Answer to Question #142016 in Chemistry for Kosalu

Question

In an electrolyte cell, solution of CuSO4 and dilute H2SO4 were electrolysed on series. If 1.27g of Cu was deposited, calculate the volume of oxygen liberated at S.T.P. (molar volume of gas at S.T.P.=22400cm³, O=16, Cu=63.5)

Answer 1

The positive anode reaction in the solution of CuSO₄ and H₂SO₄ is:

2H₂O $\rightarrow$ O₂ + 4H⁺ + 4e^-.

In this reaction, 2 water molecules are oxidized, 1 oxygen molecule is formed and 4 electrons are lost by 2 water molecules.

The negative cathode reaction is:

Cu²⁺ + 2e^- $\rightarrow$ Cu.

Here, 2 electrons are gained by copper ion and 2 copper atoms are formed.

As one can see, when one oxyhen molecule is formed, 2 copper ions are reduced. The number of the moles of copper is:

$n(Cu) = \frac{m(Cu)}{M(Cu)} = \frac{1.27\text{g}}{63.55 \text{g/mol}} = 0.020$ mol.

Therefore, the number of the moles of oxygen molecules formed is:

$n(O_2)=\frac{ n(Cu)}{2} = 0.010$ mol.

Finally, the volume of oxygen liberated is:

$V(O_2) = n\cdot V_m = 0.010\text{mol}\cdot 22400\text{cm}^3 = 447.6$ cm³, or 447.6 mL.

Answer: if 1.27g of Cu was deposited, the volume of oxygen liberated at S.T.P. is 447.6 mL.