pH of the anodic solution of the following cell is `Pt, H_2(1atm)|H^+(xM)||H^(+)(1M)|H_2(1 atm ),Pt` if `E_("cell") =0.2364V.`

Answer

To find the pH of the anodic solution in the given cell, we can follow these steps:

### Step 1: Identify the cell reactions
In the given cell, we have:
- Anode reaction: \( \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \)
- Cathode reaction: \( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \)

### Step 2: Write the expression for the cell potential
The cell potential (\(E_{cell}\)) can be expressed using the Nernst equation:
\[ E_{cell} = E^0_{cell} - \frac{0.059}{n} \log Q \]
where \(n\) is the number of moles of electrons transferred (which is 2 for hydrogen), and \(Q\) is the reaction quotient.

### Step 3: Determine the reaction quotient \(Q\)
For the given cell, the reaction quotient \(Q\) can be expressed as:
\[ Q = \frac{[\text{H}^+]^2}{P_{\text{H}_2}} \]
In this case, the pressure of hydrogen gas \(P_{\text{H}_2}\) is 1 atm for both sides of the cell. Therefore, we can simplify \(Q\) to:
\[ Q = [\text{H}^+]^2 \]

### Step 4: Substitute values into the Nernst equation
Given that \(E_{cell} = 0.2364 \, V\) and \(E^0_{cell} = 0 \, V\) (for hydrogen electrodes), we can substitute these values into the Nernst equation:
\[ 0.2364 = 0 - \frac{0.059}{2} \log Q \]
Substituting \(Q = [\text{H}^+]^2\):
\[ 0.2364 = -\frac{0.059}{2} \log([\text{H}^+]^2) \]

### Step 5: Simplify the equation
This simplifies to:
\[ 0.2364 = -0.0295 \log([\text{H}^+]^2) \]
Taking the logarithm out:
\[ 0.2364 = -0.059 \log([\text{H}^+]) \]

### Step 6: Solve for \([\text{H}^+]\)
Rearranging gives:
\[ \log([\text{H}^+]) = -\frac{0.2364}{0.059} \]
Calculating this gives:
\[ \log([\text{H}^+]) = -4 \]
Thus:
\[ [\text{H}^+] = 10^{-4} \, M \]

### Step 7: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log([\text{H}^+]) \]
Substituting in the concentration:
\[ \text{pH} = -\log(10^{-4}) = 4 \]

### Final Answer
The pH of the anodic solution is **4**.

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