Redox Reactions and Oxidation Number

Redox Reactions:

Redox Reactions is reaction in which oxidation and reduction both are going to happen, called redox reaction. It may be in any medium (acidic, neutral, and basic).

Redox Reaction in terms of electron transfer reactions:

We have already learnt that the reactions

2Na(s) + Cl₂ (g) → 2NaCl (s)

4Na(s) + O₂ (g) → 2Na₂O(s)

2Na(s) + S(s) → Na₂S(s)

These are redox reactions because in each of these reactions sodium is oxidised due to the addition of either oxygen or more electronegative element to sodium. Simultaneously, chlorine, oxygen and sulphur are reduced because to each of these, the electropositive element sodium has been added. We know that sodium chloride, sodium oxide and sodium sulphide are ionic compounds and perhaps better written as Na⁺Cl^–(s), (Na⁺ )₂O^2–(s), and (Na⁺ )₂ S^2–(s).

For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and the other the gain of electrons.

2 Na(s) → 2 Na⁺ (g) + 2e^–

Cl² (g) + 2e^– → 2 Cl^– (g)

Each of the above steps is called a half reaction. First is called half oxidation reaction other is called half reduction reaction. The final reactions that involve loss of electrons are called oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reaction.

Definitions:

Oxidation: Loss of electrons(s) by any species.

Reduction: Gain of electron(s) by any species.

Oxidising agent: It accepts electron to each of the elements interacting with it and thus help in oxidising them.

Reducing agent: It donates electron to each of the element interacting with it and thus help in reducing them.

Oxidation Number:

Oxidation number denotes the oxidation state of an element in a compound. The oxidation number of an atom is the charge that atom would have if the compound was composed of ions.

We have seen electron transfers in a reaction which is ionic reaction. But if a reaction has covalent properties then they can transfer their whole electron and that method we can’t apply. So, we have to find out the oxidation state of the element to understand the partial sharing electron in covalent reaction.

There are some rules to find out the oxidation state of an element which are given below: -

1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O₂, O₃, P₄, S₈, and aluminium metal all have an oxidation number of 0.

2. The oxidation number of simple ions is equal to the charge on the ion. The oxidation number of sodium in the Na⁺ ion is +1, for example, and the oxidation number of chlorine in the Cl^- ion is -1.

3. The oxidation number of hydrogen is +1 when it is combined with a non-metal as in CH₄, NH₃, H₂O, and HCl.

4. The oxidation number of hydrogen is -1 when it is combined with a metal as in. LiH, NaH, CaH₂, and LiAlH₄.

5. The metals in Group IA form compounds (such as Li₃N and Na₂S) in which the metal atom has an oxidation number of +1

6. The elements in Group IIA form compounds (such as Mg₃N₂ and CaCO₃) in which the metal atom has a +2-oxidation number.

7.  Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O₂, O₃, H₂O₂, and the ion.

8.  The elements in Group VIIA often form compounds (such as AlF₃, HCl, and ZnBr₂) in which the non-metal has a -1 oxidation number.

9. The sum of the oxidation numbers in a neutral compound is zero.

H₂O: 2(+1) + (-2) = 0

10. The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of the sulfur atom in the SO₄^2- ion must be +6, for example, because the sum of the oxidation numbers of the atoms in this ion must equal -2.

SO₄^2-: (+6) + 4(-2) = -2

11. Elements toward the bottom left corner of the periodic table are more likely to have positive oxidation numbers than those toward the upper right corner of the table. Sulphur has a positive oxidation number in SO₂, for example, because it is below oxygen in the periodic table.

SO₂: (+4) + 2(-2) = 0

Definitions:

Oxidation: An increase in the oxidation number of the electron in the given substance.

Reduction: A decrease in the oxidation number of the element in the given substance.

Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.

Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents ae also called as reducants.

Redox reactions: Reactions which involve change in oxidation number of the interacting species.

