`E^(@)` for the electrochemical cell
`Zn(s)|Zn^(2+) 1 M (Aq.)||Cu^(2+) 1 M (aq.)|Cu(s)`
is 1.10 V at `25^(@)C`. The equilibrium constant for the cell reaction,
`Zn(s) +Cu^(2+) (aq.) hArr Zn^(2+) (aq.)+Cu(s)`
Will be :

Options

A. (a) `10^(37)`
B. (b) `10^(-28)`
C. (c) `10^(18)`
D. (d) `10^(17)`

Explanation

To find the equilibrium constant (K) for the reaction given the standard cell potential (E°), we can use the relationship between the cell potential and the equilibrium constant. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data**:
The reaction is:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{Cu(s)} \]
The standard cell potential (E°) is given as 1.10 V at 25°C.

2. **Determine the Number of Electrons Transferred (n)**:
In this reaction, zinc (Zn) is oxidized to Zn²⁺, losing 2 electrons, while Cu²⁺ is reduced to Cu, gaining 2 electrons. Therefore, the number of electrons transferred (n) is:
\[ n = 2 \]

3. **Use the Nernst Equation**:
The relationship between the standard cell potential (E°) and the equilibrium constant (K) is given by the equation:
\[ E° = \frac{0.0591}{n} \log K \]
Rearranging this equation to solve for K gives:
\[ \log K = \frac{n \cdot E°}{0.0591} \]

4. **Substitute the Values**:
Substitute E° and n into the equation:
\[ \log K = \frac{2 \cdot 1.10}{0.0591} \]

5. **Calculate the Value**:
Performing the calculation:
\[ \log K = \frac{2.20}{0.0591} \approx 37.225 \]

6. **Find K by Taking the Antilog**:
To find K, we take the antilog:
\[ K = 10^{37.225} \approx 1.67 \times 10^{37} \]
Since 1.67 is negligible compared to 10, we can approximate:
\[ K \approx 10^{37} \]

7. **Conclusion**:
The equilibrium constant for the reaction is:
\[ K \approx 10^{37} \]
Therefore, the correct answer is option A: \(10^{37}\).