Which of the following represents the reduction potential of silver wire dipped into `0.1M AgNO` solution at `25^(@)C`:-

Options

A. `E_(red)^(@)`
B. `(E_(red)^(@) +0.059)`
C. `(E_(OX)^(@)-0.059)`
D. `(E_(red)^(@)-0.059)`

Explanation

To determine the reduction potential of a silver wire dipped into a 0.1 M AgNO₃ solution at 25°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reduction half-reaction for silver ions (Ag⁺) gaining electrons to form solid silver (Ag) is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

2. **Standard Reduction Potential (E⁰)**:
The standard reduction potential (E⁰) for the above reaction is +0.80 V. This value is typically found in standard electrochemical tables.

3. **Determine the Number of Electrons (N)**:
In this reaction, one electron is involved in the reduction of one silver ion. Therefore, N = 1.

4. **Use the Nernst Equation**:
The Nernst equation relates the cell potential (E) to the standard reduction potential (E⁰) and the concentration of the ions involved:
\[ E = E^0 - \frac{0.059}{N} \log \left(\frac{1}{[\text{Ag}^+]}\right) \]
Here, [Ag⁺] = 0.1 M.

5. **Substituting Values into the Nernst Equation**:
Plugging in the values, we have:
\[ E = 0.80 \, \text{V} - \frac{0.059}{1} \log \left(\frac{1}{0.1}\right) \]

6. **Calculating the Logarithm**:
\[ \log \left(\frac{1}{0.1}\right) = \log(10) = 1 \]

7. **Final Calculation**:
Now substituting back into the equation:
\[ E = 0.80 \, \text{V} - 0.059 \times 1 \]
\[ E = 0.80 \, \text{V} - 0.059 \, \text{V} = 0.741 \, \text{V} \]

8. **Conclusion**:
The reduction potential of the silver wire dipped into a 0.1 M AgNO₃ solution at 25°C is approximately 0.741 V.