Redox Reaction : Reduction and Oxidation

Chemical reactions involve transfer of electrons from one chemical substance to another. These electron – transfer reactions are termed as oxidation-reduction or redox-reactions.

Redox reactions play an important role in our daily life. These reactions are accompanied by energy changes in the form of heat, light, electricity etc. Generation of electricity in batteries and many industrial processes such as production of caustic soda, KMnO₄, extraction of metals like sodium, iron and aluminium are common examples of redox reactions.

Molecular and Ionic equations

(1) Molecular equations : When the reactants and products involved in a chemical change are written in molecular forms in the chemical equation, it is termed as molecular equation.

Examples :

(i) MnO₂ + 4HCl → MnCl₂ + 2H₂O +Cl₂

(ii) 2FeCl₃ + SnCl₂ → SnCl₄

In above examples, the reactants and products have been written in molecular forms, thus the equation is termed as molecular equation.

(2) Ionic equations : When the reactants and products involved in a chemical change are ionic compounds, these will be present in the form of ions in the solution. The chemical change is written in ionic forms in chemical equation, it is termed as ionic equation.

Examples :

(i) MnO₂ + 4H⁺ + 4Cl⁻ → Mn²⁺ + 2Cl⁻ + 2H₂O + Cl₂

(ii) 2Fe³⁺ + 6Cl⁻ + 2Cl⁻ → 2Fe²⁺ + 4Cl⁻ + Sn⁴⁺ + 4Cl⁻

In above examples, the reactants and products have been written in ionic forms, thus the equation is termed as ionic equation.

(3) Spectator ions : In ionic equations, the ions which do not undergo any change and equal in number in both reactants and products are termed spectator ions and are not included in the final balanced equations.

Example :

Zn + 2H⁻ + 2Cl⁻ → Zn²⁺ + H₂ + 2Cl⁻ (Ionic equation)

Zn + 2H⁺ → Zn²⁺ + H₂ (Final ionic equation)

In above example, the Cl⁻ ions are the spectator ions and hence are not included in the final ionic balanced equation.

(4) Rules for writing ionic equations

(i) All soluble ionic compounds involved in a chemical change are expressed in ionic symbols and covalent substances are written in molecular form. H₂O, NH₃, NO₂ , NO, SO₂, CO, CO₂, etc., are expressed in molecular form.

(ii) The ionic compound which is highly insoluble is expressed in molecular form.

(iii) The ions which are common and equal in number on both sides, i.e., spectator ions, are cancelled.

(iv) Besides the atoms, the ionic charges must also be balanced on both the sides.

The rules can be explained by following examples,

Example : Write the ionic equation for the reaction of sodium bicarbonate with sulphuric acid, The molecular equation for the chemical change is,

NaHCO₃+ H₂SO₄ → Na₂SO₄+ H₂O + CO₂

NaHCO₃, H₂SO₄ and Na₂SO₄ are ionic compounds, so these are written in ionic forms.

$\[N{{a}^{+}}+HCO_{3}^{-}+2{{H}^{+}}+SO_{4}^{2-}\to 2N{{a}^{+}}+SO_{4}^{2-}+{{H}_{2}}O+C{{O}_{2}}$

Na⁺ and $\[SO_{4}^{2-}$ ions are spectator ions; hence these shall not appear in the final equation.

$\[HCO_{3}^{-}+2{{H}^{+}}\to {{H}_{2}}O+C{{O}_{2}}$

To make equal charges on both sides, $\[HCO_{3}^{-}$ should have a coefficient 2.

$\[2HCO_{3}^{-}+2{{H}^{+}}\to {{H}_{2}}O+C{{O}_{2}}$

In order to balance the hydrogen and carbon on both sides, the molecules of H₂O and CO₂ should have a coefficient 2 respectively.

$\[2HCO_{3}^{-}+2{{H}^{+}}=2{{H}_{2}}O+2C{{O}_{2}}$ or $\[HCO_{3}^{-}+{{H}^{+}}={{H}_{2}}O+C{{O}_{2}}$

This is the balanced ionic equation.

Conversion of ionic equation in molecular form can be explained by following example,

Example : Write the following ionic equation in the molecular form if the reactants are chlorides.

2Fe³⁺+ Sn²⁺ → 2Fe²⁺+ Sn⁴⁺

For writing the reactants in molecular forms, the requisite number of chloride ions are added.

$\[\underbrace{2F{{e}^{3+}}+C{{l}^{-}}}_{{}}\,\,+\,\,\underbrace{S{{n}^{2+}}+2C{{l}^{2+}}}_{{}}$ or 2FeCl₃+ SnCl₂

Similarly 8 Cl⁻ ions are added on R.H.S. to neutralise the charges.

$\[\underbrace{2F{{e}^{2+}}+4C{{l}^{-}}}_{{}}\,\,+\,\,\underbrace{S{{n}^{4+}}+4C{{l}^{-}}}_{{}}$ 2FeCl₂+ SnCl₄

Thus, the balanced molecular equation is, 2FeCl₃+ SnCl₂= 2FeCl₂+ SnCl₄

Oxidation-reduction and Redox reactions

(1) Oxidation : Oxidation is a process which involves; addition of oxygen, removal of hydrogen, addition of non-metal, removal of metal, Increase in +ve valency, loss of electrons and increase in oxidation number.

