1. Higher Chemistry Unit 3 – Chemistry in Society Section 16 – Oxidising and Reducing Agents

2. 3.16 – Oxidising and Reducing Agents
Learn that many chemical reactions are reversible
Learning Outcomes/Success Criteria Unit 3 Higher Chemistry
I can identify substances as oxidising and reducing agents
I can combine ion-electron equations and write balanced redox equations
I can state that the strongest reducing agents are found in group 1
I can state that the strongest oxidising agents are found in group 7
I can use the electrochemical series to identify highly effective reducing and oxidising agents
I can state uses of oxidising agents and why they are used
I can write ion-electron equations for more complex oxidations and reductions involving compounds

3. Redox Reactions
A redox reaction involves two half reactions – oxidation and reduction
These half equations can be written as ion-electron equations. Ion-electron equations are found in the data booklet.
Oxidation involves the loss of electrons (OIL)
The ion-electron equation for oxidation must be written in reverse when compared to the data booklet entries.
Reduction involves the gain of electrons (RIG)

4. Redox Equations
Redox reactions include reactions which involve the loss or gain of electrons.
The reactant giving away (donating) electrons is called the reducing agent (which is oxidised)
The reactant taking (accepting) electrons is called the oxidising agent (which is reduced)
Both oxidation and reduction happen simultaneously, however each is considered separately using ion-electron equations.
O.I.L. R.I.G.
Oxidation is loss, reduction is gain of electrons 4

5. Redox Reactions Oxidising and Reducing Agents
In a redox reaction the species that is oxidised is described as a reducing agent – a species that allows reduction to occur.
Similarly, a species which is reduced is described as an oxidising agent – a species that allows oxidation to occur.
e.g. Mg(s) + 2Ag+  Mg2+ + 2Ag(s)
Using the Data Booklet
When writing redox equations you must first identify the oxidation and reduction half reactions and then combine them to give the redox reaction. Use the Electrochemical Series in the data booklet.

6. Cells and Redox
magnesium(s) + silver nitrate(aq)  magnesium nitrate(aq) + silver(s).
The reducing agent in this reaction is the Mg as it will
donate electrons to the silver ions .
The oxidising agent is the Ag+ ions as they accept electrons
from the Mg
Oxidation:
Mg(s)  Mg2+(aq) e-
Half equations
or ion-equations
Reduction:
2Ag+(aq) + 2e-  2Ag(s)
Mg (s) + 2Ag+ (aq)  Mg2+ (aq) + 2Ag (s)
Redox equation, electrons cancel out 6

7. Redox Reactions
Remember, oxidation and reduction are two halves of the same chemical reaction. The combined reaction is called a REDOX reaction.
To form the overall redox reaction, the ion-electron equations for the oxidation and reduction must be combined, ensuring that the electrons cancel.

8. Combining Ion- Electron Half Equations
If magnesium is added to a solution containing Fe(III) ions, a displacement reaction occurs.
Write the equation for the oxidation of magnesium metal
Write the equation for the reduction of iron (III) ions to Fe atoms
Multiply through so the number of electrons is the same on both sides. Write the overall redox reaction and then remove the spectators.
How many electrons were transferred?
Mg(s)  Mg2+(aq) e-
Fe3+(aq) e-  Fe (s)
3Mg(s) Fe3+(aq) e-  3Mg2+(aq) Fe (s) e-
3Mg(s) Fe3+(aq)  3Mg2+(aq) Fe (s)
6e-

9. Displacement reactions
Some metals are more reactive than others. In this
experiment, a strip of metal is added to a solution of
another. If the metal is more reactive than the metal
in solution, this metal displaces (pushes out) the less
reactive metal from the solution. 9

10. Single Displacement reactions
In a single displacement reaction, one element takes the place of (displaces) another element in a compound.
There are two general forms of equations for a single displacement reaction:

11. Reactivity Series

12. Electrochemical Series – Page 12

13. Displacement Reactions13

14. X - √ 14

15. What is the order of reactivity of these metals (from most to least)?
Write the formula then ionic formula equation for the reaction of magnesium and zinc (II) nitrate.
Rewrite the equation omitting the spectator ions.
Write a half equation for the reducing agent in the above reaction
Magnesium, zinc, lead, copper
Mg(s) + Zn(NO3)2(aq) → Mg(NO3)2(aq) + Zn(s)
Mg(s) + Zn2+(NO3-)2(aq) → Mg2+(NO3-)2(aq) + Zn(s)
Mg(s) + Zn2+(aq) + 2NO3-(aq) → Mg2+(aq) + 2NO3-(aq) + Zn(s)
Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)
Mg(s) → Mg2+(aq) + 2e- 15

