Oxidation numbers and balancing equations

1
Oxidation numbersandbalancing equations
2

• Oxidation was originally named for dealing with
  processes where oxygen was transferred during
  reactions

3

• Oxidation was originally named for dealing with
  processes where oxygen was transferred during
  reactions, as in
• H2 CuO ? Cu H2O

4

• Oxidation was originally named for dealing with
  processes where oxygen was transferred during
  reactions, as in
• H2 CuO ? Cu H2O
• Oxidation is a process, which always occurs
  simultaneously with Reduction, and which
  characterises many chemical reactions.

5

• As chemistry developed, it was found that many
  reactions, not necessarily involving oxygen,
  shared very similar characteristics.

6

• As chemistry developed, it was found that many
  reactions, not necessarily involving oxygen,
  shared very similar characteristics.
• The words oxidation and reduction have remained.

7

• As chemistry developed, it was found that many
  reactions, not necessarily involving oxygen,
  shared very similar characteristics.
• The words oxidation and reduction have remained.
• Reactions of this type are abbreviated to REDOX
  reactions. They all involve a transfer of
  electrons from one particle to another.

8

• As chemistry developed, it was found that many
  reactions, not necessarily involving oxygen,
  shared very similar characteristics.
• The words oxidation and reduction have remained.
• Reactions of this type are abbreviated to REDOX
  reactions. They all involve a transfer of
  electrons from one particle to another.
• We have developed the idea of oxidation number
  to help in dealing with these reactions.

9
It is easiest to start by looking at ionic
reactions where there is an obvious transfer of
electrons from one atom to another.
10
It is easiest to start by looking at ionic
reactions where there is an obvious transfer of
electrons from one atom to another.E.g. The
reaction in which ordinary salt, sodium chloride,
is formed from the elements sodium and chlorine.
11
It is easiest to start by looking at ionic
reactions where there is an obvious transfer of
electrons from one atom to another.E.g. The
reaction in which ordinary salt, sodium chloride,
is formed from the elements sodium and
chlorine.We write the reaction formally
as 2Na Cl2 2NaCl
12
It is easiest to start by looking at ionic
reactions where there is an obvious transfer of
electrons from one atom to another.E.g. The
reaction in which ordinary salt, sodium chloride,
is formed from the elements sodium and
chlorine.We write the reaction formally
as 2Na Cl2 2NaClbut we also realise that
sodium chloride contains only sodium and chloride
ions, there are no separate sodium chloride
molecules.
13

• We can then write ?2e-? 2Na Cl2
  2Na 2Cl-
• atoms molecules ions ions

14

• We can then write ?2e-? 2Na Cl2
  2Na 2Cl-
• atoms molecules ions ions
• where the sodium atoms have lost electrons to
  become Na ions and these electrons have been
  transferred to the chlorine atoms, making them
  into Cl- ions.

15

• We can then write ?2e-? 2Na Cl2
  2Na 2Cl-
• atoms molecules ions ions
• where the sodium atoms have lost electrons to
  become Na ions and these electrons have been
  transferred to the chlorine atoms, making them
  into Cl- ions.
• In the reaction the sodium atoms had an increase
  of one unit in their positive charge (from 0 to
  1) and the chlorine a one unit decrease (from 0
  to -1). Overall change is 2 for the two sodiums
  and -2 for the two chlorines 0 total

16

• For ionic reactions like this one, we associate
  the loss of electrons as oxidation and the
  corresponding gain of electrons as reduction.

17

• For ionic reactions like this one, we associate
  the loss of electrons as oxidation and the
  corresponding gain of electrons as reduction.
• There is a useful mnemonic for remembering this
  idea, namely OILRIG. Which comes from the
  initial letters of the words in the sentence-
• Oxidation is the loss of electrons, reduction is
  the gain of electrons.

18

• We choose this definition because of the
  similarity between, for example, the sodium in
  its reaction with chlorine and its behaviour in
  the reaction-

19

• We choose this definition because of the
  similarity between, for example, the sodium in
  its reaction with chlorine and its behaviour in
  the reaction-
• ?4e-?
• 4Na O2 ? 2Na2O ( 4Na 2O2- )

20

• We choose this definition because of the
  similarity between, for example, the sodium in
  its reaction with chlorine and its behaviour in
  the reaction-
• ?4e-?
• 4Na O2 ? 2Na2O ( 4Na 2O2- )
• where the sodium oxide is also an ionic compound
  like sodium chloride.

