1. 9.1 OXIDATION AND REDUCTION
TOPIC 9 REDOX PROCESSES
9.1
OXIDATION AND REDUCTION

2. ESSENTIAL IDEA
Redox (reduction-oxidation) reactions play a key role in many chemical and biochemical processes.
NATURE OF SCIENCE (1.9)
How evidence is used – changes in the definition of oxidation and reduction from one involving specific elements (oxygen and hydrogen), to one involving electron transfer, to one invoking oxidation numbers is a good example of the way that scientists broaden similarities to general principles.

3. INTERNATIONAL-MINDEDNESS
Access to a supply of clean drinking water has been recognized by the United Nations as a fundamental human right, yet it is estimated that over one billion people lack this provision. Disinfection of water supplies commonly uses oxidizing agents such as chlorine or ozone to kill microbial pathogens.

4. THEORY OF KNOWLEDGE
Chemistry has developed a systematic language that has resulted in older names becoming obsolete. What has been lost and gained in this process? Oxidation states are useful when explaining redox reactions. Are artificial conversions a useful or valid way of clarifying knowledge?

5. UNDERSTANDING/KEY IDEA 9.1.A
Oxidation and reduction can be considered in terms of oxygen gain/hydrogen loss, electron transfer or change in oxidation number.

6. Early definitions for oxidation and reduction were based upon observations of the gain and loss of oxygen and hydrogen during chemical change.
Oxidation: gain of oxygen or loss of hydrogen
Reduction: loss of oxygen or gain of hydrogen
It is now recognized that oxidation/reduction occur whenever there is a shift in electron density from one atom to another, whether complete or partial.

7. Oxidation is the loss of electrons.
Atom gets more positive.
Reduction is the gain of electrons.
Atom gets more negative.
Reactions that involve oxidation and reduction are called redox reactions.
You can never have one without the other.

8. Lions roar http://www.youtube.com/watch?v=_22gJ5kB31k
Lion is also Know as LEO

10. Oxidation Reduction Oxidation is the loss of electrons;
Reduction is the gain of electrons
Oxidation and reduction go together.
Whenever a substance loses electrons and another substance gains electrons

11. Oxidation Numbers
Oxidation Numbers are a system that we can use to keep track of electron transfers

12. Oxidation Numbers Oxidation numbers always refer to single atoms
The oxidation number of an uncombined element is always 0
O2, H2, Ne Zn
The oxidation number of Hydrogen is usually +1 Hydrides are an exception They are -1
HCl, H2SO4
The oxidation number of Oxygen is usually -2 Peroxides are an exception They are –1 In OF2 oxygen is a +2
H2O, NO2, et
Oxidation numbers of monatomic ions follow the charge of the ion
O2-, Zn2+
The sum of oxidation numbers is zero for a neutral compound. It is the charge on a polyatomic ion
LiMnO4
SO42-

13. Practice Assigning Oxidation Numbers
NO2
N2O5
HClO3
HNO3
Ca(NO3)2
KMnO4

14. Practice Assigning Oxidation Numbers
NO2
N= +4, O = -2
N2O5
N = +5, O = -2
HClO3
H=+1, Cl=+5, O = -2
HNO3
H=+1, N = +5, O = -2
Ca(NO3)2
Ca=+2, N =+5, O= -2
KMnO4
K=+1, Mn=+7, O= -2

15. Practice Assigning Oxidation Numbers
Fe(OH)3
K2Cr2O7
CO32-
CN-
K3Fe(CN)6

16. Practice Assigning Oxidation Numbers
Fe(OH)3
Fe =+3, O=-2, H=+1
K2Cr2O7
K=+1, Cr=+6, O=-2
CO32-
C=+4, O =-2
CN-
C=+4, N=-5
K3Fe(CN)6
K=+1, Fe=+3, C=+4, N=-5

17. Naming compound using oxidation number
More Correctly now named as
SO2 Sulfur dioxide
SO3 Sulfur trioxide
Sulfur(IV) oxide
Sulfur(VI) oxide
Roman Numerals are used the oxidation numbers in the name of the compounds

18. Names of some compounds
H2SO3
H2SO4
PCl3
PCl5
N2O
NO2
CuSO4
Cu2SO4
Sulfuric(IV) acid
Sulfuric(VI) acid
Phosphorous (III) chloride
Phosphorous (V) chloride
Nitrogen (I) oxide
Nitrogen (IV) oxide
Copper(II) Sulfate
Copper(I) Sulfate

19. Names of some ions SO42- Cr2O72- CrO42- PO42- MnO42- Sulfate(VI) ions
Dichromate (VI) ion
Chromate (VI) ion
Phosphate(V) ions
Manganate (VII) ion

