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. `1xx10^(-10)`
B. `29.5xx10^(-2)`
C. `10`
D. `1xx10^(10)`

Explanation

To find the equilibrium constant (K) for the given electrochemical reaction, we can use the relationship between the standard electromotive force (emf) of the cell and the equilibrium constant. The equation we will use is derived from the Nernst equation:

1. **Write the Nernst Equation**:
The Nernst equation at standard conditions (25°C) is given by:
\[ E^\circ = -\frac{0.0591}{n} \log K \]
where \(E^\circ\) is the standard emf, \(n\) is the number of electrons transferred in the reaction, and \(K\) is the equilibrium constant.

2. **Substitute the Given Values**:
From the problem, we know:
- \(E^\circ = 0.295 \, \text{V}\)
- \(n = 2\) (since the reaction involves a two-electron change)

Plugging these values into the equation:
\[ 0.295 = -\frac{0.0591}{2} \log K \]

3. **Rearranging the Equation**:
To isolate \(\log K\), we rearrange the equation:
\[ \log K = -\frac{2 \times 0.295}{0.0591} \]

4. **Calculate the Right-Hand Side**:
First, calculate the numerator:
\[ 2 \times 0.295 = 0.590 \]
Now divide by \(0.0591\):
\[ \log K = -\frac{0.590}{0.0591} \approx -10 \]

5. **Finding K**:
Now, we can find \(K\) by taking the antilogarithm:
\[ K = 10^{-10} \]

Thus, the equilibrium constant \(K\) for the reaction is \(1 \times 10^{-10}\).

### Final Answer:
The equilibrium constant \(K\) is \(1 \times 10^{-10}\).

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