Page 512 - Advanced thermodynamics for engineers
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21.2 THEORY OF FUEL CELLS        505




                  Hence
                            dQ oc ¼ 214852 kJ=kmol Zn ¼ u Cu þ u       u Zn   u             (21.4)
                                                             ZnSO 4 ðaqÞ     CuSO 4 ðaqÞ
                  The reaction described above is basically an open circuit reaction for a Daniell cell, and d Q oc
               represents the open circuit energy released. If, as is more usual, the reaction takes place while work is
               being taken from the cell then supposing a current I flows for time, t, then, from Faraday’s laws of
               electrolysis, the ratio of the amount of substance (Zn) dissolved to the valency of the element, n/z,is
               proportional to the electric charge passed, i.e.

                                              n=z f Q; i:e: n=z ¼ FQ                        (21.5)
                  When n/z ¼ 1 kmol, Q ¼ 96485 kC (where kC denotes kilo-coulomb), and hence F ¼ 96485 kC/
               kmol. This is known as the Faraday constant, and is the product of Avogadro’s number and the charge
               on a proton. Now, if the potential between the electrodes is E oc , then the work done is
                                     dW ¼ 2E oc F volt:kC=kmol ¼ 2E oc F kJ=kmol            (21.6)
                  In a Daniell cell, the potential at zero current (i.e. on open circuit), which is called the electromotive
               force (emf),E oc ¼ 1.107 Vat 25 C. If it is assumed that the cell can maintain this potential at low

               currents, then
                                     dW ¼ 2   1:107   96485 ¼ 213618 kJ=kmol:
                  If the reaction described above takes place isothermally in a closed system, then it must obey the
               First Law which this time is applied to the closed circuit system and gives

                                      dU ¼ dQ   dW
                                                                                            (21.7)
                                         ¼ dQ R   dW ¼ dQ R   213618 kJ=kmol
                  Now the change in internal energy is simply due to the chemical changes taking place, and for an
               isothermal reaction must be equal to dQ oc defined in Eqn (21.4), giving
                                     dQ R ¼ 214852 þ 213618 ¼ 1234 kJ=kmol:                 (21.8)

                  This indicates that heat must be transferred from the cell to maintain the temperature constant. This
               heat transfer is a measure of the change of entropy contained in the bonds of the product (ZnSO 4 )
               compared to the reactant (CuSO 4 ), and

                                          DS ¼ 1230=298 ¼ 4:141 kJ=K:                       (21.9)
                  In this approach, the Daniell cell has been treated as a thermodynamic system – a black box. It is
               possible to develop this approach further to evaluate the electrical performance of the cell. Suppose
               that an amount of substance of Zn, dn, enters the solution at the negative pole, then it will carry with it a
               charge of zFdn, where z is the valency (or charge number of the cell reaction) of the Zn, and is 2 in this
               case. The Coulombic forces in the cell are such that an equal and opposite charge has to be absorbed by
               the copper electrode, and this is achieved by the copper ions absorbing electrons which have flowed
               around the external circuit. The maximum work that can be done is achieved if the cell is reversible,
               and the potential is equal to the open circuit potential; thus

                                                   dW ¼ zFEdn                              (21.10)
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