`E_(cell)^(@)` and `Delta G^(@)` are related as :

Options

A. `DeltaG^(@) = n FE_(cell)^(@)`
B. `DeltaG = - n F E_(cell)^(@)`
C. `Delta G^(@) = - n F E_(cell)^(@)`
D. `Delta G^(@) = n FE_(cell)^(@) = 0`

Answer

To solve the problem of how \( E_{\text{cell}}^\circ \) and \( \Delta G^\circ \) are related, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Terms**:
- \( E_{\text{cell}}^\circ \) refers to the standard cell potential, which can be either reduction or oxidation potential of the electrochemical cell.
- \( \Delta G^\circ \) represents the change in Gibbs free energy under standard conditions (298 K).

2. **Identify the Relationship**:
- The relationship between \( E_{\text{cell}}^\circ \) and \( \Delta G^\circ \) is given by the equation:
\[ \Delta G^\circ = -nFE_{\text{cell}}^\circ \]
- Here, \( n \) is the number of moles of electrons transferred in the electrochemical reaction, and \( F \) is Faraday's constant, approximately \( 96500 \, \text{C/mol} \).

3. **Interpret the Equation**:
- The negative sign indicates that a positive cell potential (\( E_{\text{cell}}^\circ \)) corresponds to a spontaneous reaction (negative \( \Delta G^\circ \)).
- Conversely, a negative cell potential indicates a non-spontaneous reaction (positive \( \Delta G^\circ \)).

4. **Conclusion**:
- Therefore, the relationship shows that the Gibbs free energy change is directly related to the cell potential, and it helps predict the spontaneity of the electrochemical reaction.

### Final Expression:
\[ \Delta G^\circ = -nFE_{\text{cell}}^\circ \]