At `25^(@)C`, the standard emf of a cell having reaction involving two electrons change is found to be 0.295 V. The equilibrium constant of the reaction is :

Options

A. `29.5 xx 10^(-2)`
B. 10
C. `1 xx 10^(10)`
D. `29.5 xx 10^(10)`

Answer

To find the equilibrium constant (K_eq) of a cell reaction given the standard emf (E°) of the cell, we can use the relationship between E° and K_eq derived from the Nernst equation. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Standard emf (E°) = 0.295 V
- Number of electrons transferred (n) = 2
- Temperature (T) = 25°C = 298 K

2. **Use the Nernst Equation**:
The relationship between the standard emf and the equilibrium constant is given by:
\[ E° = \frac{RT}{nF} \ln K_{eq} \]
where:
- R = universal gas constant = 8.314 J/(mol·K)
- F = Faraday's constant = 96485 C/mol

3. **Convert the Natural Logarithm to Base 10**:
To convert from natural logarithm (ln) to base 10 logarithm (log), we use:
\[ \ln K_{eq} = 2.303 \cdot \log K_{eq} \]
Thus, we can rewrite the equation as:
\[ E° = \frac{2.303RT}{nF} \log K_{eq} \]

4. **Substitute the Known Values**:
Substitute R, T, n, and F into the equation:
\[ E° = \frac{2.303 \times 8.314 \times 298}{2 \times 96485} \log K_{eq} \]

5. **Calculate the Constant**:
First, calculate the constant:
\[ \frac{2.303 \times 8.314 \times 298}{2 \times 96485} = 0.0591 \]
Therefore, the equation simplifies to:
\[ E° = 0.0591 \log K_{eq} \]

6. **Rearrange to Find K_eq**:
Rearranging gives:
\[ \log K_{eq} = \frac{E°}{0.0591} \]
Substitute E°:
\[ \log K_{eq} = \frac{0.295}{0.0591} \]

7. **Calculate log K_eq**:
Performing the calculation:
\[ \log K_{eq} \approx 5.0 \]

8. **Find K_eq**:
To find K_eq, we take the antilogarithm:
\[ K_{eq} = 10^{\log K_{eq}} = 10^{5.0} = 100000 \]

### Final Answer:
The equilibrium constant (K_eq) of the reaction is approximately \( 10^5 \).