16. Write redox equations for
a. Displacement of silver ions by Zinc
b. Displacement of copper (II) ions by Aluminium
Zn(s)  Zn2+(aq) e-
Oxidation x2
2Ag+(aq) e-  2Ag(s)
Reduction
2Ag+(aq) Zn(s)  2Ag(s) Zn2+(aq)
Combined Redox x2 2
Al(s)  Al3+(aq) e- 2 6
Oxidation x3 3
Cu2+(aq) e-  Cu(s) 6 3
Reduction
3Cu2+(aq) Al(s)  2Al3+(aq) Cu(s)
Combined Redox

17. Identify the 2 half equations and say which is oxidation and which is reduction for
Zn + 2Ag+  2Ag +Zn2+
I SO32- +H20  2I- + SO H+
2I- + 2Fe3+  I2 +2Fe2+
Zn(s)  Zn2+(aq) e-
Oxidation
2Ag+(aq) e-  2Ag(s)
Reduction
SO32- (aq) H2O(l)  SO42-(aq) H+(aq) + 2e-
Oxidation
I2(s) e-  2I-(aq)
Reduction
2I-(aq)  I2(s) e-
Oxidation
2Fe3+(aq) e-  2Fe2+ (aq)
Reduction

18. Oxidising and reducing agents
The elements with high electronegativities tend to form ions by gaining electrons (reduction) and so act as oxidising agents.
The elements with low electronegativities tend to form ions by losing electrons (oxidation) and so act as reducing agents
½O2 + 2e-  O2-
Note that, in general,
Non-Metal Oxidising Agents
½Cl2 + e-  Cl-
Metal Reducing Agents
Mg  Mg2+ + 2e-
Al  Al e- 18

19. e.g. e.g. Al  Al3+ + 3e- ½O2 + 2e-  O2- Note that, in general,
½Cl2 + e-  Cl-
e.g.
Mg  Mg2+ + 2e-
Al  Al e-
Metals on the LHS of the Periodic Table ionise by electron loss and are called reducing agents
Non-metals on the RHS of the Periodic Table ionise by electron gain and are called oxidising agents 19

20. Redox and the Electrochemical Series
Oxidising agents
Li+(aq) + e  Li(s)
Na+(aq) + e  Na(s)
Mg2+(aq) + 2e  Mg(s)
Pb2+(aq) + 2e  Pb(s)
2H+(aq) +2e  H2(g)
Cu2+(aq) + 2e  Cu(s)
Ag+(aq) + e  Ag(s)
Increasing powerful reducing agent
(write the reaction backwards)
Hydrogen reference
Increasing powerful oxidising agent
(write the reaction as it appears)
Considering the two ion-equations,
Mg 2+ (aq ) + 2e-  Mg (s) and Ag + (aq) + e-  Ag ,
Mg, being higher up the electrochemical series, would act as the reducing agent. (i.e. the ion-electron equation would be written backwards). While Ag would be written as it appears in the electrochemical series.
Mg(s) + 2Ag+(aq)  Mg2+(aq) + 2Ag(s) 20

21. Electrochemical Series – Page 12
Increasing powerful reducing agent
(write the reaction backwards)
Hydrogen reference
Increasing powerful oxidising agent
(write the reaction as it appears)

22. Reactions of halogens
The Group 7 elements are called the halogens. This experiment involves some reactions of the halogens. 22

23. reaction (colour change)
Bleached (fast)
Bleached (slow)
Bleached (very slow)
No reaction
reaction (colour change) 23

24. Which halogen is the most reactive.
From the results which halogen solution is the strongest bleaching agent?
Which halogen is the most reactive.
Write an ionic formula equation for the reaction of chlorine and potassium bromide then eliminate the spectators.
Write a half equation for the oxidising agent in the above reaction
Chlorine
Chlorine
Cl2(g) + 2K+Br-(aq) → 2K+Cl-(aq) + Br2(aq)
Cl2(g) + 2Br-(aq) → 2Cl-(aq) + Br2(aq)
Cl2(g) + 2e- → 2Cl- (aq) 24

