The equivalent constant of the reaction:
`Cu(s)+2Ag^(+)(aq.) rarr Cu^(2+)(aq.)+2Ag(s)`
`E^(@)=0.46 V` at `298 K`,is:

Options

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

Answer

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 \( E^0 = 0.46 \, \text{V} \) at \( 298 \, \text{K} \), we can use the Nernst equation. Here are the steps to solve the problem:

### Step 1: Write the Nernst Equation
The Nernst equation relates the cell potential to the equilibrium constant:

\[ E_{cell} = E^0 - \frac{2.303RT}{nF} \log K_{equilibrium} \]

### Step 2: Set Up the Equation at Equilibrium
At equilibrium, the cell potential \( E_{cell} \) becomes 0. Thus, we can rewrite the equation as:

\[ 0 = E^0 - \frac{2.303RT}{nF} \log K_{equilibrium} \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:

\[ E^0 = \frac{2.303RT}{nF} \log K_{equilibrium} \]

### Step 4: Substitute Known Values
We know:
- \( E^0 = 0.46 \, \text{V} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( F = 96485 \, \text{C/mol} \)
- \( n = 2 \) (since 2 moles of electrons are transferred in the reaction)

Substituting these values into the equation:

\[ 0.46 = \frac{2.303 \times 8.314 \times 298}{2 \times 96485} \log K_{equilibrium} \]

### Step 5: Calculate the Coefficient
Calculating the coefficient:

\[ \frac{2.303 \times 8.314 \times 298}{2 \times 96485} \approx 0.0591 \]

### Step 6: Substitute Back into the Equation
Now we have:

\[ 0.46 = 0.0591 \log K_{equilibrium} \]

### Step 7: Solve for \( \log K_{equilibrium} \)
Rearranging gives:

\[ \log K_{equilibrium} = \frac{0.46}{0.0591} \]

Calculating this:

\[ \log K_{equilibrium} \approx 7.79 \]

### Step 8: Find \( K_{equilibrium} \)
To find \( K_{equilibrium} \), we take the antilogarithm:

\[ K_{equilibrium} = 10^{7.79} \]

### Step 9: Calculate the Final Value
Calculating \( 10^{7.79} \):

\[ K_{equilibrium} \approx 6.17 \times 10^7 \]

### Final Answer
Thus, the equilibrium constant \( K \) for the reaction is approximately:

\[ K \approx 6.17 \times 10^7 \]