Step 1: Understanding the Concept:
In a Latimer diagram or a sequence of electrochemical steps, the standard Gibbs free energy changes are additive ($\Delta G^{\circ}_{total} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2}$). Standard potentials ($E^{\circ}$) are not directly additive. Step 2: Key Formula or Approach:
Relation: $\Delta G^{\circ} = -nFE^{\circ}$.
For the process $Fe \xrightarrow{E_{1}^{\circ}} Fe^{2+} \xrightarrow{E_{2}^{\circ}} Fe^{3+}$, let the total potential for $Fe \to Fe^{3+}$ be $E_{3}^{\circ}$.
The formula is: $n_{3}E_{3}^{\circ} = n_{1}E_{1}^{\circ} + n_{2}E_{2}^{\circ}$. Step 3: Detailed Explanation:
Given from the diagram:
1. $Fe \to Fe^{2+}$: $n_{1} = 2$ electrons, $E_{1}^{\circ} = 0.439\text{ V}$.
2. $Fe^{2+} \to Fe^{3+}$: $n_{2} = 1$ electron, $E_{2}^{\circ} = ?$ (This is what we need to find).
3. $Fe \to Fe^{3+}$: $n_{3} = 3$ electrons, $E_{3}^{\circ} = 0.036\text{ V}$.
Substitute the values:
\[ 3 \times 0.036 = (2 \times 0.439) + (1 \times E_{2}^{\circ}) \]
\[ 0.108 = 0.878 + E_{2}^{\circ} \]
\[ E_{2}^{\circ} = 0.108 - 0.878 = -0.77\text{ V} \]
Ignoring the negative sign as requested: $0.77\text{ V}$. Step 4: Final Answer:
The value of $E_{2}^{\circ}$ is 0.77.
