NEET Chemistry Notes Chapter 13 Redox Reactions PDF

60 Chapter E3 13 Redox Reactions (ii) Removal of hydrogen : H S+Cl 2HCl + S Molecular and Ionic equations (iv) Removal of metal : 2KI+H O 2KOH+I ID Chemical reactions involve transfer of electrons from one chemical substance to another. These electron – transfer reactions are termed as oxidation-red...

60 Chapter E3 13 Redox Reactions (ii) Removal of hydrogen : H S+Cl 2HCl + S Molecular and Ionic equations (iv) Removal of metal : 2KI+H O 2KOH+I ID Chemical reactions involve transfer of electrons from one chemical substance to another. These electron – transfer reactions are termed as oxidation-reduction or redox-reactions. 2 1 0 1 2 3 4 M M M M M M D YG U ST (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 as spectator ions and are not included in the final balanced equations. Example, Zn  2 H  2Cl Zn  2 H  Zn  H 2  2Cl Zn 2  H 2 In above example, the Cl  ions are the spectator ions and hence are not included in the final ionic balanced equation. 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. 2 2MgO (vi) Loss of electrons (also known as de-electronation) –e –e –e –e –e –e –e –e – – – – – – – – Loss of electrons (a) H 0 H   e  (Formation of proton) (b) MnO42  MnO4  e  (De-electronation of MnO42 ) (c) 2 Fe0 2 Fe 3   6 e  (De-electronation of iron) (vii) Increase in oxidation number (a) Mg 0 Mg 2  (b) Fe 2 (CN )6  4 (From 0 to +2)  Fe 3 (CN )6  3 (From +2 to +3) (Ionic equation) (Final ionic equation) (i) Addition of oxygen : 2Mg + O (v) Increase in +ve valency : Fe 2  Fe3   e  M In above example the reactants and products have been written in ionic forms, thus the equation is termed as ionic equation.  2 3 Mn2   2Cl   2 H 2 O  Cl 2 2 2 M (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. Example,  2 FeS 4 In above example the reactants and products have been written in molecular forms, thus the equation is termed as molecular equation.  (iii) Addition of Non-metal : Fe + S M Example : MnO2  4 HCl MnCl2  2 H 2 O  Cl 2 MnO2  4 H   4 Cl  2 U (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. 2 (c) 2Cl  Cl 20 (From –1 to 0) (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 : CuO  C Cu  CO (ii) Addition of hydrogen : Cl 2  H 2 2 HCl (iii) Removal of non-metal 2 HgCl2  SnCl 2 Hg 2 Cl 2  SnCl 4 (iv) Addition of metal : HgCl2  Hg Hg2 Cl 2 (v) Decrease in +ve valency (+ve valency decreases) 3 4 (b) [Fe(CN )6 ] [Fe(CN )6 ] (–ve valency increases) (vi) Gain of electrons (also known as electronation) 4 3 2 1 0 1 2 3 4 M M M M M M M M M +e +e +e +e +e +e +e +e – – – – – – – – Gain of electrons 2  (a) Zn (aq)  2e Zn(S ) (b) Pb 2   2e Pb 0 (iii) Oxides of elements, MgO, CuO, CrO3 , CO 2 , P4 O10 , 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, H , S etc. (iii) Hydracids : HCl, HBr, HI, H S etc. (iv) A few compounds containing anelement in the lower oxidation state (ous). 2 (Electronation of Zn 2 (Electronation of Pb 2 ) ) (c) [Fe(CN )6 ]3   e  [Fe(CN )6 ]4  (Electronation of [Fe(CN )6 ]3  ) (vii) Decrease in oxidation number (a) Mg 2 Mg 0 (From +2 to 0) (b) Fe(CN )6 3  Fe(CN )6 4  (From +3 to +2)  (c) (From 0 to –1) 2Cl (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) Types of redox reaction (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, 2 Al  Fe2O3 Al2O3  2 Fe 2 Example : FeCl2 , FeSO4 , SnCl 2 , Hg2 Cl 2 , Cu 2 O etc. (v) Metallic hydrides e.g. NaH, LiH etc. (vi) Organic compounds like HCOOH and (COOH) 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 Na 2 S 2 O3 and KI. 2 (ix) Hypo prefix