A hydrogen electrode is dipped in a solution at `25^(@)C`. The potential of cell is `-0.177 V`. Calcualte the `pH` of the solution.

Explanation

To calculate the pH of the solution in which a hydrogen electrode is dipped, we can use the Nernst equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[products]}{[reactants]} \right) \]
For the hydrogen electrode, the half-reaction is:
\[ \text{H}^+ + e^- \leftrightarrow \frac{1}{2} \text{H}_2 \]
Here, \( n = 1 \) (one electron is transferred).

### Step 2: Identify Given Values
From the problem, we have:
- The potential of the cell, \( E = -0.177 \, V \)
- The standard electrode potential, \( E^\circ = 0 \, V \) (for the standard hydrogen electrode)

### Step 3: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ -0.177 = 0 - \frac{0.059}{1} \log \left( \frac{1}{[\text{H}^+]} \right) \]
This simplifies to:
\[ -0.177 = -0.059 \log \left( \frac{1}{[\text{H}^+]} \right) \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ 0.177 = 0.059 \log \left( [\text{H}^+] \right) \]

### Step 5: Solve for \([\text{H}^+]\)
Dividing both sides by \( 0.059 \):
\[ \log \left( [\text{H}^+] \right) = \frac{0.177}{0.059} \]
Calculating the right side:
\[ \log \left( [\text{H}^+] \right) \approx 3 \]

### Step 6: Convert Logarithm to Concentration
To find \([\text{H}^+]\):
\[ [\text{H}^+] = 10^{-3} \, \text{mol/L} \]

### Step 7: Calculate pH
The pH is defined as:
\[ \text{pH} = -\log [\text{H}^+] \]
Substituting the concentration:
\[ \text{pH} = -\log(10^{-3}) = 3 \]

### Final Answer
The pH of the solution is **3**.

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