`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe.`
The `EMF` of the above cell is `0.2905`. The equilibrium constant for the cell reaction is

Options

A. `10^(0.32//0.0591)`
B. `10^(0.32//0.0295)`
C. `10^(0.26//0.0295)`
D. `e^(0.32//0.2995)`

Explanation

To solve the problem step by step, we will follow the outlined process to find the equilibrium constant for the given electrochemical cell reaction.

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The given cell is represented as:
\[ \text{Zn} | \text{Zn}^{2+} (a = 0.1 \, \text{M}) || \text{Fe}^{2+} (a = 0.01 \, \text{M}) | \text{Fe} \]
The oxidation and reduction half-reactions can be written as:
- Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- Reduction: \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \)

The overall cell reaction is:
\[ \text{Zn} + \text{Fe}^{2+} \rightarrow \text{Zn}^{2+} + \text{Fe} \]

2. **Use the Nernst Equation**:
The Nernst equation relates the EMF of the cell to the standard EMF (\(E^\circ\)) and the concentrations of the ions:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Fe}^{2+}]} \right) \]
Here, \(E\) is the EMF of the cell, \(n\) is the number of moles of electrons transferred (which is 2 in this case), and the concentrations are given as \( [\text{Zn}^{2+}] = 0.1 \, \text{M} \) and \( [\text{Fe}^{2+}] = 0.01 \, \text{M} \).

3. **Substitute Known Values**:
We know:
- \(E = 0.2905 \, \text{V}\)
- \(n = 2\)
- \([\text{Zn}^{2+}] = 0.1\)
- \([\text{Fe}^{2+}] = 0.01\)

Substituting these values into the Nernst equation:
\[ 0.2905 = E^\circ - \frac{0.059}{2} \log \left( \frac{0.1}{0.01} \right) \]

4. **Calculate the Logarithm**:
Calculate the logarithm:
\[ \log \left( \frac{0.1}{0.01} \right) = \log(10) = 1 \]

5. **Substitute and Solve for \(E^\circ\)**:
Substitute the logarithm back into the equation:
\[ 0.2905 = E^\circ - \frac{0.059}{2} \cdot 1 \]
Simplifying gives:
\[ 0.2905 = E^\circ - 0.0295 \]
Therefore:
\[ E^\circ = 0.2905 + 0.0295 = 0.32 \, \text{V} \]

6. **Calculate the Equilibrium Constant \(K\)**:
At equilibrium, the cell potential \(E\) becomes 0. Thus, we can use the relationship:
\[ 0 = E^\circ - \frac{0.059}{n} \log K \]
Rearranging gives:
\[ E^\circ = \frac{0.059}{n} \log K \]
Substituting \(E^\circ = 0.32 \, \text{V}\) and \(n = 2\):
\[ 0.32 = \frac{0.059}{2} \log K \]
Simplifying:
\[ 0.32 = 0.0295 \log K \]
Dividing both sides by \(0.0295\):
\[ \log K = \frac{0.32}{0.0295} \approx 10.84 \]

7. **Find \(K\)**:
Taking the antilogarithm gives:
\[ K = 10^{10.84} \]

### Final Answer:
The equilibrium constant \(K\) for the cell reaction is approximately \(10^{10.84}\).