Redox Notes

Page # 1 REDOX REDOX 1. CLASSICAL CONCEPT OF OXIDATION REDUCTION A. Oxidation : According to this concept, oxidation is considered as the addition of oxygen or removal of hydrogen in an ion, in a compound or in a species. Or the addition of an electronegative element or removal of electropositive element in a group, in a ion, in a species or in a compound is called oxidation. For example : (a) 2Mg + O2  2MgO  Addition of Oxygen. (b) C + O2  CO2 (c) H 2S + Cl2  2HCl + S  Removal of Hydrogen (d) MnO2 + 4HCl  MnCl 2 + Cl2 + 2H2O B. Reduction : According to this concept, reduction is considered as addition of hydrogen or removal of oxygen in an atom, in a group, in an ion, in a species or in a compound. Or addition of an electropositive element or removal of an electronegative element in a group, in an ion, in a species or in a compound is called reduction. For example : (a) H2S + Cl2  2HCl + S  Addition of Hydrogen (b) H2 + Cl2  2HCl (C) ZnO + C  Zn + CO  Removal of Oxygen (d) Fe2O3 + 3CO  2Fe + 3CO2 2. VALENCY CONCEPT OF OXIDATION REDUCTION A. Oxidation: According to this concept, increase in (+)ve valency or decrease in (–)ve valency in a reaction is called oxidation. For example : 2Mg + O2  2MgO (0) (+2) B. Reduction : According to this concept, It is the process in which (+) ve valency decreases whereas (–)ve valency increases in a reaction is called reduction. For example : 2HgCl2 + SnCl2  Hg2Cl 2 + SnCl 4 (+2) (+1) 3. OXIDATION NUMBER CONCEPT OF OXIDATION-REDUCTION A. Oxidation : According to this concept, increase in oxidation no. in an element in a reaction is called oxidation. B. Reduction : According to this concept, decrease in oxidation no. in an element in a reaction is called reduction. For example : KMnO4 + FeSO4  MnO + Fe2(SO4)3 + 7 + 2 + 2 + 3 4. MODERN OR ELECTRONIC CONCEPT OF OXIDATION REDUCTION A. Oxidation : According to this concept, the process in which involves the loss of e– in an element or species is called oxidation. It is also called deelectronation. For example : Fe  Fe2+ + 2e– (O.N. of Fe, 0  2) Fe2+  Fe3+ + e– (O.N. of Fe, 2  3) 2Cl–  Cl2 + 2e– (O.N. of Cl, –1  0) B. Reduction : According to this concept, the process which involves the gain of electrons by an element or an atom or an ion is called reduction. It is also called electronation. For example : I2 + 2e–  2I – (O.N. of I,0  –1) Fe3+ + e–  Fe+2 (O.N. of Fe, + 3  + 2) Sn+4 + 2e–  Sn+2 (O.N. of Sn, + 4  + 2) 5. OXIDANT OR OXIDISING AGENT Species, which oxidise other species, which is present in a reaction and reduce it self. This type of species is called oxidant or oxidising agent. Or species, which accepts electron in a reaction by another species and show decreases in its oxidation no. in the reaction is called oxidant or oxidising agent. 5.1Some Important oxidising agent or oxidant 1. All elements with high electronegative character like N, O, F, Cl, etc. 2. All metallic oxides like Li 2O, Na2O, Na2O2, CO2, CaO, MgO, BaO2 etc. 3. Some nonmetallic oxides like CO2, SO2, H2O2, O3. 4. All neutral compound or ion in which element shows their higher oxidation no. or state are act as oxidant or oxidising agent like KMnO4, H2SO4, SnCl4, H3PO4, K2Cr2O7, HClO4, CuCl2, HNO3, H2SO5, FeCl 3, HgCl2, etc. Page # 2 REDOX 6. REDUCTANT OR REDUCTION AGENT Species which reduce other element in a reaction and oxidise itself to donate electrons and show increase in its oxidation no. is called reductant or reducing agent. 