25. Molecules and Group Ions can act as Oxidising and Reducing agents
Compounds can also act as oxidising or reducing agents.
The electrochemical series contain a number of ions and molecules.
Oxidising and reducing agents can be selected using an electrochemical series found in your data book. 25

26. 1. Permanganate MnO4- is a strong oxidising agent
Oxidising agents such as potassium permanganate (KMnO4) are
used in explosives and fire works
MnO4-(aq) + 8H+ +5e- → Mn2+(aq) + 4H2O(l)
Mn(VII) Mn(II)
The potassium permanganate oxidises the glycerol to carbon dioxide and water (hence the steam) and is itself reduced. 26

27. 2.Dichromate and permanganate in acidic solutions are strong oxidising agents.
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
Orange green
MnO4-(aq) + 8H+ + 5e- → ) + 4H2O(l)
Purple colourless
Both half equations contain H+ ions showing they are strong oxidising agents in the presence of acid 27

28. 3. An Oxidation and Reduction reaction – Blue Bottle experiment
An alkaline solution of glucose acts as a reducing agent and reduces added methylene blue from a blue to a colourless form. Shaking the solution raises the concentration of oxygen in the mixture and this oxidises the methylene blue back to its blue form. When the dissolved oxygen has been consumed, the methylene blue is slowly reduced back to its colourless form by the remaining glucose, and the cycle can be repeated many times by further shaking.
A conical flask contains a colourless solution. When shaken, a blue colour forms. After a few seconds, the blue colour fades and the solution again becomes colourless.
The process can be repeated. It is an oxidation followed by a reduction process 28

29. 2. Put 100 cm3 of potassium hydroxide solution into a conical flask.
What to do
1. Put some water in the conical flask. Put in the stopper. Shake vigorously to check for leaks. If there are none, pour the water away and proceed.
2. Put 100 cm3 of potassium hydroxide solution into a conical flask.
3. Add 3.3 g dextrose.
4. Add 3–4 drops of methylene blue indicator.
5. Put a stopper on the flask.
6. Shake vigorously.
7. When the solution clears, repeat the process.
8. It is necessary periodically to remove the stopper. 29

30. The stopper must be taken off to allow more oxygen in.
Methylene blue is reduced by the alkaline dextrose solution to produce a colourless solution.
When the solution is shaken, it is oxidised by the oxygen in the flask to produce the blue dye.
The stopper must be taken off to allow more oxygen in. 30

31. 4. Hydrogen peroxide is a strong oxidising agent
The overall equation for this reaction is:
2 H2O2(aq) --> 2 H2O(l) + O2(g)
Oxidising agents such as hydrogen peroxide are used
in many bleaching products
H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)
2I-(aq)  I2(s) e-
Reduction
Oxidation
H2O2(aq) + 2H+(aq) + 2I-(aq) → I2(s) + 2H2O(l)
Redox 31

32. Everyday uses for strong oxidising agents
Oxidising agents are widely employed because of the effectiveness with which they can kill fungi and bacteria, and can inactivate viruses.
The oxidation process is also an effective means of breaking down coloured compounds making oxidising agents ideal for use as “bleach” for clothes and hair.
The oxidising agents include bleaches and chemicals such as potassium permanganate, KMnO4. 32

33. Potassium permanganate
Potassium permanganate solution will react with any
organic matter in a pond including algae, bacteria, fish,
particulate and dissolved organic, and organic bottom
sediments. It has been used in fish ponds to treat
common fish pathogens such as gill parasites
and external bacterial and fungal infections.
Athlete's foot, another fungal infection in
humans can be treated with very dilute
potassium permanganate solution. 33

34. Hydrogen peroxide
The oxidation process is also an effective means of breaking down coloured compounds making oxidising agents ideal for use as "bleach" for clothes and hair. Hydrogen peroxide, H2O2(aq) can be used to bleach hair and it behaves here as an oxidising agent. In the reaction, coloured pigments in the hair, related to the pigment melanin are oxidised to colourless substances.
H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)     34

