1. 1.5a Learning Outcomes define oxidation number, oxidation state
define oxidation and reduction in terms of change of oxidation numbers
define oxidising agent and reducing agent
state the rules for oxidation numbers (exclude peroxides, except for hydrogen peroxide)
calculate oxidation numbers of transition metals in their compounds and of other elements
use oxidation numbers in nomenclature of transition metal compounds

2. Learning Outcomes
arrange the electrochemical series of metals in order of their ease of oxidation (reactions, other than displacement reactions, not required)

3. 1.5 Oxidation & Reduction Also known as Redox
Means something in terms of:
Addition or Removal Oxygen / Hydrogen
Electrons
Oxidation Numbers
Introduction to topic via three simple examples

4. Redox in Terms of O2 / H2
Oxidation = addition of oxygen to a substance (or removal of hydrogen)
C + O2 CO2
Reduction is loss of oxygen (or addition of hydrogen)
CuO + H2  Cu + H2O

5. Summary of Addition /Removal of Oxygen & Hydrogen
C2H5OH
Oxidation = -H2
CH3CHO
Ox. Agent
CH3CHO
Reduction = +H2
C2H5OH
Red. Agent
Oxidation = +O
CH3CHO
CH3COOH
Ox. Agent
Reduction = -O
CH3COOH
CH3CHO
Red. Agent
Ox. Agent = Na2Cr2O7 / H+ Red. Agent = H2 / Ni

6. Oxidation in Terms of Electrons
OILRIG
Oxidation involves Loss of electrons
Reduction Involves Gain of Electrons
Species Oxidised is Reducing Agent
Species Reduced is the Oxidising Agent

7. Example 1 of 3 Sodium + chlorine  sodium chloride Na + Cl  Na+ + Cl-
OILRIG
Oxidn involves loss / Redn involves gain
Na loses an electron [oxidised]
Cl gains an electron [reduced]

9. Example 2 of 3 Magnesium + Oxygen  magnesium oxide Mg + O  MgO
=> Oxidn / Redn by inspection?
Mg  Mg+2 loses 2 electrons [oxidation]
O  O-2 gains 2 electrons [reduction]

10. Example 3 of 3 Zinc loses electrons (oxidised)
Zinc +copper sulphate  Zinc sulphate+ Copper
Zn + CuSO4  ZnSO4 + Cu
Zn + Cu+2  Zn+2 + Cu
Zinc loses electrons (oxidised)
Copper gains electrons (reduced)

11. Oxidising Agent
A substance that causes oxidation in another substance – the oxidising agent is itself reduced

12. Reducing agent
A substance that causes reduction in another substance.

13. Summary of Redox in terms of Electrons
Oxidation is loss of electrons;
Reduction is gain of electrons
CuO + H2  Cu + H2O
CuO is made up of Cu+2 and O-2
Overall: Cu+2  Cu [gains 2 electrons] reduced
Overall: H2  H2+2[loses 2 electrons] oxidised
Overall: O-2  O-2 [ no change]

14. Summary of Redox in terms Oxidation numbers
Oxidation Number: The charge that an atom has or appears to have assuming that the compound is ionic.
Electrons always go to the most electronegative element
Oxidation is an increase on oxidation number
Reduction is a decrease in oxidation number.

15. Oxidation Number Rule 1 Elements on their own = 0 H2 = 0 Zn = 0
Cl2 = 0

16. Oxidation Number Rule 2 Ions = same as charge Cu +2 = +2 O-2 = -2
Cl-1 = -1

17. Oxidation Number Rule 3 Charges of all elements in a compound = 0
CuSO4
Cu = +2
S = +6
O4 = -8 [O = -2]
Total = –8 = 0

18. Oxidation Number Rule 4 Oxygen = -2 Exceptions are
peroxides O = -1 [H2O2, Na2O2 ]
OF2 O = +2, F = -1
(Oxygen diflouride)

19. Oxidation Number Rule 5 Hydrogen = +1
Exceptions are the metal hydrides
NaH Na = +1, H = -1
(Sodium hydride)
MgH2 Mg = +2 each H = -1
Magnesium Hydride

20. Oxidation Number Rule 6
Halogens [ F, Cl, Br, I] are always –1 except when joined to a more electronegative element
Cl2O
Cl = +1, O = -2

21. Oxidation Number Rule 7
In a complex ion the sum of all the charges = the charge on the ion.
SO4-2 (Sulfate anion)
S = +6, O4 = -8 [O = -2]
+6 –8 = -2
What about Sulfite anion = SO3-2?

