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

Step 1: See why we cannot just subtract potentials.
Standard potentials are intensive, so when half-reactions involve different electron counts we must first turn them into Gibbs free energies, which do add up. The link is \[ \Delta G^\circ = -nFE^\circ \]
Step 2: Write the known half-reactions.
Reaction A: \(Fe^{3+} + 3e^- \rightarrow Fe\), \(E^\circ = -0.04\ V\), \(n=3\).
Reaction B: \(Fe^{2+} + 2e^- \rightarrow Fe\), \(E^\circ = -0.44\ V\), \(n=2\).
Step 3: Express the target.
We want \(Fe^{3+} + e^- \rightarrow Fe^{2+}\) with \(n=1\). This equals reaction A minus reaction B.
Step 4: Combine the free energies.
\[ \Delta G^\circ_{target} = \Delta G^\circ_A - \Delta G^\circ_B = -3F(-0.04) - [-2F(-0.44)] \] \[ = (0.12F) - (0.88F) = -0.76F \]
Step 5: Convert back to a potential.
For the target \(n=1\), so \[ E^\circ = -\frac{\Delta G^\circ}{nF} = -\frac{-0.76F}{1\cdot F} = +0.76\ V \]
Step 6: State the answer.
The potential for \(Fe^{3+}/Fe^{2+}\) is \(+0.76\ V\), option 3.
\[ \boxed{E^\circ = +0.76\ \text{V}} \]