Oxidation%20and%20Reduction

1
Chapter 8

• Oxidation and Reduction

2
Redox Terminology

• Oxidation Generally,
• Involves Oxidation is an
• Loss of electrons increase in
• oxidation number.
• Reduction Reduction is a
• Involves decrease in
• Gain of electrons oxidation number.

3
Oxidation Numbers

• Specify the charge on an atom
• also called oxidation states
• Characterized by a set of 5 rules
• The oxidation number, Nox, of an atom as an
  element is zero.
• Example Na(s) and F2(g)

4
Oxidation Numbers

• Specify the charge on an atom
• also called oxidation states
• Characterized by a set of 5 rules
• The oxidation number of a monatomic ion is the
  same as its ion charge
• Example Na (Nox1) and F- (Nox-1)

5
Oxidation Numbers

• Specify the charge on an atom
• also called oxidation states
• Characterized by a set of 5 rules
• The algebraic sym of the oxidation numbers in a
  neutral, polyatomic compound is zero in a
  polyatomic ion, it is equal to the ion charge.
• Example HCl, H (Nox1), Cl- (Nox-1)
• (OH)-, O2- (Nox-2), H (Nox1)

6
Oxidation Numbers

• Specify the charge on an atom
• also called oxidation states
• Characterized by a set of 5 rules
• In combinations of elements, the more
  electronegative element has a negative oxidation
  number.
• Example HCN, H- (Nox-1), C4 (Nox4), N3-
  (Nox-3)

7
Oxidation Numbers

• Specify the charge on an atom
• also called oxidation states
• Characterized by a set of 5 rules
• Hydrogen usually has an oxidation number of 1,
  except with more electropositive elements, when
  it is 1.
• Example HF, H (Nox1), F- (Nox-1)
• NH3, N3 (Nox3), H- (Nox-1)

8
Oxidation Numbers

• Determining from relative electronegativities
• draw a Lewis structure for the desired molecule
• assume the shared electrons are completely
  owned by the more electronegative element

9
Oxidation Numbers

• Determining from relative electronegativities
• the difference in the number of valence electrons
  in the free ion and the number of valence
  electrons in the molecular ion is the oxidation
  number
• Cl, 7 8 -1 (Cl-)
• H, 1 0 1 (H)

10
Oxidation Numbers

• Polyatomic ions
• thiosulfate, S2O32-

O, 6 8 -2, (O2-) S, 6 1 5, (S5) S, 6
7 -1, (S-)
11
Oxidation Number vs. Formal Charge

• Formal charge
• electrons are shared equally
• the favored structures are those with the lowest
  formal charges
• Oxidation number
• electrons are shared unequally
• can have larger oxidation numbers

12
Periodic Variations of Oxidation Numbers

• An atoms maximum positive oxidation number is
  equal to the number of valence electrons
• Al, Ne3s23p1, maximum Nox 3
• Br, Ar4s23d103p5, maximum Nox 7

Ion Nox
ClO- 1
ClO2- 3
ClO3- 5
ClO4- 7
13
Redox Equations

• One substance is oxidized (increase in oxidation
  number) and one substance is reduced (decrease in
  oxidation number)
• Cu(s) 2Ag(aq) ? Cu2(aq) 2Ag(s)
• Cu, 0 ? 2, oxidized
• Ag, 1 ? 0, reduced

14
Redox Equations

• Can be put into two half-reactions
• Cu(s) 2Ag(aq) ? Cu2(aq) 2Ag(s)
• Cu(s) ? Cu2(aq) 2e-
• 2Ag(aq) 2e- ? 2Ag(s)

15
Redox Equations

• H2S(g) 2Fe3(aq) ? S(s) 2Fe2(aq) 2H(aq)
• H2S(g) ? S(s) 2H(aq) 2e-
• 2Fe3(aq) 2e- ? 2Fe2(aq)

16
Balancing Redox Equations

• Acidic reaction of permanganate (MnO4-) oxidizing
  Fe2 to produce Mn2 and Fe3
• Write the unbalanced equation
• MnO4-(aq) Fe2(aq) ? Mn2(aq) Fe3(aq)

17
Balancing Redox Equations

• Acidic reaction of permanganate (MnO4-) oxidizing
  Fe2 to produce Mn2 and Fe3
• Write the half-reactions
• Fe2(aq) ? Fe3(aq)
• MnO4-(aq) ? Mn2(aq)

