Building and balancing chemical equations (5) Redox equations using oxidation numbers

This post is the last for now in a series providing examples of building and balancing chemical equations for different types of inorganic reactions such as neutralisation, combustion, ionic precipitation and thermal decomposition.  

So I’m going to finish this short series of posts right now on building and balancing chemical equations by discussing how to use oxidation numbers in redox equations.

Basic Definitions of Oxidation and Reduction

We'd better recap some basic inorganic chemistry of reduction and oxidation i.e. redox chemistry.

You first think of redox as removal of oxygen or gain of oxygen by a another substance.

So for example the reaction between copper (II) oxide and methane (natural gas is a redox reaction

CH₄    +   4CuO       =      2H₂O    +     4Cu      +    CO₂

The copper oxide loses oxygen: it is reduced and the carbon and hydrogen of the methane gain oxygen: they are oxidised.

Its all fairly straightforward and easy to follow that is until you ask yourself what is oxidised and what is reduced in this reaction:

Cl₂   +   2NaOH    =   NaCl    +   NaClO    +    H₂O

It is impossible to say using the basic definitions given above.

So what do we do? 

We use Oxidation Numbers and apply the oxidation number rules to our reaction equation. 

We’ll come back to this unusual equation later in this blog.

Using Oxidation Numbers to define Oxidation and Reduction

First, what are the oxidation number rules?

Now you can glean a copy of these rules from any good college /Alevel chemistry text.

Here is a version of them in no particular order

1.    The oxidation number of any uncombined element is zero

2.    The oxidation number of a monatomic ion is the charge on that ion

3.    The oxidation numbers of the elements in a compound add up to zero

4.    The oxidation numbers of the elements in a polyatomic ion add up to the charge on that ion

5.    All Group 1 metals have a fixed oxidation number +1, Group 2 metals have oxidation number +2 and aluminium +3.

6.    Oxygen has oxidation number -2 except in compounds where it is combined with fluorine where it can be +2 e.g. in difluorine oxide F₂O and in peroxides where it is -1 e.g. in hydrogen peroxide  H₂O₂

7.    Hydrogen has oxidation number +1 except in metal hydrides e.g. NaH where it is -1

8.    All Group 7 halogens have oxidation number -1, except in compounds with oxygen.

So let’s just illustrate a few examples

a) the sulphate ion is SO₄ ^2-

If oxygen has to be -2 then sulphur’s oxidation number is +6 since

+6  +   4(–2)  = –2

b) the compound copper chloride CuCl₂

In this case chlorine has to be –1 therefore copper is +2  since

+2  +  2(–1)  =  0  the total oxidation number of a neutral compound

c) the compound hydrated iron (III) nitrate  Fe(NO₃)₃.9H₂O 

It is pictured below as pale violet crystals and very beautiful it is too. 

We treat the water of crystallisation separate from the rest of the compound.

So there are three nitrate ions each with a charge of –1 which means the iron has the oxidation number +3 to ensure the compound is neutral.

Each nitrate ion is NO₃ ^—

Oxygen has oxidation number –2 therefore each nitrogen must have oxidation number +5  since    +5   +   3(–2)  =   –1

Now you can begin to see where the Roman numerals come from that sometimes appear in the names of inorganic compounds.

They refer to the element's oxidation number and form the Stock notation named after Edwin Stock the American chemist who developed these ideas before WW2. 

But the real power of oxidation number hasn’t yet been revealed.

The Analytical Power of Oxidation Numbers

Oxidation numbers make the decision about whether a reaction is a redox reaction easy to make and also make it easier to balance such equations especially complicated examples. 

So let’s return to our first example:

CH₄    +   4CuO       =      2H₂O    +     4Cu      +    CO₂

And let’s add the oxidation numbers to the elements that do not have fixed oxidation values i.e. copper and carbon

CH₄    +   4CuO       =      2H₂O    +     4Cu      +    CO₂

–4               4(+2)                                      0              +4

Then lets look at what happens to each elements oxidation number

For copper the oxidation number falls from +8   to  0  a change of –8

For carbon its oxidation number increases from -4 to +4 a change of +8

Notice three things from this example:

i) Oxidation can now be defined as an increase in a species oxidation number look at carbon

ii) Reduction now means what it says on the tin it can be defined now as a decrease in oxidation number look at copper

iii) And finally the changes in oxidation number match +8 with –8.

And there is strategy for balancing redox equation as we shall see:—

But first what about that other difficult equation?

Cl₂   +   2NaOH    =   NaCl    +   NaClO    +    H₂O

Let’s add to the equation the oxidation numbers of elements whose oxidation numbers are not fixed but change in a reaction like this i.e. chlorine.

