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:
Wel = - DG =nFE
So theoretical electrical potential E= = - DG/ nF
=- (-237*1000)/ (2*96500) = 1.23 V
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