Introduction
Electrolysis of copper(II) sulfate (CuSO4) using platinum electrodes involves the transfer of ions and the liberation of gases. When oxygen is liberated at the anode, it indicates the oxidation of water or sulfate ions.

Determining Moles of Oxygen
- The molar mass of O₂ is 32 g/mol.
- Given mass of O₂ liberated: 1.6 g.
- Moles of O₂ = Mass / Molar Mass
Moles of O₂ = 1.6 g / 32 g/mol = 0.05 mol

Faraday's Laws of Electrolysis
- According to Faraday’s first law, the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
- The reaction at the anode can be represented as:
2H₂O → O₂ + 4H⁺ + 4e⁻
- For every mole of O₂ produced, 4 moles of electrons are transferred.

Calculating Charge (Q)
- Total moles of electrons = 4 × moles of O₂
Total moles of electrons = 4 × 0.05 mol = 0.20 mol
- Total charge (Q) can be calculated using Faraday's constant (F = 96500 C/mol):
Q = moles of electrons × Faraday’s constant
Q = 0.20 mol × 96500 C/mol = 19300 C

Current and Deposition at Cathode
- The total current (I) passed is related to charge and time (t) by the equation:
I = Q/t
- At the cathode, copper ions (Cu²⁺) are reduced:
Cu²⁺ + 2e⁻ → Cu (s)
- For every 2 moles of electron, 1 mole of copper is deposited.

Conclusion
To summarize, 1.6 g of O₂ evolved at the anode correlates with 0.20 mol of electrons, leading to the deposition of copper at the cathode. The precise amount of copper deposited can be calculated from the total charge and the reaction stoichiometry.