21

• We choose this definition because of the
  similarity between, for example, the sodium in
  its reaction with chlorine and its behaviour in
  the reaction-
• ?4e-?
• 4Na O2 ? 2Na2O ( 4Na 2O2- )
• where the sodium oxide is also an ionic compound
  like sodium chloride.
• Or, our original example
• ? 2e- ?
• H2 CuO ? Cu H2O
• where copper ions become copper metal

22

• We define Oxidation Number (ON) in this way

23

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions

24

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions
• e.g. Fe3 has ON 3 and S2- has ON -2

25

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions
• e.g. Fe3 has ON 3 and S2- has ON -2
• In most compounds H has ON 1 and O has ON -2.

26

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions
• e.g. Fe3 has ON 3 and S2- has ON -2
• In most compounds H has ON 1 and O has ON -2.
  (Exceptions are metal hydrides where ON H -1
  and peroxides where ON O -1)

27

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions
• e.g. Fe3 has ON 3 and S2- has ON -2
• In most compounds H has ON 1 and O has ON -2.
  (Exceptions are metal hydrides where ON H -1
  and peroxides where ON O -1)
• The total ON for a compound is always zero.

28

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions
• e.g. Fe3 has ON 3 and S2- has ON -2
• In most compounds H has ON 1 and O has ON -2.
  (Exceptions are metal hydrides where ON H -1
  and peroxides where ON O -1)
• The total ON for a compound is always zero.
• 4 The ON of an unreacted element is zero.

29

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions
• e.g. Fe3 has ON 3 and S2- has ON -2
• In most compounds H has ON 1 and O has ON -2.
  (Exceptions are metal hydrides where ON H -1
  and peroxides where ON O -1)
• The total ON for a compound is always zero.
• 4 The ON of an unreacted element is zero.
• 5 The overall ON change in a reaction is zero.

30

• We define Oxidation Number (ON) in this way
• Equal to the ionic charge for simple ions
• e.g. Fe3 has ON 3 and S2- has ON -2
• In most compounds H has ON 1 and O has ON -2.
  (Exceptions are metal hydrides where ON H -1
  and peroxides where ON O -1)
• The total ON for a compound is always zero.
• 4 The ON of an unreacted element is zero.
• The overall ON change in a reaction is zero.
• Some atoms with variable ON have the value shown
  by a Roman numeral, e.g. FeIIICl3

31

• So why is this idea useful in constructing
  equations?

32

• So why is this idea useful in constructing
  equations?
• Since the overall ON change is zero, this fixes
  the ratio of the numbers of the reduced atoms to
  those of the oxidised atoms.

33

• So why is this idea useful in constructing
  equations?
• Since the overall ON change is zero, this fixes
  the ratio of the numbers of the reduced atoms to
  those of the oxidised atoms.
• In our first example, there had to be equal
  numbers of sodium and chlorine atoms involved
  since the ON of Na went up by 1 unit (0?1) and
  that of Cl went down by 1 unit (0?-1).
• 2Na Cl2 2NaCl

34

• Here is another ionic reaction
• FeIICl2 Cl2 FeIIICl3

35

• Here is another ionic reaction
• FeIICl2 Cl2 FeIIICl3
• ON 2 -1 0 3 -1

36

• Here is another ionic reaction
• FeIICl2 Cl2 FeIIICl3
• ON 2 -1 0 3 -1
• Each iron atom increases its ON by 1 (2?3),
  whilst the elemental chlorine atoms decrease by 1
  (0?-1).

37

• Here is another ionic reaction
• FeIICl2 Cl2 FeIIICl3
• ON 2 -1 0 3 -1
• Each iron atom increases its ON by 1 (2?3),
  whilst the elemental chlorine atoms decrease by 1
  (0?-1).
• For an ON change of zero, this means that there
  must be equal numbers of FeIICl2 units and
  reacting Cl atoms. Since these atoms come in
  pairs (Cl2) the balanced equation MUST start with
  2 in front of the FeCl2-

38

• 2FeIICl2 1Cl2 ...FeIIICl3

39

• 2FeIICl2 1Cl2 ...FeIIICl3
• and this means that the balanced equation has to
  be 2FeIICl2 Cl2 2FeIIICl3

40

• 2FeIICl2 1Cl2 ...FeIIICl3
• and this means that the balanced equation has to
  be 2FeIICl2 Cl2 2FeIIICl3
• This was a long-winded solution to an easy
  problem, but it shows how the method works.