indicates that central atom of compound has the minimum oxidation state so it will act as a reducing agent. ID Cl 20 HNO3 , NaNO 3 , FeCl3 , HgCl2 , KClO4 , SO 3 , CO 2 , H 2 O2 etc. 60 Fe Example : KMnO4 , K 2 Cr2 O7 , Na 2 Cr2 O7 , CrO3, H 2 SO 4 , 2 E3 (a) Fe 3 Example : H 3 PO2 (hypophosphorous acid). (4) Substances which act as oxidising as well as reducing agents Examples : H O ,SO ,H SO ,HNO ,NaNO ,Na SO ,O etc. 2 2 2 2 3 2 2 2 3 3 D YG U (5) Tips for the identification of oxidising and reducing agents Here, Al is oxidised to Al2 O3 while Fe2 O3 is reduced to Fe. (d) Intramolecular redox reactions : In which one element of a compound is oxidised while the other is reduced.  For example, 2 KClO3  2 KCl  3 O2 U Here, Cl 5 in KClO3 is reduced to Cl 1 in KCl while O 2  in KClO3 is oxidised to O 20. Oxidising and Reducing agents ST (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. Or 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. Example : O , O and X (halogens). (ii) Compounds containing an element which is in the highest oxidation state. 2 3 2 (i) If an element is in its highest possible oxidation state in a compound, the compound can function as an oxidising agent. Example : KMnO4 , K2Cr2O7 , HNO3 , H 2 SO 4 , 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. Example : H 2 S , H 2 C2 O4 , FeSO4 , Na 2 S 2O3 , SnCl 2 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. Example : H 2 O2 , H 2 SO 3 , HNO 2 , SO 2 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. Example : KClO4 , KClO3 , KBrO3 , KIO3 etc. (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. Example : I  , Br  , N 3  etc. (6) Equivalent weight of oxidising and reducing agents 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. Eq. wt. of O. A. = Molecular weight No. of electrons gained by one molecule = Molecular weight Change in O. N. per mole Eq. wt. of R. A. = Molecular weight No. of electrons lostby one molecule Molecular weight Change in O. N. per mole = Table : 13.1 Equivalent weight of few oxidising/reducing agents O. N. Product O. N. Change in O. N. per atom Total Change in O. N. per mole Eq. wt. Cr2 O7 2  +6 Cr 3  +3 3 3×2=6 Mol. wt./6 C 2 O4 2  +3 CO 2 +4 1 1×2=2 Mol. wt./2 S 2 O3 2  +2 S 4 O6 2  + 2.5 0.5 0.5 × 2 = 1 Mol. wt./1 H 2 O2 –1 H 2O –2 1 H 2 O2 –1 O2 0 1 MnO4 +7 Mn2  +2 5 MnO4 (Neutral medium) +7 MnO2 +4 MnO4 (Alkaline medium) +7 MnO42  +6 V2O5 Mol. wt./2 1×2=2 Mol. wt./2 5×1=5 Mol. wt./5 3 3×1=3 Mol. wt./3 1 1×1=1 Mol. wt./1 Vanadium (V) oxide CuO Copper (II) oxide Tin (IV) oxide FeCl3 Iron (III) chloride ID Oxidation number or Oxidation state 1×2=2 E3 (Acidic medium) 60 Agents SnO2 (4) Rules for the determination of oxidation number of an atom : The following rules are followed in ascertaining the oxidation number of an atom, D YG U (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 Oxidation number Valency 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. O.N. of an element may have different values. It depends on the nature of compound in which it is present. O.N. of the element may be a whole number or fractional. O.N. of the element may be zero. It is the combining capacity of the element. No plus or minus sign is attached to it. Valency of an element is usually fixed. Valency is always a whole number. U Valency of the element is never zero except of noble gases. ST oxidation state and the suffix – ic is used for the cation with higher oxidation state. For example : Cu (oxidation number +1) cuprous + Cu (oxidation number +2) cupric 2+ (ii) Albert