The equilibrium constent of the reaction.
`Cu(s)+2Ag(aq). Leftrightarrow Cu^(2+) (aq.)+2Ag(s)`
`E^(@)=0.46" V at 298 K is"`

Options

A. `2.0 xx 10^(10)`
B. `4.0 xx 10^(10)`
C. `4.0 xx 10^(15)`
D. `2.4 xx 10^(10)`

Explanation

To find the equilibrium constant (K_eq) for the reaction:

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

given that the standard cell potential (E°) is 0.46 V at 298 K, we can follow these steps:

### Step 1: Write the Nernst Equation
The Nernst equation relates the cell potential to the concentrations of the reactants and products:

\[ E = E^\circ - \frac{0.0591}{n} \log K_{eq} \]

### Step 2: Understand the Reaction
In the given reaction:
- Copper (Cu) is oxidized to copper ions (Cu²⁺).
- Silver ions (Ag⁺) are reduced to solid silver (Ag).
- The number of electrons (n) transferred in this reaction is 2.

### Step 3: Set Up the Equation at Equilibrium
At equilibrium, the cell potential (E) is 0. Therefore, we can set up the equation as follows:

\[ 0 = E^\circ - \frac{0.0591}{n} \log K_{eq} \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:

\[ 0 = 0.46 - \frac{0.0591}{2} \log K_{eq} \]

### Step 5: Rearrange the Equation
Rearranging the equation to solve for \(\log K_{eq}\):

\[ \frac{0.0591}{2} \log K_{eq} = 0.46 \]

\[ \log K_{eq} = \frac{0.46 \times 2}{0.0591} \]

### Step 6: Calculate \(\log K_{eq}\)
Calculating the right-hand side:

\[ \log K_{eq} = \frac{0.92}{0.0591} \approx 15.57 \]

### Step 7: Convert Logarithm to Exponential Form
To find \(K_{eq}\), we convert from logarithmic form:

\[ K_{eq} = 10^{15.57} \]

### Step 8: Calculate \(K_{eq}\)
Calculating \(K_{eq}\):

\[ K_{eq} \approx 3.68 \times 10^{15} \]

### Conclusion
Thus, the equilibrium constant \(K_{eq}\) for the reaction is approximately:

\[ K_{eq} \approx 3.68 \times 10^{15} \]