If the `E_(cell)^(@)` for a given reaction has a positive value, then
which of the following gives the correct relationship for the values
of `DeltaG^(@)` and `K_(eq)` :-

Options

A. `DeltaG^(@)lt0,K_(eq)lt1`
B. `DeltaG^(@)gt0,K_(eq)gt1`
C. `DeltaG^(@)gt0,K_(eq)lt1`
D. `DeltaG^(@)lt0,K_(eq)gt1`

Explanation

To solve the question regarding the relationship between \(\Delta G^\circ\) and \(K_{eq}\) when \(E_{cell}^\circ\) is positive, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between \(\Delta G^\circ\) and \(E_{cell}^\circ\)**:
The relationship is given by the equation:
\[ \Delta G^\circ = -nFE_{cell}^\circ \]
where:
- \(n\) = number of moles of electrons transferred in the reaction (always positive),
- \(F\) = Faraday's constant (approximately 96500 C/mol, also positive),
- \(E_{cell}^\circ\) = standard cell potential.

2. **Analyze the Sign of \(\Delta G^\circ\)**:
Since \(E_{cell}^\circ\) is given as positive, and both \(n\) and \(F\) are positive, the product \(nFE_{cell}^\circ\) is also positive. Therefore:
\[ \Delta G^\circ = -nFE_{cell}^\circ < 0 \]
This means that \(\Delta G^\circ\) is negative.

3. **Understand the Relationship Between \(\Delta G^\circ\) and \(K_{eq}\)**:
The relationship is given by:
\[ \Delta G^\circ = -2.303RT \log K_{eq} \]
where:
- \(R\) = universal gas constant (positive),
- \(T\) = temperature in Kelvin (positive).

4. **Analyze the Sign of \(K_{eq}\)**:
Since we have already established that \(\Delta G^\circ < 0\), we can substitute this into the equation:
\[ -2.303RT \log K_{eq} < 0 \]
Since \(-2.303RT\) is negative (as \(R\) and \(T\) are positive), for the entire expression to be negative, \(\log K_{eq}\) must be positive. Thus:
\[ \log K_{eq} > 0 \]
This implies that:
\[ K_{eq} > 1 \]

5. **Conclusion**:
Therefore, when \(E_{cell}^\circ\) is positive, \(\Delta G^\circ\) is negative, and \(K_{eq}\) is greater than 1.

### Final Answer:
The correct relationship is:
\[ \Delta G^\circ < 0 \quad \text{and} \quad K_{eq} > 1 \]