The equilibrium constant of the reaction :
`Zn(s)+2Ag^(+)(aq)toZn(aq)+2Ag(s),E^(@)=1.50V` at
298 K is

Options

A. `2.6xx10^(49)`
B. `8.7xx10^(51)`
C. `6.1xx10^(30)`
D. `5.37xx10^(50)`

Explanation

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

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

given that the standard cell potential \( E^\circ = 1.50 \, V \) at \( 298 \, K \), we can follow these steps:

### Step 1: Use the relationship between Gibbs free energy and cell potential
We know that the change in Gibbs free energy (\( \Delta G^\circ \)) is related to the standard cell potential (\( E^\circ \)) by the equation:

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

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: Identify \( n \)
In the given reaction, 2 electrons are transferred (from 2 \( \text{Ag}^+ \) ions to form 2 \( \text{Ag} \) atoms). Therefore, \( n = 2 \).

### Step 3: Substitute values into the equation
Now we can substitute the values into the equation:

\[ \Delta G^\circ = -2 \times 96500 \, C/mol \times 1.50 \, V \]

### Step 4: Calculate \( \Delta G^\circ \)
Calculating this gives:

\[ \Delta G^\circ = -2 \times 96500 \times 1.50 = -289500 \, J/mol \]

### Step 5: Use the relationship between Gibbs free energy and equilibrium constant
We also know that:

\[ \Delta G^\circ = -2.303RT \log K \]

where:
- \( R \) = universal gas constant (\( 8.314 \, J/(mol \cdot K) \))
- \( T \) = temperature in Kelvin (298 K)

### Step 6: Substitute values into the equation
Setting the two expressions for \( \Delta G^\circ \) equal to each other gives:

\[ -289500 = -2.303 \times 8.314 \times 298 \log K \]

### Step 7: Calculate the right-hand side
Calculating the right-hand side:

\[ 2.303 \times 8.314 \times 298 \approx 5730.5 \]

### Step 8: Solve for \( \log K \)
Now we can solve for \( \log K \):

\[ 289500 = 5730.5 \log K \]

\[ \log K = \frac{289500}{5730.5} \approx 50.5 \]

### Step 9: Calculate \( K \)
To find \( K \), we take the antilog:

\[ K = 10^{50.5} \approx 3.16 \times 10^{50} \]

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

\[ K \approx 3.16 \times 10^{50} \]