Types of Redox Reactions:

1. Combination reaction:

A combination reaction may be denoted in the manner:

A + B → C

Either A and B or both A and B must be in the 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:

C(s) + O₂ (g) CO₂ (g)

CH₄ (g) + 2O₂ (g) CO₂ (g) + 2H₂O(l)

2. Decomposition reactions

Decomposition reactions are the opposite of combination reactions. 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.

2H₂O (l) 2H₂ (g) + O₂ (g)

2KClO₃ (s) 2KCl (s) + 3O₂ (g)

All decomposition reactions are not redox reaction. For example, decomposition of calcium carbonate is not a redox reaction

CaCO₃ (s) CaO(s) + CO₂ (g)

3. Displacement reaction

Displacement reactions, also known as replacement reactions, involve compounds and the “replacing” of elements. They occur as single and double replacement reactions. In a displacement reaction, an ion in a compound is replaced by an ion 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 uncombine state. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores.

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

V₂O₅ (s) + 5Ca (s) 2V (s) + 5CaO (s)

Cr₂O₃ (s) + 2 Al (s) Al2O3 (s) + 2Cr(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.

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

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

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

Mg(s) + 2H₂O(l) Mg(OH)₂ (s) + H₂ (g)

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.

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

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

· 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:

2H₂O (l) + 2F₂ (g) → 4HF(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:

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

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

· 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:

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

Cl₂ (g) + 2I ^– (aq) → 2Cl ^– (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 oxidising 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.

4. Disproportionation reaction

In some redox reactions, substances can be both oxidized and reduced. These are known as disproportionation reactions. One real-life example of such a process is the reaction of hydrogen peroxide, H₂O₂, when it is poured over a wound. At first, this might look like a simple decomposition reaction, because hydrogen peroxide breaks down to produce oxygen and water:

2H₂O₂(aq) → 2 H₂O(l) + O₂(g)

The key to this reaction lies in the oxidation states of oxygen, however. Notice that oxygen is present in the reactant and both products. In H₂O₂, oxygen has an oxidation state of -1. In H₂O, its oxidation state is -2, and it has been reduced. In O₂ however, its oxidation state is 0, and it has been oxidized. Oxygen has been both oxidized and reduced in the reaction, making this a disproportionation reaction. The general form for this reaction is as follows:

2A → A’ + A”

The paradox of Fractional Oxidation Number:

We have discussed that oxidation number can be positive, negative and zero also. But sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction not in integer. There are some Examples:

C₃O₂ [where oxidation number of carbon is ],

Br₃O₈ [where oxidation number of bromine is ]

Na₂S₄O₆ (where oxidation number of sulphur is 2.5)

As we know that electrons are never shared/transferred in fraction. It is done by integers only. In actual, this fraction 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 realized is present in different oxidation states.

We can explain this rule by their structure which are given below-

O = C = C^*= C = O

(structure of carbon suboxide)

Carbon oxide has three carbon and among them two carbon have similar properties and other (starred carbon) has different properties. (Here, properties in a sense, we are talking about oxidation number)

Oxidation state of without star carbon = +2

Oxidation sate of starred carbon = 0

The average oxidation number of carbon = =

(structure of tribromooctaoxide)

It contains three bromine and in them two are same oxidation number and other has different oxidation number.

Oxidation state of first and last bromine in structure = +6

Oxidation state of middle bromine in structure = +4

The average oxidation state of bromine = =

Structure of sodium tetrathionate has four Sulphur but they are two type of Sulphur in that. First and fourth Sulphur have same properties, and second and third have different properties from first and fourth Sulphur.

Oxidation state of first and fourth Sulphur is = +5

Oxidation state of second and third Sulphur is = 0

The average oxidation state of Sulphur is = = 2.5

We can conclude that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structure only. 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. However, the oxidation state may be in fraction as in O₂⁺ and O₂^- where it is +1/2 and -1/2 respectively.