Question

Calculate the weight (in grams) of copper that will be deposited at the cathode in the electrolysis of a 0.4 M solution of copper sulphate when quantity of electricity is equal to the quantity of electricity required to liberate 4.48 L of hydrogen at 273 K, 1 atm pressure from a 0.2 M aqueous sulphuric acid, is passed through the copper sulphate solution. (atomic mass of Cu = 63.5)

Answer 1

12.7

Answer 2

To calculate the weight of copper deposited, we need to follow these steps:

Step 1: Calculate the moles of hydrogen gas liberated. At STP (273 K and 1 atm pressure), 1 mole of any gas occupies 22.4 L. Given volume of hydrogen (H₂) = 4.48 L Number of moles of H₂ = $\frac{\text{Volume of H₂}}{\text{Molar volume at STP}}$ Number of moles of H₂ = $\frac{4.48 \text{ L}}{22.4 \text{ L/mol}}$ Number of moles of H₂ = 0.2 mol

Step 2: Determine the quantity of electricity (in Faradays) required to liberate 0.2 mol of hydrogen. The reaction for the liberation of hydrogen at the cathode from aqueous sulphuric acid is: $2\text{H}^+(\text{aq}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g})$ From the stoichiometry, 1 mole of H₂ gas requires 2 moles of electrons. Since 1 mole of electrons carries 1 Faraday (F) of charge (96500 C), 1 mole of H₂ requires 2 Faradays of electricity. Quantity of electricity required for 0.2 mol of H₂ = $0.2 \text{ mol H}_2 \times \frac{2 \text{ F}}{1 \text{ mol H}_2}$ Quantity of electricity = 0.4 F

Step 3: Calculate the mass of copper deposited using the same quantity of electricity. The reaction for the deposition of copper at the cathode from copper sulphate solution is: $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$ From the stoichiometry, 1 mole of copper (Cu) requires 2 moles of electrons for deposition. The atomic mass of Cu is 63.5 g/mol. So, 63.5 g of Cu requires 2 Faradays of electricity for deposition.

We have 0.4 F of electricity passed through the copper sulphate solution. Using the relationship: Mass of substance deposited = $\frac{\text{Equivalent weight} \times \text{Quantity of electricity (in Coulombs)}}{\text{Faraday's constant}}$ Or, more simply using Faradays: Mass of Cu deposited = $\frac{\text{Quantity of electricity (in F)}}{\text{Electrons per mole of Cu}} \times \text{Molar mass of Cu}$ Mass of Cu deposited = $\frac{0.4 \text{ F}}{2 \text{ F/mol Cu}} \times 63.5 \text{ g/mol}$ Mass of Cu deposited = $0.2 \text{ mol Cu} \times 63.5 \text{ g/mol}$ Mass of Cu deposited = 12.7 g

Alternatively, using Faraday's second law of electrolysis: $\frac{\text{Mass of H}_2}{\text{Equivalent weight of H}_2} = \frac{\text{Mass of Cu}}{\text{Equivalent weight of Cu}}$ Mass of H₂ = $0.2 \text{ mol} \times 2 \text{ g/mol} = 0.4 \text{ g}$ Equivalent weight of H₂ = $\frac{\text{Molar mass of H}_2}{\text{n-factor}} = \frac{2 \text{ g/mol}}{2} = 1 \text{ g/eq}$ Equivalent weight of Cu = $\frac{\text{Molar mass of Cu}}{\text{n-factor}} = \frac{63.5 \text{ g/mol}}{2} = 31.75 \text{ g/eq}$ $\frac{0.4 \text{ g}}{1 \text{ g/eq}} = \frac{\text{Mass of Cu}}{31.75 \text{ g/eq}}$ Mass of Cu = $0.4 \times 31.75 \text{ g}$ Mass of Cu = 12.7 g

The concentrations of the solutions (0.4 M CuSO₄ and 0.2 M H₂SO₄) are not needed for this calculation, as long as the solutions are sufficiently concentrated to provide the necessary ions.

The final answer is $\boxed{\text{12.7}}$.

Explanation of the solution:

1. Calculate moles of H₂ from its volume at STP (4.48 L / 22.4 L/mol = 0.2 mol).
2. Determine the Faradays of electricity required for H₂ liberation (0.2 mol H₂ × 2 F/mol H₂ = 0.4 F).
3. Use this quantity of electricity to find the mass of Cu deposited (0.4 F / 2 F/mol Cu × 63.5 g/mol = 12.7 g).

Answer: 12.7 g