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. `6.6xx10^(50)`

Explanation

To find the equilibrium constant \( K_{eq} \) for the reaction

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

with a given standard cell potential \( E^\circ = 1.50 \, \text{V} \) at 298 K, we can use the Nernst equation and the relationship between the standard cell potential and the equilibrium constant.

### Step-by-Step Solution:

1. **Identify the reaction and the number of electrons transferred (N)**:
- The reaction involves the oxidation of Zn to Zn²⁺ and the reduction of Ag⁺ to Ag.
- Each Zn atom loses 2 electrons, so \( N = 2 \).

2. **Use the Nernst equation at equilibrium**:
- At equilibrium, the cell potential \( E_{cell} \) is 0 V. The Nernst equation simplifies to:
\[ 0 = E^\circ - \frac{0.0591}{N} \log K_{eq} \]
Rearranging gives:
\[ E^\circ = \frac{0.0591}{N} \log K_{eq} \]

3. **Substitute the known values**:
- Substitute \( E^\circ = 1.50 \, \text{V} \) and \( N = 2 \):
\[ 1.50 = \frac{0.0591}{2} \log K_{eq} \]

4. **Solve for \( \log K_{eq} \)**:
- Multiply both sides by 2:
\[ 3.00 = 0.0591 \log K_{eq} \]
- Divide by 0.0591:
\[ \log K_{eq} = \frac{3.00}{0.0591} \approx 50.8 \]

5. **Calculate \( K_{eq} \)**:
- Convert from logarithmic form to exponential form:
\[ K_{eq} = 10^{50.8} \]

6. **Final Calculation**:
- Using a calculator, we find:
\[ K_{eq} \approx 6.31 \times 10^{50} \]

### Final Answer:
\[ K_{eq} \approx 6.31 \times 10^{50} \]