41

• 2FeIICl2 1Cl2 ...FeIIICl3
• and this means that the balanced equation has to
  be 2FeIICl2 Cl2 2FeIIICl3
• This was a long-winded solution to an easy
  problem, but it shows how the method works.
• It is more useful in balancing this reactions
  equation-
• Copper nitric acid ? Copper nitrate nitric
  oxide water

42

• 2FeIICl2 1Cl2 ...FeIIICl3
• and this means that the balanced equation has to
  be 2FeIICl2 Cl2 2FeIIICl3
• This was a long-winded solution to an easy
  problem, but it shows how the method works.
• It is more useful in balancing this reactions
  equation-
• Copper nitric acid ? Copper nitrate nitric
  oxide water
• Cu HNO3 ? Cu(NO3)2 NO H2O

43

• Cu HNO3 ? Cu(NO3)2 NO H2O
• ON 0 15-2 2 5-2 2-2 1 -2

44

• Cu HNO3 ? Cu(NO3)2 NO H2O
• ON 0 15-2 2 5-2 2-2 1 -2
• For HNO3, the ON of nitrogen is 5 since the H
  is 1 and the O is -2.
• Total ON 0 1 3(-2) ON of nitrogen
• 0 -5 ON of nitrogen
• ON of nitrogen 5

45

• Cu HNO3 ? Cu(NO3)2 NO H2O
• ON 0 15-2 2 5-2 2-2 1 -2
• For HNO3, the ON of nitrogen is 5 since the H
  is 1 and the O is -2.
• Total ON 0 1 3(-2) ON of nitrogen
• 0 -5 ON of nitrogen
• ON of nitrogen 5
• We can see that the oxidation number of nitrogen
  changes from 5?2 (3 steps down) and that of
  copper from 0?2(two steps up).

46

• This means that for every three copper atoms
  oxidised, two nitrogen atoms are reduced and we
  can start the balancing-
• 3Cu ...HNO3 ? 3Cu(NO3)2 2NO ...H2O

47

• This means that for every three copper atoms
  oxidised, two nitrogen atoms are reduced and we
  can start the balancing-
• 3Cu ...HNO3 ? 3Cu(NO3)2 2NO ...H2O
• The next step might be to check the nitrate
  groups, six on the right hand side giving eight
  nitrogens all together, coming from nitric acid.
• 3Cu 8HNO3 ? 3Cu(NO3)2 2NO ...H2O

48

• This means that for every three copper atoms
  oxidised, two nitrogen atoms are reduced and we
  can start the balancing-
• 3Cu ...HNO3 ? 3Cu(NO3)2 2NO ...H2O
• The next step might be to check the nitrate
  groups, six on the right hand side giving eight
  nitrogens all together, coming from nitric acid.
• 3Cu 8HNO3 ? 3Cu(NO3)2 2NO ...H2O
• Now we just need to organise the water formed
  from the eight hydrogens
• 3Cu 8HNO3 ? 3Cu(NO3)2 2NO 4H2O

49

• Although this took some time, we knew that the
  Cu/NO ratio was fixed at 3/2 and this left only a
  few options for the remainder.

50

• Although this took some time, we knew that the
  Cu/NO ratio was fixed at 3/2 and this left only a
  few options for the remainder.
• We can apply this method to a very similar
  reaction-
• Cu HNO3 ? Cu(NO3)2 NO2 H2O

51

• Although this took some time, we knew that the
  Cu/NO ratio was fixed at 3/2 and this left only a
  few options for the remainder.
• We can apply this method to a very similar
  reaction-
• Cu HNO3 ? Cu(NO3)2 NO2 H2O
• The only difference being in the different
  nitrogen oxide produced.

