Standard Reduction Potential Calculation for Iron System

To find the standard electrode potential $E^{\ominus}_{\text{Fe}^{3+}/\text{Fe}^{2+}}$ from the given values, we use the relationship between the standard reduction potential ($E^{\ominus}$) and the standard Gibbs free energy change ($\Delta G^{\ominus}$):

$\Delta G^{\ominus} = -nFE^{\ominus}$

where $n$ is the number of electrons transferred and $F$ is Faraday's constant.

1. Half-reaction for $\text{Fe}^{2+}/\text{Fe}$: $\text{Fe}^{2+}_{\text{(aq)}} + 2e^- \longrightarrow \text{Fe}_{\text{(s)}} \quad \text{--- (1)}$ The standard reduction potential is $E^{\ominus}_1 = X\text{ V}$ and $n_1 = 2$. $\Delta G^{\ominus}_1 = -2FX$

2. Half-reaction for $\text{Fe}^{3+}/\text{Fe}$: $\text{Fe}^{3+}_{\text{(aq)}} + 3e^- \longrightarrow \text{Fe}_{\text{(s)}} \quad \text{--- (2)}$ The standard reduction potential is $E^{\ominus}_2 = Y\text{ V}$ and $n_2 = 3$. $\Delta G^{\ominus}_2 = -3FY$

3. Target half-reaction for $\text{Fe}^{3+}/\text{Fe}^{2+}$: $\text{Fe}^{3+}_{\text{(aq)}} + e^- \longrightarrow \text{Fe}^{2+}_{\text{(aq)}} \quad \text{--- (3)}$ Let the standard reduction potential be $E^{\ominus}_3$ and $n_3 = 1$. $\Delta G^{\ominus}_3 = -1FE^{\ominus}_3$

4. Combining the reactions: Subtracting reaction (1) from reaction (2) yields reaction (3): $\text{Reaction (3)} = \text{Reaction (2)} - \text{Reaction (1)}$

Since Gibbs free energy is an state function and an extensive property, we can write: $\Delta G^{\ominus}_3 = \Delta G^{\ominus}_2 - \Delta G^{\ominus}_1$

Substituting the expressions for Gibbs free energy into the equation: $-1FE^{\ominus}_3 = -3FY - (-2FX)$

$-FE^{\ominus}_3 = -3FY + 2FX$

Dividing both sides by $-F$: $E^{\ominus}_3 = 3Y - 2X$

Thus, the value of $E^{\ominus}_{\text{Fe}^{3+}/\text{Fe}^{2+}}$ in Volt is $3Y - 2X$.

Correct Option: B