35. How does Chlorine in swimming pools work?
Chlorine kills bacteria though a fairly simple chemical reaction. The chlorine solution you pour into the water breaks down into many different chemicals, including hypochlorous acid (HOCl) The hypochlorus acid is a strong oxidising agent and kills microorganisms by removing electrons from the bacteria destroying the cell walls and enzymes and structures inside the cell, rendering them harmless. 35

36. Balancing Redox equations
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 reduced 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 36

37. SO2(g) + 2H2O(l)  SO42-(aq) + 4H+(aq)
1. Write down what you know….
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 37

38. Redox Reactions

39. IO3-  I2 Pb2+  PbO2 ClO-  Cl- Mn2+  MnO2 Sb  SbO+ VO2+  V3+
Write a balanced ion-electron equation for each of the following reactions 1.
IO  I2
Pb  PbO2 2. 3.
ClO-  Cl- 4.
Mn  MnO2
Sb  SbO+ 5.
VO  V3+ 6.
HOBr  Br2 7.
C2 H5OH  CH3CHO 8. 39

40. 2 IO3-  I2 + 12 H+ + 10e - + 6H2O Pb2+  PbO2 + 2H2O + 4 H+ + 2e -1. 2
IO  I2
+ 12 H+
+ 10e -
+ 6H2O
Pb  PbO2 2.
+ 2H2O
+ 4 H+
+ 2e -
ClO-  Cl-
+ 2 H+
+ 2e -
+ H2O 3. 4.
Mn  MnO2
+ 2H2O
+ 4 H+
+ 2e -
Sb  SbO+
+ H2O 5.
+ 2 H+
+ 3e - 40

41. VO2+  V3+ + 2 H+ + e - + H2O 2 HOBr  Br2 + 2 H+ +2 e - + 2 H2O6. 2
HOBr  Br2 7.
+ 2 H+
+2 e -
+ 2 H2O
C2 H5OH  CH3CHO
+ 2 H+
+2 e - 8. 41

42. 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 42

43. No spectator ions shown
Mg(s) +
2 H+(aq) →
Mg 2+(aq)
H2(g) +
No spectator ions shown
Mg(s) +
2 H+(aq) →
Mg 2+(aq)
H2(g) +
oxidising agent H+(aq)
Mg(s) →
Mg 2+(aq) +
2e-
reducing agent Mg(s)
2 H+(aq) +
2e- →
H2(g) 43

44. No spectator ions shown
2I -(aq)
2 Fe3+(aq) →
I2(aq)
2 Fe2+(aq) + +
No spectator ions shown
2I -(aq) →
I2(aq)
2e- +
2 Fe3+(aq)
2e- → +
2 Fe2+(aq)
Oxidising agent Fe3+(aq)
Reducing agent I-(aq) 44

45. oxidising agent Cl 2 (aq) reducing agent Br - (aq)
2NaBr(aq)
Cl2 (g) →
2 NaCl(aq)
Br2(aq) + +
Show charges on ions
2Na+Br -(aq)
Cl2 (g) →
2 Na+Cl -(aq)
Br2(aq) + +
Omit spectator ions
2Br -(aq)
Cl2 (g) + →
2 Cl -(aq) +
Br2(aq) →
Br2(aq)
2e- +
2Br -(aq)
Cl2 (g)
2e- + →
2 Cl -(aq)
oxidising agent Cl 2 (aq)
reducing agent Br - (aq) 45

46. oxidising agent Ag+ (aq) reducing agent Cu (s)
2AgNO3 (aq)
Cu(s) →
2 Ag (s)
Cu(NO3)2 (aq) + +
Show charges on ions
2 Ag (s)
Cu2+(NO3-)2 (aq)
2Ag+NO3- (aq)
Cu(s) → + +
Omit spectator ions
2Ag+(aq)
Cu(s) →
2 Ag (s)
Cu2+ (aq) + +
Cu(s)
2e- →
Cu2+ (aq) +
2e-
2 Ag (s)
2Ag+(aq) + →
oxidising agent Ag+ (aq)
reducing agent Cu (s) 46

47. (K+)2SO32- + Br2 + H2O (K+)2SO42- + 2H+Br
Show charges on ions
(K+)2SO32- + Br2 + H2O (K+)2SO H+Br
Omit spectator ions
SO32- + Br2 + H2O  SO H+Br -
SO32-(aq)
H2O(l) →
SO42-(aq)
2H+ (aq)
2e- + + +
Br2(l) →
2e-
2 Br - (aq) +
Oxidising agent Br2 (l)
Reducing agent SO32-(aq) 47

