Question

The standard electrode potential (\(E^\circ\)) for the half-cell reaction \(\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}\) at \(298\text{ K}\) is (Given: \(E^\circ(\text{Fe}^{3+}/\text{Fe}) = -0.04\text{ V}\) and \(E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V}\) at \(298\text{ K}\))

• A. \(+0.92\text{ V}\)
• B. \(+0.40\text{ V}\)
• C. \(+0.76\text{ V}\)
• D. \(-0.48\text{ V}\)

Hint:

Never add or subtract $E^\circ$ values directly unless the number of electrons transferred in all reactions is exactly identical. Always convert to $\Delta G^\circ$ components using the formula $\Delta G^\circ = -nFE^\circ$, or use the shortcut formula for consecutive states: $E_3^\circ = \frac{n_1E_1^\circ - n_2E_2^\circ}{n_3}$.

Answer 1

The Correct Option is C

Concept: Standard electrode potentials (\(E^\circ\)) are intensive thermodynamic properties and cannot be added or subtracted directly when combining half-cell reactions with different numbers of transferred electrons. To find the unknown potential of a combined half-reaction, we must convert the potentials into their corresponding standard Gibbs free energy changes (\(\Delta G^\circ\)), which is an extensive property and can be linearly combined. The mathematical connection between the standard Gibbs free energy change and the standard reduction potential is given by: \[ \Delta G^\circ = -nFE^\circ \] Where:

• \(n\) is the number of moles of electrons transferred in the specific half-reaction.

• \(F\) is the Faraday constant (\(96485\text{ C mol}^{-1}\)).

• \(E^\circ\) is the standard reduction potential of the half-cell.
Let us define the given half-cell reactions along with their respective numbers of electrons and reduction potentials:

Step 1: Write down the thermodynamic components for the first given half-reaction.
The reduction of \(\text{Fe}^{3+}\) to metallic iron Fe is represented by: \[ \text{Reaction 1:} \quad \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \] For this half-cell, the number of electrons transferred is \(n_1 = 3\), and the standard reduction potential is given as \(E_1^\circ = -0.04\text{ V}\). The corresponding change in standard Gibbs free energy is: \[ \Delta G_1^\circ = -n_1FE_1^\circ = -3 \times F \times (-0.04) = +0.12F \quad \cdots (1) \]

Step 2: Write down the thermodynamic components for the second given half-reaction.
The reduction of \(\text{Fe}^{2+}\) to metallic iron Fe is represented by: \[ \text{Reaction 2:} \quad \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \] For this half-cell, the number of electrons transferred is \(n_2 = 2\), and the standard reduction potential is given as \(E_2^\circ = -0.44\text{ V}\). The corresponding change in standard Gibbs free energy is: \[ \Delta G_2^\circ = -n_2FE_2^\circ = -2 \times F \times (-0.44) = +0.88F \quad \cdots (2) \]

Step 3: Establish the target half-reaction using algebraic combination.
Our target half-cell reaction is: \[ \text{Target Reaction:} \quad \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \] Let its unknown reduction potential be denoted as \(E_3^\circ\) and its number of electrons be \(n_3 = 1\). The target Gibbs free energy expression is: \[ \Delta G_3^\circ = -n_3FE_3^\circ = -1 \times F \times E_3^\circ = -FE_3^\circ \quad \cdots (3) \] Now, observe how we can algebraically combine Reaction 1 and Reaction 2 to yield the target equation: \[ \text{Reaction 1:} \quad \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \] \[ \text{Reverse of Reaction 2:} \quad \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \] Adding these two expressions gives: \[ \text{Fe}^{3+} + 3e^- + \text{Fe} \rightarrow \text{Fe} + \text{Fe}^{2+} + 2e^- \] Canceling out Fe and \(2e^-\) from both sides leads directly to our target equation: \[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \] Because the target reaction is obtained by subtracting Reaction 2 from Reaction 1, we can apply the exact same linear combination to their extensive Gibbs free energy values: \[ \Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ \]

Step 4: Substitute the expressions to calculate the final potential \(E_3^\circ\).
Substitute equations (1), (2), and (3) into the free energy relationship: \[ -FE_3^\circ = 0.12F - 0.88F \] We can divide the entire equation by the common factor \(-F\): \[ E_3^\circ = -(0.12 - 0.88) \] \[ E_3^\circ = -(-0.76) = +0.76\text{ V} \] Thus, the standard electrode potential for the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell is \(+0.76\text{ V}\).