52

• Although this took some time, we knew that the
  Cu/NO ratio was fixed at 3/2 and this left only a
  few options for the remainder.
• We can apply this method to a very similar
  reaction-
• Cu HNO3 ? Cu(NO3)2 NO2 H2O
• The only difference being in the different
  nitrogen oxide produced.
• Cu HNO3 ? Cu(NO3)2 NO2 H2O
• ON 0 15-2 2 5-2 4-2 1 -2

53

• Now the nitrogen changes only by one unit, from
  5 to 4.
• This fixes the Cu/ NO2 ratio as 1Cu / 2NO2
• So now we can start the balancing
• Cu ...HNO3 ? Cu(NO3)2 2NO2 ...H2O

54

• Now the nitrogen changes only by one unit, from
  5 to 4.
• This fixes the Cu/ NO2 ratio as 1Cu / 2NO2
• So now we can start the balancing
• Cu ...HNO3 ? Cu(NO3)2 2NO2 ...H2O
• There are now 4 nitrogens on the right and so
• Cu 4HNO3 ? Cu(NO3)2 2NO2 ...H2O

55

• Now the nitrogen changes only by one unit, from
  5 to 4.
• This fixes the Cu/ NO2 ratio as 1Cu / 2NO2
• So now we can start the balancing
• Cu ...HNO3 ? Cu(NO3)2 2NO2 ...H2O
• There are now 4 nitrogens on the right and so
• Cu 4HNO3 ? Cu(NO3)2 2NO2 ...H2O
• and finally
• Cu 4HNO3 ? Cu(NO3)2 2NO2 2H2O

56

• We could have written this equation in its ionic
  form

57

• We could have written this equation in its ionic
  form as-
• Cu 2NO3- 4H ? Cu2 2NO2 H2O

58

• We could have written this equation in its ionic
  form as-
• Cu 2NO3- 4H ? Cu2 2NO2 H2O
• which makes it clear that the Cu/N ratio is 1/2

59

• We could have written this equation in its ionic
  form as-
• Cu 2NO3- 4H ? Cu2 2NO2 H2O
• which makes it clear that the Cu/N ratio is 1/2
• remember that the charges have to balance, as
  well as the numbers of atoms, in an ionic
  reaction. Here we have a net 2 charge on each
  side.

60

• As a last example we can look at a reaction where
  more than one atom is oxidised
• As2S3 Cl2 ? AsCl5 S8

61

• As a last example we can look at a reaction where
  more than one atom is oxidised
• As2S3 Cl2 ? AsCl5 S8
• In this reaction there are two complications.
• Firstly, both arsenic and sulphur atoms are
  oxidised by chlorine

62

• As a last example we can look at a reaction where
  more than one atom is oxidised
• As2S3 Cl2 ? AsCl5 S8
• In this reaction there are two complications.
• Firstly, both arsenic and sulphur atoms are
  oxidised by chlorine and, secondly, the sulphur
  atoms in elemental sulphur come in rings of
  eight.

63

• As a last example we can look at a reaction where
  more than one atom is oxidised
• As2S3 Cl2 ? AsCl5 S8
• In this reaction there are two complications.
• Firstly, both arsenic and sulphur atoms are
  oxidised by chlorine and, secondly, the sulphur
  atoms in elemental sulphur come in rings of
  eight.
• It is easiest to balance the reaction using
  single sulphur atoms and then multiply everything
  by eight.

64

• As2S3 Cl2 ? AsCl5 S
• ON 3 -2 0 5 -1 0

65

• As2S3 Cl2 ? AsCl5 S
• ON 3 -2 0 5 -1 0
• The overall oxidation number increase for As is
  2 for the each of the two atoms ( 4) and for
  sulphur it is 2 for each of the three atoms (
  6).

