An aqueous solution of CuSO₄ measuring 1 L was electrolysed till the pH of the solution was found to be 1. Then, the amount of Cu deposited and the volume of gases evolved (at STP) are respectively (Cu = 63.5 g/mol)

To determine the amount of copper deposited and the volume of gases evolved during the electrolysis of a 1 L aqueous solution of CuSO4 until the pH reaches 1, we need to analyze the electrochemical reactions taking place and apply some stoichiometric principles.

Understanding the Electrolysis of CuSO4

When you electrolyze copper(II) sulfate (CuSO4) in water, the solution contains Cu²⁺ ions and SO₄²⁻ ions. The electrolysis process involves the reduction of Cu²⁺ ions at the cathode and the oxidation of water at the anode. The overall reactions can be summarized as follows:

• At the cathode: Cu²⁺ + 2e⁻ → Cu (s)
• At the anode: 2H₂O → O₂ (g) + 4H⁺ + 4e⁻

Calculating the Change in pH

Initially, the pH of the CuSO4 solution is around 7. When the electrolysis continues until the pH drops to 1, it indicates a significant increase in hydrogen ion concentration. A pH of 1 corresponds to a hydrogen ion concentration of 0.1 M. This change in pH suggests that a considerable amount of water has been oxidized, producing H⁺ ions.

Determining the Amount of Copper Deposited

To find out how much copper is deposited, we first need to calculate the total charge passed through the solution. The change in pH from 7 to 1 means that the concentration of H⁺ ions increased from 10⁻⁷ M to 0.1 M. This change indicates that a total of 0.1 moles of H⁺ ions were produced in 1 L of solution.

Since 4 moles of electrons are required to produce 2 moles of H₂O, we can calculate the moles of electrons needed:

• For 0.1 moles of H⁺ produced, we need 0.1 moles of H₂ (which comes from 0.1 moles of H⁺).
• Thus, the moles of electrons required = 0.1 moles H⁺ × 4 = 0.4 moles of electrons.

Now, using Faraday's law, we know that 1 mole of electrons can deposit 1 mole of Cu. Therefore, the moles of Cu deposited will also be 0.4 moles.

Next, we can calculate the mass of copper deposited:

Mass of Cu = moles of Cu × molar mass of Cu = 0.4 moles × 63.5 g/mol = 25.4 g.

Calculating the Volume of Gases Evolved

During the electrolysis, oxygen gas is produced at the anode. From the reaction at the anode, we know that for every 4 moles of electrons, 1 mole of O₂ is produced. Since we calculated that 0.4 moles of electrons were used, the moles of O₂ produced can be calculated as follows:

• Moles of O₂ = 0.4 moles of electrons / 4 = 0.1 moles of O₂.

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of oxygen gas evolved is:

Volume of O₂ = moles of O₂ × volume per mole = 0.1 moles × 22.4 L/mol = 2.24 L.

Final Results

In summary, during the electrolysis of 1 L of CuSO4 solution until the pH reaches 1, the amount of copper deposited is approximately 25.4 grams, and the volume of oxygen gas evolved is about 2.24 liters. This process illustrates the principles of electrochemistry and stoichiometry in a practical context.