For a cell involving two electron change , `E_(cell)^(@) = 0.3 V ` at `25^(@) C` . The equilibrium constant of the reaction is

Options

A. `10^(-10)`
B. `3 xx 10^(-2)`
C. `10`
D. `10^(10)`

Explanation

To find the equilibrium constant (K_eq) for a cell reaction involving a two-electron change with a standard cell potential (E°_cell) of 0.3 V at 25°C, we can use the Nernst equation and the relationship between the standard cell potential and the equilibrium constant.

### Step-by-Step Solution:

1. **Understand the Nernst Equation**:
The Nernst equation is given by:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q \]
where:
- \(E_{cell}\) is the cell potential under non-standard conditions.
- \(E^{\circ}_{cell}\) is the standard cell potential.
- \(n\) is the number of moles of electrons transferred in the reaction.
- \(Q\) is the reaction quotient.

2. **Set Up for Equilibrium**:
At equilibrium, the cell potential \(E_{cell}\) is 0 V, and the reaction quotient \(Q\) is equal to the equilibrium constant \(K_{eq}\). Thus, we can rewrite the Nernst equation for equilibrium:
\[ 0 = E^{\circ}_{cell} - \frac{0.0591}{n} \log K_{eq} \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq} \]

4. **Substituting Values**:
Given:
- \(E^{\circ}_{cell} = 0.3 \, \text{V}\)
- \(n = 2\) (since it is a two-electron change)

Substitute these values into the equation:
\[ 0.3 = \frac{0.0591}{2} \log K_{eq} \]

5. **Solving for \(\log K_{eq}\)**:
Multiply both sides by 2:
\[ 0.6 = 0.0591 \log K_{eq} \]
Now, divide both sides by 0.0591:
\[ \log K_{eq} = \frac{0.6}{0.0591} \approx 10.14 \]

6. **Finding \(K_{eq}\)**:
To find \(K_{eq}\), we take the antilogarithm:
\[ K_{eq} = 10^{10.14} \approx 1.38 \times 10^{10} \]

### Final Answer:
The equilibrium constant \(K_{eq}\) for the reaction is approximately \(1.38 \times 10^{10}\).