48.    5CrCl2 + KMnO4+ 8HCl 5CrCl3 + KCl +MnCl2 + 4H2O
Show charges on ions 
5Cr2+(Cl-)2 + K+MnO4-+ 8H+Cl -
5Cr3+(Cl-)3 + K+Cl- +Mn2+(Cl-)2 + 4H2O
Omit spectator ions 
5Cr MnO H+
5Cr Mn H2O
5Cr2+ 
+ 5e-
5Cr3+ 
Mn H2O
MnO H+
+ 5e-
Oxidising agent MnO4- / H+
Reducing agent Cr2+ 48

49. 2Cr 3+ + 7H2O + 3Cl2 Cr2O72- + 14H+ + 6Cl-
2Cr H2O  Cr2O H+
+ 6e -
3Cl  Cl-
+ 6e -
oxidising agent Cl2 (l)
Reducing agent Cr3+(aq) 49

50. b) MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
For each of the following reactions, combine the oxidation and reduction step to form a balanced equation.
a) Cu(s)  Cu2+(aq) + 2e-
Ag+(aq) + e-  Ag(s)
b) MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+(aq) + e-
c) Ce4+(aq) + e-  Ce3+(aq)
2Br-(aq)  Br2(l) + 2e- 50

51. Sodium hypochlorite
Sodium hypochlorite has a wide application as a bleaching agent for textiles and textile laundering. It also acts as a powerful disinfectant. In solution the hypochlorite ions form hypochlorous acid, HClO. This is the oxidising agent.
2HClO-(aq) + 2H+(aq) + 2e– → Cl2(g) + 2H2O(l)  
Household bleach is, in general, a solution containing 4-6% sodium hypochlorite 51

52. Experiment
4 drops of yellow food colouring (E102) or 4 drops of blue food colouring (E 124) are dissolved in 40 cm3 of water.
A solution containing 4 drops of household bleach in 20 cm3 of water is added to the solution and the mixture stirred.
The hypochlorite oxidises the colourings taking the solution through a number of colour changes. 52

53. Luminol Chemilumenescence
Luminol can be oxidised by a solution of bleach to an aminophthalate ion which is produced in an excited state and emits light as it drops back into a ground state. 53

54. A bleach which is a reducing agent
Aran sweaters used to be bleached using sulphur dioxide gas on the wet sweaters. The reaction involves the dissolving of the gas to make a solution containing hydrogen sulphite. This produces some sulphite ions.
SO2(g) + H2O(l) → H2SO3 (aq) 54

55. SO32-(aq) + H2O(l) → SO42-(aq) + 2H+ + 2e- Oxidation
   
This is not an example of a bleach which is an oxidising agent. As the reaction is oxidation, SO2(g) in water and SO32-(aq) are examples of a bleach which is a reducing agent.
They are described as anti-oxidants and are used to sterilise glass bottles in wine-making and are present in some foods. If you see E220 amongst the products of a food stuff, this represents sulphur dioxide. 55

56. Soundbite molecules: Potassium permanganate This Royal Society of Chemistry article looks at the uses of potassium permanganate, a compound used by chemists as a strong oxidising agent.
Chemical functional definitions - Bleach systems This Procter and Gamble article looks at what bleach is, why it is used in cleaning products and how it improves the laundry process. 56

57. Iodide ions can be oxidised using acidified potassium permanganate solution. The equations are: 2I–(aq) → I2(aq) + 2e– MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) How many moles of iodide ions are oxidised by one mole of permanganate ions? A 1.0 B 2.0 C 2.5 D 5.0 D 57

58. In which reaction is hydrogen gas acting as an oxidising agent
In which reaction is hydrogen gas acting as an oxidising agent? A H2 + CuO → H2O + Cu B H2 + C2H4 → C2H6 C H2 + Cl2 → 2HCl D H2 + 2Na → 2NaH D 58

59. The following reactions take place when nitric acid is added to zinc.
NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O(l)
Zn(s) → Zn2+(aq) + 2e–
How many moles of NO3– (aq) are reduced by
one mole of zinc?
A 2/3
B 1
C 3/2
D 2 A 59