(i) Addition of oxygen

(a) 2Mg + O₂→ 2MgO (Oxidation of magnesium)

(b) S + O₂ → SO₂ (Oxidation of sulphur)

(d) Na₂SO₃+ H₂O₂ → Na₂SO₄+ H₂O (Oxidation of sodium sulphite)

(ii) Removal of hydrogen

(a) H₂S + Cl₂ → 2HCl + S (Oxidation of hydrogen sulphide)

(b) 4HI + O₂→ 2H₂O + 2I₂ (Oxidation of hydrogen iodide)

(d) 4HCl + MnO₂ → MnCl₂ + 2H₂O + Cl₂(Oxidation of hydrogen chloride)

(iii) Addition of an electronegative element or addition of Non-metal

(a) Fe + S → FeS (Oxidation of iron)

(b) SnCl₂ + Cl₂→ SnCl₄(Oxidation of stannous chloride)

(iv) Removal of an electropositive element or removal of metal

(a) 2KI + H₂O₂ 2KOH + I₂(Oxidation of potassium iodide)

(b) 2K₂MnO₄+ Cl₂ 2KCl + 2KMnO₄(Oxidation of potassium manganate)

(v) Increase in +ve valency and Decrease in – ve valency

Increase in +ve valency

–4, –3, –2, –1, 0, +1, +2, +3, +4

Decrease in –ve valency

(a) Fe²⁺ → Fe³⁺ + e⁻ (+ve valency increases)

(b) Sn²⁺ → Sn⁴⁺ + 2e⁻ (+ve valency increases)

(d) $\[MnO{{_{4}^{2-}}^{{}}}\to MnO_{4}^{-}+{{e}^{-}}$ (–ve valency decreases)

(vi) Loss of electrons (also known as de-electronation)

(a) H^o → H⁺ + e⁻ (Formation of proton)

(b) $\[H_{2}^{0}\to H_{2}^{+}+{{e}^{-}}{{\sin }^{-1}}\theta$ (De-electronation of hydrogen)

(d) Mg → Mg²⁺ + 2e⁻ (De-electronation of Magnesium)

(e) $\[MnO_{4}^{2-}\to MnO_{4}^{-}+{{e}^{-}}$ (De-electronation of $\[MnO_{4}^{2-}$ )

(f) 2Cl⁻ → Cl₂ + 2e⁻ (De-electronation of chloride ion)

(g) 2Fe^o → 2Fe³⁺ + 6e⁻ (De-electronation of iron)

(vii) Increase in oxidation number

(a) Mg^o → Mg²⁺ (From 0 to +2)

(b) [Fe⁺²(CN)₆]⁴⁻→ [Fe⁺³(CN)₆]³⁻ (From +2 to +3)

(2) Reduction : Reduction is just reverse of oxidation. Reduction is a process which involves; removal of oxygen, addition of hydrogen, removal of non-metal, addition of metal, decrease in +ve valency, gain of electrons and decrease in oxidation number.

(i) Removal of oxygen

(a) CuO + C → Cu +Co (Reduction of cupric oxide)

(b) $\[\underset{Steam}{\mathop{{{H}_{2}}O}}\,\,\,\,+\,\,\,\underset{Coke}{\mathop{C}}\,\to \,\,\,\underbrace{CO+{{H}_{2}}}_{Water\stackrel{\scriptscriptstyle\to}{\leftarrow}gas}$ (Reduction of water)

(d) C₆H₅OH + Zn → C₆H₅OH + ZnO (Reduction of phenol)

(ii) Addition of hydrogen

(a) Cl₂ + H₂ → 2HCl (Reduction of chlorine)

(b) S + H₂ → H₂S (Reduction of sulphur)

(iii) Removal of an electronegative element or removal of Non-metal

(a) 2HgCl₂ + SnCl₂ → HgCl₂ + SnCl₄

(Reduction of mercuric chloride)

(b) 2FeCl₃ + H₂ → 2FeCl₂ + 2HCl

(Reduction of ferric chloride)

(iv) Addition of an electropositive element or addition of metal

(a) HgCl₂ + Hg → Hg₂Cl₂

(Reduction of mercuric chloride)

(b) CuCl₂ + Cu → Cu₂Cl₂

(Reduction of cupric chloride)

(v) Decrease in +ve valency and Increase in –ve valency

(a) Fe³⁺ → Fe²⁺ (+ve valency decreases)

(b) Sn⁴⁺ → Sn²⁺ (+ve velency decreases)

(d) $\[MnO_{4}^{-}\to MnO_{4}^{2-}$ (–ve valency increases)

(vi) Gain of electrons (also known as electronation)

(a) Zn²⁺(aq) + 2e⁻ → Zn(s) (Electronation of Zn²⁺)

(b) Pb²⁺ + 2e⁻ → Pb^o (Electronation of Pb²⁺)

(d) Fe³⁺ + e⁻ → Fe²⁺ (Electronation of Fe³⁺)

(e) Sn⁴⁺ + 2e⁻ → Sn²⁺ (Electronation of Sn⁴⁺)

(f) Cl + e⁻ → Cl⁻ (Formation of chloride ion)

(g) [Fe(CN)₆]³⁻ + e⁻ → [Fe(CN)₆]⁴⁻ (Eectronation of [Fe(CN)₆]³⁻)

(vii) Decrease in oxidation number

(a) Mg²⁺ → Mg⁰ (From +2 to 0)

(b) [Fe(CN)₆]³⁻ → [Fe(CN)₆]⁴⁻ (From +3 to +2)

(3) Redox-reactions

(i) An overall reaction in which oxidation and reduction takes place simultaneously is called redox or oxidation-reduction reaction. These reactions involve transfer of electrons from one atom to another. Thus every redox reaction is made up of two half reactions; One half reaction represents the oxidation and the other half reaction represents the reduction.

