If `E^(o)""_(Fe^(+2)//Fe)` is `x_(1),E^(o)""_(Fe ^(+3)//Fe)`is `x_(2),` then `E^(o) ""_(Fe^(+3)//Fe^(2+))` will be :

Options

A. `3x_(2) - 2x_(1)`
B. `x_(2) - x_(1)`
C. `x_(2) + x_(1)`
D. `2x_(1) + 3x_(2)`

Answer

To find the standard electrode potential \( E^{\circ}_{Fe^{3+} // Fe^{2+}} \) in terms of \( E^{\circ}_{Fe^{2+} // Fe} \) (denoted as \( x_1 \)) and \( E^{\circ}_{Fe^{3+} // Fe} \) (denoted as \( x_2 \)), we can use the relationship between Gibbs free energy and electrode potentials.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Potentials**:
- The reduction half-reaction for \( Fe^{2+} \) to \( Fe \) is:
\[ Fe^{2+} + 2e^- \rightarrow Fe \quad (E^{\circ} = x_1) \]
- The reduction half-reaction for \( Fe^{3+} \) to \( Fe \) is:
\[ Fe^{3+} + 3e^- \rightarrow Fe \quad (E^{\circ} = x_2) \]

2. **Set Up the Reaction for \( Fe^{3+} \) to \( Fe^{2+} \)**:
- The reduction half-reaction for \( Fe^{3+} \) to \( Fe^{2+} \) is:
\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \quad (E^{\circ} = y) \]

3. **Use the Gibbs Free Energy Relation**:
- The relationship between Gibbs free energy (\( \Delta G^{\circ} \)) and electrode potential (\( E^{\circ} \)) is given by:
\[ \Delta G^{\circ} = -nFE^{\circ} \]
- For the overall reaction from \( Fe^{3+} \) to \( Fe \):
\[ \Delta G^{\circ}_{Fe^{3+} \rightarrow Fe} = \Delta G^{\circ}_{Fe^{3+} \rightarrow Fe^{2+}} + \Delta G^{\circ}_{Fe^{2+} \rightarrow Fe} \]

4. **Express the Gibbs Free Energies in Terms of Potentials**:
- This gives us:
\[ -3FE^{\circ}_{Fe^{3+} // Fe} = -FE^{\circ}_{Fe^{3+} // Fe^{2+}} - 2FE^{\circ}_{Fe^{2+} // Fe} \]
- Substituting the known potentials:
\[ -3x_2 = -y - 2x_1 \]

5. **Rearranging the Equation**:
- Rearranging the equation to solve for \( y \):
\[ y = 3x_2 - 2x_1 \]

6. **Final Result**:
- Therefore, the standard electrode potential \( E^{\circ}_{Fe^{3+} // Fe^{2+}} \) is:
\[ E^{\circ}_{Fe^{3+} // Fe^{2+}} = 3x_2 - 2x_1 \]

### Final Answer:
\[ E^{\circ}_{Fe^{3+} // Fe^{2+}} = 3x_2 - 2x_1 \]