Page 104 - Chemical equilibria Volume 4
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80     Chemical Equilibria
                           3.3.2.6. Redox reactions and inversion temperature in a metal-oxide
                           Ellingham diagram
                             Ellingham diagrams  enable us to discover, at a given temperature, the
                           direction of the reaction between a metal and an oxide of a different metal.
                           By way of example, let us consider the redox reaction between silicon and
                           chrome oxide, which would be written as:

                                Si + 2/3Cr 2O 3  = SiO 2 + 4/3Cr                        [3R.13]

                             To this end, let us consider the following two reactions of chromium and
                           silicon with oxygen (reactions [3R.14] and [3R.15]), each represented by an
                           Ellingham line (Figure 3.9(a)):


                                4/3Cr + O 2  = 2/3Cr 2O 3                               [3R.14]
                                Si + O 2  = SiO 2                                       [3R.15]

                             We can see that the balance equation for reaction [3R.13] is the
                           difference between the two balance  equations for reactions  [3R.14] and
                           [3R.15], so according to relation [3.22], its affinity will be:

                                 A 13  = A 15  − A 14                                    [3.52]

                             Thus, as a function of the standard Gibbs energies for the two reactions
                           [3R.14] and [3R.15]:

                                      −
                                                        −
                                                          −
                                                                          )
                                 A 13  = ( Δ g°+  RT ln P O 2 ) ( Δ g°+ RT ln P O 2       [3.53]
                                         15
                                                             14
                             Hence, this affinity is only a function of the standard Gibbs energies of
                           the two reactions [3R.14] and [3R.15]:
                                    =  Δ A  Δ g° −  g°                                   [3.54]
                                  13   14     15
                             Thus, this affinity will be positive, so reaction [3R.13] is possible from
                           left to right if:

                                 Δ  14  Δ g°>  15 g°                                     [3.55]


                             This means that the Ellingham line of reaction [3R.15] must be placed
                           below that of reaction [3R.14], as shown by Figure 3.9(a).
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