/mol e^-

Using the formula:
Charge (Coulombs) = Current (Amperes) × Time (seconds)

Charge = 0.5 A × 10,800 seconds = 5,400 C

Number of moles of electrons = Charge (Coulombs) / Faraday's constant
Number of moles of electrons = 5,400 C / 96,500 C/mol e^-
Number of moles of electrons = 0.05598 moles of electrons

Step 2: Calculate the number of moles of copper produced.

According to the balanced equation, for every 2 moles of electrons, 1 mole of copper is produced.

Number of moles of copper = (0.05598 moles of electrons) / (2 moles of electrons/mole of copper)
Number of moles of copper = 0.02799 moles of copper

Step 3: Calculate the mass of copper produced.

Given:
Molar mass of copper (Cu) = 63.55 g/mol

Mass of copper = Number of moles of copper × Molar mass of copper
Mass of copper = 0.02799 moles × 63.55 g/mol
Mass of copper = 1.78 g (rounded to two decimal places)

Therefore, the mass of copper produced by passing a current of 0.5 A through a CuSO4 solution for 3 hours would be approximately 1.78 grams.

Now, let's calculate the volume of oxygen produced.

The balanced equation for the production of oxygen in the electrolysis of water is:

2H2O (l) → 4H^+ (aq) + O2 (g) + 4e^-

According to the balanced equation, for every 4 moles of electrons, 1 mole of oxygen gas is produced.

Step 1: Calculate the number of moles of electrons passed.

We already calculated the number of moles of electrons as 0.05598 moles in the previous calculation.

Step 2: Calculate the number of moles of oxygen produced.

Number of moles of oxygen = (0.05598 moles of electrons) / (4 moles of electrons/mole of oxygen)
Number of moles of oxygen = 0.013995 moles of oxygen

Step 3: Calculate the volume of oxygen produced using the ideal gas law.

Given:
Temperature (T) = 25°C = 25 + 273.15 = 298.15 K (Convert to Kelvin)
Pressure (P) = 760 mmHg (Convert to atm)

Ideal gas law equation: PV = nRT

Where:
P = Pressure (atm)
V = Volume (L)
n = Number of moles
R = Ideal gas constant (0.0821 L × atm/mol × K)
T = Temperature (K)

Volume of oxygen = (Number of moles of oxygen × R × T) / P
Volume of oxygen = (0.013995 mol × 0.0821 L × atm/mol × K × 298.15 K) / 760 mmHg
Volume of oxygen = 0.005708 L or 5.7 mL (rounded to two decimal places)

Therefore, the volume of oxygen produced by passing a current of 0.5 A through a CuSO4 solution between Pt electrodes for 3 hours would be approximately 5.7 mL at 25°C and 760 mmHg.