(ii) The redox reactions are of following types

(a) Direct redox reaction : The reactions in which oxidation and reduction takes place in the same vessel are called direct redox reactions.

(b) Indirect redox reaction : The reactions in which oxidation and reduction takes place in different vessels are called indirect redox reactions. Indirect redox reactions are the basis of electro-chemical cells.

(c) Intermolecular redox reactions : In which one substance is oxidised while the other is reduced. For example,

2Al + Fe₂O₃ → Al₂O₃ + 2Fe

Here, Al is oxidised to Al₂O₃ while Fe₂O₃ is reduced to Fe.

(d) Intramolecular redox reactions : In which one element of a compound is oxidised while the other is reduced. For example,

$\[2KCl{{O}_{3}}\xrightarrow{\Delta }2KCl+3{{O}_{2}}$

Here, Cl⁺⁵ in KClO₃ is reduced to Cl⁻¹ in KCl while O²⁻ in KClO₃ is oxidised to .

(iii) To see whether the given chemical reaction is a redox reaction or not, the molecular reaction is written in the form of ionic reaction and now it is observed whether there is any change in the valency of atoms or ions. If there is a change in valency, the chemical reaction will be a redox reaction otherwise not. For example,

(a) BaO₂ + H₂SO₄ → BaSO₄ + H₂O₂

(b) CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄

In above examples there is no change in the valency of any ion or atom, thus these are not redox reactions.

(iv) Some examples of redox reactions are,

(a)

Here mercuric ion is reduced to mercurous ion and stannous ion is oxidised to stannic ion, i.e., mercuric ion acts as an oxidising agent while stannous ion acts as a reducing agent.

(b)

Here ferric ion is reduced to ferrous ion by gain of one electron while stannous ion is oxidised to stannic ion by loss of two electrons. The ferric ion acts as an oxidising agent while stannous ion acts as a reducing agent.

(c)

Here thiosulphate ion is oxidised to tetrathionate ion by loss of electrons while iodine is reduced to iodide ion by gain of electrons. Thiosulphate ion acts as a reducing agent and iodine acts as an oxidising agent.

(d)

(e) (Where X = F, Cl, Br, I)

(f)

(g)

(h)

Oxidizing and Reducing agents (Oxidants and Reductants)

(1) Definition : The substance (atom, ion or molecule) that gains electrons and is thereby reduced to a low valency state is called an oxidising agent, while the substance that loses electrons and is thereby oxidised to a higher valency state is called a reducing agent.

An oxidising agent is a substance the oxidation number of whose atom or atoms decreases while a reducing agent is a substance the oxidation number of whose atom increases.

(2) Important oxidising agents

(i) Molecules made up of electronegative elements, e.g. O₂, O₃ and X₂ (halogens).

(ii) Compounds containing an element which is in the highest oxidation state e.g.

KMnO₄, K₂Cr₂O₇, Na₂cr₂O₇, CrO₃, H₂SO₄

HNO₃, NaNO₃, FeCl₃, HgCl₂, KClO₄, SO₃, CO₂, H₂O₂ etc.

(iii) Oxides of elements e.g. MgO, CuO, CrO₃, CO₂, P₄O₁₀ etc.

(iv) Fluorine is the strongest oxidising agent.

(3) Important reducing agents

(i) All metals e.g. Na, Zn, Fe, Al, etc.

(ii) A few non-metals e.g. C, H₂, S etc.

(iii) Hydracids : HCl, HBr, HI, H₂S etc.

(iv) A few compounds containing an-element in the lower oxidation state (ous), e.g. FeCl₂, FeSO₄, SnCl₂, Hg₂Cl₂, Cu₂O etc.

(v) Metallic hydrides e.g. NaH, LiH etc.

(vi) Organic compounds like HCOOH and (COOH)₂ and their salts, aldehydes, alkanes etc.

(vii) Lithium is the strongest reducing agent in solution.

(viii) Cesium is the strongest reducing agent in absence of water. Other reducing agents are Na₂S₂O₃ and KI.

(ix) Hypo prefix indicates that central atom of compound has the minimum oxidation state so it will act as a reducing agent. e.g., H₃PO₂ (hypophosphorous acid).

(4) Substances which act as oxidising as well as reducing agents

Example :

H₂O₂, SO₂, H₂SO₃, HNO₃, HNO₂, NaNO₂, Na₂SO₃, O₃ etc.