Stock proposed a new system known as Stock system. In this system, Roman numeral written in parentheses immediately after the name of the element indicates the oxidation states. For example, SnO * * For example, Oxidation number of Cl in Cl , O in O and N in N is 2 2 2 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 : + – +2 (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 Copper (I) oxide A : A or A – A A + A A – B  A + : B : (3) Oxidation number and Nomenclature Cu2O (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. Tin (II) oxide FeCl2 Iron (II) chloride Mn2O7 Manganess (VII) oxide K2Cr2O7 Potassium dichromate (VI) Na2CrO4 Sodium chromate (VI) –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  A + :B : 2+ -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 N 2 , Cl2 , H 2 , P4 , S 8 , 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  , Mg 2  and Al3  ions are + 1, + 2 and + 3 respectively while those of Cl  , S 2  and N 3  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, MgH , CaH etc. 2 IA ns1 +1 II A ns2 +2 III A ns2np1 +3, +1 IV A ns np +4,+3,+2,+1, –1, –2, –3, –4 VA 3 ns np +5,+3,+1, –1, –3 VI A 4 ns np +6,+4,+2,–2 VII A ns2np5 +7,+5,+3, +1, –1 2 (vii) The oxidation number of oxygen is – 2 in most of its compounds, except in peroxides like H 2O2 , BaO2 etc. where it is –1. Another interesting exception is found in the compound OF (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. 2 2 2 (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. Example : Ni has zero oxidation state in NiCO 4 . 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. 2 60 (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. 2 (xv) Those compounds which have only C, H and O the oxidation number of carbon can be calculated by following formula, (b) The oxidation number of halogens (F, Cl, Br, I) is always –1 in metal halides such as KF, AlCl , MgBr , CdI. etc. Oxidation number of ' C '  (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. Where, nO is the number of oxygen atom, n H is the number of 2 (nO  2 - nH ) nC E3 3 2 hydrogen atom, nC is the number of carbon atom. For example, * For example, (a) CH 3 OH ; nH  4, nC  1, nO  1 3 ID (a) N is given an oxidation number of –3 when it is bonded to less electronegative atom as in NH and NI , but is given an oxidation number of + 3 when it is bonded to more electronegative atoms as in NCl. 3 3 Oxidation number of ‘C’ = (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. (1  2  4 )  2 1 * U (b) HCOOH ; nH  2, nO  2, nc  1 (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 BrCl , the oxidation number of Cl is –1 while that of Br is +3. D YG 3 (x) For neutral molecule, the sum of the oxidation numbers of all the atoms is equal to zero. For example, in NH 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 Oxidation number of carbon = (2  2  2)  2 1 (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. 3 (i) Write down the formula of the given molecule/ion leaving some space between the atoms. charge on the ion. For example, in SO 42  ion, the sum of the oxidation numbers of sulphur atom and 4 oxygen atoms must be equal to –2. (ii) Write oxidation number on the top of each atom. In case of the atom whose oxidation number has to be calculated write x. (xi) It may be noted that oxidation number is also frequently called as oxidation state. For example, in H O, 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. (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. (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, (iv) Equate