Calculate equilibrium constant for the reaction at `25^(@)C`
`Cu(s)+2Ag^(+)(aq) hArr Cu^(2+)(aq)+2Ag(s)`
`E^(@)` value of the cell is 0.46`" V "`.

Explanation

To calculate the equilibrium constant \( K \) for the reaction

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

at \( 25^\circ C \) given that the standard cell potential \( E^\circ \) is \( 0.46 \, V \), we can follow these steps:

### Step 1: Use the relationship between Gibbs free energy and cell potential
The relationship between the Gibbs free energy change (\( \Delta G^\circ \)) and the standard cell potential (\( E^\circ \)) is given by:

\[ \Delta G^\circ = -nFE^\circ \]

where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (\( 96500 \, C/mol \))
- \( E^\circ \) = standard cell potential

### Step 2: Determine the number of moles of electrons transferred
In the given reaction, 2 moles of electrons are transferred (from 2 Ag\(^+\) to Cu). Therefore, \( n = 2 \).

### Step 3: Substitute values into the Gibbs free energy equation
Now, substituting the values into the equation:

\[ \Delta G^\circ = -2 \times 96500 \, C/mol \times 0.46 \, V \]

Calculating this gives:

\[ \Delta G^\circ = -2 \times 96500 \times 0.46 = -88940 \, J/mol \]

### Step 4: Use the relationship between Gibbs free energy and equilibrium constant
The relationship between Gibbs free energy change and the equilibrium constant (\( K \)) is given by:

\[ \Delta G^\circ = -RT \ln K \]

where:
- \( R \) = gas constant (\( 8.314 \, J/(K \cdot mol) \))
- \( T \) = temperature in Kelvin (for \( 25^\circ C \), \( T = 298 \, K \))

### Step 5: Rearrange the equation to solve for \( K \)
We can rearrange the equation to solve for \( K \):

\[ \ln K = -\frac{\Delta G^\circ}{RT} \]

Substituting \( \Delta G^\circ \) into the equation:

\[ \ln K = -\frac{-88940}{8.314 \times 298} \]

Calculating the right-hand side:

\[ \ln K = \frac{88940}{2478.572} \approx 35.88 \]

### Step 6: Calculate \( K \) from \( \ln K \)
Now, to find \( K \):

\[ K = e^{35.88} \]

Calculating this gives:

\[ K \approx 3.63 \times 10^{15} \]

### Final Answer
The equilibrium constant \( K \) for the reaction at \( 25^\circ C \) is approximately:

\[ K \approx 3.63 \times 10^{15} \]

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