Page 233 - Natural Antioxidants, Applications in Foods of Animal Origin
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212                Natural Antioxidants: Applications in Foods of Animal Origin
  VetBooks.ir  TABLE 6.1  (Continued)



                          Potentio metric determination of peroxide values
            Titrant  [Analyte]  [Analyte]  [Titrant]  [Titrant]  Eo  Constant  n  E system
            (Liters)  Ox.   Red.    Red.    Ox.                        V
            0.0253                  0.0336  0.003  0.08  0.0592    2   0.050
            0.0260                  0.0342  0.003  0.08  0.0592    2   0.090
            0.0300                  0.0375  0.003  0.08  0.0592    2   0.050
            0.0350                  0.0412  0.004  0.08  0.0592    2   0.050
            0.0400                  0.0444  0.004  0.08  0.0592    2   0.050

               Components for potentiometric titration are:

               1.  Buret containing the titrant (sodium thiosulfate).
               2.  Known volume of the sample containing the analyte (extracted fat
                   or oil).
               3.  Reference electrode (saturated calomel or hydrogen electrode).
               4.  Electrode for measuring the potential, working electrode.
               5.  Meter to record the change during titration.

               To demonstrate the  application  of the  Nernst equation, titration  of
            hydrogen peroxide  (H O ) with sodium thiosulfate  (Na S O ) in a model
                                                                  4
                                  2
                                                              2 2
                                2
            system was calculated with Microsoft  Excel  2010 spreadsheet, E  value
                                                                        o
                                              ®
                                                    ™
            sourced from CRC Handbook of Chemistry and Physics 71st edition.
             Rxn. in aqueous system 2S O   –2  + H O  – 2e  <–> 2H O + 2e  + 2S O  –2
                                                     –
                                                                   –
                                     4  6    2  2            2           2  3
                         RT
                                ( )
            K eq    =    E cell  =  nF  × ln K eq  (6.6a); solving for K  (rxn driven strongly to the right)
                     o
            E     = (E  Red.) + (E  Ox.)   (6.6b)
                                o
                      o
             (Redox)
            E   (analyte at T=0)  = 0.281 V
             o
                       ne E o   ne E o
                            +
                    =
            Eq. point             =  0.200;vol. to eq. pt.   M ×   M ×    V 2  (6.6c)
                                                            V =
                                                   =
                         ne ne                         1    1    2
                            +
            K  = 1.43 × 10 57
             eq
            E  (Redox)    1.69 V
            E eq. pt.     0.200 V (estimate)
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