the sum of the oxidation numbers to zero for neutral molecule and equal to charge on the ion. ST U 2 Group Outer shell configuration (v) Solve for the value of x. Common oxidation numbers (states) except zero in free state Table : 13.2 Oxidation number of some elements in compounds, ions or chemical species Element Sulphur (S) Oxidation Number –2 Compounds, ions or chemical species H S, ZnS, NaHS, (SnS ) , BaS, CS 2– 2 3 2 0 S, S , S , SCN +1 S F , S Cl +4 SO , H SO , (SO ) , SOCl , NaHSO , Ca[HSO ] , [HSO ] SF – 4 2 2 8 2 2 2– 2 2 3 3 – 2 3 3 2 3 , 4 Phosphorus (P) Oxygen (O) NH , (NH ) , AlN, Mg N , (N) , Ca N , CN –2 N H , (N H ) –1 NH OH –1/3 NaN , N H 4 2– , 4, + 3 3 4, 3– 4 3 2 3 6 2 2 7 2 7 – 2 + 2 4 2 5 2 3 0 N +1 NO 3 2 2 NO HNO , (NO ) , NaNO , N O , NF +4 NO +5 HNO , (NO ) , KNO , N O –1 HCl, NaCl, CaCl , AlCl , ICl, ICl , SOCl , CrO Cl , KCl, K PtCl , HAuCl , CCl 0 Cl, Cl +1 HOCl, NaOCl, (OCl) , Cl O +3 KClO , (ClO ) , HClO +4 ClO +5 (ClO ) , KClO , NaClO , HClO +7 HClO , Cl O , KClO , (ClO ) –1 NaH, CaH , LiAlH , LiH +1 NH , PH , HF –3 PH , (PH ) , Ca P 60 +2 +3 – 2 2 2 2 3 3 2 – 3 3 2 2 5 3 5 2 2 2 2 – 2 – 2 2 2 2 – 3 3 3 3 – 4 2 7 4 2 3 4 3 + 3 4 4 3 E3 3 2 6 2 0 P +1 H PO , KH PO , BaH P O +3 PI , PBr , PCl , P O , H PO +5 (PO ) , H PO , Ca (PO ) , H P O , P O , PCl , (P O ) , Mg P O , ATP –2 H O, PbO , (CO ) , (PO ) , SO , (C O ) , HOCl, (OH) , (O) –1 Na O , BaO , H O , (O ) , Peroxides 4 3 2 2 3 2 4 3 3 2 3 2 4 3 3 3– 4 3 4 3 4 2 4 2– 2 2 2 7 4 2– 3 4– 10 5 2 7 2 2– 4 2 2 – 7 2– 4 KO 2 2 2 2 2 2 0 O, O , O +1 OF +2 OF –4 CH –3 CH –2 CH Cl, C H –1 CaC , C H 0 Diamond, Graphite, C H O , C H O , HCHO, CH Cl ST 2 2– 2 U – 1/2 Carbon (C) – 4 2 2 3 2 2 2 4 6 3 2 2 2 4 2 6 12 6 2 4 2 2 2 +2 CO, CHCl , HCN +3 H C O , (C O ) +4 CO , H CO , (HCO ) , CCl , Na CO , Ca CO , CS , CF , (CO ) 3 2– 2 2 4 2 4 – 2 2 3 3 –2 4 2 4 ID Hydrogen (H) –3 2- 4 2 U Chlorine (Cl) H SO , (SO ) , [HSO ] BaSO KHSO SO , SF , H S O , (S O ) D YG Nitrogen (N) +6 3 2 3 2 4 3 4 Cr (SO ) , CrCl , Cr O , [Cr(H O) Cl ] +6 K CrO , (CrO ) , K Cr O , (Cr O ) , KCrO Cl, CrO Cl , Na Cr O , CrO +2 MnO, MnSO , MnCl , Mn(OH) 2 2 3 2 4 2 4 3 2– 4 2 4 2 7 2 2 7 3 2 2 2 3 3 4 Mn(OH) +4 MnO , K MnO +6 K MnO , (MnO ) +7 KMnO , (MnO ) , HMnO –4 SiH , Mg Si +4 SiO , K SiO , SiCl 3 2 2 3 2– 2 4 4 – 4 4 4 4 2 2 2 3 4 Fe O 8 3 3 4 +2 FeSO. (NH ) SO (Ferrous ammonium sulphate), K Fe(CN) , FeCl +3 K [Fe(CN) ], FeCl +7 H 4 IO 6 , KIO 4 4 3 2 4 6 Osmium (Os) +8 OsO Xenon(Xe) +6 XeO , XeF 3 4 3 4 6 6 D YG 2 5 2 × (+1) + 2 × (x) (for H) U 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 H 2 SO 5 shows peroxide linkage as shown below, or ST 2 2 8 (for S) (for O–O) 2 + 2x – 12 – 2 = 0 or x = + 6. (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 CrO are in peroxide linkage. 5 The chemical structure of CrO is 5 O Peroxide linkage O Peroxide linkage Cr O By usual method ; H 2 S 2O8 1 × 2 + 2x + 8 (–2) = 0 2x = + 16 – 2 = 14 or (for O) 5 Peroxide linkage 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 H S O (Peroxidisulphuric acid or Marshall's acid) + 6 × (–2) + 2 × (–1) = 0 (iii) Oxidation number of Cr in CrO H Therefore the evaluation of o.n. of sulphur here should be made as H Therefore the evaluation of oxidation state of sulphur should be made as follow, x  8 follows, Peroxide linkage O O O S O O S O O O H By usual method; H 2 SO 5 2  1  x  5  (2)  0 or 2 Similarly Caro's acid, Marshall's acid also has a peroxide linkage so that in which S shows +6 oxidation state. U (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 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 H SO (Permonosulphuric acid or Caro's acid) H 10 Mn O 4 O O S O O O 3 2 +3  Iodine (I) 3 60 Iron (Fe) 3 2– + 8/3 Silicon (Si) 4 E3 Manganese (Mn) +3 ID Chromium (Cr) O O x=+7 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) Type II. In molecules containing covalent and coordinate bonds, following rules are used for evaluating the oxidation numbers of atoms. 