Find the value of E° in the missing link in following diagram.
\(Fe^{3+} \xrightarrow{E_1^\circ = 0.77\ V} Fe^{2+} \xrightarrow{E_2^\circ = -0.44\ V} Fe\)
\(Fe^{3+} \xrightarrow{\hspace{2em} E_3^\circ = \ ? \hspace{2em}} Fe\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → E3° = −0.037 V
Standard electrode potentials of half-reactions cannot be added directly when the number of electrons changes, because E° is an intensive property. The Gibbs energies are additive, so each E° is first converted into ΔG° = −nFE°:
\( \ce{Fe^{3+} + e- -> Fe^{2+}} \qquad \Delta G_1^\circ = -1 \times F \times 0.77 = -0.77F \)
\( \ce{Fe^{2+} + 2e- -> Fe} \qquad \Delta G_2^\circ = -2 \times F \times (-0.44) = +0.88F \)
Adding the two half-reactions gives the missing link:
\( \ce{Fe^{3+} + 3e- -> Fe} \qquad \Delta G_3^\circ = \Delta G_1^\circ + \Delta G_2^\circ = -0.77F + 0.88F = +0.11F \)
\( \displaystyle -3F E_3^\circ = +0.11F \;\Rightarrow\; E_3^\circ = -\frac{0.11}{3} = -0.0367 \approx \mathbf{-0.037\ V} \)
Shortcut: E° of the combined step is the electron-weighted average of the steps:
\( \displaystyle E_3^\circ = \frac{n_1E_1^\circ + n_2E_2^\circ}{n_1 + n_2} = \frac{1(0.77) + 2(-0.44)}{3} = -0.037\ \text{V} \)
NCERT (2026-27 edition):
Electrochemical Cell and Gibbs Energy of the Reaction
