1. (D) OXIDISING AND REDUCING AGENTSHC
CHEMISTRY HC
CHEMISTRY
CHEMISTRY IN SOCIETY
(D) OXIDISING AND REDUCING AGENTS

2. Ion-electron equation and Redox with oxyanionsHC
CHEMISTRY HC
CHEMISTRY
(D) OXIDISING AND REDUCING AGENTS
Ion-electron equation and Redox with oxyanions

3. (D) CHEMISTRY IN SOCIETY Ion-electron equation and RedoxHC
CHEMISTRY HC
CHEMISTRY
(D) CHEMISTRY IN SOCIETY
Ion-electron equation and Redox
After completing this lesson you should be able to :
The electrochemical series is a series of reduction reactions. It indicates the effectiveness of oxidising and reducing agents. Oxidising and reducing agents can be selected using an electrochemical series from a data booklet or can be identified in the equation showing a redox reaction.
Oxidation and reduction reactions can be represented by ion-electron equations. When molecules or group ions are involved, if the reactant and product species are known, a balanced ion-electron equation can be written by adding appropriate numbers of water molecules, hydrogen ions and electrons.
Learners would not be expected to complete and balance ion-electron equations for reactions occurring in alkaline solutions.

4. Writing REDOX equations
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Consider the reaction between sodium and water:
Na(s) + H2O(l)  NaOH(aq) + ½H2(g)
Consider how the ions are formed in this reaction
Na(s)  Na+(aq) + e-
H2O(l) + e-  OH-(aq) + ½H2(g)
Na(s)  Na+(aq)+ e-
H2O(l) + e-  OH-(aq) + ½H2(g)
and, we could say that a water
molecule must be accepting
the electron
A sodium atom loses
an electron 4

5. OIL RIG These are called ion-electron equations
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Na(s)  Na+(aq) + e-
OIL
RIG
H2O(l) + e-  OH-(aq) + ½H2(g)
These are called ion-electron equations
(or ionic half equations). 5

6. Na(s) + H2O(l)  NaOH(aq) + ½H2(g)
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Na(s)  Na+(aq) + e-
Electrons cancel!
H2O(l) + e-  OH-(aq) + ½H2(g)
Reduction and oxidation occur simultaneously. Adding the two equations together gives us the overall equation for a reaction.
Na(s) + H2O(l)  NaOH(aq) + ½H2(g) 6

7. COPY a) Cu(s)  Cu2+ + 2e- Ag+(aq) + e-  Ag(s)
For each of the following reactions, combine the oxidation and reduction step to form a balanced equation.
a) Cu(s)  Cu2+ + 2e-
Ag+(aq) + e-  Ag(s)
b) MnO4- + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+ + e-
c) Ce4+(aq) + e-  Ce3+(aq)
2Br-(aq)  Br2(aq) + 2e- 7

8. Balancing Redox equations
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Most redox reaction you will come across will occur in
neutral or acidic conditions.
1. Make sure there are the same number of atoms of each
element being oxidised or reduce on each side of the
half equation.
2. If there are any oxygen atoms present, balance them by
adding water molecules to the other side of the half-equation.
3. If there are any hydrogen atoms present, balance them
by adding hydrogen ions on the other side of the
half-equation.
4. Make sure the half-reactions have the same overall
charge on each side by adding electrons.
For basic solutions H atoms are balanced using H2O and then the same
number of OH- ions to the opposite side to balance the oxygen atoms 8

9. SO2(g) + 2H2O(l)  SO42-(aq) + 4H+(aq)
1. Write down what you know….
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sulphur dioxide is oxidised to sulphate ions
SO2(g)  SO42-(aq)
2. Balance the oxygen atoms by adding water
2H2O(l)
SO2(g)  SO42-(aq)
3. Balance the hydrogen atoms by adding hydrogen ions
SO2(g) + 2H2O(l)  SO42-(aq) +
4H+(aq)
4. Balance the charges by adding electrons
SO2(g) + 2H2O(l)  SO42-(aq) + 4H+(aq) +
2e-
charge is zero
4 - and 4 + equals zero 9