(5) Tips for the identification of oxidising and reducing agents

(i) If an element is in its highest possible oxidation state in a compound, the compound can function as an oxidising agent, e.g. KMnO₄, K₂Cr₂O₇, HNO₃, H₂SO₄, HClO₄ etc.

(ii) If an element is in its lowest possible oxidation state in a compound, the compound can function only as a reducing agent, e.g.

H₂S, H₂C₂O₄, FeSO₄, Na₂S₂O₃, SnCl₂ etc.

(iii) If an element is in its intermediate oxidation state in a compound, the compound can function both as an oxidising agent as well as reducing agent, e.g.

H₂O, H₂SO₃, HNO₂, SO₂ etc.

(iv) If a highly electronegative element is in its highest oxidation state in a compound, that compound can function as a powerful oxidising agent, e.g.

KClO₄, KClO₃, KBrO₃, KIO₃etc.

(v) If an electronegative element is in its lowest possible oxidation state in a compound or in free state, it can function as a powerful reducing agent, e.g.

I⁻, Br⁻, N³⁻etc.

(6) Tests for oxidising agents

(i) Aqueous solutions of oxidising agents react with,

(a) Hydrogen sulphide to give a milky yellow precipitate of sulphur.

H₂S + Oxidising agent → S (milky yellow ppt.)

(b) Potassium iodide solution and evolve iodine which gives intense blue colour with starch solution

KI + Oxidising agent → I₂ $\[{{I}_{2}}\xrightarrow{Starch\,\,solution}$ Intense blue colour

(c) Freshly prepared solution of ferrous ammonium sulphate in presence of il. H₂SO₄. Ferric ions (Fe³⁺) can be detected by adding ammonium thiocyanate solution when a deep red colouration is produced.

Oxidising agent + $\[F{{e}^{2+}}\to F{{e}^{3+}}\xrightarrow{CN{{S}^{-}}}Fe{{(CNS)}_{3}}$ (deep red clouration)

(ii) Insoluble oxidising agents on,

(a) Strong heating evolve oxygen which relights a glowing splinter.

(b) Warming with concentrated hydrochloric acid evolve chlorine which bleaches the moist litmus paper.

(7) Tests for reducing agents

(i) Aqueous solutions of reducing agents react with,

(a) Acidified potassium permanganate solution and decolourise it.

(b) Few drops of acidified potassium dichromate solution, green colouration is produced.

(c) Few drops of ferric chloride solution. The ferrous ions thus formed give a deep blue colouration with potassium ferricyanide (K₃[Fe(CN)₆]).

(d) Insoluble reducing agents on, Heating with concentrated nitric acid, evolve brown fumes of nitrogen dioxide.

(e) Heating with powdered cupric salt, form a red deposit of copper which does not dissolve in warm dilute sulphuric acid.

(8) Equivalent weight of oxidising and reducing agents

(i) Equivalent weight of a substance (oxidant or reductant) is equal to molecular weight divided by number of electrons lost or gained by one molecule of the substance in a redox reaction.

Equivalent weight of oxidizing agent = $\[\frac{\text{Molecular}\,\,\text{weight }}{\text{No}\text{.}\,\,\text{of}\,\,e\text{lectrons}\,\,\text{gained}\,\,\text{by}\,\,\text{one}\,\,\text{molecule}}$

Equivalent weight of reducing agent = $\[\frac{\text{Molecular}\,\,\text{weight}}{\text{No}\text{.}\,\,\text{of}\,\,\text{electrons}\,\,\text{lost}\,\,\text{by}\,\,\text{one}\,\,\text{molecule}}$

(ii) In other words, it is equal to the molecular weight of oxidant or reductant divided by the change in oxidation number.

Equivalent weight of oxidising agent = $\[\frac{\text{Molecular}\,\,\text{weight}}{\text{Change}\,\,\text{in}\,\,\text{O}\text{.N}\text{.}\,\,\text{per}\,\,\text{mole}}$

Equivalent weight of reducing agent = $\[\frac{\text{Molecular}\,\,\text{weight}}{\text{Change}\,\,\text{in}\,\,\text{O}\text{.N}\text{.}\,\,\text{per}\,\,\text{mole}}$

Equivalent weight of few oxidising/reducing agents

Agents	O. N.	Product	O. N.	Change in O. N. per atom	Total Change in O. N. per mole	Eq. wt.
$\[C{{r}_{2}}O_{7}^{2-}$	+ 6	Cr³⁺	+ 3	3	3 × 2 = 6	Mol. wt./6
$\[{{C}_{2}}O_{4}^{2-}$	+ 3	CO₂	+ 4	1	1 × 2 = 2	Mol. wt./2
$\[{{S}_{2}}O_{3}^{2-}$	+ 2	$\[{{S}_{4}}O_{6}^{2-}$	+ 2.5	0.5	0.5 × 2 = 1	Mol. wt./1
H₂O₂	– 1	H₂O	– 2	1	1 × 2 = 2	Mol. wt./2
H₂O₂	– 1	O₂	0	1	1 × 2 = 2	Mol. wt./2
$\[MnO_{4}^{-}$ (Acidic medium)	+ 7	Mn²⁺	+ 2	5	5 × 1 = 5	Mol. wt./5
$\[MnO_{4}^{-}$ (Neutral medium)	+ 7	MnO₂	+ 4	3	3 × 1 = 3	Mol. wt./3
(Alkaline medium)	+ 7	$\[MnO_{4}^{2-}$	+ 6	1	1 × 1 = 1	Mol. wt./1

Oxidation number or Oxidation state

(1) Definition : Charge on an atom produced by donating or accepting electrons is called oxidation number or oxidation state. It is the number of effective charges on an atom.

