To solve the problem step by step, we will follow these calculations:
### Step 1: Calculate the moles of silver deposited
Given:
- Mass of silver (Ag) = 108 g
- Molar mass of silver (Ag) = 108 g/mol
Using the formula for moles:
\[
\text{Moles of Ag} = \frac{\text{Mass of Ag}}{\text{Molar mass of Ag}} = \frac{108 \, \text{g}}{108 \, \text{g/mol}} = 1 \, \text{mol}
\]
### Step 2: Determine the charge required to deposit silver
The reaction for the deposition of silver is:
\[
\text{Ag}^+ + e^- \rightarrow \text{Ag}
\]
From this reaction, we see that 1 mole of silver requires 1 mole of electrons (1 Faraday, F).
### Step 3: Relate the charge to the production of oxygen gas
The reaction for the electrolysis of water is:
\[
2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4 e^-
\]
From this reaction, we can see that:
- 4 moles of electrons produce 1 mole of oxygen gas (O₂).
- Therefore, 1 mole of electrons will produce \(\frac{1}{4}\) mole of O₂.
### Step 4: Calculate the moles of oxygen produced
Since we have 1 mole of electrons (from the deposition of 1 mole of Ag), the moles of O₂ produced will be:
\[
\text{Moles of O}_2 = \frac{1 \, \text{mol of electrons}}{4} = \frac{1}{4} \, \text{mol}
\]
### Step 5: Use the ideal gas law to find the volume of oxygen
Using the ideal gas law:
\[
PV = nRT
\]
Where:
- \(P = 1 \, \text{bar} = 1 \, \text{atm}\) (since 1 bar ≈ 1 atm)
- \(n = \frac{1}{4} \, \text{mol}\)
- \(R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
- \(T = 273 \, \text{K}\)
Substituting the values into the ideal gas equation:
\[
V = \frac{nRT}{P} = \frac{\left(\frac{1}{4} \, \text{mol}\right) \times (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (273 \, \text{K})}{1 \, \text{atm}}
\]
Calculating this gives:
\[
V = \frac{(0.25) \times (0.0821) \times (273)}{1} = 5.675 \, \text{L}
\]
### Final Answer
The volume of oxygen gas produced at 273 K and 1 bar pressure is **5.675 L**.
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