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+))^(@)`.

Options

A. 1.84 V
B. 1.42 V
C. 1.3 V
D. 2.0 V

Answer

To determine the value of \( E_{FeO_4^{2-} // Fe^{2+}}^{\circ} \), we will use the given standard electrode potentials and the relationships between the reactions. Here's a step-by-step solution:

### Step 1: Write the half-reactions and their standard potentials

1. **Half-reaction for \( FeO_4^{2-} \) to \( Fe^{3+} \)**:
\[ FeO_4^{2-} + 3e^- \rightarrow Fe^{3+} \quad (E^{\circ} = +2.20 \, V) \]

2. **Half-reaction for \( Fe^{3+} \) to \( Fe^{2+} \)**:
\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \quad (E^{\circ} = +0.77 \, V) \]

3. **Half-reaction for \( Fe^{2+} \) to \( Fe^{0} \)**:
\[ Fe^{2+} + 2e^- \rightarrow Fe^{0} \quad (E^{\circ} = -0.445 \, V) \]

### Step 2: Write the Gibbs free energy changes for each reaction

The Gibbs free energy change (\( \Delta G \)) for each half-reaction can be expressed as:
\[ \Delta G = -nFE^{\circ} \]
where \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( E^{\circ} \) is the standard electrode potential.

1. For the first reaction:
\[ \Delta G_1 = -3F \cdot 2.20 \]

2. For the second reaction:
\[ \Delta G_2 = -1F \cdot 0.77 \]

3. For the third reaction:
\[ \Delta G_3 = -2F \cdot (-0.445) \]

### Step 3: Combine the reactions

To find \( E_{FeO_4^{2-} // Fe^{2+}}^{\circ} \), we need to combine the first two reactions:

\[ FeO_4^{2-} + 3e^- \rightarrow Fe^{3+} \quad (1) \]
\[ Fe^{3+} + e^- \rightarrow Fe^{2+} \quad (2) \]

Adding these two reactions gives:
\[ FeO_4^{2-} + 3e^- + e^- \rightarrow Fe^{2+} \]
This simplifies to:
\[ FeO_4^{2-} + 4e^- \rightarrow Fe^{2+} \]

### Step 4: Write the equation for the overall Gibbs free energy change

The overall Gibbs free energy change for the combined reaction is:
\[ \Delta G_{total} = \Delta G_1 + \Delta G_2 \]
Substituting the values:
\[ \Delta G_{total} = -3F \cdot 2.20 - 1F \cdot 0.77 \]

### Step 5: Relate to the standard potential for the combined reaction

Now, we can relate this to the standard potential for the overall reaction:
\[ \Delta G_{total} = -4F \cdot E_{FeO_4^{2-} // Fe^{2+}}^{\circ} \]

### Step 6: Set the equations equal and solve for \( E_{FeO_4^{2-} // Fe^{2+}}^{\circ} \)

Setting the two expressions for \( \Delta G_{total} \) equal gives:
\[ -3F \cdot 2.20 - 1F \cdot 0.77 = -4F \cdot E_{FeO_4^{2-} // Fe^{2+}}^{\circ} \]

Dividing through by \( -F \) (and canceling \( F \)):
\[ 3 \cdot 2.20 + 0.77 = 4E_{FeO_4^{2-} // Fe^{2+}}^{\circ} \]

Calculating the left side:
\[ 6.60 + 0.77 = 7.37 \]

Thus:
\[ 4E_{FeO_4^{2-} // Fe^{2+}}^{\circ} = 7.37 \]

Dividing by 4:
\[ E_{FeO_4^{2-} // Fe^{2+}}^{\circ} = \frac{7.37}{4} = 1.8425 \, V \approx 1.84 \, V \]

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

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