STATEMENT 1: `DeltaG^(@)=-nFE^(@)` and
STATEMENT 2: `E^(@)` should be positive for a spontaneous reaction

Options

A. Statement 1 is true , Statement 2 is true Statement 2 is correct explanation for Statement 1
B. Statement 1 is true Statement 2 is true Statement 2 is NOT a correct explanation for Statement 1
C. Statement 1 is true statement 2 is true
D. Statement 1 is false Statement 2 is true

Answer

To solve the question regarding the statements about Gibbs free energy and cell potential, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
**Statement 1:** \(\Delta G^\circ = -nFE^\circ\)

1. **Understanding Gibbs Free Energy:** Gibbs free energy (\(\Delta G^\circ\)) is a thermodynamic potential that can be used to predict the spontaneity of a reaction. The relationship given in Statement 1 connects Gibbs free energy with the cell potential.

2. **Nernst Equation:** The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln K_{\text{equilibrium}} \]
At equilibrium, \(E_{\text{cell}} = 0\). Thus, we can rearrange this equation to express \(E^\circ_{\text{cell}}\):
\[ E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K_{\text{equilibrium}} \]

3. **Relating \(\Delta G^\circ\) and \(K_{\text{equilibrium}}\):** We also know that:
\[ \Delta G^\circ = -RT \ln K_{\text{equilibrium}} \]

4. **Substituting \(K_{\text{equilibrium}}\):** If we substitute \(K_{\text{equilibrium}}\) from the rearranged Nernst equation into the expression for \(\Delta G^\circ\), we get:
\[ \Delta G^\circ = -RT \ln \left( e^{\frac{nFE^\circ_{\text{cell}}}{RT}} \right) \]
This simplifies to:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Thus, Statement 1 is **true**.

### Step 2: Analyze Statement 2
**Statement 2:** \(E^\circ_{\text{cell}}\) should be positive for a spontaneous reaction.

1. **Understanding Spontaneity:** A reaction is spontaneous if \(\Delta G^\circ < 0\). From Statement 1, we have:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

2. **Condition for Spontaneity:** For \(\Delta G^\circ\) to be negative, the term \(-nFE^\circ_{\text{cell}}\) must be negative. Since \(n\) (the number of moles of electrons) and \(F\) (Faraday's constant) are both positive, this means that \(E^\circ_{\text{cell}}\) must be positive:
\[ E^\circ_{\text{cell}} > 0 \implies \Delta G^\circ < 0 \]
Therefore, Statement 2 is also **true**.

### Conclusion
Both statements are true, but Statement 2 does not serve as a direct explanation for Statement 1; rather, it explains the condition for spontaneity in terms of cell potential.

### Final Answer
- **Statement 1 is true.**
- **Statement 2 is true.**
- **Statement 2 is not a correct explanation for Statement 1.**

Thus, the correct option is that both statements are true, but Statement 2 does not explain Statement 1.

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