The cell `Pt|H_(2)(g,01` bar) `|H^(+)(aq),pH=x||Cl^(-)(1M)|Hg_(2)Cl_(2)|Hg|Pt` has emf of 0.5755 V at `25^(@)C` the SOP of calomel electrode is `-0.28V` then pH of the solution will be
(a)11
(b)4.5
(c)5
(d)None of these

Options

A. 11
B. 4.5
C. 5
D. None of these

Answer

To solve the problem, we need to find the pH of the solution given the cell and its EMF. Let's break down the steps to find the solution.

### Step-by-Step Solution:

1. **Identify the Components of the Cell**:
The cell is given as:
\[ \text{Pt} | \text{H}_2(g, 1 \text{ bar}) | \text{H}^+(aq), pH = x || \text{Cl}^-(1M) | \text{Hg}_2\text{Cl}_2 | \text{Hg} | \text{Pt} \]
Here, the left side is the hydrogen electrode and the right side is the calomel electrode.

2. **Given Data**:
- EMF of the cell, \( E_{\text{cell}} = 0.5755 \, \text{V} \)
- Standard oxidation potential of the calomel electrode, \( E^{\circ}_{\text{calomel}} = -0.28 \, \text{V} \)

3. **Calculate the Standard Electrode Potential of the Hydrogen Electrode**:
The EMF of the cell can be expressed as:
\[ E_{\text{cell}} = E^{\circ}_{\text{H}_2} - E^{\circ}_{\text{calomel}} \]
Rearranging gives:
\[ E^{\circ}_{\text{H}_2} = E_{\text{cell}} + E^{\circ}_{\text{calomel}} \]
Substituting the values:
\[ E^{\circ}_{\text{H}_2} = 0.5755 \, \text{V} + 0.28 \, \text{V} = 0.8555 \, \text{V} \]

4. **Apply the Nernst Equation**:
The Nernst equation for the hydrogen electrode is:
\[ E = E^{\circ} - \frac{0.059}{n} \log \left( \frac{1}{[\text{H}^+]} \right) \]
Here, \( n = 2 \) (for the reaction \( \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \)).
Thus, the equation becomes:
\[ E_{\text{H}_2} = E^{\circ}_{\text{H}_2} - \frac{0.059}{2} \log \left( \frac{1}{[\text{H}^+]} \right) \]

5. **Substituting Known Values**:
We know \( E_{\text{H}_2} = 0.8555 \, \text{V} \) and \( E_{\text{cell}} = 0.5755 \, \text{V} \):
\[ 0.5755 = 0.8555 - \frac{0.059}{2} \log \left( \frac{1}{[\text{H}^+]} \right) \]

6. **Rearranging to Find \([\text{H}^+]\)**:
Rearranging gives:
\[ \frac{0.059}{2} \log \left( \frac{1}{[\text{H}^+]} \right) = 0.8555 - 0.5755 \]
\[ \frac{0.059}{2} \log \left( \frac{1}{[\text{H}^+]} \right) = 0.28 \]
\[ \log \left( \frac{1}{[\text{H}^+]} \right) = \frac{0.28 \times 2}{0.059} \]
\[ \log \left( \frac{1}{[\text{H}^+]} \right) = 9.49 \]
Therefore,
\[ [\text{H}^+] = 10^{-9.49} \approx 3.24 \times 10^{-10} \, \text{M} \]

7. **Calculating pH**:
The pH is calculated as:
\[ \text{pH} = -\log [\text{H}^+] = -\log (3.24 \times 10^{-10}) \approx 9.49 \]

### Conclusion:
The calculated pH value does not match any of the provided options. Therefore, the answer is:
**(d) None of these**