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74   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    75


                      4.8.1  E-pH Diagram of Water
                      The following example illustrates how the stability or predominance
                      diagram of water can be constructed from its basic thermodynamic
                      information.  Equation  (4.33)  describes  the  equilibrium  between
                      hydrogen ions and hydrogen gas in an aqueous environment:

                                            +
                                                −
                                         2H +  2e   H (g)                 (4.33)
                                                      2
                      Adding sufficient OH  to both sides of reaction in Eq. (4.33) results in
                                        −
                      Eq. (4.34) in neutral or alkaline solutions:
                                          +
                                              −
                                                            −
                                     2H O 2e     H (g) +  2OH             (4.34)
                                        2
                                                   2
                         At  higher  pH  than  neutral,  Eq.  (4.34)  is  a  more  appropriate
                      representation.  However,  since  the  concentrations  of  H   and  OH
                                                                      +
                                                                               −
                      ions are related by the dissociation constant of water expressed in
                      Eq. (4.35), Eq. (4.33) and (4.34) basically represent the same reaction
                      for which the thermodynamic behavior can be expressed by Nernst
                      equation.
                                                      +
                                                           −
                                                       ]
                           H O   H +  OH with K eq  =  [H [OH ]  =  10 − 14  at 25 C    (4.35)
                                          −
                                    +
                                                                      °
                            2
                                                     [H O]
                                                        2
                                                    RT    [H ] 2
                                                            +
                                    E  +  =  E 0  +  +  ln                 (4.36)
                                     H /H 2   H /H 2  nF  p
                                                           H 2
                      that becomes Eq. (4.37) at 25°C and the partial pressure of hydrogen
                      (p ) of value unity.
                        H 2
                                      E  +  =  E 0  +  −  0.059 pH         (4.37)
                                       H /H 2  H /H 2
                         Equation (4.33) and its alkaline or basic form, Eq. (4.34), delineate
                      the stability of water in a reducing environment and are represented
                      in a graphical form by the sloping line (a) on the Pourbaix diagram in
                      Fig.  4.10.  Below  the  equilibrium  reaction  shown  as  line  (a)  in  this
                      figure,  the  decomposition  of  H O  into  hydrogen  is  favored  while
                                                 2
                      water  is  thermodynamically  stable  above  the  same  line  (a).  As
                      potential becomes more positive or noble, water can be decomposed
                      into  its  other  constituent,  oxygen,  as  illustrated  in  Eqs.  (4.38)  and
                      (4.39) for respectively the acidic form and neutral or basic form of the
                      same process.
                                       O +  4H +  4e −    2 H O           (4.38)
                                             +
                                        2
                                                         2
                                      O +  2H O 4e −    4OH               (4.39)
                                                           −
                                               +
                                        2    2
                         And again these equations are equivalent and only reflect the pH
                      condition of the environment. The corresponding Nernst expression
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