Page 80 - Corrosion Engineering Principles and Practice
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58   C h a p t e r   4                           C o r r o s i o n   T h e r m o d y n a m i c s    59


                         There is also a parasitic reaction at the aluminum anode that has
                      to be considered because it has serious safety implications, that is,
                      the  production  of  hydrogen  gas  from  the  reduction  of  water
                      described in Eq. (4.17):

                                      2H O +  2e −  →  H +  2OH −          (4.17)
                                        2
                                                     2
                         There is however only one reaction on the cathode, that is, the
                      reduction of oxygen shown in Eq. (4.18):

                                      O + 2H O + 4e −  →  4OH              (4.18)
                                                           −
                                        2
                                             2
                         The overall cell voltage can be calculated from thermodynamic
                      data  by  computing  Gibbs  free  energy  for  the  individual  species
                      involved  in  the  global  reaction  described  in  Eq.  (4.19)  using  the
                      coefficients expressed in that equation:

                                          −
                                  4Al + 4OH + 3O →  4AlO 2 −  + 2H O       (4.19)
                                                2
                                                              2
                      4.4.2  Detailed Calculations
                      Calculate  G   for  each  species.  The  free  energy  of  a  substance,  for
                                0
                      which heat capacity data are available, can be calculated as a function
                      of temperature using Eq. (4.20):

                                                       T2 2  C 0  T  2
                              G 0  =  G 0  − S 0    −  T 2 ∫  p  dT + C dT     (4.20)
                                                               ∫
                                                                 0
                                (T )
                                      (T )
                                 2
                                       1
                                            (T ) 1   T − T    T  p
                                                 2  1    T1  T  1
                         For  pure  substances  (i.e.,  solids,  liquids,  and  gases)  the  heat
                      capacity C  is often expressed, as in Table 4.3, as a function of the
                               0
                               p
                      absolute temperature:
                                          C = A BT CT                      (4.21)
                                                   +
                                           0
                                               +
                                                        2 −
                                           p
                         For  ionic  substances,  one  has  to  use  another  method,  such  as
                      proposed by Criss and Cobble in 1964 [1], to obtain the heat capacity,
                      provided the temperature does not rise above 200°C. The expression
                      of  the  ionic  capacity  in  equation  (Eq.  4.22)  makes  use  of  absolute
                      entropy values and the parameters a and b contained in Table 4.3.
                                                                 .
                                                              298 16     (4.22)
                                0
                               C = ( .  a bS 0 ( 298  K) )( T −  298 16)/ln    T T 2   
                                                       .
                                   4 186 +
                                p
                                                  2
                         By combining Eqs. (4.21) or (4.22) with Eq. (4.20) one can obtain
                      the free energy in Eq. (4.23) at any given temperature by using the
                      fundamental data contained in Table 4.3 and Table 4.4:
                                                                  T
                         G 0  T ( )  =  G 0 (298 K)  + ( C −  S 0 (298 K)  ⋅ ) ( T −  298 16)− T ln    298 16. 2      ⋅C   (4.23)
                                                             ⋅
                                                          −
                                                                         0
                                                       .
                                        0
                                                                         p
                                                  2
                                        p
                                                             2
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