Sustainable Energy Science and Engineering CenterBasic Fuel Cell ReactionsReference: PEM fuel cells: theory and Practice, Frano Barber, Elsevier Academic Press, 2005The overall reaction of a PEM fuel cell is:This reaction is the same as the reaction of hydrogen combustion, which is an exothermic process (energy is released): The heat, typically given in terms of enthalpy, of a chemical reaction is the difference between the heats of formation of products and reactants:Heat of formation of liquid water: -286 kJ/mol at 25oC and at atmospheric pressure.H2+12O2⇒ H2OH2+12O2⇒ H2O+ heatΔH = hf()H2O− hf()H2−12hf()O2=−286kJ /g − 0− 0 =−286kJmolH2+12O2⇒ H2Ol()+ 286kJ /molSustainable Energy Science and Engineering CenterHydrogen HHV and LHV Hydrogen heating value is used as a measure of energy input in a fuel cell.Hydrogen heating value: the amount of heat that may be generated by a complete combustion of 1 mol of hydrogen = the enthalpy of hydrogen combustion reaction = 286 kJ/mol The result of combustion is liquid water at 25oC and the value of 286 kJ/mol is considered as Higher Heating Value (HHV). If the combustion is done with excess oxygen and allowed to cool down to 25oC, the product will be in the form of vapor mixed with unburned oxygen. The resulting heat release is measured to be 241 kJ/mol, known as Lower Heating Value (LHV).The difference between HHV and LHV is the heat of evaporation of water at 25oC:H2+12O2⇒ H2Og()+ 241kJ /molHfg=286−241=45kJ/molSustainable Energy Science and Engineering CenterTheoretical Electrical WorkNot all the hydrogen’s energy can be converted into electricity. The portion of the reaction enthalpy that can be converted to electricity corresponds to Gibbs free energy: ΔS is the difference between entropies of products and reactants:ΔG=ΔH−TΔSΔS = Sf()H2O− Sf()H2−12Sf()O2hf(kJ/mol) sf(kJ/mol)Hydrogen 0 0.13066Oxygen 0 0.20517Water (liquid) -286.02 0.06996Water (Vapor) -241.98 0.18884At 25oC and at one atmosphereΔG =−286.02 − 298 × 0.06996−0.13066−0.5×0.20517()()=−237.36kJ /mol48.68 kJ/mol is converted into heat.Sustainable Energy Science and Engineering CenterTheoretical Fuel Cell PotentialWe=neFE=−ΔGElectrical work: ne= 2 (two electron per molecule); F =96,485 Coulombs/electron-mol.The theoretical potential of fuel cell at 25oC and at one atmosphere:Temperature Effect:a, b and c are empiricalCoefficients, different for each gasE =−ΔGneF=237,340J/mol2 × 96,485As/mol=1.23Volts215.29815.29815.29815.2981cTbTaCdtCTSSdtChhFnSTFnHESTHGPTPTTPTee++=+=+=⎟⎟⎠⎞⎜⎜⎝⎛Δ−Δ−=Δ−Δ=Δ∫∫Sustainable Energy Science and Engineering CenterTheoretical Fuel Cell PotentialabcH228.91404 -0.00084 2.01E-06O225.84512 0.012987 -3.9E-06H2O (g) 30.62644 0.009621 1.18E-06ΔH (kJ/mol) ΔS (kJ/mol) ΔG (kJ/mol)-286.02 -0.1633 -237.34-241.98 -0.0444 -228.74H2+12O2⇒ H2Ol()H2+12O2⇒ H2Og()ΔHT=ΔH298.15+ΔaT− 298.15()+ΔbT − 298.15()22+ΔcT − 298.15()33ΔST=ΔS298.15+ΔalnT298.15⎛ ⎝ ⎜ ⎞ ⎠ ⎟ +ΔbT− 298.15()+ΔcT − 298.15()22Δa = aH2O− aH2−12aO2Δb = bH2O− bH2−12bO2Δc = cH2O− cH2−12cO2For T=298.15, E = 1.23 VoltsFor T=373.15, E = 1.167 VoltsSustainable Energy Science and Engineering CenterTheoretical Fuel Cell