Final answer:
The volume of oxygen gas liberated during the electrolysis when 1.27g of copper is deposited is approximately 0.224 liters at STP, calculated using stoichiometry and molar volume of gas at STP.
Step-by-step explanation:
The question deals with the amount of oxygen gas liberated during the electrolytic process involving solutions of CuSO₄ and dilute H₂SO₄ when 1.27g of copper was deposited. To calculate the volume of oxygen liberated at STP, we need to use the stoichiometry of the electrochemical reaction involved in the electrolysis process.
Here's the basic approach to solving the problem:
- Determine the number of moles of copper deposited using the molar mass of copper (63.55 g/mol).
- Since the deposition of copper occurs at the cathode, for each mole of copper deposited, there will be half a mole of oxygen gas evolved at the anode (according to the balanced equation for the electrolysis of water).
- Use Avogadro's law to convert moles of oxygen to volume at STP (Standard Temperature and Pressure), remembering that 1 mole of any gas at STP occupies 22.4 liters.
Molar mass of Cu = 63.55 g/mol
Moles of Cu deposited = 1.27 g / 63.55 g/mol = 0.02 moles (rounded to two significant figures)
Moles of O₂ liberated = 0.02 moles Cu × (1/2 moles O₂/moles Cu) = 0.01 moles O₂
Volume of O₂ at STP = 0.01 moles × 22.4 L/mol = 0.224 L
Therefore, the volume of oxygen gas liberated during the electrolysis when 1.27 g of copper is deposited is approximately 0.224 liters at STP.