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 ($E^\ominus$) are intensive properties and cannot be added or subtracted directly. We must convert them to standard Gibbs free energy changes ($\Delta G^\ominus$), which are extensive thermodynamic properties and can be algebraically manipulated according to Hess's Law.
Step 2: Key Formula or Approach:
1. $\Delta G^\ominus = -n F E^\ominus$, where $n$ is the number of electrons transferred.
2. Find the target half-reaction by adding/subtracting the given half-reactions.
3. Add/subtract the corresponding $\Delta G^\ominus$ values to find $\Delta G^\ominus$ of the target reaction, then convert back to $E^\ominus$.
Step 3: Detailed Explanation:
Let's write down the given half-reactions as reductions to elemental iron:
Reaction 1: $Fe^{2+}(aq) + 2e^- \to Fe(s)$
$E^\ominus_1 = X$ Volt, $n_1 = 2$.
$\Delta G^\ominus_1 = -2FX$.
Reaction 2: $Fe^{3+}(aq) + 3e^- \to Fe(s)$
$E^\ominus_2 = Y$ Volt, $n_2 = 3$.
$\Delta G^\ominus_2 = -3FY$.
We want the standard electrode potential for the couple $Fe^{3+} / Fe^{2+}$.
The target half-reaction is:
Reaction 3: $Fe^{3+}(aq) + e^- \to Fe^{2+}(aq)$
This reaction involves $n_3 = 1$ electron. Let its potential be $E^\ominus_3$.
$\Delta G^\ominus_3 = -1 \cdot F \cdot E^\ominus_3$.
To obtain Reaction 3, we can subtract Reaction 1 from Reaction 2:
$(Fe^{3+} + 3e^- \to Fe) - (Fe^{2+} + 2e^- \to Fe) \implies Fe^{3+} + e^- \to Fe^{2+}$.
Applying Hess's law for free energy:
$\Delta G^\ominus_3 = \Delta G^\ominus_2 - \Delta G^\ominus_1$.
Substitute the free energy terms:
$-F E^\ominus_3 = (-3FY) - (-2FX)$.
$-F E^\ominus_3 = -3FY + 2FX$.
Divide the entire equation by $-F$:
$E^\ominus_3 = 3Y - 2X$.
Step 4: Final Answer:
The standard potential is $3Y - 2X$.