Question

Given at 298 K:
\(E^\circ_{\rm Fe^{2+}/Fe} = X\) Volt
\(E^\circ_{\rm Fe^{3+}/Fe} = Y\) Volt
The \(E^\circ_{\rm Fe^{3+}/Fe^{2+}}\) in Volt at 298 K is given by:

• A. \(2X - 3Y\)
• B. \(3Y - 2X\)
• C. \(3Y + 2X\)
• D. \(Y + X\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Electrode potentials are not additive, but Gibbs free energy changes (\(\Delta G^\circ\)) are. We use the relation \(\Delta G^\circ = -nFE^\circ\) to combine the half-reactions.
Step 2: Key Formula or Approach:
1. \(Fe^{2+} + 2e^- \to Fe \quad \Delta G^\circ_1 = -2FX\)
2. \(Fe^{3+} + 3e^- \to Fe \quad \Delta G^\circ_2 = -3FY\)
The target reaction is: \(Fe^{3+} + e^- \to Fe^{2+}\).
Step 3: Detailed Explanation:
The target reaction can be obtained by subtracting reaction (1) from reaction (2): \[ (Fe^{3+} + 3e^- \to Fe) - (Fe^{2+} + 2e^- \to Fe) = (Fe^{3+} + e^- \to Fe^{2+}) \] Therefore: \[ \Delta G^\circ_3 = \Delta G^\circ_2 - \Delta G^\circ_1 \] \[ -1 \cdot F \cdot E^\circ_3 = -3FY - (-2FX) \] \[ -F \cdot E^\circ_3 = -3FY + 2FX \] Divide both sides by \(-F\): \[ E^\circ_3 = 3Y - 2X \]
Step 4: Final Answer:
The value of \(E^\circ_{\rm Fe^{3+}/Fe^{2+}}\) is \(3Y - 2X\).