8.3.1 Types of Redox Reactions

1.Combination reactions

A combination reaction may be denoted in the manner:

A + B → C

elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions.

Some important examples of this category are:

0         0            +4 -2

C(s) + O2 (g) →  CO2 (g)


0             0              +2   -3

3Mg(s) + N2 (g) →  Mg3N2 (s)


+4+1         0               +4 -1         +1 -1

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)


2.Decomposition reactions

Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state. Examples of this class of reactions are:

+1 -2               0              0

2H2O (l) → 2H2 (g) + O2 (g)

+1 -1               0              0

2NaH (s) → 2Na (s) + H2 (g)

+1 +5 -2           +1 -1           0

2KClO3 (s) → 2KCl (s) + 3O2 (g)

It may carefully be noted that there is no change in the oxidation number of hydrogen in methane under combination reactions and that of potassium in potassium chlorate in reaction.

This may also be noted here that all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction.

+2 +4 -2           +2 -2       +4 -2

CaCO3 (s) → CaO (s) + CO2 (g)


3.Displacement reactions

In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:

X + YZ → XZ + Y

Displacement reactions fit into two categories: metal displacement and non-metal displacement.

(a) Metal displacement: A metal in a compound can be displaced by another metal in the uncombined state. We have already discussed about this class of the reactions under section 8.2.1. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores. A few such examples are:

+2 +6 -2           0              0           +2 +6 -2

CuSO4 (aq) + Zn (s) → Cu (s) + ZnSO4 (aq)


+5 -2           0                 0          +2 -2

V2O5 (s) + 5Ca (s) → 2V (s) + 5CaO (s)


+4 -1           0                 0          +2 -1

V2O5 (s) + 5Ca (s) → 2V (s) + 5CaO (s)

+3 -2              0                 +3 -2         0

Cr2O3 (s) + 2Al (s) → Al2O3 (s) + 5Cr (s)

In each case, the reducing metal is a better reducing agent than the one that is being reduced which evidently shows more capability to lose electrons as compared to the one that is reduced.

(b) Non-metal displacement: The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement.

All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, will displace hydrogen from cold water.


0               +1 -2          +1 -2 +1           0

2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g)


0           +1 -2          +2 -2 +1            0

Ca (s) + 2H2O (l) → Na(OH)2 (aq) + H2 (g)


Less active metals such as magnesium and iron react with steam to produce dihydrogen gas:


0               +1 -2          +2 -2 +1           0

Mg (s) + 2H2O (l) → Mg(OH)2 (s) + H2 (g)


0               +1 -2          +3 -2           0

2Fe (s) + 3H2O (l) → Fe2O3 (s) + 3H2 (g)


Many metals, including those which do not react with cold water, are capable of displacing hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals. A few examples for the displacement of hydrogen from acids are:

0           +1 -1              +2 -1             0

Zn (s) + 2HCl (aq) →  ZnCl2 (aq) + H2 (aq)


0           +1 -1                  +2 -1             0

Mg (s) + 2HCl (aq) →  MgCl2 (aq) + H2 (aq)

0           +1 -1               +2 -1             0

Fe (s) + 2HCl (aq) →  FeCl2 (aq) + H2 (aq)

Above three reactions are used to prepare dihydrogen gas in the laboratory. Here, the reactivity of metals is reflected in the rate of hydrogen gas evolution, which is the slowest for the least active metal Fe, and the fastest for the most reactive metal, Mg.

Very less active metals, which may occur in the native state such as silver (Ag), and gold (Au) do not react even with hydrochloric acid.

In section (8.2.1) we have already discussed that the metals – zinc (Zn), copper (Cu) and silver (Ag) through tendency to lose electrons show their reducing activity in the order Zn> Cu>Ag.

Like metals, activity series also exists for the halogens. The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table.

This implies that fluorine is so reactive that it can replace chloride, bromide and iodide ions in solution. In fact, fluorine is so reactive that it attacks water and displaces the oxygen of water :

0               +1 -2          +1 +1          0

2Fe2 (g) + 2H2O (l) → 2HF (aq) + O2 (g)

It is for this reason that the displacement reactions of chlorine, bromine and iodine using fluorine are not generally carried out in aqueous solution.