20. Using Oxidation Numbers
Careful examination of the oxidation numbers of atoms in an equation allows us to determine what is oxidized and what is reduced in an oxidation-reduction reaction

21. Using Oxidation Numbers
An increase in the oxidation number indicates that an atom has lost electrons and therefore oxidized.
A decrease in the oxidation number indicates that an atom has gained electrons and therefore reduced
Example
Zn CuSO4  ZnSO4 + Cu
Zn: 0   Oxidized
Cu: +2   Reduced

23. UNDERSTANDING/KEY IDEA 9.1.B
An oxidizing agent is reduced and a reducing agent is oxidized.

24. Redox reactions ALWAYS involve the simultaneous oxidation of one reactant with the reduction of another through the transfer of electrons.
The reactant causing the oxidation of the other reactant is called the oxidizing agent. (It is the one that was reduced.)
The reactant causing the reduction of the other reactant is called the reducing agent. (It is the one that is oxidized.)

25. APPLICATION/SKILLS
Be able to identify the species oxidized and reduced and the oxidizing and reducing agents, in redox reactions.

26. If you can determine which one was reduced (got more negative), then the other three are easy to find.
Remember that if it got reduced, then it is the oxidizing agent and vice versa.

27. Oxidizing agent Vs Reducing agent
Also called as Oxidant
An oxidizing agent takes electron away from other substance .
Also called as reductants.
A reducing agent give electrons to other substance

28. Example
Identify the oxidizing and reducing agents in the following equation.
2Al + 3PbCl2 → 2AlCl Pb
First write down the oxidation numbers.
Al went from 0 to +3 – oxidized (red agent)
Pb went from +2 to 0 – reduced (ox agent)
*Note the oxidizing agent is the whole compound (PbCl2), not just the element.*

29. Identify the substances being oxidized and reduced.
Cu + 2Ag+ → Ag + Cu2+
Copper goes from a zero to a +2.
It got more positive so it was oxidized.
Silver goes from +1 to zero.
It got more negative so it was reduced.

31. Exercise Cl2 + KBr  KCl + Br2 Cu + HNO3  Cu(NO3)2 + NO2 + H2O
For each of the following reactions find the element oxidized and the element reduced
Cl KBr  KCl Br2
Cu HNO3  Cu(NO3) NO2 + H2O
HNO I  HIO NO2

32. Exercise Cl2 + KBr  KCl + Br2 Br increases from –1 to 0 -- oxidized
For each of the following reactions find the element oxidized and the element reduced
Cl KBr  KCl Br2  
Br increases from –1 to oxidized
Cl decreases from 0 to – Reduced
K remains unchanged at +1

33. Exercise
For each of the following reactions find the element oxidized and the element reduced
Cu HNO3  Cu(NO3) NO2 + H2O –
Cu increases from 0 to It is oxidized
Only part of the N in nitric acid changes from +5 to +4. It is reduced
The nitrogen that ends up in copper nitrate remains unchanged

34. Exercise HNO3 + I2  HIO3 + NO2 1 +5 -2 0 +1+5-2 +4-2
For each of the following reactions find the element oxidized and the element reduced
HNO I  HIO NO2
N is reduced from +5 to +4. It is reduced
I is increased from 0 to +5 It is oxidized
The hydrogen and oxygen remain unchanged.

35. APPLICATION/SKILLS
Be able to deduce redox reactions using half-equations in acidic or neutral solutions.

36. ½ EQUATIONS
Since oxidation cannot occur without reduction, you can identify each type of reaction within a redox reaction.
These are called the ½ equations.
Include the electrons in your equations.

37. EXAMPLE Deduce the two ½ equations for the following reaction:
Zn + Cu2+ → Zn Cu
Zinc goes from 0 to 2+ (more pos – oxidized)
Cu goes from 2+ to 0 (more neg – red)
Oxidation: Zn → Zn e-
Reduction: Cu e- → Cu

38. Disproportionation Reactions
when the same element is both oxidized and reduced.
3HNO HNO3 + 2NO + H2O
3ClO Cl ClO31-
Cu2O(s)+H2SO4(aq)→Cu(s)+CuSO4(aq)+H2O(l)

39. RULES FOR BALANCING ½ REACTIONS
WRITE DOWN THE ½ REACTIONS.
BALANCE ALL ELEMENTS BESIDES OXYGEN AND HYDROGEN.
BALANCE OXYGEN WITH WATER.
BALANCE HYDROGENS WITH H+
DETERMINE OVERALL CHARGES ON BOTH SIDES.
BALANCE THE CHARGES ON EACH SIDE WITH e-
MULTIPLY BY A FACTOR IF NEEDED SO THE e- CANCEL.
ADD EQUATIONS AND CANCEL OUT COMMON FACTORS ON BOTH SIDES.
FOR BASIC SOLUTIONS, ADD OH- TO BOTH SIDES TO GET RID OF H+’S.

