Calculation of the theoretical heat generated in the Proton Exchange Membrane fuel cell & Electrical potential Difference

Calculation of the theoretical heat generated in the fuel cell.

Basic reaction at fuel cell electrodes-

                 At the anode:       H₂ →  2H⁺ + 2e^-                                                                                                   

                 At the cathode:    ½ O₂  + 2H+ + 2e- →  H₂O                                            

                 Overall:               H₂ + ½ O₂ →  H₂O                                                                      

The overall reaction is same as the hydrogen combustion. Since combustion is exothermic process so heat will be released -

                                  H₂ + ½ O₂ →  H₂O + heat

Standard enthalpy of reaction can be calculated by difference of heat of formation of products and reactant.

       ΔH= (hf)_H2O(l)- (hf)_H2- (hf)_0.5O2

Heat of formation of liquid water is -

(hf)_H2O = -286 kJmol^-1 (at 25°C)

Heat of formation of reactant gases 

(hf)H₂₌ (hf)0.5O₂ =0 kJ/mol

So,                                                  ΔH= -286-0-0= -286 kJ/mol.

-ve sign indicates that reaction is exothermic.

But if products H2O is not cooled enough to become liquid, and it remains as vapour then the enthalpy of reaction reduced by enthalpy of vaporisation of water ie. 44 kJ/mol

     ΔH= (hf)_H2O(g)- (hf)_H2- (hf)_0.5O2

              = -286+44 -0-0= -242 kJ/mol.

All the generated energy cannot be converted into electricity due the reaction entropy. The portion of the reaction enthalpy (or hydrogen's higher heating value) that can be converted to electricity in a fuel cell corresponds to Gibbs free energy and is given by the following equation:

                                                       DG = DH- TDS    .

So Gibbs free energy in the liquid water conversion at 25⁰ C

                                                      DG= -286-(25+273)*(-0.1663)= -237 kJ/mol.

Heat produced theoretically in the reaction is = -237-(-286)= 49 kJ/mol

And Gibbs free energy in the vapour water conversion at 25⁰ C

                                                     DG= -242- (25+273)*(-0.0444) = -229 kJ/mol.

Heat produced theoretically in the reaction is = -229- (-242)= 13kJ/mol.

Electrical work:                           W _el= nFE,       

 n= no of electrons per mole. E= electric potential

Maximum amount of electrical energy generated in a fuel cell corresponds to Gibbs free energy, DG:                                              

W_el = - DG =nFE

So theoretical electrical potential   E= = - DG/ nF

                                                            =- (-237*1000)/ (2*96500) = 1.23 V 

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