10. COPY MnO4-  Mn2+ IO3-  I2 Pb2+  PbO2 ClO-  Cl- Mn2+  MnO2
Write a balanced ion-electron equation for each of the following reactions
MnO  Mn2+
IO  I2
Pb  PbO2
ClO-  Cl-
Mn  MnO2
Sb  SbO+
VO  V3+
HOBr  Br2
C2 H5OH  CH3CHO 10

11. COPY MnO4-  Mn2+ + 8 H+ + 5e - + 4H2O 2 IO3-  I2 + 12 H+ + 10e -
Pb  PbO2
+ 2H2O
+ 4 H+
+ 2e -
ClO-  Cl-
+ 2 H+
+ 2e -
+ H2O 11

12. COPY
Mn  MnO2
+ 2H2O
+ 4 H+
+ 2e -
Sb  SbO+
+ H2O
+ 2 H+
+ 3e - 12

13. COPY VO2+  V3+ + 2 H+ + e - + H2O 2 HOBr  Br2 + 2 H+ +2 e - + 2 H2O
C2 H5OH  CH3CHO
+ 2 H+
+2 e - 13

14. Redox Reactions From a reaction you should be able to identify
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Redox Reactions
From a reaction you should be able to identify
a) spectator ions if present
b) half equations for reduction/oxidation
c) the reducing and oxidising agents 14

15. No spectator ions shown
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Mg(s) +
2 H+(aq) →
Mg 2+(aq)
H2(g) +
No spectator ions shown 15

16. COPY
Mg(s) +
2 H+(aq) →
Mg 2+(aq)
H2(g) + 16

17. COPY
Mg(s) +
2 H+(aq) →
Mg 2+(aq)
H2(g) + 17

18. COPY
Mg(s) →
Mg 2+(aq)
2 H+(aq) →
H2(g) 18

19. COPY
Mg(s) →
Mg 2+(aq)
2 H+(aq) →
H2(g) 19

20. COPY
Mg(s) →
Mg 2+(aq)
2e- +
2 H+(aq)
2e- →
H2(g) + 20

21. oxidising agent H+(aq)
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Mg(s) →
Mg 2+(aq)
2e- +
oxidation
oxidising agent H+(aq)
2 H+(aq)
2e- →
H2(g) +
reduction
reducing agent Mg(s) 21

22. No spectator ions shown
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2I -(aq)
2 Fe3+(aq) →
I2(aq)
2 Fe2+(aq) + +
No spectator ions shown 22

23. COPY
2I -(aq)
I2(aq)
2 Fe2+(aq)
2 Fe3+(aq) → + + 23

24. COPY
2I -(aq) →
I2(aq)
2e- +
2 Fe3+(aq)
2e- → +
2 Fe2+(aq) 24

25. oxidising agent Fe3+(aq)
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2I -(aq) →
I2(aq)
2e- +
oxidation
oxidising agent Fe3+(aq)
2 Fe3+(aq)
2e- → +
2 Fe2+(aq)
reduction
reducing agent I-(aq) 25

26. COPY 2NaBr(aq) Cl2 (g) → 2 NaCl(aq) Br2(aq) Show charges on ions + +26

27. COPY 2Na+Br -(aq) Cl2 (g) → 2 Na+Cl -(aq) Br2(aq) Omit spectator ions
Two Na+ ions 27

28. COPY
2Br -(aq)
Cl2 (g) →
2 Cl -(aq)
Br2(aq) + + 28

29. COPY
2Br -(aq)
Cl2 (g) →
2 Cl -(aq)
Br2(aq) + + 29

30. COPY
2Br -(aq) →
Br2(aq)
2 Cl -(aq)
Cl2 (g) → 30

31. COPY
2Br -(aq) →
Br2(aq)
2e- +
2 Cl -(aq)
Cl2 (g)
2e- → + 31

32. oxidising agent Cl 2 (aq)
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2Br -(aq) →
Br2(aq)
2e- +
oxidation
oxidising agent Cl 2 (aq)
2 Cl -(aq)
Cl2 (g)
2e- → +
reduction
reducing agent Br - (aq) 32

