For a reaction, `A + B^(2+) rarr B + A^(2+)`, at `25^(@)C`
`E^(@)= 0.2955V`. The value of `K_(eq)` is

Options

A. 10
B. `10^(10)`
C. `-10`
D. `10^(-10)`

Explanation

To find the equilibrium constant \( K_{eq} \) for the reaction \( A + B^{2+} \rightleftharpoons B + A^{2+} \) at \( 25^\circ C \) with a standard cell potential \( E^\circ = 0.2955 \, V \), 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 Standard Cell Potential:**
The reaction is given as:
\[ A + B^{2+} \rightleftharpoons B + A^{2+} \]
The standard cell potential \( E^\circ \) is provided as \( 0.2955 \, V \).

2. **Use the Nernst Equation:**
The Nernst equation relates the standard cell potential to the equilibrium constant:
\[ E^\circ = \frac{0.0591}{n} \log K_{eq} \]
where \( n \) is the number of moles of electrons transferred in the reaction.

3. **Determine \( n \):**
In the given reaction, \( A \) is oxidized to \( A^{2+} \) and \( B^{2+} \) is reduced to \( B \). Since \( A \) loses 2 electrons to become \( A^{2+} \), we have:
\[ n = 2 \]

4. **Substitute Values into the Nernst Equation:**
Rearranging the Nernst equation to solve for \( K_{eq} \):
\[ K_{eq} = 10^{\frac{n \cdot E^\circ}{0.0591}} \]
Substituting \( n = 2 \) and \( E^\circ = 0.2955 \, V \):
\[ K_{eq} = 10^{\frac{2 \cdot 0.2955}{0.0591}} \]

5. **Calculate \( K_{eq} \):**
First, calculate the exponent:
\[ \frac{2 \cdot 0.2955}{0.0591} \approx \frac{0.591}{0.0591} \approx 10.01 \]
Therefore,
\[ K_{eq} \approx 10^{10.01} \approx 10^{10} \]

6. **Final Result:**
Thus, the value of \( K_{eq} \) is approximately:
\[ K_{eq} \approx 10^{10} \]