On the other hand, chlorine can displace bromide and iodide ions in an aqueous solution as shown below:


0             +1 -1          +1 +1          0

Cl2 (g) + 2KBr (aq) → 2KCl (aq) + Br2 (l)                             (8.41)


0             +1 -1          +1 +1          0

Cl2 (g) +  2KI (aq) → 2KCl (aq) + I2 (s)                                  (8.42)


As Br2 and I2 are coloured and dissolve in CCl4, can easily be identified from the colour of the solution. The above reactions can be written in ionic form as:


0             -1                -1                0

Cl2 (g) + 2Br- (aq) → 2Cl- (aq) + Br2 (l)


0            -1               -1               0

Cl2 (g) +  2I- (aq) → 2Cl- (aq) + I2 (s)

Reactions (8.41) and (8.42) form the basis of identifying Br– and I– in the laboratory through the test popularly known as ‘Layer Test’. It may not be out of place to mention here that bromine likewise can displace iodide ion in solution:

0            -1                 -1             0

Br2 (l) +  2I- (aq) → 2Br- (aq) + I2 (s)

The halogen displacement reactions have a direct industrial application. The recovery of halogens from their halides requires an oxidation process, which is represented by:

2X- → X2 + 2e-

here X denotes a halogen element. Whereas chemical means are available to oxidise Cl, Br and I, as fluorine is the strongest oxidizing agent; there is no way to convert F – ions to F2 by chemical means. The only way to achieve F2 from F is to oxidise electrolytically, the details of which you will study at a later stage.


Disproportionation reactions

Disproportionation reactions are a special type of redox reactions. In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced.

One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction.

The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation.

+1 -1                 +1 -2         0

2H2O2 (aq) → 2H2O (l) + O2 (g)

Here the oxygen of peroxide, which is present in –1 state, is converted to zero oxidation state in O2 and decreases to –2 oxidation state in H2O.

Phosphorous, sulphur and chlorine undergo disproportionation in the alkaline medium as shown below :

0                                                        -3                 +1

P4 (s) +  3OH- (aq) + 3H2O (l) → PH3 (g) + 3H2PO2- (aq)


0                                       -2               +2

S8 (s) +  12 OH- (aq) → 2S2- (aq) + 2S2O3 (aq) + 6H2O (l)


0                                   +1              -1

Cl2 (g) +  2OH- (aq) → CIO- (aq) + Cl- (aq) +  H2O (l)                          (8.48)


The reaction (8.48) describes the formation of household bleaching agents. The hypochlorite ion (ClO–) formed in the reaction oxidises the colour-bearing stains of the substances to colourless compounds.

It is of interest to mention here that whereas bromine and iodine follow the same trend as exhibited by chlorine in reaction (8.48), fluorine shows deviation from this behaviour when it reacts with alkali. The reaction that takes place in the case of fluorine is as follows:

2F2 (g) +  2OH- (aq) → 2 F- (aq) + OF2 (g) +  H2O (l)                          (8.49)

(It is to be noted with care that fluorine in reaction (8.49) will undoubtedly attack water to produce some oxygen also).

This departure shown by fluorine is not surprising for us as we know the limitation of fluorine that, being the most electronegative element, it cannot exhibit any positive oxidation state. This means that among halogens, fluorine does not show a disproportionation tendency.

The Paradox of Fractional Oxidation Number

Sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction. Examples are:


C3O2[where oxidation number of carbon is (4/3)],

Br3O8 [where oxidation number of bromine is (16/3)]

And Na2S4O6  (where oxidation number of sulphur is 2.5)


We know that the idea of fractional oxidation number is unconvincing to us, because electrons are never shared/transferred in fraction. Actually this fractional oxidation state is the average oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is present in different oxidation states. Structure of the species C3O2, Br3O8 and S4O62– reveal the following bonding situations:



The element marked with asterisk in each species is exhibiting the different oxidation state (oxidation number) from rest of the atoms of the same element in each of the species.

This reveals that in C3O2, two carbon atoms are present in +2 oxidation state each, whereas the third one is present in zero oxidation state and the average is 4/3.

However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br3O8, each of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine is present in +4 oxidation state.

Once again the average, that is different from reality, is 16/3. In the same fashion, in the species S4O62– , each of the two extreme sulphurs exhibits oxidation state of +5 and the two middle sulphurs as zero.

The average of four oxidation numbers of sulphurs of the S4O62– is 2.5, whereas the reality being + 5,0,0 and +5 oxidation number respectively for each sulphur.

We may thus, in general, conclude that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structures only.

Further, whenever we come across with fractional oxidation state of any particular element in any species, we must understand that this is the average oxidation number only.

In reality (revealed by structures only), the element in that particular species is present in more than one whole number oxidation states.

Fe3O4, Mn3O4, Pb3O4 are some of the other examples of the compounds, which are mixed oxides, where we come across with fractional oxidation states of the metal atom.

However, the oxidation states may be in fraction as in O2+ and O where it is +½ and –½ respectively.

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