22. Redox Equations MnO4- + H2S + H+  Mn+2 + S + H2O 2-8 +2
MnO H2S + H+  Mn+2 + S + H2O
+ 7 +1 -2 +2 -2
Ox => Loses 2e-
L.C.M. = 10
Red => Gains 5e- 2
MnO H2S + H+  Mn+2 + S + H2O 5 2
MnO H2S + H+  Mn S + H2O 5 6 2 5 8 -2 +6 +4

23. Redox Equations Zn + NO3- + H+  Zn+2 + N2H4 + H2O 7+5 -6 +4
Zn + NO3- + H+  Zn+2 + N2H4 + H2O +5 -2 +2 -2 +1
Ox. => Loses 2e-
Red. => Gains 7e-
L.C.M. = 14 7
Zn + NO3- + H+  Zn+2 + N2H4 + H2O 2
7Zn + 2NO3- + H+  Zn+2 + N2H4 + H2O 16 7 6
+16
+14 -2

24. Redox Equations MnO4- + H+ + Cl-  Mn+2 + Cl2 + H2O-8
MnO4- + H+ + Cl-  Mn+2 + Cl2 + H2O -1 +2 +7 -2
Red => Gains 5e-
Ox => Loses 1e-
L.C.M. = 5
MnO4- + H+ + Cl-  Mn+2 + Cl2 + H2O 5
MnO4- + H+ + 5Cl-  Mn+2 + Cl2 + H2O 8
5/2 4 -1 +8 -5 +2

25. Redox Equations ClO3- + I- + H+  I2 + Cl- + H2O-6
ClO I- + H+  I2 + Cl- + H2O +5 -2 -1 -1
Red => Gains 6e-
Ox => Loses 1e-
L.C.M. = 6
ClO I- + H+  I2 + Cl- + H2O 6
ClO3- + 6I- + H+  I2 + Cl- + H2O 6 3 3 -1 -1 -6 +6

26. Redox Equations Cr2O72- + Cl- + H+  Cr+3 + H2O + Cl2
-12
-14
Cr2O Cl- + H+  Cr H2O + Cl2 +6 -2 -1 +3
Red => EACH Gains 3e-
=> 2 x 3e- = 6
Ox => Loses 1e-
L.C.M. = 6
Cr2O Cl- + H+  Cr H2O + Cl2 6
Cr2O Cl H+  Cr H2O + Cl2 14 2 7 3 -2 -6
+14 +6

27. Redox Equations Cr2O72- + H+ + S  SO2 + OH- + Cr2O3 2
-12
-14 -4 -6
Cr2O H+ + S  SO OH Cr2O3 -2 +6 -2 +4 -2 +3
Red => EACH Gains 3e-
=> 2 x 3e- = 6
Ox => Loses 4e-
L.C.M. = 12 2
Cr2O H+ + S  SO OH Cr2O3 3
2Cr2O H S  SO OH Cr2O3 2 3 2 2 -2 -4 +2

28. Redox Equations MnO4- + H2O2 + H+  Mn+2 + O2 + H2O 2-8 +2 -2
MnO H2O H+  Mn O H2O +7 -2 +1 -1 +2
Red => Gains 5e-
Ox => EACH loses 1e-
=> 2 x 1e- = 2
L.C.M. = 10 2
MnO H2O H+  Mn O H2O 5
2MnO H2O H+  Mn O H2O 6 2 5 8 +4 -2 +6

29. Exam Q’s
Q.4 ’02
(e) What is the oxidation number of sulfur in Na2S2O3?
(e) + 2 / 2 / 0 and 4 (but only if both are given – no marks for 0 or 4 on its own) (6)
10.
(a) Define oxidation number. (4)
(i) Using oxidation numbers, identify which species is being oxidised and which species is being reduced in the following reaction. (12)
MnO4− + Cl− + H+ → Mn2+ + Cl2 + H2O
oxidised: Cl¯ / Cl (–1) (6) [OR Cl (–1 to 0) (3)
reduced: MnO4¯/ Mn(+7) / Mn(7) / Mn(VII) (6) [OR Mn (7 to 2) (3) reduced (3)]