40. EXAMPLE
Balance the following redox reaction: NO3- + Cu → NO + Cu2+ Step 1: Write the ½ reactions Ox: Cu → Cu2+ Red: NO3- → NO Step 2: Balance all elements besides H and O Cu and N are balanced already. Step 3: Balance oxygen with water Ox: Cu → Cu2+ Red: NO3- → NO + 2H2O

41. Step 4: Balance hydrogens with H+ Ox: Cu → Cu2+ Red: 4H+ + NO3- → NO + 2H2O Step 5: Determine overall charges on both sides Ox: (zero) Cu → Cu2+ (+2) Red: (+3) 4H+ + NO3- → NO + 2H2O (zero) Step 6: Balance the charges on each side with e- Ox: Cu → Cu2+ + 2e- Red: 3e- + 4H+ + NO3- → NO + 2H2O

42. Step 7: Multiply by a factor if needed to cancel out the e- 3(Cu → Cu2+ + 2e-) 2(3e- + 4H+ + NO3- → NO + 2H2O) 3Cu → 3Cu2+ + 6e- 6e- + 8H+ + 2NO3- → 2NO + 4H2O Step 8: Add both equations and cancel out anything common on both sides. 8H+ + 3Cu + 2NO3- → 2NO + 3Cu2+ + 4H2O

43. Balancing Redox Equations 1
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms
Check your work. Make sure that both the atoms and charges balance
Cu + HNO3  Cu(NO3)2 + NO + H2O

44. Balancing Redox Equations 2
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms
Check your work. Make sure that both the atoms and charges balance
HNO3 + I2  HIO3 + NO2 + H2O

45. Balancing Ionic Redox Equations 3
Fe2+ +MnO4-  Mn2+ +Fe3+ (acidic)
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms. You may need to add H+ or OH- and H2O in ionic equations
Check your work. Make sure that both the atoms and charges balance

46. Balancing Ionic Redox Equations 5
VO2+ + Zn VO2+ + Zn2+ (Acidic)
Assign oxidation numbers to the species in the reaction
Find the substance oxidized and the substance reduced
Write half reactions for the oxidation and reduction
Balance the atoms that change in the half reaction
Determine the electrons transferred and balance the electrons between the half reactions
Combine the half reactions and balance the remaining atoms. You may need to add H+ or OH- and H2O in ionic equations
Check your work. Make sure that both the atoms and charges balance

47. UNDERSTANDING/KEY IDEA 9.1.C
Variable oxidation numbers exist for transition metals and for most main-group non-metals.

48. APPLICATION/SKILLS
Be able to deduce the oxidation states of an atom in an ion or a compound.

49. GUIDANCE
Oxidation states should be represented with the sign before the given number, not after like as in ions.

50. SIGN CONVENTION
Oxidation numbers are shown with the +or- in front of the number such as +7.
Ions with their charges are shown with the number first followed by the charge such as 2+.

51. OXIDATION NUMBERS
The concept of oxidation numbers provides a way to keep track of electrons in redox reactions.
They are not really charges, but we will use them to assign numbers in covalent compounds to show that the overall charge in a compound is zero.
It is essentially a convention to assign which element has electron control.

52. OXIDATION NUMBER RULES
1. Elements by themselves are zero.
2. In simple ions, the oxidation number is the same as its charge.
3. Oxidation numbers in a neutral compound must add up to zero.
4. Oxidation numbers in a polyatomic ion must add up to the charge of the ion.
You can predict many oxidation numbers from the periodic table.

53. GUIDANCE
Know that the oxidation state of hydrogen can be -1 in metal hydrides and oxygen can be -1 in peroxides.

54. COMMON EXAMPLES Fluorine is always (-1).
Oxygen is usually (-2) except:
Peroxides (-1)
OF2 (+2)
Hydrogen is (+1) except:
Metal hydrides NaH (-1)
Chlorine is (-1) except:
When combined with O or F, then it is (+1).

55. EXAMPLES
Find the oxidation states for the elements in H2SO4 and Na2C2O4.
H2SO4: H +1 (2 of them for a total of +2)
O -2 (4 of them for a total of -8)
S +6 (make the overall charge 0)
Na2C2O4: Na +1 (2 of them for a total of +2)
O -2 (4 of them for a total of -8)
C +3 (2 of them to equal +6)

56. APPLICATION/SKILLS
Be able to deduce the name of a transition metal compound from a given formula, applying oxidation numbers represented by Roman numerals.