6.1Some Important reducing agent or reductant 1. All metals like, K, Mg, Ca, etc. 2. All metallic hydrides like NaH, CaH2, LiAlH4, NaBH4, AlH3, etc. 3. All hydroacids like HF, HCl, HBr, H2S etc. 4. Some organic compounds like Aldehyde, formic acid, oxalic acid, tartaric acid. 5. All neutral compounds or ions, which show their lower oxidation state. MnO, HClO, HClO2, H3PO2, HNO2, H2SO3, FeCl2, SnCl2, Hg2Cl 2, CH2Cl 2 etc. Some Important compound which can act as oxidant and reductant both HNO2, SO2, H2O2, O3, Al2O3, CrO2, MnO2, ZnO, CuO, NOTE : Al2O3, CrO2, MnO2, ZnO, CuO are called as amphoteric oxide. 7. OXIDATION NUMBER A. Definition : It represents the number of electron gained or lost by atom when it changes in compound from a free state. (a) If electron are lose by an atom in the formation of compound, oxidation number is given (+)ve sign. (b) If electrons are gain by an atom in the formation of compound oxidation number given is (–)ve sign. (c) It represents the real charge in case of ionic compounds and represents the imaginary charge in case of covalent compounds. (d) Maximum oxidation no. of an element is equal to group no. in the periodic table (e) Minimum oxidation no. of an element is equal to group no. – 8. I group elements always shows +1 oxidation no. II group elements always shows +2 oxidation no. III group elements show +3 oxidation no. but +1 becomes more stable going down the group (due to inert pair effect) IV group shows –4 to +4 oxidation no. V group shows –3 to +5 oxidation no. VI group shows –2 to +6 oxidation no. VII group shows –1 to +7 oxidation no. Inert gases show zero oxidation no. 7.1Oxidation no. for Coordinate bond (a) When coordinate bond is formed from low electronegative element to high electronegative element then the e– donor element shows +2 oxidation number whereas e– acceptor element shows –2, oxidation no. in this type of bonded compounds. For example in H2SO4. O A H O  S O H || O Here 'S' is low electronegative element than O. therefore, number of S = +2 and O. N. of O = –2 (b) When coordinate bond is formed between the two same electronegative elements then the e– donor element shows +2 oxidation number where e– acceptor element shows –2 oxidation number in this type of bonded compound. For example :– In Na2S2O3 S A B O NaO  S  ONa Here O.N. of 'S' is +2, Because it is e– donor and the other 'S' is –2, Because it is e– acceptor. (c) When coordinate bond is formed from high electronegative element to low electronegative element then no change will be shown by both the elements, which is bonded by coordinate bond. eg. CH3NC 8. OXIDATION STATE Oxidation state of an atom is defined as oxidation number per atom for all practical purposes. Oxidation state is often expressed as oxidation number. 8.1The rules to derive oxidation number or oxidation state (a) The O.S. of an element in its free state is zero. Example O.S.’ s of Na, Cu, I2, Cl2, O2 etc. are zero (b) Sum of O.S.’ s of all the atoms in neutral molecule is zero. (c) Sum of O.S.’ s of all the atoms in a complex ion is equal number of charge present on it. (d) In complex compounds, O.S. of some neutral molecules (ligands) is zero. Example CO, NO, NH3, H2O. (e) Generally O.S. of oxygen is –2 but in H2O2 it is –1 and in OF2 it is +2. (f) Generally O.S. of Hydrogen is +1 but in metallic hydrides it is –1. (g) Generally O.S. of halogen atoms is –1 but in interhalogen compounds it changes. NOTE : Some times same atom in a compound has different O.S. For example, structure of Na2S2O3 is S2 A B Na  O  S1  O  Na O Page # 3 REDOX Here S1 and S2 both are sulphur atoms but they have different O.S. O.S. of S1 = 6 O.S. of S2 = –2 (it is accepting two electrons from S1) 62 Average O.S. of S = = 2 2 (h) Generally, O.S. of alkali metals is +1 and that of alkaline earth metals is +2. (i) O.S. of transition elements very high from compound to compound. Mn has O.S. from +1 to +7. Mn2O  +1, MnO  +2, Mn3O4  8/3, MnO2  +4, Mn2O5  +5, MnO42–  +6, MnO4–  +7 (j) O.S. of an atom may be fractional, negative, zero as well as Positive. 