Cl₂   +   2NaOH    =   NaCl    +   NaClO    +    H₂O

0                                       –1              +1

The interesting thing here is that only the chlorine oxidation number changes increasing and decreasing

So chlorine is both oxidised and reduced in this reaction!!

This is an example of a disproportionation reaction—a reaction in which one species in this example chlorine is both oxidised and reduced and by the same value of oxidation numbers +/-1.

Can you explain now why these reactions are disproportionation reactions?

2H2O2    =     2H2O     +     O2

Na2S2O3    +     2HCl     =      2NaCl    +    H2O    +     S    +   SO2

Balancing Redox Equations using Oxidation Numbers

Let’s look at a simple then a very much more complex example in order to illustrate how to balance redox equations using oxidation numbers

The simple example is the reaction between magnesium and hydrochloric acid producing hydrogen and magnesium chloride

Magnesium + hydrochloric acid  =  magnesium chloride  +  hydrogen

Let’s add the symbols

Magnesium + hydrochloric acid  =  magnesium chloride  +  hydrogen

Mg     +          HCl                     =    MgCl₂       +         H₂

Now add the oxidation numbers of the elements so

Magnesium + hydrochloric acid  =  magnesium chloride  +  hydrogen

Mg     +          HCl                     =    MgCl₂       +         H₂

0                   +1  -1                        +2   -2                      0

Next note how the oxidation numbers change:

Mg goes from 0 to +2   and increase of 2

Hydrogen goes from +1   to 0  a decrease of 1

But to balance the equation the increase and decrease in oxidation number must be the same so we have to increase the hydrogens and double their number in the equation so

Magnesium + hydrochloric acid  =  magnesium chloride  +  hydrogen

Mg     +          2HCl                     =    MgCl₂       +         H₂

0                 2(+1  -1)                        +2   -2                      0

Now the increase and decrease in oxidation number are the same and the equation is balanced.

Here is a more complex example

The reaction between potassium manganate(VII) and iron(II) sulphate solution in sulphuric acid.

Here are the words first

Pot. managanate(VII) + iron(II)sulfate + sulphuric acid = pot. sulfate + manganese sulfate + water + iron(III)sufate

Adding symbols we get:

Pot. managanate(VII) + iron(II)sulfate + sulphuric acid = pot. sulfate + manganese sulfate + water + iron(III)sufate

KMnO₄  + FeSO₄ + H₂SO₄ = K₂SO₄ + MnSO₄ + H₂O + Fe₂(SO₄)₃

Now look for the oxidation numbers that change which in this example are iron and manganese both transition metals and transition metals are known to have variable oxidation states.

KMnO4 + FeSO4 + H2SO4 = K2SO4 + MnSO4 + H2O + Fe2(SO4)3

   +7          +2                                                   +2                            +3

The change to iron is from +2 to +3 but the numbers of iron already double so we must first double the iron on the left hand side thus

KMnO4 + 2FeSO4 + H2SO4 = K2SO4 + MnSO4 + H2O + Fe2(SO4)3

   +7          +2                                                   +2                            +3

Similarly the potassiums double so we do the same with them like so

2KMnO4 + 2FeSO4 + H2SO4 = K2SO4 + 2MnSO4 + H2O + Fe2(SO4)3

   +7             +2                                                     +2                            +3

Now we can examine the changes in oxidation number

First manganese is reduced from a total of +14 to +4 a reduction of -10

Therefore the iron species must be oxidised by a total of +10 oxidation numbers

AS written they increase from a total of +4 to +6 an increase of +2 so if we multiply each species by 5 then the increase in oxidation number will balance that of the reduction like so

2KMnO4 + 10FeSO4 + H2SO4 = K2SO4 + 2MnSO4 + H2O + 5Fe2(SO4)3

   +14             +20                                                     +4                            +30

The numbers indicate the total oxidation number values of the species in the equation that change oxidation number.

All that is left to do is to balance the number of sulfates first.

There are 11 on he left hand side and 18 on the right hand side so adding 8 amounts of sulphuric acid to the left hand side balances them up like so

2KMnO4 + 10FeSO4 + 8H2SO4 = K2SO4 + 2MnSO4 + H2O + 5Fe2(SO4)3

   +14             +20                                                     +4                            +30

Now the last thing we do is we balance up the hydrogen of the acid and the oxygen of the manganate(VII) with water on the right hand side so

2KMnO4 + 10FeSO4 + 8H2SO4 = K2SO4 + 2MnSO4 + 8H2O + 5Fe2(SO4)3

   +14             +20                                                     +4                            +30

And so the equation is now balanced.

You might like to try another example such as the reaction (at about 70^oC between ethandioic acid, sulphuric acid and potassium manganate(VII) which produces potassium sulfate, water, carbondioxide and manganese sulphate.