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 silver wire dipped into a 0.1 M AgNO₃ solution at 25°C, we will follow these steps:

### Step 1: Understand the Reduction Reaction
The reduction of silver ions (Ag⁺) from the AgNO₃ solution can be represented as:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag (s)} \]

### Step 2: Identify the Standard Reduction Potential
The standard reduction potential (E°) for the half-reaction of silver ions is typically given as +0.80 V. This value is essential for our calculations.

### Step 3: Apply the Nernst Equation
The Nernst equation relates the standard reduction potential to the actual reduction potential under non-standard conditions:
\[ E = E° - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

Where:
- \( E \) = cell potential
- \( E° \) = standard reduction potential
- \( n \) = number of electrons transferred (which is 1 for this reaction)
- \([\text{products}]\) = concentration of solid Ag (which is 1 for pure solids)
- \([\text{reactants}]\) = concentration of Ag⁺ (which is 0.1 M)

### Step 4: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ E = E° - \frac{0.0591}{1} \log \left( \frac{1}{0.1} \right) \]

### Step 5: Calculate the Logarithm
Calculate the logarithm:
\[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

### Step 6: Complete the Calculation
Now substitute this back into the equation:
\[ E = E° - 0.0591 \times 1 \]
\[ E = E° - 0.0591 \]

### Step 7: Final Expression
Thus, the reduction potential of the silver wire in the 0.1 M AgNO₃ solution at 25°C is:
\[ E = E° - 0.0591 \]

### Conclusion
The correct option that represents the reduction potential of silver wire dipped into a 0.1 M AgNO₃ solution at 25°C is:
\[ E = E° - 0.0591 \]