8.2Oxidation State As A periodic Property Oxidation state of an atom depends upon the electronic configuration of atom it is periodic properties. (a) I A group or alkali metals shows +1 oxidation state. (b) II A group or alkaline earth metals show +2 O.S. (c) The maximum normal oxidation state, show by III A group elements is +3. These elements also show +2 to +1 oxidation states also. (d) Elements of IVA group show their max & min. oxidation states +4 & - 4 respectively. (e) Non metals shows number of oxidation states, the relation between max & min. oxidation states for non metals is equal to maximum O.S. –minimum O.S. = 8 For example sulphur has maximum oxidation number +6 as being in VI A group element. 8.3Fractional Oxidation States Lot of elements shows fractional oxidation states. For example oxidation state of oxygen in superoxides of alkali metals (KO2, SO2, RbO2) is -1/2. eg. In Fe3O4, Fe shows its oxidation state as 8/3 as it is a mixed oxide and can be written as FeII Fe2IIIO4 . Oxidation number and oxidation state S.No. Reaction Oxidant (Getting Reduced) 1. 2. 3. 4. 5. 6. 7. 8. C + O2  CO2 PbS + 4O3  PbSO4 + 4O2 PbS + 4H2O2  PbSO4 + 4H2O Sn + 2F2  SnF4 SO2 + 2H2O +Cl2  2HCl + H2SO 4 I2 + 10HNO3  2HIO3 + 10NO2 + 4H2O CuO + H2  Cu + H2O 2KMnO4 + 3H2SO4 + 5H2S  K2SO4 + 2MnSO4 + 8H2O + 5S H2O2 + Ag2O  2Ag + H2O + O2 (Oxygen of H2O2) H2SO4 + 2HI  SO2 + I2 + 2H2O O O O F Cl N Cu Mn 9. 10. 9. REDOX REACTIONS (a) The reactions in whi ch oxidation and reduction both occur simultaneously are called redox reactions. (b) Most of the chemical reactions are redox because if one el ement i s there to lose electrons, other element has to be there to accept them. (c) Any redox reaction may be divided in two parts: (i) Oxidation half reaction (ii) Reduction half reaction Now, we will study some reaction. [0   – 2] [+ 2  0] [–1  –2] [0  –1] [0  –1] [+ 5  + 4] [+ 2  0] [+ 7  + 2] Ag [+ 1  0] S [+6  +4] Reductant (Getting Oxidised) C S S Sn S I H S [0  + 4] [ – 2  + 6] [– 2  + 6] [0 + 4] [+ 4  + 6] [0  + 5] [ 0  + 1] [–2  0] O [–1  0] I [–1  0] NOTE : In reaction 2 oxygens of ozone have different OS. Structure of ozone is O3  O1=O2 OS of O1 = + 2 OS of O2 = 0 Here O1 is getting reduced in reaction 2 OS of O3 = – 2 (d) Redox reactions may be intramolecular or disproportionation reactions. It depends upon whether the migration of electron takes place in the atoms of the same compound or different compounds . Page # 4 REDOX (i) Intermolecular redox reaction +6 +2 +4 2H2SO4(conc.) + Cu  CuSO4 + SO2 + 2H2O (ii) Intramolecular redox reaction +5–2 –1 0 2KClO3  2KCl + 3O2 (iii) Disproportions Redox Reaction : In this type of redox reactions same element acts as both oxidsing & reducing agent. 0 +1 Cl2 + H2O  HClO + HCl – 10.BALANCING OF REDOX EQUATIONS BY ION ELECTRON METHOD 10.1 Acidic Medium (a) consider the example, H  Cr3+ + CO2 Cr2O72– + C2O42–  (b) Write both the half reactions. (c) Balance O– atoms by adding H2O. H2O + Cr(OH)3  CrO42– IO3–  I– + 3H2O (d) Balance H–atoms by adding H2O to side having deficiency and OH – to side having deficiency of H–atoms. 