(2) Valency and oxidation number : Valency and oxidation number concepts are different. In some cases (mainly in the case of electrovalent compounds), valency and oxidation number are the same but in other cases they may have different values. Points of difference between the two have been tabulated below

Valency	Oxidation number
It is the combining capacity of the element. No plus or minus sign is attached to it.	O.N. is the charge (real or imaginary) present on the atom of the element when it is in combination. It may have plus or minus sign.
Valency of an element is usually fixed.	O.N. of an element may have different values. It depends on the nature of compound in which it is present.
Valency is always a whole number.	O.N. of the element may be a whole number or fractional.
Valency of the element is never zero except of noble gases.	O.N. of the element may be zero.

(3) Oxidation number and Nomenclature

(i) When an element forms two monoatomic cations (representing different oxidation states), the two ions are distinguished by using the ending-ous and ic. The suffix – ous is used for the cation with lower oxidation state and the suffix – ic is used for the cation with higher oxidation state.

For example : Cu⁺ (oxidation number +1) cuprous : Cu²⁺ (oxidation number +2) cupric

(ii) Albert Stock proposed a new system known as Stock system. In this system, the oxidation states are indicated by Roman numeral written in parentheses immediately after the name of the element. For example,

Cu₂O	Copper (I) oxide	SnO	Tin (II) oxide
FeCl₂	Iron (II) chloride	Mn₂O₇	Manganess (VII) oxide
K₂Cr₂O₇	Potassium dichromate (VI)	Na₂CrO₄	Sodium chromate (VI)
V₂O₅	Vanadium(V) oxide	CuO	Copper (II) oxide
SnO₂	Tin (IV) oxide	FeCl₃	Iron (III) chloride

Note : Stock system is not used for non-metals.

(4) Rules for the determination of oxidation number of an atom

The following rules are followed in ascertaining the oxidation number of an atom,

(i) If there is a covalent bond between two same atoms then oxidation numbers of these two atoms will be zero. Bonded electrons are symmetrically distributed between two atoms. Bonded atoms do not acquire any charge. So oxidation numbers of these two atoms are zero.

A : A or A – A A^*+ A^*

For e.g. Oxidation number of Cl in Cl₂, O in O₂ and N and N₂ is zero.

(ii) If covalent bond is between two different atoms then electrons are counted towards more electronegative atom. Thus oxidation number of more electronegative atom is negative and oxidation number of less electronegative atom is positive. Total number of charges on any element depends on number of bonds.

A – B A⁺+ B : ^–

A – B A⁺²+ : B : ^–2

The oxidation number of less electronegative element (A) is + 1 and + 2 respectively.

(iii) If there is a coordinate bond between two atoms then oxidation number of donor atom will be + 2 and of acceptor atom will be – 2.

A → B → A²⁺+ B⁻²

(iv) The oxidation number of all the atoms of different elements in their respective elementary states is taken to be zero. For example, in N₂, Cl₂, H₂, P₄,S₈, O₂, Br₂, Na, Fe, Ag etc. the oxidation number of each atom is zero.

(v) The oxidation number of a monoatomic ion is the same as the charge on it. For example, oxidation numbers of Na⁺, Mg²⁺ and Al³⁺ ions are + 1, + 2 and + 3 respectively while those of Cl⁻, S²⁻ and N³⁻ ions are –1, –2 and –3 respectively.

(vi) The oxidation number of hydrogen is + 1 when combined with non-metals and is –1 when combined with active metals called metal hydrides such as LiH, KH, MgH₂, CaH₂ etc.

(vii) The oxidation number of oxygen is – 2 in most of its compounds, except in peroxides like H₂O₂, BaO₂ etc. where it is –1. Another interesting exception is found in the compound OF₂ (oxygen difluoride) where the oxidation number of oxygen is + 2. This is due to the fact that fluorine being the most electronegative element known has always an oxidation number of –1.

(viii)In compounds formed by union of metals with non-metals, the metal atoms will have positive oxidation numbers and the non-metals will have negative oxidation numbers. For example,

(a) The oxidation number of alkali metals (Li, Na, K etc.) is always +1 and those of alkaline earth metals (Be, Mg, Ca etc) is + 2.

(b) The oxidation number of halogens (F, Cl, Br, I) is always –1 in metal halides such as KF, AlCl₃, MgBr₂, CdI₂. etc.

(ix) In compounds formed by the union of different elements, the more electronegative atom will have negative oxidation number whereas the less electronegative atom will have positive oxidation number. For example,

(a) N is given an oxidation number of –3 when it is bonded to less electronegative atom as in NH₃ and NI₃, but is given an oxidation number of + 3 when it is bonded to more electronegative atoms as in NCl₃.

(b) Since fluorine is the most electronegative element known so its oxidation number is always –1 in its compounds i.e. oxides, interhalogen compounds etc.