18
Balancing Redox Equations

• Acidic reaction of permanganate (MnO4-) oxidizing
  Fe2 to produce Mn2 and Fe3
• Balance the half-reactions for mass
• Fe2(aq) ? Fe3(aq)
• MnO4-(aq) 8H(aq) ? Mn2(aq) 4H2O(l)

19
Balancing Redox Equations

• Acidic reaction of permanganate (MnO4-) oxidizing
  Fe2 to produce Mn2 and Fe3
• Balance the half-reactions for charge
• Fe2(aq) ? Fe3(aq) e-
• MnO4-(aq) 8H(aq) 5e- ? Mn2(aq) 4H2O(l)

20
Balancing Redox Equations

• Acidic reaction of permanganate (MnO4-) oxidizing
  Fe2 to produce Mn2 and Fe3
• Equivalate the charge in each half-reaction
• (Fe2(aq) ? Fe3(aq) e-)
• 5Fe2(aq) ? 5Fe3(aq) 5e-
• MnO4-(aq) 8H(aq) 5e- ? Mn2(aq) 4H2O(l)

21
Balancing Redox Equations

• Acidic reaction of permanganate (MnO4-) oxidizing
  Fe2 to produce Mn2 and Fe3
• Add the two half-reactions
• 5Fe2(aq) MnO4-(aq) 8H(aq) ? 5Fe3(aq)
  Mn2(aq) 4H2O(l)

22
Balancing Redox Equations

• Basic disproportionation reaction of Cl2 to
  chloride and chlorate ions
• Write the unbalanced equation
• Cl2(aq) ? Cl-(aq) ClO3-(aq)

23
Balancing Redox Equations

• Basic disproportionation reaction of Cl2 to
  chloride and chlorate ions
• Write the half-reactions
• Cl2(aq) ? Cl-(aq)
• Cl2(aq) ? ClO3-(aq)

24
Balancing Redox Equations

• Basic disproportionation reaction of Cl2 to
  chloride and chlorate ions
• Balance the half-reactions for mass
• Cl2(aq) ? 2Cl-(aq)
• Cl2(aq) 12OH-(aq) ? 2ClO3-(aq) 6H2O(l)

25
Balancing Redox Equations

• Basic disproportionation reaction of Cl2 to
  chloride and chlorate ions
• Balance the half-reactions for charge
• Cl2(aq) 2e- ? 2Cl-(aq)
• Cl2(aq) 12OH-(aq) ? 2ClO3-(aq) 6H2O(l) 10e-

26
Balancing Redox Equations

• Basic disproportionation reaction of Cl2 to
  chloride and chlorate ions
• Equivalate the charge in each reaction
• 5Cl2(aq) 10e- ? 10Cl-(aq)
• Cl2(aq) 12OH-(aq) ? 2ClO3-(aq) 6H2O(l) 10e-

27
Balancing Redox Equations

• Basic disproportionation reaction of Cl2 to
  chloride and chlorate ions
• 6. Add the two half-reactions
• 5Cl2(aq) Cl2(aq) 12OH-(aq) ? 10Cl-(aq)
  2ClO3-(aq) 6H2O(l)

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Half-reaction Quantitative Aspects

• Half-cell potential (reduction potentials)
• relative oxidizing or reducing power
• potential of a half-reaction relative to hydrogen
• 2H(aq) 2e- ? H2(g) E 0.00V
• Cu2(aq) 2e- ? Cu(s) E 0.34V

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Half-reaction Quantitative Aspects

• Half-cell potential (reduction potentials)
• the more positive E is what is being reduced
• the more negative E is what is being oxidized
• Cu2(aq) 2e- ? Cu(s) E 0.34V
• 2Ag(aq) 2e- ? 2Ag(s) E 0.80V
• 2Ag(aq) Cu(s) ? 2Ag(s) Cu2(aq)

30
Half-reaction Quantitative Aspects

• Half-cell potential (reduction potentials)
• If the sum of the reduction potentials for a
  redox reaction is positive, then the reaction is
  spontaneous
• Cu(s) ? Cu2(aq) 2e- E -0.34V
• (Ag(aq) e- ? Ag(s)) E 0.80V
• 2Ag(aq) 2e- ? 2Ag(s) E 0.80V
• 2Ag(aq) Cu(s) ? 2Ag(s) Cu2(aq)
• E 0.46V

