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Potentiometry [LAB] - PowerPoint

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Reference   Indicator
Electrode   Electrode




                        3
NHE
PPE
SHE   Ag/AgCl/Cl-   SCE

                     4
              1st Kind




             Indicator
             Electrodes


Inert Kind                2nd Kind

                                     5
       Metal rod immersed in its solution
                   M0 / M n+(xM)

      Cu / CuSO4
      Cd / CdSO4
      Zn / ZnSO4


ُ = E0 + (0.0591 / n) log [Mn+] / [M0]
E

                                            6
Metal rod / sparingly soluble salt / anion solution.



   Ag / AgCl (sat’d) / KCl (xM)
   Hg / Hg2Cl2 (sat’d) / KCl (xM).

     Ac Bd + ne- ↔ cA0 + dB-


  ُ = E0 + (0.0591 / n) log [AcBd ] / [A0]c* [B-]d
  E
                                                       7
Pt, Au, C electrodes are used for electrical
     contact in the following systems:

          System            Example
     Solution/solution      Fe3+/Fe2+
       Gas/solution         H2/H+
      Liquid/solution       Br2/Br-

                                               8
            Electrochemical
                  cells



  Galvanic                Electrolytic
or voltaic cell               cell


                                         9
                Determination
Determination
                     of
     of
                  unknown
    E cell
                concentration


                                10
11
Cu0 / Cu 2 +(0.1M) // Ag+ (0.2M) / Ag 0
         Anode                       Cathod




        Cu0   ↔ Cu   2 +   +   2e-    E0 Cu2+/Cu0 = 0.337volt



E left = E0 Cu2+/Cu0+ 0.0591 / n log [Cu+2] / [Cu0]
   = 0.337 + 0.0591 / 2 log (0.1 / 1)
   = 0.307 volt.

                                                                12
  Cu0 / Cu 2 +(0.1M) // Ag+ (0.2M) / Ag 0
        Anode                   Cathod




       Ag+   +e-   ↔ Ag   0       E0 Ag+ / Ag0 = 0.799 volt



E right = E0 Ag+/Ag0 + (0.0591 / n) log [Ag+] / [Ag0]


 = 0.799 + 0.0591 / 1 log (0.2 / 1)
 = 0.757 volt.
                                                              13
     Cu0 / Cu 2 + // Ag+ (0.2M) / Ag 0
         Anode                  Cathod




   E cell = 0.757 – 0.307 = + 0.45 volts



The reaction proceeds in the written direction.
                                                  14
Cu0 / Cu 2 + // Ag+ (0.2M) / Ag 0

        Cu0 ↔ Cu 2 + + 2e-
        2Ag+ +2e- ↔ 2Ag 0


    Cu 0 + 2 Ag + ↔ Cu2+ + 2 Ag0



                                    15
   Ag 0 / Ag+ (0.2M) // Cu 2 +(0.1M) / Cu0
         Anode                  Cathod


      E cell = 0.307 –0.757 = - 0.45 volts




The reaction proceeds in the opposite direction.

          Cu 0 + 2 Ag + ↔ Cu2+ + 2 Ag0
                                                   16
    Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M)              //
           Fe2+ (0.01M) , Fe3+(0.025M) / Pt0




   U4+ + 2H2O ↔ UO22+ + 4H+ + 2e-       E0 UO22+,H+/U4+ = 0.334volt



E left = E0 UO22+,H+/U4+ + 0.0591 / n log [UO22+] [H+]4 / [U4+]
    = 0.334 + 0.0591 / 2 log [0.015* (0.03)4/ 0.015]
    = 0.12 volt.

                                                                 17
Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M)             //
       Fe2+ (0.01M) , Fe3+(0.025M) / Pt0




Fe3+ +e- ↔ Fe2+          E0 Fe3+/Fe2+ = 0.771volt




E right = E0 Fe3+,Fe2+ + 0.0591 / n log [Fe3+] / [Fe2+]
     = 0.771 + 0.0591 / 1 log [0.025/ 0.01]
     = 0.795 volt.

                                                         18
Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M)       //
        Fe2+ (0.01M) , Fe3+(0.025M) / Pt0




   E cell = 0.795 – 0.12 = + 0.675 volts



The reaction proceeds in the written direction.
                                                   19
Pt0 / UO22+(0.015M),U4+(0.2M),H+(0.03M)   //
      Fe2+ (0.01M) , Fe3+(0.025M) / Pt0

       U4+ + 2H2O ↔ UO22+ + 4H+ + 2e-
       2Fe3+ + 2e- ↔ 2Fe2+


      U4+ +2Fe3+ +2H2O ↔ UO22+ + 2Fe2+ +4H+



                                               20
Pt0/VO2+(0.25M) , V3+ (0.1M), H+ (0.001M)
    // Tl+(0.05M), Tl3+(0.1M),Pt0




                                       21
     Pb0 / PbSO4 (sat’d),SO42-(0.2M) //
             Sn4+(0.25M) / Sn2+(0.15M)


               PbSO4 ↔ Pb2+ + SO42-
                   Pb2+ + 2e- ↔ Pb0
            PbSO4+ 2e- ↔ Pb0 + SO42-
E left = E0 PbSO4/Pb0,SO4 +
             0.0591 / n log [PbSO4] / [Pb0][SO42-]
     =-0.35 + 0.0591 / 2 log [1/1*0.2]
                                                 22
     = -0.33 volt.
     Pb0 / PbSO4 (sat’d),SO42-(0.2M) //
             Sn4+(0.25M) / Sn2+(0.15M)



            Sn4++2e- ↔ Sn2+

E right = E0Sn4+/Sn2+ + 0.0591 / n log [Sn4+] / [Sn2+]
       = 0.154 + 0.0591 / 2 log [0.25/ 0.15]
       = 0.16 volt.



