An electric current is passed through electrolytic cells in series one containing `Ag(NO_(3))`(eq) and other `H_(2)SO_(4)` (aq). What volume of `O_(2)` measured at `25^(@)C` and `750`mm Hg pressure would be liberated form `H_(2)SO_(4)` if
(a) one mole of `Ag^(+)` is deposited from `AgNO_(3)` solution
(b) `8 xx 10^(22)` ions of `Ag^(+)` are deposited from `AgNO_(3)` solution.

Explanation

To solve this problem, we need to analyze the electrochemical reactions occurring in the electrolytic cells and apply the ideal gas law to find the volume of oxygen produced.

### Step-by-Step Solution:

**Step 1: Understand the Electrochemical Reactions**

In the electrolytic cell containing `AgNO3`, silver ions (`Ag^+`) are reduced at the cathode:
\[ Ag^+ + e^- \rightarrow Ag \]

At the anode, water is oxidized to produce oxygen:
\[ 2H_2O \rightarrow 4H^+ + O_2 + 4e^- \]

**Step 2: Determine the Relationship Between Silver and Oxygen Production**

From the balanced reactions, we see that:
- 4 moles of `Ag^+` produce 1 mole of `O_2`.
- Therefore, 1 mole of `Ag^+` produces \( \frac{1}{4} \) mole of `O_2`.

**Step 3: Calculate Volume of `O_2` for Part (a)**

For part (a), we are given that 1 mole of `Ag^+` is deposited:
- Moles of `O_2` produced = \( \frac{1}{4} \) moles.

Using the ideal gas law \( PV = nRT \):
- Rearranging gives us \( V = \frac{nRT}{P} \).

Substituting the values:
- \( n = \frac{1}{4} \) moles,
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \),
- \( T = 25^\circ C = 298 \, K \),
- \( P = 750 \, \text{mm Hg} = \frac{750}{760} \, \text{atm} \approx 0.9868 \, \text{atm} \).

Now, calculate the volume:
\[ V = \frac{\left(\frac{1}{4}\right) \times 0.0821 \times 298}{0.9868} \]
\[ V \approx \frac{6.116}{0.9868} \approx 6.20 \, \text{L} \]

**Step 4: Calculate Volume of `O_2` for Part (b)**

For part (b), we are given \( 8 \times 10^{22} \) ions of `Ag^+`:
- First, convert ions to moles:
\[ \text{Moles of } Ag^+ = \frac{8 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.133 \, \text{moles} \]

Now, calculate moles of `O_2` produced:
\[ \text{Moles of } O_2 = \frac{0.133}{4} \approx 0.03325 \, \text{moles} \]

Using the ideal gas law again:
\[ V = \frac{0.03325 \times 0.0821 \times 298}{0.9868} \]
\[ V \approx \frac{0.815}{0.9868} \approx 0.826 \, \text{L} \]

### Final Answers:
(a) Volume of `O_2` liberated = **6.20 L**
(b) Volume of `O_2` liberated = **0.826 L**