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21.2 THEORY OF FUEL CELLS        513





                  Now, from the Van’t Hoff equation
                                                 d           Q p
                                                         Þ¼                                (21.44)
                                                   ðln K p r   2
                                                dT          <T
                  Substituting from Eqn (21.44) into Eqn (21.43) gives

                                          vE     <T   Q p   E   Q p   E
                                               ¼          þ   ¼     þ                      (21.45)
                                          vT      z i F <T 2  T  z i FT  T
                                              p
                  Equation (21.45) shows that the change of emf is related to the heat of reaction, Q p .Now,for
               exothermic reactions Q p is negative, and this means that the emf of a cell based on an exothermic
               reaction decreases with temperature. This is in line with the effect of temperature on the degree of
               dissociation in a combustion process, as discussed in Chapter 12.

               21.2.4.1 Examples

               Example 1
                  A Pb–Hg fuel cell operates according to the following equation
                                            Pb þ Hg Cl 2 /PbCl 2 þ 2Hg
                                                   2
               and the heat of reaction, Q p ¼ 95,200 kJ/kmol Pb. The cell receives heat transfer of 8300 kJ/
               kmol Pb from the surroundings. Calculate the potential of the cell, and evaluate the electrical
               work produced per kilogram of reactants, assuming the following atomic weights: Pb – 207; Hg –
               200; Cl – 35.
               Solution
                  This problem can be solved by a macroscopic approach to the cell as a closed system. Applying the
               First Law gives
                                         zFEdn ¼ dU   dQ trans þ TdS   pdV
                                                ¼ dG   dQ trans
                                                                                   0
                  Now this value of energy transfer has to be used in Eqn (21.34) in place of DG , giving
                                                                                   T
                                          0

                                       DG   dQ trans     95200   8300
                                          T
                                 E ¼                ¼                 ¼ 0:5363 V
                                            zF            2   96487
                  The quantity of work obtained per kg of reactants is given in the following way. Assume the
               potential of the cell is not reduced by drawing a current, then the charge transferred through the cell per
               kmol Pb is 2F (because the valency of lead is 2). Hence the work done per kmol Pb is
                                    dW ¼ 2   96485   0:5363 ¼ 103500 kJ=kmol Pb
               giving the work per unit mass of reactants as
                                      dW          103500
                                          ¼                    ¼ 152:9kJ=kg
                                     m react  207 þ 2  ð200 þ 35Þ
               Example 2
                  A hydrogen–oxygen fuel cell operates at a constant temperature of 227 C and the hydrogen and

               oxygen are fed to the cell at 40 bar, and the water is taken from it at the same pressure. Evaluate the
               emf at this condition, and the heat transfer from the cell.
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