57. GUIDANCE
Oxidation number and oxidation state are often used interchangeably, though IUPAC does formally distinguish between the two terms. Oxidation numbers are represented by Roman numerals according to IUPAC.

58. IONIC NAMING RULES You know how to do this – Yeah!!!
Use Roman numerals when you have more than one choice of ion.
You may see covalent compounds using Roman numerals.
NO – nitrogen monoxide or
nitrogen II oxide
NO2 – nitrogen dioxide or
nitrogen IV oxide

59. UNDERSTANDING/KEY IDEA 9.1.D
The activity series ranks metals according to the ease with which they undergo oxidation.

60. Not all oxidizing and reducing agents are the same strength.
Their strength depends upon how easily they lose or gain electrons.

61. Reducing Agents
Metals tend to give up electrons forming positive ions so they cause other elements to become more negative or to be reduced.
This is why metals are commonly reducing agents.
More reactive metals lose their electrons more readily so they are stronger reducing agents.

62. Reactivity Series
A sample reactivity series Mg strongest reducing agent Al (most readily oxidized) Zn Fe Pb Cu Ag weakest reducing agent (least readily oxidized)

63. Oxidizing Agents
Nonmetals tend to gain electrons forming negative ions so they cause other elements to become more positive or to be oxidized.
This is why nonmetals are commonly oxidizing agents.
More reactive nonmetals gain their electrons more readily so they are stronger oxidizing agents.

64. Reactivity Series
A sample reactivity series F2 strongest oxidizing agent Cl2 (most readily reduced) Br2 I2 weakest oxidizing agent (least readily reduced)

65. You do not have to memorize the activity series, but you will have to interpret information from one.
Remember that any metal or nonmetal above another will cause a displacement reaction.
If it is not higher on the activity series, the reaction will not occur.
If you were given a series of viable reactions, you should be able to determine the activity series.

66. SAMPLE REDOX TITRATIONS
Redox titrations are commonly used in the food and beverage industry.
They are very similar to acid-base titrations, but sometimes do not need an indicator as a color change can naturally occur at the equivalence point.

67. Analysis of Iron with Manganate VII
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
This reaction used potassium permanganate in an acidic solution as the oxidizing agent, which oxidizes Fe2+ ions to Fe3+ ions. The manganese is reduced from Mn7+ to Mn2+.
The reaction does not need an indicator as it goes from a deep purple to colorless at equivalence.

68. Iodine-thiosulfate reaction
Several different redox titrations use an oxidizing agent to react with excess iodine ions to form iodine.
2I- + oxidizing agent → I2 + reduced product
Examples of oxidizing agents are: KMnO4, KIO3, K2Cr2O7 and NaOCl.
The I2 is then titrated with sodium thiosulfate using starch as an indicator.
The starch indicator is added during the titration (not at the start) and forms a deep blue color by forming a complex with the free I2. As the I2 is reduced to I- ions, the blue color disappears marking equivalence.

69. Redox equations:
oxidation: 2S2O32- → S4O e-
reduction: I2 + 2e- → 2I-
Overall equation:
2S2O32 + I2 → 2I- + S4O62-

70. UNDERSTANDING/KEY IDEA 9.1.E
The Winkler Method can be used to measure biochemical oxygen demand (BOD), used as a measure of the degree of pollution in a water sample.

71. The dissolved oxygen content of water is one of the most important indicators of its quality.
As pollution increases, the dissolved oxygen content decreases as the oxygen is used by bacteria in decomposition reactions.
The BOD (biological oxygen demand) is used as a means of measuring the degree of pollution.
BOD is defined as the amount of oxygen used to decompose the organic matter in a sample of water over a specified time period, usually 5 days at a specified temperature.

72. APPLICATION/SKILLS
Be able to apply the Winkler Method to calculate BOD.

73. WINKLER METHOD
The Winkler Method uses redox titrations to measure the dissolved oxygen in water to calculate the BOD.
The dissolved oxygen in the water is “fixed” by the addition of a manganese II salt such as MnSO4.
Reaction of this salt with oxygen in basic solution causes oxidation of Mn(II) to higher oxidation states such as Mn(IV).
2Mn2+ + O2 + 4OH- → 2MnO2 + 2H2O

74. Acidified iodide ions are added to the solution
and are oxidized by the Mn(IV) to I2.
MnO2 + 2I- + 4H+ → Mn2+ + I2 + 2H2O
4. The iodine produced is then titrated with sodium
thiosulfate as described earlier.
2S2O32 + I2 → 2I- + S4O62-
So we can see that for every 1 mole of O2 in the water, 4 moles of S2O32- are used.
Work the sample problem page 384

75. Citations
International Baccalaureate Organization. Chemistry Guide, First assessment Updated Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.