(c) In interhalogen compounds of Cl, Br, and I; the more electronegative of the two halogens gets the oxidation number of –1. For example, in BrCl₃, the oxidation number of Cl is –1 while that of Br is +3.

(x) For neutral molecule, the sum of the oxidation numbers of all the atoms is equal to zero. For example, in NH₃ the sum of the oxidation numbers of nitrogen atom and 3 hydrogen atoms is equal to zero. For a complex ion, the sum of the oxidation numbers of all the atoms is equal to charge on the ion. For example, in $\[SO_{4}^{2-}$ ion, the sum of the oxidation numbers of sulphur atom and 4 oxygen atoms must be equal to –2.

(xi) It may be noted that oxidation number is also frequently called as oxidation state. For example, in H₂O, the oxidation state of hydrogen is +1 and the oxidation state of oxygen is – 2. This means that oxidation number gives the oxidation state of an element in a compound.

(xii) In the case of representative elements, the highest oxidation number of an element is the same as its group number while highest negative oxidation number is equal to (8 – Group number) with negative sign with a few exceptions. The most common oxidation states of the representative elements are shown in the following table,

Group	Outer shell configuration	Common oxidation numbers (states) except zero in free state
I A	ns¹	+1
II A	ns²	+2
III A	ns²np¹	+3, +1
IV A	ns²np²	+4, +3, +2, +1, –1, –2, –3, –4
V A	ns²np³	+5, +3, +1, –1, –3
VI A	ns²np⁴	+6, +4, +2, –2
VII A	ns²np⁵	+7, +5, +3, +1, –1

(xiii) Transition metals exhibit a large number of oxidation states due to involvement of (n –1) d electron besides ns electron.

(xiv) Oxidation number of a metal in carbonyl complex is always zero, e.g. Ni has zero oxidation state in [Ni(CO₄)].

(xv) Those compounds which have only C, H and O the oxidation number of carbon can be calculated by following formula,

$\[\text{Oxidation number of }\!\!'\!\!\text{ }C\text{ }\!\!'\!\!\text{ }=\frac{\text{(}{{n}_{O}}\times \text{2-}{{n}_{H}}\text{)}}{{{n}_{C}}}$

Where, n₀ is the number of oxygen atom, n_H is the number of hydrogen atom, n_c is the number of carbon atom.

For example,

(a) $\[{{\overset{*}{\mathop{CH}}\,}_{3}}OH\,\,;\,\,{{n}_{H}}=4,\,\,{{n}_{c}}=1,\,\,{{n}_{o}}=1$

Oxidation number of ‘C’ = $\[\frac{(1\times 2-4)}{1}=-2$

(b) $\[\overset{*}{\mathop{HCOOH}}\,\,\,;\,\,{{n}_{H}}=2,\,\,\,{{n}_{0}}=2,\,\,\,{{n}_{c}}=1$

Oxidation number of carbon = $\[\frac{(2\times 2-2)}{1}=+2$

(5) Procedure for calculation of oxidation number: By applying the above rules, we can calculate the oxidation numbers of elements in the molecules/ions by the following steps.

(i) Write down the formula of the given molecule/ion leaving some space between the atoms.

(ii) Write oxidation number on the top of each atom. In case of the atom whose oxidation number has to be calculated write x.

(iii) Beneath the formula, write down the total oxidation numbers of each element. For this purpose, multiply the oxidation numbers of each atom with the number of atoms of that kind in the molecule/ion. Write the product in a bracket.

(iv) Equate the sum of the oxidation numbers to zero for neutral molecule and equal to charge on the ion.

(v) Solve for the value of x.

