`E^(o)` for the reaction `Fe + Zn^(2+) rarr Zn + Fe^(2+)` is `– 0.35` V. The given cell reaction is :

Options

A. feasible
B. not feasible
C. in equilibrium
D. can't say anything

Answer

To solve the problem regarding the feasibility of the cell reaction \( \text{Fe} + \text{Zn}^{2+} \rightarrow \text{Zn} + \text{Fe}^{2+} \) with a given standard cell potential \( E^\circ = -0.35 \, \text{V} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
The reaction involves iron (Fe) and zinc ions (\( \text{Zn}^{2+} \)). In this reaction, iron is oxidized to \( \text{Fe}^{2+} \) and zinc ions are reduced to zinc metal (Zn).

2. **Write the Half-Reactions**:
- Oxidation half-reaction:
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \]
- Reduction half-reaction:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \]

3. **Determine the Number of Electrons Transferred (n)**:
From the half-reactions, we can see that 2 electrons are transferred in the overall reaction.

4. **Use the Gibbs Free Energy Relation**:
The relationship between Gibbs free energy (\( \Delta G \)) and cell potential (\( E \)) is given by:
\[ \Delta G = -nFE \]
where \( F \) is Faraday's constant (approximately \( 96485 \, \text{C/mol} \)).

5. **Substitute the Values**:
Given \( E^\circ = -0.35 \, \text{V} \) and \( n = 2 \):
\[ \Delta G = -2 \times 96485 \, \text{C/mol} \times (-0.35 \, \text{V}) \]
\[ \Delta G = 2 \times 96485 \times 0.35 \]
\[ \Delta G = 67639.75 \, \text{J/mol} \quad (\text{or } 67.64 \, \text{kJ/mol}) \]

6. **Interpret the Result**:
Since \( \Delta G \) is positive (\( \Delta G > 0 \)), this indicates that the reaction is non-spontaneous.

7. **Conclusion**:
Therefore, the cell reaction \( \text{Fe} + \text{Zn}^{2+} \rightarrow \text{Zn} + \text{Fe}^{2+} \) is not feasible under standard conditions.