30. (ii) Hence, or otherwise, balance the equation. (9)
MnO4¯ + 5Cl¯ + 8H+ Mn Cl2 + 4H2O (9)

31. 2004 Q.4
(g) What is the oxidation number (i) of oxygen in H2O2 and (ii) of bromine in KBrO3?
(g) (i) – 1 (3) (ii) + 5 / 5 / V (3)
2005 Q.11
(a) (i) Define oxidation in terms of change in oxidation number.
(a) (i) DEFINE: increase in oxidation number (4)
(ii) What is observed when chlorine gas is bubbled into an aqueous solution of sodium bromide?
Explain your answer in terms of oxidation and reduction. (9)
(ii) WHAT: solution turns red-brown (red, orange, yellow) (3
EXPLAIN: bromide ions oxidised to bromine / Br¯ to Br2 / Br(-1) to Br(0) (3)
chlorine reduced to chloride ions / Cl2 to Cl¯ /
Cl(0) to Cl(-1) (3)

32. 2006 q.10
(b) Define oxidation in terms of change in oxidation number. (4)
What is the oxidation number of (i) chlorine in NaClO and (ii) nitrogen in NO3¯? (6)
DEFINE: increase (4)
WHAT: (i) +1 [Accept 1] (3)
(ii) +5 [Accept 5] (3)
State and explain the oxidation number of oxygen in the compound OF2. (6)
EXPLAIN: oxygen is more electropositive / less electronegative / fluorine is more
electronegative / fluorine is less electropositive [Allow even if ox. no. incorrect.] (3)

33. Using oxidation numbers or otherwise, identify the reducing agent in the reaction between acidified potassium manganate(VII) and potassium iodide solutions represented by the balanced equation below. Use your knowledge of the colours of the reactants and products to predict the colour change you would expect to see if you carried out this reaction. (9)
2MnO4¯ + 10I¯ + 16H+ 2Mn2+ + 5I2 + 8H2O
IDENTIFY: potassium iodide (KI) solution / potassium iodide (KI) / iodide (I¯) / I( –1 to 0) (3)
COLOURS: purple / violet / maroon (3)
to brown / red / orange/ yellow (3)

34. 2007 Q.10 (c) The halogens are good oxidising agents.
(i) How does the oxidation number of the oxidising agent change during a redox reaction? (4)
(c) (i) HOW: it decreases (4)
(ii) Assign oxidation numbers in each of the following equations to show clearly that the halogen is the oxidising agent in each case. (12)
Br2 + 2Fe2+ → 2Br– + 2Fe 3+
oxidation number of Br in Br2 = 0 (3)
oxidation number of Br in Br¯ = – 1 (3) …reduced => ox. agent
Cl2 + SO32– + H2O → Cl– + SO4 2– + H+
oxidation number of Cl in Cl2 = 0 (3)
oxidation number of Cl in Cl¯ = – 1 (3) …reduced => ox. agent

35. Q (b) Define oxidation in terms of (i) electron transfer, (ii) change in oxidation number. (7)
DEFINE: (i) loss (decrease) of electrons
(ii) increase (rise, gain) in oxidation number (4 + 3)
(iii) For the redox reactions shown below, use oxidation numbers to identify the species oxidised in the first reaction and the oxidising reagent in the second reaction. (6)
ClO¯ + I¯ + H+  Cl¯ + I2 + H2O
I2 + S2O32–  I¯ + S4O62–
1st I¯ (iodide ion) / I (–1) (3) ….oxidised itself
2nd I2 (iodine molecule) / I (0) / I2 /(3).. reduced itself

36. (iv) Using oxidation numbers or otherwise balance both equations. (12)
ClO¯ + I¯ + H+  Cl¯ + I2 + H2O
Balanced: ClO¯ + 2I¯ + 2H+ → Cl¯ + I2 + H2O
I2 + S2O32–  I¯ + S4O62–
Balanced: I2 + 2S2O32– → 2I¯ + S4O62–

37. Electrochemical Series
Electrochemical Series – Metals listed in order of Decreasing ability as Reducing Agents (themselves oxidised)
Ca  Ca+2 + 2e- (Oxidation)

38. Metals Electrochemical Series
[K] => Best Reducing Agent – becomes easily oxidized
[Na]
[Ca]
[Mg]
[Al]
[Zn]
[Fe]
[Sn]
[Pb]
[H]
[Cu]
[Ag] => Worst Reducing Agent – most difficult in list to oxidize