The emf of the cell,
`Zn|Zn^(2+)(0.01M)|| Fe^(2+)(0.001M) | Fe`
at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is:

Options

A. `e^(0.32/0.0295)`
B. `1-^(0.32/0.0295)`
C. `10^(0.26/0.0295)`
D. `10^(0.32/0.0591)`

Answer

To find the equilibrium constant for the cell reaction given the EMF of the cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- EMF of the cell (E_cell) = 0.2905 V
- Concentration of Zn²⁺ = 0.01 M
- Concentration of Fe²⁺ = 0.001 M
- Temperature (T) = 298 K

2. **Use the Nernst Equation:**
The Nernst equation relates the cell potential (E_cell) to the standard cell potential (E°_cell) and the concentrations of the reactants and products:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Fe^{2+}]} \right) \]
where \( n \) is the number of moles of electrons transferred in the reaction. For the reaction between Zn and Fe²⁺, \( n = 2 \).

3. **Substitute Known Values into the Nernst Equation:**
\[ 0.2905 = E^{\circ}_{cell} - \frac{0.059}{2} \log \left( \frac{0.01}{0.001} \right) \]

4. **Calculate the Logarithm:**
\[ \log \left( \frac{0.01}{0.001} \right) = \log(10) = 1 \]

5. **Substitute the Logarithm Value:**
\[ 0.2905 = E^{\circ}_{cell} - \frac{0.059}{2} \cdot 1 \]
\[ 0.2905 = E^{\circ}_{cell} - 0.0295 \]

6. **Solve for E°_cell:**
\[ E^{\circ}_{cell} = 0.2905 + 0.0295 = 0.32 \, V \]

7. **Determine the Equilibrium Constant (K_eq):**
At equilibrium, E_cell = 0. Thus, we set up the equation:
\[ 0 = E^{\circ}_{cell} - \frac{0.059}{n} \log(K_{eq}) \]
\[ 0 = 0.32 - \frac{0.059}{2} \log(K_{eq}) \]

8. **Rearranging the Equation:**
\[ \frac{0.059}{2} \log(K_{eq}) = 0.32 \]
\[ \log(K_{eq}) = \frac{0.32 \times 2}{0.059} \]
\[ \log(K_{eq}) = \frac{0.64}{0.059} \approx 10.84 \]

9. **Calculate K_eq:**
\[ K_{eq} = 10^{10.84} \]

### Final Answer:
The value of the equilibrium constant for the cell reaction is approximately \( K_{eq} \approx 10^{10.84} \).