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) (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. Examples, (for O) (for S) (for coordinated S) x – 2 = 0 or x = + 6 Thus two sulphur atoms in Na S O have oxidation number of – 2 or + 2 – 6 + 2 (ii) Oxidation number of chlorine in CaOCl 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. – Therefore the oxidation number of N in HN  C remains it has three covalent bonds. – 3 as D YG 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 in HC  N is taken as – 3 4 3 (wrong) No doubt NH NO 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 4 3 NH 4 x + 4 × (+1) = +1 or x = – 3 NO 3 x + 3 (– 2) = –1 or x = + 5. (iv) Oxidation number of Fe in Fe O (i) Oxidation number of S in Na S O U 3 By usual method Na 2 S 2 O3 3 4 3 4 2 3 3 4 2 4 6 Its structure can be represented as follows, O Na O S S O  S O S O Na  O The two S-atoms which are linked to each other have oxidation number zero. The oxidation number of other S-atoms can be calculated as follows Let oxidation number of S = x. 2×x (for S) x = 2. 4 (v) Oxidation number of S in sodium tetrathionate (Na S O )   2 × (+1) + 2 × x + 3 (–2) = 0 or 2 + 2x – 6 = 0 3 In Fe O , Fe atoms are in two different oxidation states. Fe O can be considered as an equimolar mixture of FeO [iron (II) oxide] and Fe O [iron (III) oxide]. Thus in one molecule of Fe O , two Fe atoms are in + 3 oxidation state and one Fe atom is in + 2 oxidation state.  x=+2 Type III. In a molecule containing two or more atoms of same or different elements in different oxidation states. + 2 × 0 + 6 × ( – 2) = – 2 (for S–S) (for O) ST x = + 5. But this is unacceptable as the two sulphur atoms in Na S O cannot have the same oxidation number because on treatment with dil. H SO , one sulphur atom is precipitated while the other is oxidised to SO. 2 2 3 2 2 In this case, the oxidation number of sulphur is evaluated from concepts of chemical bonding. The chemical structure of Na S O is 2 Na  2 3 Balancing of oxidation-reduction reactions 4 Na 2 S 2 O3  H 2 SO 4 Na 2 SO 4  SO 2  S  H 2 O O 2 2x + 4 – 3 = 0 or 2x = + 1 U +x = 0 (for H) (for N) (for C) or 1 + x – 3 = 0 or x = + 2. (c) Oxidation number of carbon in HC  N Na  By usual method N H O ; 2x + 4 × (+1) + 3 × (–1) = 0 ID S S O O 3 for NH 4 and NO 3 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). 2 2 (bleaching powder) E3 (b) Oxidation number of carbon in H – N  C 2 3 4 In such cases, evaluation of oxidation number should be made using indirect concept or by the original concepts of chemical bonding. Now HC  N  +1 + x – 3 = 0 2 and +6. (iii) Oxidation number of N in NH NO 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. 