Oxidation number of some elements in compounds, ions or chemical species

Element	Oxidation Number	Compounds, ions or chemical species
Sulphur (s)	– 2	H₂S, ZnS, NaHS, (SnS₃)^2–, BaS, CS₂
	0	S, S₄, S₈, SCN^–
	+ 1	S₂, F₂, S₂Cl₂
	+ 4	SO₂, H₂SO₃, (SO₃)^2–, SOCl₂, NaHSO₃, Ca[HSO₃]₂, [HSO₃]^–_, SF₄
	+ 6	H₂SO₄, (SO₄)^2-, [HSO₄]^–_, BaSO_4,KHSO_4,SO₃, SF₆, H₂S₂O₇, (S₂O₇)^2–
Nitrogen (N)	– 3	NH₃, (NH₄)⁺, AlN, Mg₃N₂, (N)^3–, Ca₃N₂, CN^–
	– 2	N₂H₄, (N₂H₅)⁺
	– 1	NH₂OH
	–1/3	NaN₃, N₃H
	0	N₂
	+ 1	N₂O
	+ 2	NO
	+ 3	HNO₂, (NO₂)^–, NaNO₂, N₂O₃, NF₃
	+ 4	NO₂
	+ 5	HNO₃, (NO₃)^–, KNO₃, N₂O₅
Chlorine	– 1	HCl, NaCl, CaCl₂, AlCl₃, ICl, ICl₅, SOCl₂, CrO₂Cl₂, KCl, K₂PtCl₆, HAuCl₄, CCl₄
(Cl)	0	Cl, Cl₂
	+ 1	HOCl, NaOCl, (OCl)^–, Cl₂O
	+ 3	KClO₂, (ClO₂)^–, HClO₂
	+ 4	ClO₂
	+ 5	(ClO₃)^–, KClO₃, NaClO₃, HClO₃
	+ 7	HClO₄, Cl₂O₇, KClO₄, (ClO₄)^–
Hydrogen	– 1	NaH, CaH₂, LiAlH₄, LiH
(H)	+ 1	NH₃, PH₃, HF
Phosphorus	– 3	PH₃, (PH₄)⁺, Ca₃P₂
(P)	0	P₄
	+ 1	H₃PO₂, KH₂PO₂, BaH₄P₂O₄
	+ 3	PI₃, PBr₃, PCl₃, P₂O₃, H₃PO₃
	+ 5	(PO₄)^3–, H₃PO₄, Ca₃(PO₄)₂, H₄P₂O₇, P₄O₁₀, PCl₅, (P₂O₇)^4–, Mg₂P₂O₇, ATP
Oxygen	– 2	H₂O, PbO₂, (CO₃)^2–, (PO₄)^2–, SO₂, (C₂O₄)^2–, HOCl, (OH)^–, (O)^2-
(O)	– 1	Na₂O₂, BaO₂, H₂O₂, (O₂)^2–, Peroxides
	– 1/2	KO₂
	0	O, O₂, O₃
	+ 1	O₂F₂
	+ 2	OF₂
Carbon	– 4	CH₄
(C)	– 3	C₂HHHhhhhjkjkkkH₆
	– 2	CH₃Cl, C₂H₄
	– 1	CaC₂, C₂H₂
	0	Diamond, Graphite, C₆H₁₂O₆, C₂H₄O₂, HCHO, CH₂Cl₂
	+ 2	CO, CHCl₃, HCN
	+ 3	H₂C₂O₄, (C₂O₄)^2–
	+ 4	CO₂, H₂CO₃, (HCO₃)^–, CCl₄, Na₂CO₃, Ca₂CO₃, CS₂, CF₄, (CO₃)^–2
Chromium	+ 3	Cr₂(SO₄)₃, CrCl₃, Cr₂O₃, [Cr(H₂O)₄Cl₃]
(Cr)	+ 6	K₂CrO₄, (CrO₄)^2–, K₂Cr₂O₇, (Cr₂O₇)^2–, KCrO₃Cl, CrO₂Cl₂, Na₂Cr₃O₁₀, CrO₃
Manganese	+ 2	MnO, MnSO₄, MnCl₂, Mn(OH)₂
(Mn)	+ 8/3	Mn₃O₄
	+ 3	Mn(OH)₃
	+ 4	MnO₂, K₂MnO₃
	+ 6	K₂MnO₄, (MnO₄)^2–
	+ 7	KMnO₄, (MnO₄)^–, HMnO₄
Silicon	– 4	SiH₄, Mg₂Si
(Si)	+ 4	SiO₂, K₂SiO₃, SiCl₄
Iron (Fe)	$\[+\frac{8}{3}$	Fe₃O₄
	+ 2	FeSO₄ (Ferrous ammonium sulphate), K₄Fe(CN)₆, FeCl₂
	+ 3	K₃[Fe(CN)₆], FeCl₃
Iodine (I)	+ 7	H₄IO₆⁻, KIO₄
Osmium (Os)	+ 8	OsO₄
Xenon (Xe)	+ 6	XeO₃, XeF₆

(6) Exceptional cases of evaluation of oxidation numbers : The rules described earlier are usually helpful in determination of the oxidation number of a specific atom in simple molecules but these rules fail in the following cases. In these cases, the oxidation numbers are evaluated using the concepts of chemical bonding involved.

Type I. In molecules containing peroxide linkage in addition to element-oxygen bonds. For example,

(i) Oxidation number of S in H₂SO₅ (Permonosulphuric acid or Caro’s acid). By usual method; H₂SO₅

2 × 1 + x + 5 × (−2) = 0 or x = + 8

But this cannot be true as maximum oxidation number for S cannot exceed + 6. Since S has only 6 electrons in its valence shell. This exceptional value is due to the fact that two oxygen atoms in H₂SO₅ shows peroxide linkage as shown below,

Therefore the evaluation of o. n. of sulphur here should be made as follows,

2 × (+1) + x + 3 × (–2) + 2 × (–1)

(for H) (for S) (for O) (for O–O)

or 2 + x – 6 – 2 = 0 or x = + 6.

(ii) Oxidation number of S in H₂S₂O₈ (Peroxidisulphuric acid or Marshall’s acid) By usual method ; H₂S₂O₈

1 × 2 + 2x + 8 (–2) = 0

2x = + 16 – 2 = 14 or x = + 7

Similarly Caro’s acid, Marshall’s acid also has a peroxide linkage so that in which S shows +6 oxidation state.

Therefore the evaluation of oxidation state of sulphur should be made as follow,

2 × (+1) + 2 × (x) + 6 × (–2) + 2 × (–1) = 0

(for H) (for S) (for O) (for O–O)

Or 2 + 2x – 12 – 2 = 0 or x = + 6.