5OH– + H2O + Cr(OH)3  CrO42– + 5H2O or 5OH– + Cr(OH)3  CrO42– + 4H2O IO3– + 6H2O  I– + 3H2O + 6OH– or IO3– + 3H2O  I– + 6OH– (e) Balance the charges by electrons 5OH– + Cr(OH)3  CrO42– + 4H2O + 3e– IO3– + 6H2O + 6e–  I– + 3H2O + 6OH– (f) Multiply first equation by 2 and add to second to give 10OH– + 2Cr(OH)3  2CrO42– + 8H2O + 6e– IO3– + 6H2O + 6e–  I– + 3H2O +6OH– ___________________________________________________ Cr2O72–  Cr3+ (Reduction half reaction) C2O42–  CO2 (Oxidation half reaction) (c) Atoms other than H and O are balanced Cr2O7 C 2O4 2–  2– 2Cr3+  2CO2 (d) Balance O–atoms by the addition of H2O to 4OH–+2Cr(OH)3+IO3–  5H2O+2CrO42– + I– 11. EQUIVALENT WEIGHT (a)Equivalent wt. of an oxidant (get reduced) = Mol. wt. No. of electrons gained by one mole = Mol. wt. Decrease in O. S.  No. of atom undergoing reduction another side Cr2O72–  2Cr3+ C 2O4 2– + 7H2O  2CO2 (e) Balance H–atoms by the addition of H+ ions to another side Cr2O72– + 14H+  2Cr3+ + 7H2O C2O42–  2CO2 (f) Now, balance the charge by the addition of electrons (e–). Cr2O7 C 2O4 2– 2– +14H+ +6e–  2CO2 +  2Cr3+ 2e– +7H2O Example : In acidic medium 6e– + Cr2O72– + 14H+  2Cr3+ + 7H2O Here atoms undergoing reduction is Cr its O.S. is decreasing from 6 to 3 Eq. wt. of (g) Multiply equations by a constant to get number of electrons same in both side. In the above case second equation is multiplied by 3 and then added to first equation. Cr2O72– + 14H+ + 6e–  2Cr3+ + 7H2O Cr2O72– + 3C2O42– + 14 H+  2Cr3+ + 6CO2 + 7H2O 10.2 Alkaline Medium (a) Consider the reaction Cr(OH)3 + IO3 – OH    I– + CrO4 2– (b) Separate the two half reactions. Cr(OH)3  CrO42– (Oxidation half reaction) IO3–  I– (Reduction half reaction) = NOTE : electrons from the (b)Similarly oxidised) = = K2Cr2O7 = Mol. wt. of K 2Cr2O7 32 Mol. wt. 6 [6 in denominator indicates that 6 were gained by Cr2O72– as it is clear given balanced equation] equivalent wt. of a reductant (gets Mol. wt. No. of electrons lost by one mole Mol. wt. increase in O. S.  No. of atom undergoing oxidation Example : In acidic medium, C2O42–  2CO2 + 2e– Here, O.S. of carbon is changing from +1 to Page # 5 REDOX +2 i.e. each carbon atom is l osing one electron and there are 2 atoms of carbon so total electrons lost = 2 so, eq. wt = Mol.wt 2 (e)Try to balance reaction (a) (b) (c) of reduction of KMnO4 by ion electron method as you should get following – (i) MnO4– + 8H+ + 5e–  Mn2+ + 4H2O (acidic medium) (c)In different condition a compound may have different equivalent wts. Because, it depends (ii) MnO4 – + e–  MnO42– (alkaline medium) upon the number of electrons gained or lost by 12.EQUIVALENT WEIGHT IN DISPROPORTION that compound in that reaction. REACTION Example : (i) MnO4 (+7) –  Mn2+ (acidic medium) (+2) undergoing disproportion can be determined by So, Here 5 electrons are taken so eq. wt = Mol. wt. of KMn O 4 5 = 158 = 31.6 5 (ii) MnO4–  MnO2 (neutral medium) (+7) (+4) Here, only 3 electrons are gained so, eq. wt = = The equ ival ent weight of th e su bstance the following method. Consider the substance ‘X’ is undergoing disproportionation reaction. (i) Find out the equivalent weight of ‘X’ for oxidation reaction, say it is E1. (ii) Also find out the equivalent weight of ‘X’ for Mol. wt. of KMnO4 3 reduction reaction, say it is E2. (iii) The equivalent weight of ‘X’ = E1 + E2 for 158 = 52.7 3 example, Cl2 disproportionate in Cl – & ClO3– (iii) MnO4–  MnO 4–2 (alkaline medium) (+7) (+6) Here, only one electron is gained so, eq. wt = Mol. wt. of KMn O 4 1 = 158 NOTE : It is important to note that KMnO4 acts as an oxidant in every medium although with different strength which follows the order – acidic medium > neutral med. > Alkaline medium while, K2Cr2O7 acts as an oxidant only in acidic medium as follows Cr2O72–  2 Cr3+ (2 x 6)   (2 x 3) Here, 6 electrons are gained so eq. wt = Mol. wt. of K 2Cr2O7 6 = 294.21 = 49.03 6 (d)It is clear that KMnO4 is better oxidant than K2Cr2O7 . The X factor for oxidation = 2 The X factor for reduction = 10 The equivalent wt of Cl2 = 71 71 + 10 2 = 35.5 + 7.1 = 42.6 Ans.2