1 × (+ 1) + 1 × (– 3) + 1 × (– 2) = 0 (for Na) – (a) Oxidation number of C in HC  N and HN  C or + 3 × (–2) + x × 1 60 or x – 2 – 4 = 0 or x = + 6. Though there are a number of methods for balancing oxidation – reduction reactions, two methods are very important. These are, (1) Oxidation number method (2) Ion – electron method (1) Oxidation number method : The method for balancing redox reactions by oxidation number change method was developed by Johnson. In a balanced redox reaction, total increase in oxidation number must be equal to the total decrease in oxidation number. This equivalence provides the basis for balancing redox reactions. This method is applicable to both molecular and ionic equations. The general procedure involves the following steps, (a) Balance the atoms other than H and O for each half reaction using simple multiples. (i) Write the skeleton equation (if not given, frame it) representing the chemical change. (b) Add water molecules to the side deficient in oxygen and H to the side deficient in hydrogen. This is done in acidic or neutral solutions. (ii) Assign oxidation numbers to the atoms in the equation and find out which atoms are undergoing oxidation and reduction. Write separate equations for the atoms undergoing oxidation and reduction. (c) In alkaline solution, for each excess of oxygen, add one water molecule to the same side and 2OH ions to the other side. If hydrogen is still unbalanced, add one OH ion for each excess hydrogen on the same side and one water molecule to the other side. (iii) Find the change in oxidation number in each equation. Make the change equal in both the equations by multiplying with suitable integers. Add both the equations. (iv) Add electrons to the side deficient in electrons as to equalise the charge on both sides. The final balanced equation should be checked to ensure that there are as many atoms of each element on the right as there are on the left. (v) In ionic equations the net charges on both sides of the equation must be exactly the same. Use H ion/ions in acidic reactions and OH ion/ions in basic reactions to balance the charge and number of hydrogen and oxygen atoms. + – The following example illustrate the above rules, Step: II Writing the oxidation number of all the atoms. 5 2 4 2 1 2 Cu  H N O 3 Cu ( N O 3 )2  N O 2  H 2 O 2 0 D YG Step: III Change in oxidation number has occurred in copper and nitrogen. Cu Cu (NO 3 )2 5......(i) 4 H N O3 N O 2......(ii) Increase in oxidation number of copper = 2 units per molecule Cu Decrease in oxidation number of nitrogen = 1 unit per molecule HNO 3 To make increase and decrease equal, equation (ii) is U Step: IV multiplied by 2. (Ionic equation) I2  OH  IO3  H 2 O ; I2 I  (Oxidation half reaction) (Reduction half reaction) Step: III Adding OH  ions, I2  12OH  2 IO3  6 H 2O Step: IV Adding electrons to the sides deficient in electrons, (Si) I2  12OH  2 IO3  6 H 2 O  10e  ; 5 [I2  2e  2 I  ] Step: VI Adding both the half reactions. 6 I2  12OH  2 IO3  6 H 2 O  10 I  ; U 2 Step: I I2  OH  IO3  I   H 2O Step: II Splitting into two half reactions, ID (Skeleton equation) 1 5  2 (vi) Add the two balanced half reactions and cancel any term common to both sides. The following example illustrate the above rules I2  12OH  2 IO3  6 H 2 O  10e  ; I2  2e  2 I  Step: V Balancing electrons in both the half reactions. Step : I Cu  HNO3 Cu(NO 3 )2  NO 2  H 2O 0 (v) Multiply one or both the half reactions by a suitable number so that number of electrons become equal in both the equations. 