If `E_(1 )^(0)`is standard electrode potential for `Fe//Fe^(+2) , E_(2)^(0)` is for `Fe^(+2)//Fe^Fe^(+3) and E_(3)^(0) ` is for `Fe// Fe^(+3) ` then the relation between `E_(1)^(0), E_(2)^(0) and E_(3)^(0) ` will be :

Options

A. `E_(3)^(0) = E_(1) ^(0) + E_(2)^(0)`
B. `E_(3) ^(0) = (E_(1) ^(0) + E_(2) ^(0))/3`
C. `E_(3)^(0) = (2E_(1)^(0) +E_(2)^(0))/3 `
D. `E_(3)^(0) = (E_(1)^(0)+2E_(2)^(0))/3`

Answer

To solve the problem, we need to establish the relationships between the standard electrode potentials \( E_1^0 \), \( E_2^0 \), and \( E_3^0 \) for the given half-reactions involving iron. Let's break down the steps systematically.

### Step 1: Identify the Half-Reactions
1. The first half-reaction is:
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \quad (E_1^0) \]
This corresponds to the standard electrode potential \( E_1^0 \).

2. The second half-reaction is:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad (E_2^0) \]
This corresponds to the standard electrode potential \( E_2^0 \).

3. The third half-reaction is:
\[ \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \quad (E_3^0) \]
This corresponds to the standard electrode potential \( E_3^0 \).

### Step 2: Write the Gibbs Free Energy Relationships
The relationship between Gibbs free energy (\( \Delta G \)) and standard electrode potential (\( E^0 \)) is given by:
\[ \Delta G = -nFE^0 \]
where \( n \) is the number of moles of electrons transferred, and \( F \) is Faraday's constant.

### Step 3: Set Up the Equations
For each half-reaction, we can write the Gibbs free energy changes:
1. For the first half-reaction:
\[ \Delta G_1 = -2FE_1^0 \]

2. For the second half-reaction:
\[ \Delta G_2 = -1FE_2^0 \]

3. For the third half-reaction:
\[ \Delta G_3 = -3FE_3^0 \]

### Step 4: Relate the Gibbs Free Energies
Since the overall reaction from Fe to Fe³⁺ can be seen as the sum of the first two half-reactions, we can write:
\[ \Delta G_3 = \Delta G_1 + \Delta G_2 \]

Substituting the expressions for Gibbs free energy:
\[ -3FE_3^0 = -2FE_1^0 - 1FE_2^0 \]

### Step 5: Simplify the Equation
Dividing through by \(-F\) (since \( F \) is positive):
\[ 3E_3^0 = 2E_1^0 + E_2^0 \]

### Step 6: Rearranging the Equation
Rearranging gives us the final relationship:
\[ E_3^0 = \frac{2E_1^0 + E_2^0}{3} \]

### Conclusion
The relation between the standard electrode potentials is:
\[ E_3^0 = \frac{2E_1^0 + E_2^0}{3} \]