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					  REDOX

Electrochemistry
       8-1
Redox reactions involve the transfer of the
                electron.

   • Spontaneous redox reactions can
     transfer energy
     – Electrons (electricity)
     – Heat
   • Non-spontaneous redox reactions can
     be made to happen with electricity.
    Trends in Oxidation
    and Reduction
Active metals:
                 Lose electrons easily
                 Are easily oxidized
                 Are strong reducing agents



Active nonmetals:
              Gain electrons easily
              Are easily reduced
              Are strong oxidizing agents
    There are five basic rules for the
   determination of oxidation number:
• Rule 1: The oxidation number for any atom
  in its elementary state is 0.
• Rule 2: The oxidation number for any simple
  ion is the charge on the ion.
•     a.      The oxidation number of alkali
  metals in compounds is 1+
       (Li1+ , Na1+ , K1+ , Rb1+ , Cs1+ , Fr1+ ).
•     b.      The oxidation number of alkaline-
  earth metals in compounds is 2+
•        (Mg2+ , Ca2+ , and Ba2+ ).
 Find the oxidation number for
   magnesium and chlorine


Mg  Cl 2  Mg Cl2

  0       0      2   1
 Mg  Cl 2  Mg Cl 2
• Rule 3: The oxidation number for
  oxygen usually is 2-. In peroxides, it is
  1-.



      1     2                   1      1
     H2O                         H 2 O2
            water                      peroxide
• Rule 4: The oxidation number for hydrogen
  is 1+ in all its compounds except in metallic
  hydrides like NaH or BaH2 , where it is 1-.
• Rule 5: All other oxidation numbers are
  assigned so that the sum of oxidation
  numbers equals the net charge on the
  molecule or polyatomic ion.
  Find the oxidation numbers
• H2S                 Ca(OH)2


        1   2            2     2 1
        H2 S              Ca(O H ) 2
   2(+1) + (-2) = 0     (+2) + 2(-2) + 2(+1) = 0
      H     S            Ca      O       H
Find the oxidation numbers


      ? 2            ? 2
                              2
    N O3              S O4
   X + 3(-2) = -1   X + 4(-2) = -2
   N     O          S     O

        X = +5           X = +6
Find the oxidation number of Cr:
• Ex: Cr2O72-

    2X + 7(-2) = -2
    Cr     O

         X = +6
    Write the oxidation numbers for
              each atom:
                0
•   S8
                 S = +4 and O = -2
•   SO2
•   S2O32-       S = +2 and O = -2

•   SO42-        S = +6 and O = -2

•   MgSO4        Mg = +2, S = +6 and O = -2

•   H2SO4        H = +1, S = +6 and O = -2
    Find the oxidation numbers
•   P4        0

•   P2O5      P = +5 and O = -2

•   PCl5      P = +5 and Cl = -1

•   H3PO4     H = +1, P = +5 and O = -2

•   PO43-     P = +5 and O = -2
•   Na3PO4    Na = +1, P = +5 and O = -2
  A summary of terminology for oxidation-
       reduction (redox) reactions
                        e-


                X      transfer
                                  Y
                      or shift of
                      electrons


X loses electron(s)                   Y gains electron(s)

X is oxidized                         Y is reduced

X is the reducing agent               Y is the oxidizing agent

X increases its                       Y decreases its
oxidation number                      oxidation number
     Not all reactions are redox
   1 5 2              1   1      1   1           1 5 2
  Ag N O3 (aq)  Na Cl (aq)  Ag Cl ( s)  Na N O3 (aq)


  1 2 1          1    6 2            1   6 2              1   2
2 Na O H (aq)  H 2 S O 4 (aq)   Na 2 S O 4 (aq)  H 2 O(l )


             Reactions in which there has been no change in
             oxidation number are not redox rxns.
               Oxidation
  - is defined as the loss of an electron
•     (LEO - loss of electron; oxidation)
- when oxidation happens, charge
   becomes more positive
• Fe → Fe2+ + 2e-

 4 Fe (s) + 3 O2 (aq) →        2 Fe2O3 (s)
               Reduction
• is defined as the gain of an electron
• (GER- gain of electrons – reduction
• when reduction happens, charge becomes
  more negative
•     Fe2+ + 2e- → Fe

•    2 Fe2 O3 (s) → 4 Fe (s) + 3 O2 (aq)
•    N2O → NO
•    Has the nitrogen gained or lost
  electrons?
• N2O →        NO
  +1            +2
Nitrogen has lost one electron and is
  therefore being oxidized.
• Just like the Bronsted-Lowry theory of
  acids, oxidation and reduction also
  happens in pairs; a species cannot
  donate an electron unless another
  species gains the electron. Since
  oxidation and reduction always happen
  together, these are most often called
  redox reactions.
• These pairs of reactions are called half-
  reactions. For example, in the reaction
•    Cu2+ (aq) + Fe (s) → Cu (s) + Fe2+ (aq)
•    two half-reactions occur; iron is oxidized
  while the copper (II) ion is reduced. Seen
  another way, the copper (II) ion causes the
  oxidation of the iron; it is the oxidizing
  agent. Likewise, the iron is the reducing
  agent for the copper (II) ion.
Reducing and oxidizing agents

   The substance reduced is the oxidizing agent
   The substance oxidized is the reducing agent
               0             1
                                     
              Na  Na  e
    Sodium is oxidized – it is the reducing agent
               0                   1
                         
             Cl  e  Cl
    Chlorine is reduced – it is the oxidizing agent
          FIND THE OXIDATION
               NUMBERS:
•   2H2(g) + O2(g)                    2H2O(g)


    0             0                   +1   -2



• 2H2(g) + O2(g)                     2H2O(g)


        The O.N. of H increases; it is oxidized; it is the reducing agent.
        The O.N. of O decreases; it is reduced; it is the oxidizing agent.
Find the oxidation numbers:

2Al(s)+ 3H2SO4(aq)                  Al2(SO4)3(aq) + 3H2(g)

  0         +1   +6 -2                +3 +6 -2                  0



 2Al(s) + 3H2SO4(aq)                 Al2(SO4)3(aq) +3H2(g)


  The O.N. of Al increases; it is oxidized; it is the reducing agent.
  The O.N. of H decreases; it is reduced; it is the oxidizing agent.
      Find the oxidation numbers
•    PbO(s) + CO(g)               →         Pb(s) + CO2(g)

    +2 -2         +2 -2                 0          +4 -2



• PbO(s) + CO(g)                  → Pb(s) + CO2(g)


        The O.N. of C increases; it is oxidized; it is the reducing agent.

        The O.N. of Pb decreases; it is reduced; it is the oxidizing agent.
     0            0                    1   1
2 Na  Cl 2  2 Na Cl
           0              1
                                  
          Na  Na  e
Each sodium atom loses one electron:
          Lose Electrons = Oxidation
      0                            1
                      
    Cl 2  2e  2 Cl
Each chlorine atom gains one electron:
           Gain Electrons = Reduction
Using oxidation states work out what is being
    oxidized and what is being reduced




a) 6Na (s)+ N2 (g)  2Na3N (s)
b) Mg (s)+ Cl2 (g)  MgCl2 (s)
c) 4Fe (s)+ 3O2 (g)  2Fe2O3 (s)
d) Ca (s)+ C (s)  CaC2 (s)
e) MnO4-(aq) + S2O32-(aq)  Mn 2+(aq) +S2O62(aq)
f) VO2+(aq) + Zn (s)  VO 2+(aq) + Zn2+(aq)

				
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