Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?
I. `logK = (nFE^(@))/(2.303RT)`
II. `K = e^((nFE)/(RT))`
III. `log K = (-nFE^(@))/(2.303 RT)`
IV. `log K = 0.4342 (nFE^(@))/(RT)`
Choose the correct statement(s).

Options

A. I, II and III are correct
B. II and III are correct
C. I, II and IV are correct
D. I and IV are correct

Explanation

To solve the problem, we need to analyze the relationships between standard electrode potential (E°), Gibbs free energy (ΔG), and the equilibrium constant (K).

### Step-by-Step Solution:

1. **Understand the Relationship between ΔG and E°**:
The Gibbs free energy change for a reaction can be expressed in terms of the standard electrode potential as:
\[ \Delta G = -nFE^0 \]
where:
- \( n \) = number of moles of electrons transferred
- \( F \) = Faraday's constant (approximately 96485 C/mol)
- \( E^0 \) = standard electrode potential

2. **Understand the Relationship between ΔG and K**:
The Gibbs free energy change is also related to the equilibrium constant by the equation:
\[ \Delta G = -2.303RT \log K \]
where:
- \( R \) = universal gas constant (approximately 8.314 J/(mol·K))
- \( T \) = temperature in Kelvin

3. **Set the Two Expressions for ΔG Equal**:
Since both expressions represent ΔG, we can set them equal to each other:
\[ -nFE^0 = -2.303RT \log K \]

4. **Rearranging the Equation**:
By cancelling the negative signs, we can rearrange the equation to solve for \( \log K \):
\[ nFE^0 = 2.303RT \log K \]
Dividing both sides by \( 2.303RT \):
\[ \log K = \frac{nFE^0}{2.303RT} \]
This corresponds to the first statement.

5. **Convert to Natural Log (ln)**:
We can also express \( K \) in terms of the natural logarithm:
\[ \Delta G = -RT \ln K \]
Setting this equal to the first expression for ΔG:
\[ -nFE^0 = -RT \ln K \]
Rearranging gives:
\[ nFE^0 = RT \ln K \]
Dividing both sides by \( RT \):
\[ \ln K = \frac{nFE^0}{RT} \]
Exponentiating both sides gives:
\[ K = e^{\frac{nFE^0}{RT}} \]
This corresponds to the second statement.

6. **Check the Third Statement**:
The third statement is:
\[ \log K = \frac{-nFE^0}{2.303RT} \]
This is incorrect because we derived that \( \log K \) is positive, not negative.

7. **Check the Fourth Statement**:
The fourth statement is:
\[ \log K = 0.4342 \frac{nFE^0}{RT} \]
We can convert \( \frac{1}{2.303} \) to \( 0.4342 \) (since \( 0.4342 \approx \frac{1}{2.303} \)). Thus, this statement is also correct.

### Conclusion:
The correct statements are:
- I: \( \log K = \frac{nFE^0}{2.303RT} \) (Correct)
- II: \( K = e^{\frac{nFE^0}{RT}} \) (Correct)
- III: \( \log K = \frac{-nFE^0}{2.303RT} \) (Incorrect)
- IV: \( \log K = 0.4342 \frac{nFE^0}{RT} \) (Correct)

Thus, the correct options are I, II, and IV.