                                                    23
    Pb0 / PbSO4 (sat’d),SO42-(0.2M) //
            Sn4+(0.25M) / Sn2+(0.15M)




   E cell = 0.16 – (-0.337) = + 0.49 volts




The reaction proceeds in the written direction.
                                                  24
Pb0 / PbSO4 (sat’d),SO42-(0.2M) //
        Sn4+(0.25M) / Sn2+(0.15M)

    Pb0 + SO42- ↔ PbSO4+ 2e-
    Sn4++2e-    ↔ Sn2+


   Pb0 + SO42-+ Sn4+ ↔ PbSO4 + Sn2+


                                      25
    Pt0/ I2(S) , I- (0.02M)
              //
Fe2+(0.04M), Fe3+(0.04M)/Pt0



                               26
Cu0 / CuSO4 (0.05M)
         //
CdSO4 (0.02M) / Cd0



                      27
Ag0 / AgBr(sat’d), Br-(0.04M)
              //
H+(10 4-M),H2(gas)(0.9atm),Pt0




                                 28
0.09M CE // V2+(0.2M) , V3+(0.3M)/Pt0


           Calomel Electrode




Hg0l , Hg2Cl2 (sat’d) , KCl (X M = 0.09M)


                                            29
Hg0, Hg2Cl2 (sat’d) ,Cl-( 0.09M) // V2+(0.2M) , 3+(0.3M)/Pt0



                     Hg2Cl2     ↔ Hg22+ + 2Cl-
                    Hg22+ + 2e- ↔ 2Hg0
                Hg2Cl2+ 2e- ↔ 2Hg0 + 2Cl-

   E left = E0CE + 0.0591 / n log [Hg2Cl2] / [Hg0]2[Cl-]2
       =0.268 + (0.0591 / 2) log [1/1*(0.09)2]
       = 0.329 volt.
                                                       30
Hg0, Hg2Cl2 (sat’d) ,Cl-( 0.09M) // V2+(0.2M) , 3+(0.3M)/Pt0




                 V3++e- ↔ V2+

     E right = E0V3+/V2+ + (0.0591 / n) log [V3+] / [V2+]
           = - 0.256 + (0.0591 / 1) log [0.3/ 0.2]
          = - 0.245 volt.



                                                       31
Hg0, Hg2Cl2 (sat’d) ,Cl-( 0.09M) // V2+(0.2M) , 3+(0.3M)/Pt0




      E cell = -0.245 – 0.329 = - 0.574volts




  The reaction proceeds in the OPPOSITE direction.
                                                       32
 2Hg0 + 2Cl-      ↔ Hg2Cl2+ 2e-
        2V3++ 2e- ↔ 2V2+


2Hg0 + 2Cl- +2V3+↔ Hg2Cl2+ 2V2+




                                  33
34
Pt0 / Fe2+(0.05M),Fe3+(0.25) // Ag+ (xM) / Ag0



            Ecell = -0.106 volt
        E0 Fe3+,Fe2+ = 0.771 volt
        E0 Ag+ / Ag0 = 0.799 volt


                                             35
 Pt0 / Fe2+(0.05M),Fe3+(0.25) // Ag+ (xM) / Ag0




Fe3+ +e- ↔ Fe2+          E0 Fe3+/Fe2+ = 0.771volt




 E left = E0 Fe3+,Fe2+ + 0.0591 / n log [Fe3+] / [Fe2+]
     = 0.771 + 0.0591 / 1 log [0.025/ 0.05]
     = 0.8123 volt.

                                                     36
   Pt0 / Fe2+(0.05M),Fe3+(0.25) // Ag+ (xM) / Ag0


    Ag+ +e- ↔ Ag 0                   E0 Ag+ / Ag0 = 0.799 volt


   E right = E0 Ag+/Ag0 + (0.0591 / n) log [Ag+] / [Ag0]
         = 0.799 + 0.0591 / 1 log ( x / 1)
Ecell    = E right – E left
-0.106   = {0.799 + 0.0591 log x} – 0.8123
Log Ag+ = - 1.56
[Ag+]    = 0.027 M
                                                                 37
SCE // Cu2+ (xM) / Cu0




                         38
            Equimolar ratio
O                             OH


    +   2H+ + 2e-


O                             OH



                                   39
    O                                   OH


          +   2H+ + 2e-


    O                                   OH


E QE = E0QE + (0.0591/n) log [Q]*[H+]2 / [H2Q]
     = 0.699 + 0.0591/2 log [H+]2
     = 0.699 + 0.0591 log [H+]
     =0.699 - 0.0591 pH
                                                 40
SCE // QE (sat’d), H+(xM)/Pt0




       Ecell = 0.313 volt
      E0 QE = 0.699 volt
      E SCE = 0.246 volt

                                41
SCE // QE (sat’d), H+(xM)/Pt0

      E cell = E QE – E SCE
  0.313 = 0.699- 0.0591 pH -0.246
     ………………………….
            pH = 2.36


                                    42
43
44
SCE // (H+) = xM , GE

   E cell    pH

  0.2094    4.006

  -0.3011     ?

  0.0020      ?
                        45
         SCE // (H+) = xM , GE
E cell   = E GE – E SCE
         = L - 0.0591 pH -0.246
0.2094 = L – 0.0591(4.006) – 0.246
-0.3011 = +L – 0.0591 (pH) – 0.246
0.2094+0.3011=0.0591(-4.006+pH)
pH = 12.6

                                     46
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