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

Answer

To find the potential of a hydrogen electrode dipped in a solution of pH 3 at 25°C, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Identify the Nernst Equation
The Nernst equation for the half-cell reaction of the hydrogen electrode can be expressed as:

\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{P_{\text{H}_2}^{1/2}}{[H^+]}\right) \]

Where:
- \( E_{\text{cell}} \) is the cell potential.
- \( E^\circ_{\text{cell}} \) is the standard reduction potential (for hydrogen, this is 0 V).
- \( n \) is the number of electrons transferred (for hydrogen, \( n = 1 \)).
- \( P_{\text{H}_2} \) is the pressure of hydrogen gas (assumed to be 1 atm).
- \( [H^+] \) is the concentration of hydrogen ions.

### Step 2: Determine the Concentration of Hydrogen Ions
Given that the pH of the solution is 3.0, we can calculate the concentration of hydrogen ions:

\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 3: Substitute Values into the Nernst Equation
Now we can substitute the known values into the Nernst equation:

\[ E_{\text{cell}} = 0 - \frac{0.059}{1} \log \left( \frac{1^{1/2}}{10^{-3}} \right) \]

### Step 4: Simplify the Logarithmic Expression
This simplifies to:

\[ E_{\text{cell}} = -0.059 \log \left( \frac{1}{10^{-3}} \right) \]

\[ E_{\text{cell}} = -0.059 \log (10^{3}) \]

Using the property of logarithms:

\[ \log (10^{3}) = 3 \]

So we have:

\[ E_{\text{cell}} = -0.059 \times 3 \]

### Step 5: Calculate the Final Potential
Calculating this gives:

\[ E_{\text{cell}} = -0.177 \, \text{V} \]

### Conclusion
The potential of the hydrogen electrode in a solution of pH 3 at 25°C is \( -0.177 \, \text{V} \).

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