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

### Step 1: Understand the relationship between Gibbs free energy and cell potential
The relationship between the standard Gibbs free energy change (\( \Delta G^{\circ} \)) and the standard electrode potential (\( E^{\circ} \)) is given by the equation:
\[ \Delta G^{\circ} = -nFE^{\circ} \]
where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (\( 96500 \, \text{C/mol} \))
- \( E^{\circ} \) = standard electrode potential

### Step 2: Identify the values
In this case, we have:
- \( E^{\circ} = -0.35 \, \text{V} \)
- Assume \( n = 1 \) (for simplicity, as only one electron is transferred in this reaction)

### Step 3: Substitute the values into the equation
Substituting the values into the Gibbs free energy equation:
\[ \Delta G^{\circ} = - (1)(96500)(-0.35) \]

### Step 4: Calculate \( \Delta G^{\circ} \)
Calculating the above expression:
\[ \Delta G^{\circ} = 96500 \times 0.35 \]
\[ \Delta G^{\circ} = 33775 \, \text{J/mol} \quad (\text{or } 33.775 \, \text{kJ/mol}) \]

### Step 5: Analyze the result
Since \( \Delta G^{\circ} \) is positive (\( 33775 \, \text{J/mol} \)), this indicates that the reaction is not feasible under standard conditions. For a reaction to be spontaneous (feasible), \( \Delta G^{\circ} \) must be negative.

### Conclusion
The given cell reaction \( \text{Fe} + \text{Zn}^{2+} \rightarrow \text{Zn} + \text{Fe}^{2+} \) is **not feasible**.