If `E_(fe^(2+)//Fe)^(@)` is `x_(1),E_(Fe^(3+)//Fe)^(@), is `x_(2)` then what will be `E_(Fe^(3+)//Fe^(2+))^(@)`

Explanation

To solve the problem, we need to find the standard electrode potential \( E^{\circ}_{Fe^{3+}/Fe^{2+}} \) given the standard electrode potentials \( E^{\circ}_{Fe^{2+}/Fe} = x_1 \) and \( E^{\circ}_{Fe^{3+}/Fe} = x_2 \).

### Step-by-Step Solution:

1. **Write the half-reactions**:
- The reduction of \( Fe^{2+} \) to \( Fe \):
\[ Fe^{2+} + 2e^- \rightarrow Fe \quad (E^{\circ} = x_1) \]
- The reduction of \( Fe^{3+} \) to \( Fe \):
\[ Fe^{3+} + 3e^- \rightarrow Fe \quad (E^{\circ} = x_2) \]

2. **Write the desired half-reaction**:
- The reduction of \( Fe^{3+} \) to \( Fe^{2+} \):
\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \quad (E^{\circ} = E^{\circ}_{Fe^{3+}/Fe^{2+}}) \]

3. **Relate the half-reactions**:
- To find \( E^{\circ}_{Fe^{3+}/Fe^{2+}} \), we can use the relationship between the potentials of the half-reactions. We can express the conversion of \( Fe^{3+} \) to \( Fe^{2+} \) in terms of the other two half-reactions:
\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \quad \text{(reaction 1)} \]
\[ Fe^{2+} + 2e^- \rightarrow Fe \quad \text{(reaction 2)} \]
\[ Fe^{3+} + 3e^- \rightarrow Fe \quad \text{(reaction 3)} \]

4. **Use the relationship of Gibbs free energy**:
- The Gibbs free energy change for the reactions can be related as follows:
\[ \Delta G_3 = \Delta G_2 - \Delta G_1 \]
- In terms of electrode potentials:
\[ -nFE^{\circ}_{cell} = -nF E^{\circ}_{Fe^{3+}/Fe} + -nF E^{\circ}_{Fe^{2+}/Fe} \]
- For the reaction \( Fe^{3+} + e^- \rightarrow Fe^{2+} \):
\[ E^{\circ}_{Fe^{3+}/Fe^{2+}} = E^{\circ}_{Fe^{3+}/Fe} - E^{\circ}_{Fe^{2+}/Fe} \]

5. **Substituting the known values**:
- From the previous steps, we know:
\[ E^{\circ}_{Fe^{3+}/Fe^{2+}} = x_2 - x_1 \]

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