Oxidation numbers and balancing equations

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Oxidation numbers and balancing equations

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Title: Oxidation numbers and balancing equations

1
Oxidation numbersandbalancing equations
2

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

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

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.

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

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

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.

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

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.

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

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

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.

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

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- )

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.

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

We define Oxidation Number (ON) in this way

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

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

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.

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)

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.

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.

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.

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

So why is this idea useful in constructing
equations?

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.

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

Here is another ionic reaction
FeIICl2 Cl2 FeIIICl3

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

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).

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-

2FeIICl2 1Cl2 ...FeIIICl3

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

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.

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

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

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

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

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).

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

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

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

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.

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

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.

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

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

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

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

We could have written this equation in its ionic
form

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

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

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.

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

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

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.

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.

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

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).

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

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

As2S3 5Cl2 ? ...AsCl5 ...S

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

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

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

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

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.

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).

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

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.

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

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?

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.

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

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

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

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.

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

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

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.

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

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

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

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-

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-

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- ? 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

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