66

• As2S3 Cl2 ? AsCl5 S
• ON 3 -2 0 5 -1 0
• The overall oxidation number increase for As is
  2 for the each of the two atoms ( 4) and for
  sulphur it is 2 for each of the three atoms (
  6).
• I.e. each As2S3 molecule requires 10 (64) ON
  steps. Each chlorine atom decreases its ON by
  one unit, so this will require five chlorine
  molecules, each with two atoms

67

• As2S3 Cl2 ? AsCl5 S
• ON 3 -2 0 5 -1 0
• The overall oxidation number increase for As is
  2 for the each of the two atoms ( 4) and for
  sulphur it is 2 for each of the three atoms (
  6).
• I.e. each As2S3 molecule requires 10 (64) ON
  steps. Each chlorine atom decreases its ON by
  one unit, so this will require five chlorine
  molecules, each with two atoms.
• This fixes the As2S3/Cl2 ratio at 15

68

• As2S3 5Cl2 ? ...AsCl5 ...S

69

• As2S3 5Cl2 ? ...AsCl5 ...S
• Now we can balance the rest of the atoms
• As As2S3 5Cl2 ? 2AsCl5 ...S

70

• As2S3 5Cl2 ? ...AsCl5 ...S
• Now we can balance the rest of the atoms
• As As2S3 5Cl2 ? 2AsCl5 ...S
• S As2S3 5Cl2 ? 2AsCl5 3S

71

• As2S3 5Cl2 ? ...AsCl5 ...S
• Now we can balance the rest of the atoms
• As As2S3 5Cl2 ? 2AsCl5 ...S
• S As2S3 5Cl2 ? 2AsCl5 3S
• To complete the process, we have to allow the
  sulphurs to be in groups of eight and so we have
  to multiply each entry by eight-
• 8As2S3 40Cl2 ? 16AsCl5 3S8

72

• Ionic reactions in solution sometimes need you to
  realise that water molecules, H ions and/or OH-
  ions may be present.

73

• Ionic reactions in solution sometimes need you to
  realise that water molecules, H ions and/or OH-
  ions may be present.
• Potassium permanganate, KMnVIIO4 contains
  MnVIIO4- ions. These are oxidising agents in
  acidic, neutral or basic solutions and give
  different products in each.

74

• Ionic reactions in solution sometimes need you to
  realise that water molecules, H ions and/or OH-
  ions may be present.
• Potassium permanganate, KMnVIIO4 contains
  MnVIIO4- ions. These are oxidising agents in
  acidic, neutral or basic solutions and give
  different products in each.
• The ions have their strongest oxidising power in
  acidic solution, where they are reduced to Mn2
  ions, giving an ON change of -5 (from 7 to 2).

75

• Lets see how these ions oxidise iron(II),
  Fe2,ions to iron(III), Fe3, ions.

76

• Lets see how these ions oxidise iron(II),
  Fe2,ions to iron(III), Fe3, ions.
• This is a single ON step and so we will need only
  one MnO4- ion to deal with 5 Fe2ions.

77

• Lets see how these ions oxidise iron(II),
  Fe2,ions to iron(III), Fe3, ions.
• This is a single ON step and so we will need only
  one MnO4- ion to deal with 5 Fe2ions.
• 5Fe2 MnO4- ? 5Fe3 Mn2

78

• Lets see how these ions oxidise iron(II),
  Fe2,ions to iron(III), Fe3, ions.
• This is a single ON step and so we will need only
  one MnO4- ion to deal with 5 Fe2ions.
• 5Fe2 MnO4- ? 5Fe3 Mn2
• The question now is, how to deal with the
  oxygens?

79

• Lets see how these ions oxidise iron(II),
  Fe2,ions to iron(III), Fe3, ions.
• This is a single ON step and so we will need only
  one MnO4- ion to deal with 5 Fe2ions.
• 5Fe2 MnO4- ? 5Fe3 Mn2
• The question now is, how to deal with the
  oxygens?
• Since the reaction is in acidic solution, there
  will be H ions present and these will react to
  form water with the oxygen.

80

• 5Fe2 MnO4- ....H ? 5Fe3 Mn2
  ...H2O

81

• 5Fe2 MnO4- ....H ? 5Fe3 Mn2
  ...H2O
• The four oxygens will produce four water
  molecules, demanding eight hydrogens as the final
  step-

82

• 5Fe2 MnO4- ....H ? 5Fe3 Mn2
  ...H2O
• The four oxygens will produce four water
  molecules, demanding eight hydrogens as the final
  step-
• 5Fe2 MnO4- ....H ? 5Fe3 Mn2 4H2O

83

• 5Fe2 MnO4- ....H ? 5Fe3 Mn2
  ...H2O
• The four oxygens will produce four water
  molecules, demanding eight hydrogens as the final
  step-
• 5Fe2 MnO4- ....H ? 5Fe3 Mn2 4H2O
• 5Fe2 MnO4- 8H ? 5Fe3 Mn2 4H2O

84

• If we wanted a full equation, rather than the
  ionic one, we need to specify both the iron
  compound and the acid. Lets choose ironII
  sulphate and sulphuric acid.
• The equation then becomes
• 10FeSO4 2KMnO4 8H2SO4 ? 5Fe2(SO4)3
  2MnSO4 8H2O
• Where the ratio Fe/Mn is still 15 as before.

