The hydrogen electrode is dpped in a solution of pH=3 at `25^(@)C`. The potential of the cell would be (the value of 2.303RT/F is 0.059V)

Options

A. 0.177V
B. `-0.177V`
C. `0.087V`
D. `0.059V`

Explanation

To find the potential of the hydrogen electrode in a solution with a pH of 3 at 25°C, we can use the Nernst equation. The Nernst equation for the hydrogen electrode can be expressed as:

\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{1}{[H^+]} \right) \]

Where:
- \( E \) is the cell potential.
- \( E^0 \) is the standard electrode potential (for the hydrogen electrode, \( E^0 = 0.00 \, V \)).
- \( n \) is the number of electrons transferred in the half-reaction (for hydrogen, \( n = 2 \)).
- \([H^+]\) is the concentration of hydrogen ions.

### Step-by-Step Solution:

1. **Determine the concentration of \( [H^+] \)**:
- The pH of the solution is given as 3.
- The concentration of hydrogen ions can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

2. **Substitute \( [H^+] \) into the Nernst equation**:
- Since \( n = 2 \) for the hydrogen electrode, we can rewrite the Nernst equation:
\[ E = E^0 - \frac{0.0591}{2} \log \left( \frac{1}{10^{-3}} \right) \]

3. **Calculate the logarithm**:
- The logarithm can be simplified:
\[ \log \left( \frac{1}{10^{-3}} \right) = \log(10^3) = 3 \]

4. **Substitute the values into the equation**:
- Now substituting back into the Nernst equation:
\[ E = 0.00 - \frac{0.0591}{2} \times 3 \]

5. **Calculate the potential**:
- Now perform the multiplication:
\[ E = 0.00 - \frac{0.0591 \times 3}{2} \]
\[ E = 0.00 - \frac{0.1773}{2} \]
\[ E = 0.00 - 0.08865 \]
\[ E = -0.08865 \, V \]

6. **Final result**:
- The potential of the hydrogen electrode in a solution of pH 3 at 25°C is approximately:
\[ E \approx -0.0887 \, V \]