Electricity is passed through an acidic solution of $$Cu^{2+}$$ till all the Cu2+ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is __ mL. (Nearest integer)
[Given:
$$Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s) E_{red}^{\circ}=+0.34V$$
$$O_{2}(g)+4H^{+}+4e^{-}\rightarrow 2H_{2}O E_{red}^{\circ}=+1.23V$$
Molar mass of Cu= 63.54 g $$mol^{-1}$$
Molar mass of $$O_{2}$$ = 32 g $$mol^{-1}$$
Farnday Constant = 96500 C $$mol^{-1}$$
Molar volume at STP = 22.4 L]

We need to find the total volume of oxygen evolved at STP during the entire electrolysis process.

We are given that 300 mg (0.3 g) of Cu is deposited, a current of 600 mA (0.6 A) is applied for 28 minutes, the molar mass of Cu is 63.54 g/mol, and the Faraday constant is 96500 C/mol.

We start by calculating the moles of Cu deposited:

$$\frac{0.3}{63.54} = 4.72 \times 10^{-3}$$ mol

Since each Cu²⁺ requires two electrons for reduction, the reaction is:

$$Cu^{2+} + 2e^- \rightarrow Cu$$

so the moles of electrons used are:

$$2 \times 4.72 \times 10^{-3} = 9.44 \times 10^{-3}$$ mol

At the anode during this phase, water is oxidized according to:

$$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$

so the moles of O₂ produced are:

$$\frac{9.44 \times 10^{-3}}{4} = 2.36 \times 10^{-3}$$ mol

After all Cu²⁺ is exhausted, only water electrolysis continues. The total charge passed is:

$$0.6 \times 28 \times 60 = 1008$$ C

which corresponds to:

$$\frac{1008}{96500} = 0.01044$$ mol of electrons

Using the same oxidation reaction at the anode, the moles of O₂ produced in this phase are:

$$\frac{0.01044}{4} = 2.61 \times 10^{-3}$$ mol

The total moles of O₂ evolved are the sum of both phases:

$$n_{O_2} = 2.36 \times 10^{-3} + 2.61 \times 10^{-3} = 4.97 \times 10^{-3}$$ mol

At STP, one mole of gas occupies 22400 mL, so the volume of O₂ is:

$$V = n \times 22400 = 4.97 \times 10^{-3} \times 22400 = 111.3 \text{ mL}$$

Therefore, rounding to the nearest integer gives a total volume of approximately 111 mL of O₂ at STP.