The equilibrium constant (K) for the reaction
`Cu(s)+2Ag^(+) (aq) rarr Cu^(2+) (aq)+2Ag(s)`, will be
[Given, `E_(cell)^(@)=0.46 V`]

Options

A. `K_(c)=` antilog 15.6
B. `K_(c)=` antilog 2.5
C. `K_(c)=` antilog 1.5
D. `K_(c)=` antilog 12.2

Explanation

To find the equilibrium constant (K) for the reaction

\[ \text{Cu(s)} + 2\text{Ag}^+ (aq) \rightleftharpoons \text{Cu}^{2+} (aq) + 2\text{Ag(s)} \]

given that the standard cell potential \( E^\circ_{cell} = 0.46 \, \text{V} \), we can use the relationship between the Gibbs free energy change and the equilibrium constant.

### Step-by-Step Solution:

1. **Identify the Reaction and Variables:**
The reaction involves copper and silver ions. The number of electrons transferred in the reaction (n) is 2, as each silver ion gains one electron.

2. **Use the Gibbs Free Energy Equation:**
The relationship between Gibbs free energy change (\( \Delta G^\circ \)), the cell potential (\( E^\circ_{cell} \)), and the equilibrium constant (K) is given by:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
where:
- \( n = 2 \) (number of electrons transferred)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( E^\circ_{cell} = 0.46 \, \text{V} \)

3. **Calculate \( \Delta G^\circ \):**
Substitute the values into the equation:
\[ \Delta G^\circ = -2 \times 96500 \times 0.46 \]
Calculate \( \Delta G^\circ \):
\[ \Delta G^\circ = -2 \times 96500 \times 0.46 = -88940 \, \text{J/mol} \]

4. **Use the Relationship Between \( \Delta G^\circ \) and K:**
The relationship between \( \Delta G^\circ \) and the equilibrium constant K is given by:
\[ \Delta G^\circ = -2.303RT \log K \]
where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \) (assuming room temperature)

5. **Rearranging the Equation:**
We can rearrange this equation to solve for \( \log K \):
\[ \log K = \frac{-\Delta G^\circ}{2.303RT} \]

6. **Substituting Values:**
Substitute \( \Delta G^\circ \), \( R \), and \( T \) into the equation:
\[ \log K = \frac{88940}{2.303 \times 8.314 \times 298} \]

7. **Calculate \( \log K \):**
Calculate the denominator:
\[ 2.303 \times 8.314 \times 298 \approx 5730.58 \]
Now calculate \( \log K \):
\[ \log K \approx \frac{88940}{5730.58} \approx 15.5 \]

8. **Find K:**
To find K, we take the antilog:
\[ K = 10^{15.5} \approx 3.16 \times 10^{15} \]

### Final Answer:
The equilibrium constant \( K \) for the reaction is approximately \( 3.16 \times 10^{15} \).