33. COPY 2AgNO3 (aq) Cu(s) → 2 Ag (s) Cu(NO3)2 (aq) Show charges on ions +33

34. COPY 2 Ag (s) Cu2+(NO3-)2 (aq) 2Ag+NO3- (aq) Cu(s) →
Omit spectator ions
Two NO3- ions 34

35. COPY
2Ag+(aq)
Cu(s) →
2 Ag (s)
Cu2+ (aq) + + 35

36. COPY
2Ag+(aq)
Cu(s) →
2 Ag (s)
Cu2+ (aq) + + 36

37. COPY
Cu(s) →
Cu2+ (aq)
2 Ag (s)
2Ag+(aq) → 37

38. COPY
Cu(s) →
Cu2+ (aq)
2e- +
2 Ag (s)
2Ag+(aq)
2e- → + 38

39. oxidising agent Ag+ (aq)
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Cu(s) →
Cu2+ (aq)
2e- +
oxidation
oxidising agent Ag+ (aq)
2 Ag (s)
2Ag+(aq)
2e- → +
reduction
reducing agent Cu (s) 39

40. COPY
K2SO4 + Br2 + H2O K2SO4 + 2HBr
Show charges on ions 40

41. (K+)2SO32- + Br2 + H2O (K+)2SO42- + 2H+Br -
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(K+)2SO32- + Br2 + H2O (K+)2SO H+Br -
Omit spectator ions
Two K+ ions 41

42. COPY
SO32- + Br2 + H2O  SO H+Br - 42

43. COPY
SO32-(aq)
Br2(l) →
SO42-(aq) +
H2O(l) +
2H+ Br - (aq) + 43

44. COPY
SO32-(aq)
H2O(l) + →
SO42-(aq)
2H+ (aq) +
2 Br - (aq)
Br2(l) → 44

45. COPY SO32-(aq) H2O(l) → SO42-(aq) 2H+ (aq) 2e- 2 Br - (aq) Br2(l) 2e-45

46. reducing agent SO32-(aq)
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SO32-(aq)
H2O(l) + →
SO42-(aq)
2H+ (aq)
2e- + +
oxidation
oxidising agent Br2 (l)
2 Br - (aq)
Br2(l)
2e- → +
reduction
reducing agent SO32-(aq) 46

47. 2Cr 3+ + 7H2O + 3Cl2 Cr2O72- + 14H+ + 6Cl-
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2Cr H2O + 3Cl2 Cr2O H+ + 6Cl-
No spectator ions shown 47

48. 2Cr 3+ + 7H2O + 3Cl2 Cr2O72- + 14H+ + 6Cl-
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2Cr H2O + 3Cl2 Cr2O H+ + 6Cl- 48

49. COPY
2Cr H2O  Cr2O H+
3Cl  Cl- 49

50. 2Cr 3+ + 7H2O  Cr2O72- + 14H+ + 6e - 3Cl2 +6e-  6Cl- COPY oxidation
oxidising agent Cl2 (l)
3Cl e-  Cl-
reduction
reducing agent Cr3+(aq) 50

51.  5CrCl2 + KMnO4+ 8HCl 5CrCl3 + KCl +MnCl2 + 4H2O COPY
Show charges on ions 51

52.  5Cr2+(Cl-)2 + K+MnO4-+ 8H+Cl -
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5Cr2+(Cl-)2 + K+MnO4-+ 8H+Cl - 
5Cr3+(Cl-)3 + K+Cl- +Mn2+(Cl-)2 + 4H2O
Omit spectator ions
One K+ ion and eighteen Cl- ions 52

53. COPY
5Cr MnO H+ 
5Cr Mn H2O 53

54. COPY
5Cr MnO H+ 
5Cr Mn H2O 54

55. 5Cr2+ + 5e- 5Cr3+   Mn2+ + 4H2O MnO4- + 8H+ + 5e- COPY oxidation
oxidising agent MnO4- / H+ 
Mn H2O
MnO H+
+ 5e-
reduction
reducing agent Cr2+ 55