(iii) Oxidation number of Cr in CrO₅ (Blue perchromate)

By usual method CrO₅ ; x – 10 = 0 or x = + 10

This cannot be true as maximum o.n. of Cr cannot be more than + 6. Since Cr has only five electrons in 3d orbitals and one electron in 4s orbital. This exceptional value is due to the fact that four oxygen atoms in CrO₅ are in peroxide linkage. The chemical structure of CrO₅is

Therefore, the evaluation of o. n. of Cr should be made as follows

x + 1 × (– 2) + 4 (–1) = 0

(for Cr) (for O) (for O–O)

Or x – 2 – 4 = 0 or x = + 6.

Type II. In molecules containing covalent and coordinate bonds, following rules are used for evaluating the oxidation numbers of atoms.

(i) For each covalent bond between dissimilar atoms the less electronegative element is assigned the oxidation number of + 1 while the atom of the more electronegative element is assigned the oxidation number of –1.

(ii) In case of a coordinate-covalent bond between similar or dissimilar atoms but the donor atom is less electronegative than the acceptor atom, an oxidation number of + 2 is assigned to the donor atom and an oxidation number of – 2 is assigned to the acceptor atom.

Conversely, if the donor atom is more electronegative than the acceptor atom, the contribution of the coordinate bond is neglected.

Example :

(a) Oxidation number of C in HCN and HN = C

The evaluation of oxidation number of C cannot be made directly by usual rules since no standard rule exists for oxidation numbers of N and C.

In such cases, evaluation of oxidation number should be made using indirect concept or by the original concepts of chemical bonding.

(b) Oxidation number of carbon in H – N Ξ C

The contribution of coordinate bond is neglected since the bond is directed from a more electronegative N atom (donor) to a less electronegative carbon atom (acceptor).

Therefore the oxidation number of N in remains – 3 as it has three covalent bonds.

1 × (+ 1) + 1 × (– 3) + x = 0

(for H) (for N) (for C)

or 1 + x – 3 = 0 or x = + 2.

(c) Oxidation number of carbon in HC ≡ N

In HC ≡ N, N is more electronegative than carbon, each bond gives an oxidation number of –1 to N. There are three covalent bonds, the oxidation number of N is HC ≡ Nis taken as – 3

Now HC ≡ N +1 + x – 3 = 0 x = + 2

Type III. In a molecule containing two or more atoms of same or different elements in different oxidation states.

(i) Oxidation number of S in Na₂S₂O₃

By usual method Na₂S₂O₃

2 × (+1) + 2 × x + 3 (–2) = 0 or 2 + 2x – 6 = 0 or x = 2.

But this is unacceptable as the two sulphur atoms in Na₂S₂O₃ cannot have the same oxidation number because on treatment with dil. H₂SO₄, one sulphur atom is precipitated while the other is oxidised to SO₂.

Na₂S₂O₃ + H₂SO₄ → Na₂SO₄ + SO₂ + S + H₂O

In this case, the oxidation number of sulphur is evaluated from concepts of chemical bonding. The chemical structure of Na₂S₂O₃ is

Due to the presence of a co-ordinate bond between two sulphur atoms, the acceptor sulphur atom has oxidation number of – 2 whereas the other S atom gets oxidation number of + 2.

2 × (+1) + 3 × (–2) + x × 1 + 1 × (– 2) = 0

(for Na) (for O) (for S) (for coordinated S)

or + 2 – 6 + x – 2 = 0 or x = + 6

Thus two sulphur atoms in Na₂S₂O₃ have oxidation number of – 2 and +6.

(ii) Oxidation number of chlorine in CaOCl₂ (bleaching powder)

In bleaching powder, Ca(OCl)Cl, the two Cl atoms are in different oxidation states i.e., one Cl^– having oxidation number of –1 and the other as OCl^– having oxidation number of +1.

(iii) Oxidation number of N in NH₄NO₃

By usual method N₂H₄O₃ ; 2x + 4 × (+1) + 3 × (–1) = 0

2x + 4 – 3 = 0 or 2x = + 1 (wrong)

No doubt NH₄NO₃ has two nitrogen atoms but one N has negative oxidation number (attached to H) and the other has positive oxidation number (attached to O). Hence the evaluation should be made separately for $\[NH_{4}^{+}$ and $\[NO_{3}^{-}$

x + 4 × (+1) = +1 or x = – 3

x + 3 (– 2) = –1 or x = + 5.

(iv) Oxidation number of Fe in Fe₃O₄

In Fe₃O₄, Fe atoms are in two different oxidation states. Fe₃O₄ can be considered as an equimolar mixture of FeO (iron (II) oxide) and Fe₂O₃ (iron (III) oxide). Thus in one molecule of Fe₃O₄, two Fe atoms are in + 3 oxidation state and one Fe atom is in + 2 oxidation state.

(v) Oxidation number of S in sodium tetrathionate (Na₂S₄O₆). Its structure can be represented as follows :

The two S-atoms which are linked to each other have oxidation number of zero. The oxidation number of other S-atoms can be calculated as follows

Let oxidation number of S = x.

2 × x + 2 × 0 + 6 × ( – 2) = – 2

(for S) (for S–S) (for O) x = + 5.

Redox Reaction : Reduction and Oxidation

Redox Reaction : Reduction and Oxidation

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