The cell `Pt, H_(2) (1atm) H^(+) (pH =x)|` Normal calomel Electrode has an EMF of `0.67V` at `25^(@)C` .Calculate the pH of the solution. The oxidation potential of the calomel electrode on hydrogen scale is `-0.28V`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Cell Representation
The cell is represented as:
\[ \text{Pt, H}_2 (1 \text{ atm}) | \text{H}^+ (pH = x) | \text{Normal Calomel Electrode} \]

### Step 2: Identify Given Data
- EMF of the cell, \( E_{\text{cell}} = 0.67 \, \text{V} \)
- Oxidation potential of the calomel electrode on the hydrogen scale, \( E^{\circ}_{\text{ox}} = -0.28 \, \text{V} \)

### Step 3: Calculate the Reduction Potential of the Calomel Electrode
The reduction potential of the calomel electrode can be calculated as:
\[ E^{\circ}_{\text{red}} = -E^{\circ}_{\text{ox}} = -(-0.28) = 0.28 \, \text{V} \]

### Step 4: Write the Nernst Equation
The Nernst equation for the cell can be expressed as:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{H}^+]^2}{1} \right) \]
Where:
- \( n = 1 \) (number of electrons transferred)

### Step 5: Calculate the Standard Cell Potential
The standard cell potential \( E^{\circ}_{\text{cell}} \) is given by:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{red}} + E^{\circ}_{\text{ox}} = 0.28 + 0 = 0.28 \, \text{V} \]

### Step 6: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ 0.67 = 0.28 - \frac{0.0591}{1} \log \left( [\text{H}^+]^2 \right) \]

### Step 7: Rearranging the Equation
Rearranging the equation to isolate the logarithm term:
\[ 0.67 - 0.28 = -0.0591 \log \left( [\text{H}^+]^2 \right) \]
\[ 0.39 = -0.0591 \log \left( [\text{H}^+]^2 \right) \]

### Step 8: Solve for the Logarithm
Dividing both sides by -0.0591:
\[ \log \left( [\text{H}^+]^2 \right) = \frac{0.39}{-0.0591} \]
Calculating the right side:
\[ \log \left( [\text{H}^+]^2 \right) \approx -6.60 \]

### Step 9: Solve for \([H^+]\)
Using the property of logarithms:
\[ \log \left( [\text{H}^+] \right) = \frac{-6.60}{2} = -3.30 \]
Thus,
\[ [\text{H}^+] = 10^{-3.30} \approx 5.01 \times 10^{-4} \, \text{M} \]

### Step 10: Calculate the pH
Using the definition of pH:
\[ \text{pH} = -\log [\text{H}^+] = -\log (5.01 \times 10^{-4}) \approx 3.30 \]

### Final Answer
The pH of the solution is approximately **3.30**.

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