Calculate the reduction potential for the following half cell reaction at 298 K.
`Ag^(+)(aq)+e^(-)toAg(s)`
`"Given that" [Ag^(+)]=0.1 M and E^(@)=+0.80 V`

Options

A. 0.741 V
B. 0.80 V
C. `-0.80`V
D. `-0.741`V

Explanation

To calculate the reduction potential for the half-cell reaction \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \) at 298 K, we will use the Nernst equation.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Standard reduction potential, \( E^\circ = +0.80 \, \text{V} \)
- Concentration of \( \text{Ag}^+ \), \( [\text{Ag}^+] = 0.1 \, \text{M} \)
- Number of electrons transferred, \( n = 1 \)

2. **Write the Nernst Equation:**
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where \( Q \) is the reaction quotient.

3. **Determine the Reaction Quotient \( Q \):**
For the reaction \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \):
- The concentration of the solid \( \text{Ag}(s) \) is considered to be 1 M (as solids are not included in the equilibrium expression).
- Therefore, the reaction quotient \( Q \) is:
\[ Q = \frac{[\text{Ag}(s)]}{[\text{Ag}^+]} = \frac{1}{0.1} = 10 \]

4. **Substitute Values into the Nernst Equation:**
Now, substitute \( E^\circ \), \( n \), and \( Q \) into the Nernst equation:
\[ E = 0.80 \, \text{V} - \frac{0.0591}{1} \log(10) \]

5. **Calculate \( \log(10) \):**
We know that \( \log(10) = 1 \). Thus, the equation simplifies to:
\[ E = 0.80 \, \text{V} - 0.0591 \times 1 \]

6. **Perform the Final Calculation:**
\[ E = 0.80 \, \text{V} - 0.0591 \, \text{V} = 0.7409 \, \text{V} \]
Rounding this, we get:
\[ E \approx 0.741 \, \text{V} \]

### Final Answer:
The reduction potential for the half-cell reaction at 298 K is approximately \( 0.741 \, \text{V} \).