If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?

Options

A. `DeltaG^(@) gt 0,K_(eq) lt 1`
B. `DeltaG^(@) gt 0,K_(eq) gt 1`
C. `DeltaG^(@) lt 0,K_(eq) gt 1`
D. `DeltaG^(@) lt 0,K_(eq) lt 1`

Explanation

To solve the problem, we need to analyze the relationships between the standard cell potential (\(E_{cell}^{\circ}\)), the standard Gibbs free energy change (\(\Delta G^{\circ}\)), and the equilibrium constant (\(K_{eq}\)).

### Step-by-Step Solution:

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

2. **Analyzing the Sign of \(E_{cell}^{\circ}\)**:
Since it is given that \(E_{cell}^{\circ}\) is negative, we can substitute this into the equation:
\[ \Delta G^{\circ} = -nF(E_{cell}^{\circ}) < 0 \]
This implies that \(\Delta G^{\circ}\) will be greater than 0 (positive) because multiplying a negative \(E_{cell}^{\circ}\) by \(-nF\) (which is positive) results in a positive \(\Delta G^{\circ}\).

3. **Understanding the Relationship Between \(\Delta G^{\circ}\) and \(K_{eq}\)**:
The relationship between the standard Gibbs free energy change and the equilibrium constant is given by:
\[ \Delta G^{\circ} = -RT \ln K_{eq} \]
where:
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

4. **Analyzing the Sign of \(K_{eq}\)**:
Since we have established that \(\Delta G^{\circ} > 0\), we can rearrange the equation:
\[ K_{eq} = e^{-\Delta G^{\circ}/RT} \]
Because \(\Delta G^{\circ}\) is positive, \(-\Delta G^{\circ}/RT\) will be negative, which means:
\[ K_{eq} < 1 \]

5. **Conclusion**:
From the analysis, we conclude that if \(E_{cell}^{\circ}\) is negative, then:
- \(\Delta G^{\circ} > 0\) (positive)
- \(K_{eq} < 1\)

### Final Answer:
The correct relationships are:
- \(\Delta G^{\circ} > 0\)
- \(K_{eq} < 1\)