31
Half-reaction Quantitative Aspects

• Half-cell potential (reduction potentials)
• concentration dependant
• Nernst equation
• R (8.31 VC/(molK)), n (moles of electrons), F
  (9.65 x 104 C/mol)

32
Half-reaction Quantitative Aspects

• Half-cell potential (reduction potentials)
• MnO4-(aq) 8H(aq) 5e- ? Mn2(aq) 4H2O(l)
• E 1.70V

33
Half-reaction Quantitative Aspects

• Half-cell potential (reduction potentials)
• MnO4-(aq) 8H(aq) 5e- ? Mn2(aq) 4H2O(l)
• E 1.70V
• E 1.70V (5.13x10-3V)ln(1.00)/((1.00)(1.0x10-4
  )8
• E 1.70V (5.13x10-3V)ln(1.0x1032)
• E 1.70V 0.38V 1.32V

34
Electrode Potentials as Thermodynamic Functions

• Potentials are measure of the free energy of the
  process.
• ?G -nFE
• Can use this to calculate the standard potential
  of an unknown half-reaction

35
Finding E Using Free Energy

• Fe3(aq) 3e- ? Fe(s) E ???
• Obtain the E from the following two
  half-reactions (cannot simply add them together)
• Fe3(aq) e- ? Fe2(aq) E 0.77V
• Fe2(aq) 2e- ? Fe(s) E -0.44V

36
Finding E Using Free Energy

• Fe3(aq) e- ? Fe2(aq) E 0.77V
• ?G -1(F)(0.77) -0.77F
• Fe2(aq) 2e- ? Fe(s) E -0.44V
• ?G -2(F)(-0.44) 0.88F
• Fe3(aq) 3e- ? Fe(s)
• ?G (-0.77F 0.88F) 0.11F
• E -?G/nF -(0.11F)/3F -0.04V

37
Latimer Diagrams

• Diagram of reduction potentials
• Used to write the half-reactions (must be
  balanced according to either acidic or basic
  media)

38
Writing Half-reactions from Latimer Diagrams
In acidic media, Fe3(aq) e- ? Fe2(aq) E
0.77V Fe2(aq) 2e- ? Fe(s) E
-0.44V Fe3(aq) 3e- ? Fe(s) E
-0.04V FeO42-(aq) 8H(aq) 3e- ? Fe3(aq)
4H2O(l) E 2.20V
39
Frost Diagrams

• Diagram of oxidation states
• free energy on the y-axis
• oxidation state on the x-axis
• Can construct Frost diagrams from Latimer diagrams

40
Constructing Frost Diagrams

• Plot of ?G/F or -nE vs. oxidation state
• For O2, Nox 0, -nE 0 (0,0)
• For H2O2, Nox -1, -nE (-1x0.68V)
  (-1,-0.68)
• For H2O, Nox -2, -nE (-1x1.78V) 0.68
  (-2,-2.46)

41
Oxygen Frost Diagram

• Using the calculated points, plot the data
• (0,0)
• (-1, -0.68)
• (-2, -2.46)
• The lowest point on the plot is the most
  thermodynamically stable species.

42
Frost Diagram Features

1. More thermodynamically stable states will be
   found lower in the diagram (Mn2 is the most
   stable)
2. A species on a convex curve will tend to
   disproportionate (MnO42- and Mn3)
3. A species on a concave curve will not
   disproportionate (MnO2)
4. A species that is high and on the right will be
   strongly oxidizing (MnO4-)
5. A species that is high and on the left will be
   strongly reducing (Mn0 is moderate)

43
Pourbaix Diagrams

• Plot of E vs. pH
• At varying pH values,
• More oxidized species are found at the top
• More reduced species are found at the bottom

44
Ellingham Diagrams

• Plots of free energy vs. temperature
• Useful for finding useable reactants for redox
  reactions with positive free energies

45
Biological Aspects

• Many biological processes depend upon redox
  reactions
• photosynthesis
• respiration
• nitrogen fixation

46
Biological Aspects

• Pourbaix diagrams are good to show the dependance
  upon the potential and pH