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Question #44771

Difficulty: 8/10 | 🟢 Free

The Question

A novel Direct Methanol Fuel Cell (DMFC) operates at 298 K using a cathodic half-cell where \ce{MnO4^{-}(aq)} is reduced. Due to the specific electrocatalyst and a buffered local cathode pH of 4.0, the reduction strictly yields \ce{MnO2(s)} as the only manganese product. The anode oxidizes \ce{CH3OH(aq)} to \ce{CO2(g)}.

Given standard reduction potentials at 298 K: E∘(\ceMnO4−/\ceMn2+)=1.51 VE^\circ(\ce{MnO4^{-}}/\ce{Mn^{2+}}) = 1.51 \text{ V} E∘(\ceMnO2/\ceMn2+)=1.21 VE^\circ(\ce{MnO2}/\ce{Mn^{2+}}) = 1.21 \text{ V} E∘(\ceCO2/\ceCH3OH)=0.01 VE^\circ(\ce{CO2}/\ce{CH3OH}) = 0.01 \text{ V}

The cell operates with [\ceMnO4−]=0.1 M[\ce{MnO4^{-}}] = 0.1 \text{ M}, [\ceCH3OH]=0.1 M[\ce{CH3OH}] = 0.1 \text{ M}, and P\ceCO2=1 barP_{\ce{CO2}} = 1 \text{ bar}. The overall enthalpy change (ΔH\Delta H) for the balanced cell reaction (per mole of \ce{CH3OH} consumed) is −1150.76 kJ/mol-1150.76 \text{ kJ/mol}.

Assuming 2.303RTF=0.06 V\frac{2.303 RT}{F} = 0.06 \text{ V} and Faraday's constant F=96500 C/molF = 96500 \text{ C/mol}, what is the operating thermodynamic efficiency (η=ΔGΔH\eta = \frac{\Delta G}{\Delta H}) of this fuel cell?

Options

  • A.

    80.0%80.0\%

  • B.

    85.5%85.5\%

  • C.

    91.1%91.1\%

  • D.

    82.5%82.5\%

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Fuel Cell Efficiency|KMnO₄ Redox pH Dependency|E° (M³⁺/M²⁺) Redox Potential

Why This Question Is Tricky

Common Mistakes

This problem tests the integration of Latimer diagram principles, Nernstian pH-dependency, and thermodynamic efficiency in fuel cells.

The primary trap is failing to calculate the correct E∘E^\circ for the specific \ceMnO4−→\ceMnO2\ce{MnO4^{-}} \rightarrow \ce{MnO2} pathway, as students often mistakenly use the \ceMn2+\ce{Mn^{2+}} potential.

Secondary traps include ignoring the Nernst equation and calculating efficiency using standard conditions (Option B), inverting the reaction quotient QQ (Option C), or misbalancing the overall reaction, which alters the stoichiometric coefficients of \ceH+\ce{H+} and \ceMnO4−\ce{MnO4^{-}} in QQ (Option D).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterElectrochemistry - Class 12
ConceptFuel Cell Efficiency
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree