A huydrgen electrode is dipped is a solution of `pH = 3.0` at ` 25^@C` The potential fo the cell will be .

Options

A. `(0.177V)`
B. `(0.087)`
C. `(0.059V)`
D. `(0.177V)`

Explanation

To find the potential of a hydrogen electrode dipped in a solution with a pH of 3.0 at 25°C, we can follow these steps:

### Step 1: Understand the relationship between pH and [H⁺]
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
Given that pH = 3.0, we can find the concentration of hydrogen ions \([H^+]\):
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Write the Nernst equation
The Nernst equation for the hydrogen electrode is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[H_2]^{1/2}}{[H^+]} \right) \]
Where:
- \(E^\circ_{\text{cell}}\) is the standard electrode potential (0.00 V for the hydrogen electrode).
- \(n\) is the number of electrons transferred (1 for the hydrogen electrode).
- \([H_2]\) is the concentration of hydrogen gas, which we can assume to be 1 atm (or 1 M for simplicity).

### Step 3: Substitute values into the Nernst equation
Since \([H_2] = 1 \, \text{M}\) and \([H^+] = 10^{-3} \, \text{M}\), we can substitute these values into the equation:
\[ E_{\text{cell}} = 0.00 - \frac{0.0591}{1} \log \left( \frac{1^{1/2}}{10^{-3}} \right) \]

### Step 4: Simplify the logarithmic term
The logarithmic term simplifies as follows:
\[ \log \left( \frac{1^{1/2}}{10^{-3}} \right) = \log(1) - \log(10^{-3}) = 0 - (-3) = 3 \]

### Step 5: Calculate the cell potential
Now substituting back into the Nernst equation:
\[ E_{\text{cell}} = 0.00 - 0.0591 \times 3 \]
\[ E_{\text{cell}} = 0.00 - 0.1773 = -0.1773 \, \text{V} \]

### Final Answer
Thus, the potential of the cell is approximately:
\[ E_{\text{cell}} \approx -0.177 \, \text{V} \]

---