Page 125 - Hydrogeology Principles and Practice
P. 125

HYDC03  12/5/05  5:37 PM  Page 108






                 108    Chapter Three


                 Table 3.9 Examples of oxidation states for various compounds that occur in groundwater. The oxidation state of free elements, whether
                 in atomic or molecular form, is zero. Other rules for assigning oxidation states include: the oxidation state of an element in simple ionic
                 form is equal to the charge on the ion; the sum of oxidation states is zero for molecules; and for ion pairs or complexes it is equal to the
                 formal charge on the species. After Freeze and Cherry (1979).
                 Carbon compounds      Sulphur compounds      Nitrogen compounds     Iron compounds

                 Substance   C state   Substance    S state   Substance   N state    Substance   Fe state

                 HCO − 3     +IV       S            0         N 2         0          Fe          0
                 CO 2−       +IV       H S          −II       SCN −       +II       FeO          +II
                   3
                                        2
                 CO 2        +IV       HS −         −II       N O         −III      Fe(OH) 2     +II
                                                               2
                 CH O        0         FeS 2        −I        NH 4 +      +III      FeCO 3       +II
                   2
                 C H O       0         FeS          −II       NO − 2      +V         FeO 3       +III
                  6 12 6
                 CH 4        −IV       SO 2−        +IV       NO − 3      −III      Fe(OH) 3     +III
                                         3
                 CH OH       −II       SO 2−        +VI       HCN         −I         FeOOH       +III
                                         4
                   3
                 and expresses the net effect of the electron transfer  oxidized forms and electrons are written on the left
                 with the absence of free electrons, thus:   and the reduced products on the right.
                                                               Rearrangement of equation 3.33 gives the electron
                 1       2+    +     3+                             −
                 /2O + 2Fe + 2H = 2Fe + H O         eq. 3.30  activity [e ] for a half-reaction as:
                    2                     2
                 By expressing redox reactions as half-reactions, the  [ ⎧ reductants] ⎫ 1 /n
                                                               −
                                                                 =
                 concept of pe is used to describe the relative electron  []   ⎨  ⎬             eq. 3.34
                                                              e
                 activity where:                                  ⎩ [oxidants] K  ⎭
                          −
                 pe =−log [e ]                      eq. 3.31  Rewriting equation 3.34 by taking the negative log-
                        10
                                                             arithm of both sides yields:
                 pe is a dimensionless quantity and is a measure of the
                                                                   −
                 oxidizing or reducing tendency of the solution where  −log [e ] = pe
                                                                 10
                 pe and pH are functions of the free energy involved in
                 the transfer of 1 mole of electrons or protons, respect-  1  ⎧      [reductants] ⎫
                 ively, during a redox reaction.                     =  ⎨ log K  −  log 10     ⎬
                                                                            10
                                                                       n  ⎩           [oxidants]  ⎭
                   For the general half-reaction:
                                                                                                eq. 3.35
                           −
                 oxidants + ne = reductants         eq. 3.32
                                                             When a half-reaction is written in terms of a single
                 where n is the number of electrons transferred, then,  electron transfer, or n = 1, the log K term is written
                                                                                       10
                                                                 o
                 from the law of mass action:                as pe such that:
                                                               =
                                                                   o
                   =
                 K    [reductants] n                eq. 3.33  pe   pe −  log 10  [reductants]   eq. 3.36
                             −
                     [oxidants][ e ]                                       [oxidants]
                 A numerical value for such an equilibrium constant  Tabulations of thermodynamic data for redox reac-
                                                                                         o
                 can be computed using Gibbs free energy data for con-  tions are commonly expressed as pe values. A set of
                 ditions at 25°C and 1 atmosphere pressure. By con-  reduction reactions of importance in groundwater,
                                                                                        o
                 vention, the equilibrium constant for a half-reaction  together with their respective pe values, is listed
                 is always expressed in the reduction form. The   in Table 3.10. The reactions are listed on the basis
   120   121   122   123   124   125   126   127   128   129   130