Free Energy for Oxidation of Fe 2+ by MnO 2 in Acid Solution: 18-Env-B5 Industrial & Hazardous Waste Management, December 2019, Q6

Question 6: Free Energy for Oxidation of Fe²⁺ by MnO₂ in Acid Solution (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Standard reduction potentials from the supplied table (25°C): $\text{Fe}^{3+}+e^-\rightarrow\text{Fe}^{2+}$, $E^{\circ}=+0.77\ \text{V}$; $\text{MnO}_2+4\text{H}^++2e^-\rightarrow\text{Mn}^{2+}+2\text{H}_2\text{O}$, $E^{\circ}=+1.23\ \text{V}$. Faraday constant $F = 96{,}485\ \text{C/mol}$.

Find. The standard free energy change $\Delta G^{\circ}$ for the oxidation of $\text{Fe}^{2+}$ by $\text{MnO}_2$ in acid solution.

Approach. Pair the $\text{Fe}^{2+}$ oxidation (the reverse of the $\text{Fe}^{3+}/\text{Fe}^{2+}$ reduction couple) with $\text{MnO}_2$ acting as the oxidant (its own reduction half-reaction), balance electrons, find $E^{\circ}_{\text{cell}}$, then apply $\Delta G^{\circ}=-nFE^{\circ}_{\text{cell}}$.

Write and balance the overall redox reaction. $\text{MnO}_2$ is the oxidant (cathode, gains 2 electrons); $\text{Fe}^{2+}$ is oxidized to $\text{Fe}^{3+}$ (anode, loses 1 electron per Fe, so the iron half-reaction is doubled to balance electrons): $$2\text{Fe}^{2+}+\text{MnO}_2+4\text{H}^{+}\longrightarrow2\text{Fe}^{3+}+\text{Mn}^{2+}+2\text{H}_2\text{O}, \qquad n=2$$
Cell potential. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$, using each half-reaction's REDUCTION potential as tabulated (MnO₂/Mn²⁺ is the cathode, Fe³⁺/Fe²⁺ is the anode couple): $$E^{\circ}_{\text{cell}} = 1.23 - 0.77 = \boxed{0.46\ \text{V}}$$ The positive cell potential confirms the reaction is spontaneous as written, i.e. $\text{MnO}_2$ does oxidize $\text{Fe}^{2+}$ in acid solution.
Free energy. $$\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} = -(2)(96{,}485\ \text{C/mol})(0.46\ \text{V}) = \boxed{-88{,}800\ \text{J/mol}} = -88.8\ \text{kJ/mol}$$

Final results
Quantity	Value
Balanced reaction	$2\text{Fe}^{2+}+\text{MnO}_2+4\text{H}^{+}\rightarrow2\text{Fe}^{3+}+\text{Mn}^{2+}+2\text{H}_2\text{O}$
Electrons transferred, $n$	2
Cell potential, $E^{\circ}_{\text{cell}}$	$+0.46\ \text{V}$
Standard free energy, $\Delta G^{\circ}$	$-88.8\ \text{kJ/mol}$ (spontaneous)

Topic: Standard electrode potentials, cell potential, and free energy of a redox reaction

Key relations: $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}$ (both read as REDUCTION potentials); $\Delta G^{\circ}=-nFE^{\circ}_{\text{cell}}$; a positive $E^{\circ}_{\text{cell}}$ corresponds to a negative (spontaneous) $\Delta G^{\circ}$

Why this works: tabulated standard potentials are always written as reductions, so identifying which half-reaction is actually being reduced (the oxidant, here MnO₂, becomes the cathode) versus which is being oxidized (Fe²⁺, the anode couple) is the entire content of the problem; once $E^{\circ}_{\text{cell}}$ and $n$ are correctly identified, the free-energy relation is a direct application of electrochemical thermodynamics used throughout corrosion, disinfection by-product, and metal-redox engineering problems.

Common pitfall: subtracting the potentials in the wrong order (anode $-$ cathode, which silently flips the sign of both $E^{\circ}_{\text{cell}}$ and $\Delta G^{\circ}$), or forgetting to double the iron half-reaction's electron count before taking $n=2$ into $\Delta G^{\circ}=-nFE^{\circ}_{\text{cell}}$ (the potentials themselves are NOT multiplied when a half-reaction is scaled, only $n$ changes).

Source: Davis & Cornwell, Introduction to Environmental Engineering, 6th ed.; LaGrega, Buckingham & Evans, Hazardous Waste Management, 2nd ed.

Question 6 of 10: Free Energy for Oxidation of Fe 2+ by MnO 2 in Acid Solution

Question 6: Free Energy for Oxidation of Fe2+ by MnO2 in Acid Solution (10 marks)

Question 6: Free Energy for Oxidation of Fe²⁺ by MnO₂ in Acid Solution (10 marks)