The emf of the cell,
`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)|Fe`
is 0.2905 V. The equilibrium constant of 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//0.0295)`

Answer

To find the equilibrium constant \( K \) for the cell reaction given the cell EMF, we can use the Nernst equation and the relationship between the standard EMF and the equilibrium constant.

### Step-by-Step Solution:

1. **Identify the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{[Ox]}{[Red]} \right) \]
where:
- \( E \) is the cell potential (EMF),
- \( E^0 \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred in the reaction,
- \([Ox]\) is the concentration of the oxidizing agent,
- \([Red]\) is the concentration of the reducing agent.

2. **Determine the Values**:
From the question, we have:
- \( E = 0.2905 \, V \)
- Concentration of \( Zn^{2+} = 0.1 \, M \) (reducing agent)
- Concentration of \( Fe^{2+} = 0.01 \, M \) (oxidizing agent)
- The number of electrons transferred \( n = 2 \) (for the reaction \( Zn \rightarrow Zn^{2+} + 2e^- \) and \( Fe^{2+} + 2e^- \rightarrow Fe \)).

3. **Substitute into the Nernst Equation**:
We can rearrange the Nernst equation to solve for \( E^0 \):
\[ E^0 = E + \frac{0.059}{n} \log \left( \frac{[Ox]}{[Red]} \right) \]
Substituting the known values:
\[ E^0 = 0.2905 + \frac{0.059}{2} \log \left( \frac{0.01}{0.1} \right) \]

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

5. **Substitute the Logarithm Value**:
Now substitute this back into the equation:
\[ E^0 = 0.2905 + \frac{0.059}{2} \times (-1) \]
\[ E^0 = 0.2905 - 0.0295 = 0.2610 \, V \]

6. **Relate \( E^0 \) to the Equilibrium Constant \( K \)**:
The relationship between the standard cell potential and the equilibrium constant is given by:
\[ E^0 = \frac{0.059}{n} \log K \]
Rearranging gives:
\[ K = 10^{\frac{nE^0}{0.059}} \]

7. **Substitute \( E^0 \) and Calculate \( K \)**:
Substituting \( E^0 = 0.2610 \, V \) and \( n = 2 \):
\[ K = 10^{\frac{2 \times 0.2610}{0.059}} = 10^{8.853} \]
This approximates to:
\[ K \approx 10^{9} \]

### Final Answer:
The equilibrium constant \( K \) for the cell reaction is approximately \( 10^{9} \).