60 2 – – E3 (iv) Complete the balancing by inspection. First balance those substances which have undergone change in oxidation number and then other atoms except hydrogen and oxygen. Finally balance hydrogen and oxygen by putting H O molecules wherever needed. + Cu  2 HNO3 Cu (NO 3 )2  2 NO 2  H 2O ST Step: V Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained. Cu  4 HNO3 Cu (NO 3 )2  2 NO 2  2 H 2O This is the balanced equation. (2) Ion-electron method (half reaction method) Jette and LaMev developed the method for balancing redox-reactions by ion electron method in 1927. It involves the following steps (i) Write down the redox reaction in ionic form. (ii) Split the redox reaction into two half reactions, one for oxidation and other for reduction. (iii) Balance each half reaction for the number of atoms of each element. For this purpose, Dividing by 2, 3 I2  6 OH  IO3  5 I   3 H 2 O Autoxidation (1) Turpentine and numerous other olefinic compounds, phosphorus and certain metals like Zn and Pb can absorb oxygen from the air in presence of water. The water is oxidised to hydrogen peroxide. This phenomenon of formation of H O by the oxidation of H O is known as autoxidation. The substance such as turpentine or phosphorus or lead which can activate the oxygen is called activator. The activator is supposed to first combine with oxygen to form an addition compound, which acts as an autoxidator and reacts with water or some other acceptor so as to oxidise the latter. For example; 2 2 Pb  O2 PbO2 (activator) 2 ; PbO2  H 2 O PbO  H 2 O2 (autoxidator) (acceptor) (2) The turpentine or other unsaturated compounds which act as activators are supposed to take up oxygen molecule at the double bond position to form unstable peroxide called moloxide, which then gives up the oxygen to water molecule or any other acceptor. RCH  CHR  O2 RHC CHR O O RHC CHR  2 H 2 O RCH  CHR  2 H 2 O2 O O 2 KI  H 2O2 2 KOH  I2 The evolution of iodine from KI solution in presence of turpentine can be confirmed with starch solution which turns blue. (3) The concept of autoxidation help to explain the phenomenon of induced oxidation. Na 2 SO 3 solution is oxidised by air but Na 3 AsO3 solution is not oxidised by air. If a mixture of both is taken, it is observed both are oxidised. This is induced oxidation. Na 2 SO 3  O2 Na 2 SO 5 can act as both oxidising & reducing agent. For example, H2O2, H2SO3, HNO3, SO2 etc. Moloxide Na 2 SO 5  Na 3 AsO3 Na 3 AsO4  Na 2 SO 4 Na 2 SO 3  Na 3 AsO3  O2 Na 2 SO 4  Na 3 AsO4 Disproportionation 1 E3 increase 0 H 2 O2  H 2 O 2  H 2 O  O 2 (1) –1 –2 5 7 ID decrease decrease 1 4 K Cl O3 3 K Cl O4  KCl (2) U increase decrease 1 0 3 D YG 4 P  3 NaOH  3 H 2 O 3 NaH 2 PO2  PH3 (3) increase decrease 5 1 0 hot  5 Na Cl  Na ClO3  3 H 2 O (4) 3 Cl 2  6 NaOH  (conc.) U increase ST  If an element is in its highest possible oxidation state in a compound, it can act as an oxidising agent. for example, KMnO , K Cr O , HNO , H SO , HClO etc. 2 2 7 3 2 4 4 4  If an element is in its lowest oxidation state in a compound, it can act as a reducing agent. For example, H S, H C O , FeSO , Na S O , SO , SnCl , many metals etc. 2 60 One and the same substance may act simultaneously as an oxidising agent and as a reducing agent with the result that a part of it gets oxidised to a higher state and rest of it is reduced to lower state of oxidation. Such a reaction, in which a substance undergoes simultaneous oxidation and reduction is called disproportionation and the substance is said to disproportionate. Following are the some examples of disproportionation, 2 2 4 4 2 2 3 2 2  The strength of oxyacids of chlorine decrease in the order. HClO4 > HClO3> HClO2 >HClO  If highly electronegative element is in its highest oxidation state in a compound that compound can act as powerful oxidant. For example, KClO4, KClO , KBrO3, KIO3 etc. 3  If an element is in intermediate oxidation state in a compound, it