Direct Methanol Fuel Cell Thermodynamics: 21-Mat-A7 Environmental Degradation of Materials, May 2018, Q1

Question 1: Direct Methanol Fuel Cell Thermodynamics (20 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 thermodynamic data at 298 K for every species in the balanced cell reaction $CH_3OH + \tfrac32 O_2 \rightarrow CO_2 + 2H_2O$ (6 mol e$^-$ transferred, from the anode half-reaction).

Given data — standard enthalpies and entropies of formation
Species	$\Delta H_f^\circ$ (kJ/mol)	$S^\circ$ (J/mol·K)
$CH_3OH(l)$	−245.98	239.83
$O_2(g)$	0	205.14
$CO_2(g)$	−393.51	213.80
$H_2O(l)$	−285.83	69.95

Find. $\Delta H_{rxn}^\circ$, $\Delta S_{rxn}^\circ$, $\Delta G_{rxn}^\circ$, and the standard cell potential $E_{cell}^\circ$.

Approach. Take products − reactants (weighted by stoichiometric coefficients) for $\Delta H^\circ$ and $\Delta S^\circ$, combine them via $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ at $T=298$ K, then convert to a cell potential with $\Delta G^\circ=-nFE_{cell}^\circ$ using the $n=6$ electrons the anode reaction transfers.

(a) Overall enthalpy. $$\Delta H_{rxn}^\circ = \big[\Delta H_f^\circ(CO_2) + 2\,\Delta H_f^\circ(H_2O)\big] - \big[\Delta H_f^\circ(CH_3OH) + \tfrac32\,\Delta H_f^\circ(O_2)\big]$$ $$\Delta H_{rxn}^\circ = \big[-393.51 + 2(-285.83)\big] - \big[-245.98 + \tfrac32(0)\big] = -965.17-(-245.98)$$ $$\boxed{\Delta H_{rxn}^\circ \approx -719.19\ \text{kJ/mol}}$$ Both product terms (the $CO_2$ and the doubled $H_2O$) are strongly exothermic, and methanol's own formation enthalpy is comparatively small, so the reaction is strongly exothermic overall — expected for a fuel oxidation.
(b) Overall entropy. The same products − reactants combination applied to $S^\circ$: $$\Delta S_{rxn}^\circ = \big[S^\circ(CO_2)+2\,S^\circ(H_2O)\big]-\big[S^\circ(CH_3OH)+\tfrac32\,S^\circ(O_2)\big]$$ $$\Delta S_{rxn}^\circ = \big[213.80+2(69.95)\big]-\big[239.83+\tfrac32(205.14)\big] = 353.70-547.54$$ $$\boxed{\Delta S_{rxn}^\circ \approx -193.84\ \text{J/mol}\cdot\text{K}}$$ The entropy falls because 2.5 mol of gas ($O_2$) on the reactant side condenses into liquid products with no compensating increase in moles of gas — a net loss of gas-phase disorder.
(c) Overall free energy. Using $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ at $T=298$ K: $$\Delta G_{rxn}^\circ = -719.19\ \text{kJ/mol} - (298\ \text{K})(-0.19384\ \text{kJ/mol}\cdot\text{K}) = -719.19+57.78$$ $$\boxed{\Delta G_{rxn}^\circ \approx -661.4\ \text{kJ/mol}}$$ The unfavourable (negative) entropy change makes the reaction slightly less favourable than $\Delta H^\circ$ alone suggests, but the enthalpy term dominates by more than an order of magnitude, so the reaction remains strongly spontaneous.
(d) Standard cell potential. The anode half-reaction shows $n=6$ electrons transferred per mole of methanol oxidised, so $$E_{cell}^\circ = \frac{-\Delta G_{rxn}^\circ}{nF} = \frac{661\,426\ \text{J/mol}}{(6)(96\,486.7\ \text{C/mol})}$$ $$\boxed{E_{cell}^\circ \approx +1.14\ \text{V}}$$

Final results — Question 1
Quantity	Symbol	Result
Enthalpy of reaction	$\Delta H_{rxn}^\circ$	$-719.2$ kJ/mol
Entropy of reaction	$\Delta S_{rxn}^\circ$	$-193.8$ J/mol·K
Free energy of reaction	$\Delta G_{rxn}^\circ$	$-661.4$ kJ/mol
Standard cell potential	$E_{cell}^\circ$	$+1.14$ V

Concept & Theory

Topic. Building a fuel cell's thermodynamic driving force from tabulated formation data, and converting a free-energy change to a cell voltage.

Key relations. Hess's law: $\Delta H_{rxn}^\circ=\sum \nu_i \Delta H_{f,i}^\circ$(products) $-\sum \nu_i \Delta H_{f,i}^\circ$(reactants), same form for $S^\circ$; Gibbs free energy $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$; electrochemical link $\Delta G^\circ=-nFE_{cell}^\circ$.

Why this works. A fuel cell's open-circuit voltage is nothing more than the reaction's free-energy release expressed per unit charge transferred; $n$ must match the electrons actually written in the balanced half-reactions (here 6, not 2 or 4) or the voltage comes out wrong by a whole integer factor. The DMFC's real open-circuit voltage is measured somewhat below this ideal value because of methanol crossover and mixed-potential effects at the cathode — the calculation here gives the thermodynamic ceiling, not the operating voltage.

Common pitfall. Using $n=2$ (from a half-reaction glance) instead of reading the full 6-electron anode reaction, or forgetting to convert $\Delta G^\circ$ from kJ to J before dividing by $F$ (which is quoted in C/mol, i.e. J/(V·mol)).

Source. Fontana, Corrosion Engineering, 3rd ed., Ch. 2 (thermodynamics of electrochemical cells); Jones, Principles and Prevention of Corrosion, 2nd ed., Ch. 2.

Question 1 of 8: Direct Methanol Fuel Cell Thermodynamics

Question 1: Direct Methanol Fuel Cell Thermodynamics (20 marks)