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 electrochemical cell, we will follow these steps:

### Step 1: Identify the Cell Components
The cell is given as:
\[ \text{Pt, } H_2(1 \text{ atm}) | H^+(xM) || H^+(1M) | H_2(1 \text{ atm}), \text{ Pt} \]
Here, the left side (anodic) has \( H^+(xM) \) and the right side (cathodic) has \( H^+(1M) \).

### Step 2: Write the Half-Reactions
At the anode, the half-reaction is:
\[ H_2(g) \rightarrow 2H^+ + 2e^- \]
At the cathode, the half-reaction is:
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]

### Step 3: Combine the Half-Reactions
The overall cell reaction can be written as:
\[ H_2(g) + 2H^+ \rightarrow H_2(g) + 2H^+ \]
This shows that \( H_2 \) is produced and consumed, and we need to focus on the concentration of \( H^+ \).

### Step 4: Write the Reaction Quotient (Q)
The reaction quotient \( Q \) for the cell can be expressed as:
\[ Q = \frac{[H^+]^2}{P_{H_2}} \]
Where \( P_{H_2} \) is the pressure of hydrogen gas.

### Step 5: Substitute Known Values
For the anodic compartment:
- \( [H^+] = x \) (unknown concentration)
- \( P_{H_2} = 1 \text{ atm} \)

For the cathodic compartment:
- \( [H^+] = 1 \text{ M} \)
- \( P_{H_2} = 1 \text{ atm} \)

Thus, the reaction quotient becomes:
\[ Q = \frac{x^2}{1} = x^2 \]

### Step 6: Apply the Nernst Equation
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log Q \]
For hydrogen electrodes, \( E^0_{cell} = 0 \) V. Given \( E_{cell} = 0.2364 \) V and \( n = 2 \):
\[ 0.2364 = 0 - \frac{0.0591}{2} \log(x^2) \]

### Step 7: Simplify the Equation
This simplifies to:
\[ 0.2364 = -0.02955 \log(x^2) \]
\[ 0.2364 = -0.0591 \log(x) \]

### Step 8: Solve for \( \log(x) \)
Rearranging gives:
\[ \log(x) = -\frac{0.2364}{0.0591} \]
Calculating this:
\[ \log(x) \approx -4 \]
Thus:
\[ x \approx 10^{-4} \text{ M} \]

### Step 9: Calculate pH
The pH is given by:
\[ \text{pH} = -\log[H^+] \]
Substituting \( [H^+] = x \):
\[ \text{pH} = -\log(10^{-4}) = 4 \]

### Final Answer
The pH of the anodic solution is:
\[ \text{pH} = 4 \]

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