The standard `EMF` of a galvanic cell involving cell reaction with `n=2` is found to be `0.295 V` at `25^(@)C` . The equilibrium constant of the reaction would be

Options

A. `4.0 xx 10^(12)`
B. `1.0xx10^(2)`
C. `1.0 xx 10^(10)`
D. `2.0xx10^(11)`

Explanation

To find the equilibrium constant (K) for the given galvanic cell reaction, we can use the Nernst equation. The Nernst equation relates the standard electromotive force (EMF) of a cell to the equilibrium constant.

### Step-by-Step Solution:

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

2. **Use the Nernst Equation**:
The Nernst equation at equilibrium is given by:
\[ E° = \frac{RT}{nF} \ln K \]
At equilibrium, the cell potential (E) becomes 0, and we can rearrange the equation to solve for K:
\[ 0 = E° - \frac{RT}{nF} \ln K \]
Rearranging gives:
\[ \frac{RT}{nF} \ln K = E° \]
Thus,
\[ \ln K = \frac{nFE°}{RT} \]

3. **Substituting Constants**:
- R (universal gas constant) = 8.314 J/(mol·K)
- F (Faraday's constant) = 96485 C/mol
- T = 298 K
Substituting these values into the equation:
\[ \ln K = \frac{(2)(96485)(0.295)}{(8.314)(298)} \]

4. **Calculating the Right Side**:
- Calculate the numerator:
\[ 2 \times 96485 \times 0.295 = 57074.075 \]
- Calculate the denominator:
\[ 8.314 \times 298 = 2477.572 \]
- Now, calculate:
\[ \ln K = \frac{57074.075}{2477.572} \approx 23.063 \]

5. **Finding K**:
To find K, we exponentiate both sides:
\[ K = e^{23.063} \approx 9.8 \times 10^{10} \]

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