A concentration cell `Pt|H_2(g)|HCl(aq)||H_2SO_(4)(aq)|H_2(g)|Pt` is constructed using equal concentration of acids and equal number of moles of `H_2` gas in both the compartments at the same temperature.If volume of `H_2` gas at the anodic compartment is `1/9` times the volume of `H_2` gas at cathodic compartment Which of the following is/are correct for the given cell `(log 2=0.3, log 3=0.48)(2.303 RT)/F=0.06 V`

Options

A. Pressure of `H_2` gas in both the cell compartment are equal
B. Concentration of `H^(+)` ion in both the cell compartment are unequal
C. `E_(cell)^(@)=0` for the above cell
D. `E_(cell)ne0` for the above cell

Explanation

To solve the given problem regarding the concentration cell `Pt|H_2(g)|HCl(aq)||H_2SO_(4)(aq)|H_2(g)|Pt`, we need to analyze the information provided and apply the Nernst equation.

### Step-by-step Solution:

1. **Identify the Components of the Cell**:
- The cell consists of two compartments:
- Anodic compartment: `H_2(g)` in `HCl(aq)`
- Cathodic compartment: `H_2(g)` in `H_2SO_4(aq)`

2. **Understand the Given Conditions**:
- Equal concentration of acids in both compartments.
- Equal number of moles of `H_2` gas in both compartments.
- The volume of `H_2` gas at the anodic compartment is `1/9` times the volume of `H_2` gas at the cathodic compartment.

3. **Relate Volume and Pressure**:
- Since the number of moles is equal and the volume of gas at the anode is `1/9` of that at the cathode, we can use the ideal gas law to relate pressure and volume:
\[ P \propto \frac{1}{V} \]
- Therefore, if \( V_{anode} = \frac{1}{9} V_{cathode} \), then:
\[ P_{anode} = 9 P_{cathode} \]

4. **Determine Concentrations of H⁺ Ions**:
- For `HCl`, the concentration of `H⁺` ions is equal to the concentration of `HCl`, denoted as \( C \).
- For `H2SO4`, which is dibasic, the concentration of `H⁺` ions will be \( 2C \) (since it produces two `H⁺` ions per molecule).

5. **Set Up the Nernst Equation**:
- The Nernst equation for the cell can be written as:
\[ E_{cell} = E^0_{cell} - \frac{0.06}{n} \log \left( \frac{[H^+]_{anode} \cdot P_{H2, cathode}^{1/2}}{[H^+]_{cathode} \cdot P_{H2, anode}^{1/2}} \right) \]
- Here, \( E^0_{cell} = 0 \) (standard potential for hydrogen), and \( n = 1 \) (one electron is involved).

6. **Substituting Values into the Nernst Equation**:
- Substitute the concentrations and pressures:
\[ E_{cell} = 0 - 0.06 \log \left( \frac{C \cdot (P_{cathode})^{1/2}}{(2C) \cdot (P_{anode})^{1/2}} \right) \]
- Since \( P_{anode} = 9 P_{cathode} \):
\[ E_{cell} = -0.06 \log \left( \frac{C \cdot (P_{cathode})^{1/2}}{(2C) \cdot (9 P_{cathode})^{1/2}} \right) \]
- Simplifying gives:
\[ E_{cell} = -0.06 \log \left( \frac{1}{18} \right) = -0.06 \log \left( \frac{1}{2} \cdot \frac{1}{9} \right) = -0.06 \left( \log \frac{1}{2} + \log \frac{1}{9} \right) \]
- Using \( \log 2 = 0.3 \) and \( \log 3 = 0.48 \):
\[ E_{cell} = -0.06 \left( -0.3 - 0.96 \right) = -0.06 \times (-1.26) = 0.0756 \text{ V} \approx 0.076 \text{ V} \]

7. **Conclusion**:
- Based on the analysis:
- The pressure of hydrogen gas in both compartments is **not equal**.
- The concentration of `H⁺` ions in both compartments is **unequal**.
- \( E^0_{cell} = 0 \).
- \( E_{cell} \) is **not equal to 0**.

### Final Answer:
- Correct statements are:
1. Concentration of H⁺ ions in both compartments are unequal.
2. \( E^0_{cell} = 0 \).
3. \( E_{cell} \) is not equal to 0.