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Step-by-Step Solution
Step 1: Identify the Relevant Half-Reactions
The question involves electrolysis of AgNO3 solution at the cathode and electrolysis of water at the anode.
Cathode half-reaction:
\mathrm{Ag^+ (aq) + e^- \to Ag (s)}
Anode half-reaction (for water):
2\,\mathrm{H_2O} \,\to\, \mathrm{O_2} + 4\,\mathrm{H^+} + 4\,\mathrm{e^-}
Step 2: Calculate Moles of Silver Deposited
Given that 108 g of silver (molar mass = 108 g/mol) is deposited, the moles of Ag deposited are:
n_{\mathrm{Ag}} = \dfrac{108 \text{ g}}{108 \,\text{g/mol}} = 1 \text{ mol}
Step 3: Determine Moles of Electrons Used
From the cathode reaction, depositing 1 mole of Ag requires 1 mole of electrons:
1\,\mathrm{Ag^+} + 1\,\mathrm{e^-} \to 1\,\mathrm{Ag}
Hence, 1 mole of \mathrm{e^-} is consumed to deposit 1 mole of Ag.
Step 4: Use Electron Count to Find Moles of O2 Produced
The oxidation of water to O2 at the anode is:
2\,\mathrm{H_2O} \to \mathrm{O_2} + 4\,\mathrm{H^+} + 4\,\mathrm{e^-}
According to this reaction, producing 1 mole of O2 requires 4 moles of electrons. We have only 1 mole of electrons available (the same total that deposited 1 mole of Ag). Hence:
1 \text{ mole of } \mathrm{e^-} \Rightarrow \dfrac{1}{4} \text{ mole of O}_2
Step 5: Calculate the Volume of O2 at STP
At 273 K and 1 bar (approximately STP), 1 mole of any ideal gas occupies 22.4 L. Therefore,
n_{\mathrm{O_2}} = \dfrac{1}{4} \text{ mol} \quad \Rightarrow \quad V = \dfrac{1}{4} \times 22.4 \text{ L}
V = 5.6\,\text{L}
Final Answer
Therefore, the volume of O2 produced under the given conditions is
5.6 L.