85

• In neutral solution the manganese becomes MnIVO2
  and so there are three reduction steps from MnO4-
  and we can start writing the equation as

86

• In neutral solution the manganese becomes MnIVO2
  and so there are three reduction steps from MnO4-
  and we can start writing the equation as
• 3Fe2 MnO4- ....H ? 3Fe3 MnO2 ...H2O

87

• In neutral solution the manganese becomes MnIVO2
  and so there are three reduction steps from MnO4-
  and we can start writing the equation as
• 3Fe2 MnO4- ....H ? 3Fe3 MnO2 ...H2O
• Now we just need to use the hydrogen ions to
  balance as before. The two unbalanced oxygens
  need four H, always present in water.

88

• In neutral solution the manganese becomes MnIVO2
  and so there are three reduction steps from MnO4-
  and we can start writing the equation as
• 3Fe2 MnO4- ....H ? 3Fe3 MnO2 ...H2O
• Now we just need to use the hydrogen ions to
  balance as before. The two unbalanced oxygens
  need four H, always present in water.
• 3Fe2 MnO4- 4H ? 3Fe3 MnO2 2H2O

89

• In neutral solution the manganese becomes MnIVO2
  and so there are three reduction steps from MnO4-
  and we can start writing the equation as
• 3Fe2 MnO4- ....H ? 3Fe3 MnO2 ...H2O
• Now we just need to use the hydrogen ions to
  balance as before. The two unbalanced oxygens
  need four H, always present in water.
• 3Fe2 MnO4- 4H ? 3Fe3 MnO2 2H2O
• And we can notice that this has also balanced the
  charges

90

• The third condition, basic solution, reduces the
  manganese to MnVIO42-, with an ON change of just
  -1 for the manganese this time.

91

• The third condition, basic solution, reduces the
  manganese to MnVIO42-, with an ON change of just
  -1 for the manganese this time.
• So with sulphite ions being oxidised to sulphate,
  with all the changing oxidation numbers shown in
  Roman numerals
• ...MnVIIO4- SIVO32- ? ...MnVI O42- SVIO42-

92

• The third condition, basic solution, reduces the
  manganese to MnVIO42-, with an ON change of just
  -1 for the manganese this time.
• So with sulphite ions being oxidised to sulphate,
  with all the changing oxidation numbers shown in
  Roman numerals
• ...MnVIIO4- SIVO32- ? ...MnVI O42- SVIO42-
• 2MnVIIO4- SIVO32- ? 2MnVI O42- SVIO42-

93

• 2MnVIIO4- SIVO32- ? 2MnVI O42- SVIO42-
• Now we need to make use of the fact that this is
  in basic solution and use hydroxide ions to
  provide the extra oxygen, with the hydrogen
  ending up as water.

94

• 2MnVIIO4- SIVO32- ? 2MnVI O42- SVIO42-
• Now we need to make use of the fact that this is
  in basic solution and use hydroxide ions to
  provide the extra oxygen, with the hydrogen
  ending up as water.
• 2MnVIIO4- SIVO32- 2OH- ?
• 2MnVIO42- SVIO42- H2O

95

• 2MnVIIO4- SIVO32- ? 2MnVI O42- SVIO42-
• Now we need to make use of the fact that this is
  in basic solution and use hydroxide ions to
  provide the extra oxygen, with the hydrogen
  ending up as water.
• 2MnVIIO4- SIVO32- 2OH- ?
• 2MnVIO42- SVIO42- H2O
• If these had all been potassium salts, a full
  equation would be-
• 2KMnO4 K2SO3 2KOH ?
• 2K2MnO4 K2SO4 H2O

96

• There are practice equations to balance on the
  associated pdf file.
• Good luck!