Types of Redox Reactions

(i) Intermolecular redox reactions In such reactions, oxidation and reduction take place separately in two compounds. e.g.,

(ii) Intramolecular redox reactions In these reactions, oxidation and reduction take place in a single compound. e.g.,

(iii) Disproportionation reactions These reactions involve reduction and oxidation of same element of a compound. e.g.,

This reaction is also known as autoredox reaction.

Classification of Redox Reactions

1. Direct Redox Reactions

Chemical reaction in which oxidation as well as reduction is carried out simultaneously in the same container, is known as direct redox reaction In such reactions, energy is generally liberated in the form of heat energy.

2. Indirect Redox Reactions

A reaction in which oxidation and reduction are carried out separately in two separate half-cells, is known as indirect redox reaction. In such reactions, energy is generally liberated in the form of electrical energy.

oxidation and Reduction

Reductants and Oxidants

Oxidant or oxidising agent is a chemical substance which can accept one or more electrons and causes oxidation of some other species. In other words, the oxidation number of oxidant decreases in a redox reaction.

Important Oxidants

Molecules of most electronegative elements such as O₂, O₃, halogens.

Compounds having element in its highest oxidation state e.g.,

K₂Cr₂O₇, KMnO₄, HCIO₄, H₂SO₄, KCIO₃, Ce(SO₄)₂,

Oxides of metals and non-metals such as MgO, CrO₃, CO₂, etc.

Reductant or reducing agent is a chemical.substance which can give one or more electrons and causes reduction of some other species. In other words, the oxidation number of reductant increases in a redox reaction.

Important Reductants

All metals such as Na, AI, Zn, etc., and some non – metals, e.g., C, S. P, H₂, etc.

Metallic hydrides like NaH, LiH. KH, CaH₂ etc.

Oxidation Number

The oxidation number is defined as the charge in which an atom appears to have when all other atoms are removed from it as ions. It may have + or – sign.

[An element may have different values of oxidation number depending upon the nature of compound in which it is present.]

Oxidation number of an element may be a whole number (positive or negative) or fractional or zero.

Important Points for Determining Oxidation Number

1. The algebraic sum of the oxidation numbers of aU the atoms in an uncharged (neutral) compound is zero. In an ion, the algebraic sum is equal to the charge on the ion.
2.  All elements in the elementary state have oxidation number zero, e.g., He, Cl₂, S₈, P₄ etc.
3.  As fluorine is the most electronegative element, it always has an oxidation number of – 1 in all of its compounds.
4.  In compounds containing oxygen, the oxidation number of oxygen is – 2 except in peroxides (-1) such as Na₂O₂, in OF₂ and in O₂ F₂ (+2 and + 1 respectively).
5. In all compounds. except ionic metallic hydrides, the oxidation number of hydrogen is +1. In metal hydrides like NaH, MgH₂, CaH₂, LiH, etc the oxidation number of hydrogen is -1.
6. Oxidation number for alkali metals is +1 and for alkaline earth metals is + 2.
7.  Oxidation number of metal in amalgams is zero.
8.  In case of coordinate bond, it gives +2 value of oxidation number to less electronegative atom and -2 values to more electronegative atom when coordinate bond is directed formless electronegative atom to more electronegative atom .
9. If coordinate bond is directed from more electronegative to less electronegative atom then its contribution be zero for both the atoms.
10. For p-block elements [Except F and 0], the highest oxidation number is equal to their group number and lowest oxidation number is equal to the group number minus eight.
11. In transition elements the lowest oxidation number is equal to the number of ns electrons and highest oxidation number is equal to number of ‘ns’ and (n – l)d unpaired electrons.

Determination of Oxidation Number of Underlined Element

Oxidation number of Na = + 1

Oxidation number of 0 = – 2

∴ 2 (1) + 4x + 6 x – 2 = 0

a = 5 / 2, this is average oxidation number. because the compound has two types of sulphur atom.

OX of sulphur bonded with coordinate bond = 5

ON of sulphur which have S-S bond = 0

Average oxidation number = 5 + 5 + 0 + 0 / 4 = 5 / 2

(vii) NH₄ NO₃

There are two types of nitrogen atoms. Therefore. evaluation should be made separately as

Oxidation number of N in NH⁺₄

x + 4 (+ 1)= + 1

x = – 3

Oxidation number of N in NO^–₃

y + 3 x (- 2) = – 1

y = 5

Stock Notations

The oxidation states of elements exhibiting variable oxidation states are specified by Roman numerals such as I, II, III, IV, etc., within parenthesis after the symbol or name of the element. This system was introduced for the first time by German chemist, Alfred Stock and is known as Stock notation. This may be illustrated as

Balancing of Redox Chemical Equations

Every chemical equation must be balanced according to law of conservation of mass. In a balanced chemical equation the atoms of various species involved in the reactants and products must be equal in number. Redox reaction can be balanced through (i) Ion electron method (ii) Oxidation number method

Ion Electron Method

This method of balancing was developed by Jette and Lamer in 1927.

For example. balance the equation

Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O

It involves the following steps.

Step I Write the redox reaction in ionic form

Cu + H⁺ + NO^–₃ → Cu²⁺ + NO + H₂O

Step II Split the redox reaction into its oxidation-half and reduction half-reaction.

Step III Balance atoms of each half-reaction (except H and O) by using simple multiples.

Cu → Cu²⁺ and NO^–₃ → NO

(Except H and O, all atoms are balanced)

Step IV Balance H and O as

(i) For acidic and neutral solutions Add H₂O molecule to the side deficient in oxygen and H+ to the side deficient in hydrogen.

(ii) For alkaline solutions For each excess of oxygen, add one water molecule to the same side and OH^– ion to the other side to balance H.

Step V Add electrons to the side deficient in electrons.

Step VI Equalize the number of electrons in both the reactions by multiplying a suitable number

Step VII Add the two balanced half reactions and cancel common terms of opposite sides

Step VIII Convert the ionic reaction into molecular form by adding spectator ions

(Ions which are present in solution but do not take part in the redox reaction, are omitted while writing the net ionic equation of a reaction and are known as spectator ions.)

Oxidation Number Method

For example, balance the equation

Mg + HNO₃ → Mg(NO₃)₂ + N₂O + H₂O

It involves the following steps.

Step I Write the skeleton equation (if not given)

Step II Assign oxidation number of each atom

Step III Balance atoms other than H and O in two processes.

Step IV Equalize the total increase or decrease in oxidation number

4Mg + 2HNO₃ → 4Mg(NO₃)₂ + NO₂O

Step V Balance H and O

8H⁺ + 4 Mg + 2HNO₃ + 8NO₃^– → 4 Mg (NO₃)₂ + N₂O + 5H₂O

4 Mg + 10 HNO₃ → 4 Mg (NO₃)₂ + N₂O + 5H₂O

Redox Reactions in Daily Life