e.m.f. diagram for some ions is given as :
`FeO_(4)^(2-)overset(E^(@)=+2.20V)rarrFe^(3+)overset(E^(@)=+0.77V)rarrFe^(2+)overset(E^(@)=-0.445V)(rarr)Fe^(0)`
Datermine the value of `E_(FeO_(4)^(2-)//Fe^(2+))^(@)`.

Answer

To determine the value of \( E_{FeO_4^{2-}/Fe^{2+}}^{\circ} \), we will analyze the given standard reduction potentials and use them to find the desired potential through a series of steps.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Standard Potentials**:
- The given reactions with their standard reduction potentials are:
- \( FeO_4^{2-} + 6H^+ + 3e^- \rightarrow Fe^{3+} + 4H_2O \) with \( E^{\circ} = +2.20 \, V \)
- \( Fe^{3+} + e^- \rightarrow Fe^{2+} \) with \( E^{\circ} = +0.77 \, V \)
- \( Fe^{2+} + 2e^- \rightarrow Fe^{0} \) with \( E^{\circ} = -0.445 \, V \)

2. **Calculate the Gibbs Free Energy Change for Each Reaction**:
- For the first reaction:
\[ \Delta G^{\circ}_1 = -n_1 F E^{\circ}_1 = -3F(2.20) \]
- For the second reaction:
\[ \Delta G^{\circ}_2 = -n_2 F E^{\circ}_2 = -1F(0.77) \]
- For the third reaction:
\[ \Delta G^{\circ}_3 = -n_3 F E^{\circ}_3 = -2F(-0.445) \]

3. **Combine the Reactions**:
- We want to find the potential for the reaction \( FeO_4^{2-} \rightarrow Fe^{2+} \).
- We can combine the first and second reactions:
\[ FeO_4^{2-} + 6H^+ + 3e^- \rightarrow Fe^{3+} + 4H_2O \quad (1) \]
\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \quad (2) \]
- Adding these two reactions gives:
\[ FeO_4^{2-} + 6H^+ + 3e^- + Fe^{3+} + e^- \rightarrow Fe^{3+} + 4H_2O + Fe^{2+} \]
- Simplifying this, we get:
\[ FeO_4^{2-} + 6H^+ + 2e^- \rightarrow Fe^{2+} + 4H_2O \]

4. **Calculate the Total Gibbs Free Energy Change**:
- The total Gibbs free energy change for the combined reaction is:
\[ \Delta G^{\circ}_{total} = \Delta G^{\circ}_1 + \Delta G^{\circ}_2 \]
- Substituting the values:
\[ \Delta G^{\circ}_{total} = -3F(2.20) - 1F(0.77) \]

5. **Relate Gibbs Free Energy to Standard Potential**:
- The Gibbs free energy change for the overall reaction can also be expressed as:
\[ \Delta G^{\circ}_{total} = -n F E^{\circ}_{total} \]
- Here, \( n = 4 \) (total electrons transferred).

6. **Set Up the Equation**:
- Equating the two expressions for \( \Delta G^{\circ}_{total} \):
\[ -3F(2.20) - 1F(0.77) = -4F E^{\circ}_{FeO_4^{2-}/Fe^{2+}} \]
- Cancel \( F \) from both sides:
\[ -3(2.20) - 0.77 = -4 E^{\circ}_{FeO_4^{2-}/Fe^{2+}} \]

7. **Solve for the Standard Potential**:
- Calculate the left-hand side:
\[ -6.60 - 0.77 = -4 E^{\circ}_{FeO_4^{2-}/Fe^{2+}} \]
\[ -7.37 = -4 E^{\circ}_{FeO_4^{2-}/Fe^{2+}} \]
- Therefore:
\[ E^{\circ}_{FeO_4^{2-}/Fe^{2+}} = \frac{7.37}{4} = 1.8425 \, V \]

### Final Answer:
\[ E^{\circ}_{FeO_4^{2-}/Fe^{2+}} \approx 1.84 \, V \]