Efficiencyη=ΔGΔH=237.34286.02= 0.83η=ΔGΔHLHV=228.74241.98= 0.945η=−ΔG−ΔH=−ΔGneF−ΔHneF=1.231.482= 0.83Potential corresponding to hydrogen’s higher heating valueSustainable Energy Science and Engineering CenterTheoretical Fuel Cell PotentialEffect of Pressure: The change in Gibbs free energy may be shown to be:Where Vm= molar volume, m3/mol and P= pressure, PaFor an ideal gas: Therefore: Go: Gibbs free energy at standard temperature, 25oC and at one atmosphereΔG=VmdPPVm=RTdG = RTdPPG = Go+ RTlnPPoSustainable Energy Science and Engineering Center11Equilibrium of a gas mixture:For a chemical reaction occurring at constant pressure and temperature, the reactant gases A and B form products M and N. Where a, b, m and n are stoichiometric coefficients.The change in the Gibbs energyWhere g is in molar quantity (kJ/mol)In terms of standard Gibbs energy. The reference pressure is usually taken as 1 atm. aA+bB⇔mM+nNΔG=mGM+nGN− aGA−bGBTheoretical Fuel Cell PotentialΔG =ΔGo+ RTlnPMPo⎛ ⎝ ⎜ ⎞ ⎠ ⎟ mPNPo⎛ ⎝ ⎜ ⎞ ⎠ ⎟ nPAPo⎛ ⎝ ⎜ ⎞ ⎠ ⎟ aPBPo⎛ ⎝ ⎜ ⎞ ⎠ ⎟ b⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥Sustainable Energy Science and Engineering CenterEquilibrium of a gas mixture:Q: Reaction coefficient for the pressuresThe change in Gibbs energy of a reaction involving gases is:ΔG =ΔGo+ RTlnPMmPNnPAaPBbΔG =ΔGo+ RTlnQQ =PMmPNnPAaPBbΔG=ΔGo+ RTlnQTheoretical Fuel Cell PotentialSustainable Energy Science and Engineering Center11For a hydrogen-oxygen fuel cell, the overall reaction stoichiometry isThe Nernst equation becomes:H2+12O2→ H2OΔG =ΔGo+RTneFlnPH2OPH2PO21/2Theoretical Fuel Cell PotentialE = Eo+RTneFlnPH2PO21/2PH2OWhen liquid water is produced in a fuel cell:For higher reactant pressures the cell potential is higherPH2O=1Sustainable Energy Science and Engineering CenterΔE = EO2− Eair=RTneFlnPO2Pair=RTneFln10.21⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0.5Theoretical Fuel Cell PotentialAir vs oxygen:At 80oC , the voltage loss becomes 0.012V. In practice, this is much higher.Sustainable Energy Science and Engineering CenterE = Eo−RTneFlnPH2OPH2PO21/2Effect of PressureThe Nernst equation is given by OrWhere Pois the standard pressure, 0.1 MPa. If the pressure on both cathode and the anode is approximately same, P then we can writeE = Eo−RTneFlnPH2OPoPH2PoPO21/2PoPH2=αP;PO2=βP;PH2O=δPthenE = Eo−RT2Flnδαβ12⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ −RT4Fln P()Sustainable Energy Science and Engineering CenterFuel and Oxidant UtilizationAs air passes through a fuel cell, the oxygen is used and so the partial pressure will reduce. Similarly, the fuel partial pressure will often decline, as the portion of the fuel reduces and reaction products increase, i.e,α and βdecrease where as δ increase. This results in the increase of magnitude ofMaking the voltage E to fall. This will vary within the cell - worst at the fuel outlet as the fuel is used. Since it is not possible for different parts of the cell to have different voltages, the current varies. The current density will be lower nearer the exit where the fuel concentration is lower. The RT term in the equation also shows that the drop in E due to fuel