For a reaction `A (s) + B^(2+) rarr B(s) + A ^(2+) , ` at `25^(@) C E^(o) "" _(cell) = 0.2955 V. ` Hence, `K_(eq)` will be :

Options

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

Answer

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

\[ A (s) + B^{2+} \rightleftharpoons B (s) + A^{2+} \]

given that the standard cell potential \( E^{\circ}_{cell} = 0.2955 \, V \) at \( 25^{\circ}C \), we can use the Nernst equation. The steps to solve for \( K_{eq} \) are as follows:

### Step 1: Write the Nernst Equation
The Nernst equation relates the cell potential to the equilibrium constant:

\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log K_{eq} \]

### Step 2: Identify the Number of Electrons Transferred
In the given reaction, \( A \) is oxidized to \( A^{2+} \) and \( B^{2+} \) is reduced to \( B \). Each of these processes involves the transfer of 2 electrons. Thus, \( n = 2 \).

### Step 3: Set Up the Equation at Equilibrium
At equilibrium, the cell potential \( E_{cell} = 0 \). Therefore, we can set up the equation:

\[ 0 = E^{\circ}_{cell} - \frac{0.0591}{n} \log K_{eq} \]

Substituting the known values:

\[ 0 = 0.2955 - \frac{0.0591}{2} \log K_{eq} \]

### Step 4: Rearrange the Equation
Rearranging the equation gives:

\[ \frac{0.0591}{2} \log K_{eq} = 0.2955 \]

### Step 5: Solve for \( \log K_{eq} \)
Now, multiply both sides by \( \frac{2}{0.0591} \):

\[ \log K_{eq} = \frac{0.2955 \times 2}{0.0591} \]

Calculating the right side:

\[ \log K_{eq} = \frac{0.5910}{0.0591} \approx 10.00 \]

### Step 6: Calculate \( K_{eq} \)
To find \( K_{eq} \), we take the antilogarithm:

\[ K_{eq} = 10^{10.00} = 10 \]

### Final Answer
Thus, the equilibrium constant \( K_{eq} \) is:

\[ \boxed{10} \]

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