Redox Reaction

REDOX REACTION 6.0 INTRODUCTION Redox reactions show vital role in non renewable energy source. In cell reaction where oxidation and reduction both occurs simultaneously will have redox reaction for interconversion of energy. 6.1 Redox Reaction (Oxidation-Reduction) : Many chemical reaction involve transfer of electrons from one chemical substance to another. These electron-transfer reaction are termed as oxidation-reduction or redox reaction. Or Those reaction which involve oxidation and reduction both simultaneously are known as oxidation reduction or redox reactions. Or Those reactions in which increase and decrease in oxidation number of same or different atoms occurs are known as redox reactions. 6.2 Oxidation State Oxidation state of an atom in a molecule or ion is the hypothetical or real charge present on an atom due to electronegativity difference. Or Oxidation state of an element in a compound represents the number of electrons lost or gained during its change from free state into that compound. Some important points concerning oxidation number :(1) Electronegativity values of no two elements are sameP>H C>H S>C Cl > N (2) Oxidation number of an element may be positive or negative. (3) Oxidation number can be zero, whole number or a fractional value. Ex. Ni(CO)4  O.S of Ni = 0 N3H  O.S of N = –1/3 HCl  O.S of Cl = –1 (4) Oxidation state of same element can be different in same or different compounds. Ex. H2S  O.S of S = –2 H2SO3  O.S of S = +4 H2SO4  O.S of S = +6 6.3 Some helping rules for calculating oxidation number : (A) In case of covalent bond – (i) For hmoatomic molecule A–A A=A AA       0 0 0 0 0 0 (ii) For heteroatomic molecule (EN of B > A) A–A A=A AA       +1 –1 +2 –2 +3 –3 (iii) The oxidation state of an element in its free state is zero. Example – Oxidation state of Na, Cu, I, Cl, O etc. are zero. (iv) Oxidation state of atoms present in homoatomic molecules is zero. Ex. H2, O2, N2, P4, S8 = zero (v) Oxidation state of an element in any of its allotropic form is zero. Ex. CDiamond, CGraphite, SMonoclinic, SRhombic = 0 (vi) Oxidation state of all the components of an alloy are 0 (vii) In complex compounds, oxidation state of some neutral molecules (ligands) is zero. Ex. CO, NO, NH3, H2O. (viii) Oxidation state of fluorine in all its compounds is –1 (ix) Oxidation state of I A & II A group elements are +1 and +2 respectively. (x) Oxidation stte of hydrogen in most of its compounds is +1 except in metal hydrides (–1) Ex. NaH LiH CaH2 MgH2        +1 –1 +1 –1 +2 –1 +2 –1 (xi) Oxidation state of oxygen in most of its compounds is –2except in(a) Peroxides  O22   Oxidation state (O) = –1 Ex. H2O2, BaO2 (b) Super oxides  O21  Oxidation state (O) = –1/2 Ex. KO2  –1/3 (d) OF2 (Oxygen difluoride) Ex. F–O–F  +2 (e) O2F2 (Dioxygen difluoride)  +1 (xii) Oxidation state of monoatomic ions is equal to the charge present on the ion. Ex. Mg+2  Oxidation state = +2 (xiii) The algebric sum of oxidation state of all the atoms present in a polytomic neutral molecule is 0. Ex. H2SO4 If O.S. of S is x then 2(+1) + x + 4(–2) = 0 x–6=0 = +6 Ex. H2SO3 If O.S. of S is x then 2(+1) + x + 3(–2) = 0 x–4=0 = +4 (xiv) The algebraic sum of oxidation state of all the atoms in a polyatomic complex ion is equal to the charge present on the ion. Ex. S O 4 2 Ex. (B) (C) If O.S. of S is x then x + 4(–2) = –2 x–6=0 = +6 HC O 3 If O.S. of C is x then +1 + x + 3(–2) = –1 x–4=0 = +4 In case of co-ordinate bond (EN of B > A) : AA BB AB      +2 –2 +2 –2 +2 –2 In case of Ionic bond : Charge on cation = O.S. of cation Charge on anion = O.S. of anion Ex. NaCl  Na+ + Cl–    +1 –1 +2 MgCl2  Mg + 2Cl–    +2 –1 BA  (neglected) Illustrations Illustration 1. Oxidation number of cobalt in [Co(NH3)6]Cl2Br is – (1) +6 (2) Zero (3) +3 (5) +2 Solution Let the oxidation number of Co be x Oxidation number of NH3 is zero Oxidation number of Cl is –1 Oxidation number of Br is –1 Hence, x + 6(0) – (1×2) – 1 = 0  x = +3 So, the oxidation number of cobalt in the given complex compound is +3. Illustration 2. The order of increasing oxidation state of S in S8, S2 O 82 , S2 O 32 , S4 O 62 is given below(1) S8 < S2 O 82 < S2 O 32 < S4 O 62 (2) S2 O 82 < S2 O 32 < S4 O 62 < S8 (3) S2 O 82 < S8 < S4 O 62 < S2 O 32 (4) S8 < S2 O 32 < S4 O 62 < S2 O 82 Solution The oxidation number of S are shown below along with the compounds S2 O 82 S2 O 32 S4 O 62 S8 0 +6 +2 +2.5 Hence the order of increasing oxidation state of S is – S8 < S2 O 32 < S4 O 62 < S2 O 82 Illustration 3. The oxidation number of Cl in NOClO4 is (1) +11 (2) +9 (3) +7 (4) +5 Solution The compound may be written as NO+ ClO 4 . For ClO 4 , Let oxidation number of Cl = a a + 4×(–2) = –1 a = +7 Hence, the oxidation number of Cl in NOClO4 is +7 Illustration 4. The two possible oxidation number of N atoms in NH4NO3 are respectively(1) +3, +5 (2) +3, –5 (3) –3, +5 (4) –3, –5 Solution There are two N atoms in NH4NO3, but one N atom has negative oxidation number (attached to H) and the other has positive oxidation number of (attached to O).Therefore evaluation should be made separately as – Oxidation number of N is NH 4 Oxidation number of N in NO3 a + 4×(+1) = +1 a + 3(–2) = –1  a = –3  a = +5 Here the two oxidation number are –3 and +5 respectively. Illustration 5. The oxidation number of S in H2S2O8 is (1) +8 (2) –8 (3) +6 Solution In H2S2O8 two O atoms form peroxide linkage i.e. O O (4) +4 H–O–S–O–O–S–O–H O O 2 × 1 + 2a + 6(–2) + 2(–1) =0  a = +6 Thus the oxidation number of S in H2S2O8 is +6 Illustration 6. The oxidation number of S in (CH3)2 SO is (1) 1 (2) 2 (3) 0 (4) 3 Solution Let the oxidation number of S is ‘a’ Oxidation number of CH3 = +1 Oxidation number of O = –2 2(+1) + a + (–2) = 0 a=0 Hence the oxidation number of S in dimethyl sulphoixde is zero. 1. 2. 3. BEGINNER’S BOX-1 In which of the following compounds, the oxidation state of I atom is highest ? (1) KI3 (2) KIO4 (3) KIO3 (4) IF5 The oxidation number of phosphorus in Ba(H2PO2)2 is (1) +3 (2) +2 (3) +1 (4) –1 Oxidation number of Ni in Ni(CO4) is (1) 0 (2) 4 (4) 2 (3) 8 4. Positive oxidation state of an element indicates that it is (1) Elementry form (2) Oxidised (3) Reduced (4) Only reductant 5. Predict the highest and lowest oxidation state of (a) Ti and (b) Tl in combined state. (1) a[0, +3] b[0, +2] (2) a[+3, 0] b[+4, 0] (3) a[+4, 0] b[+4, 0] (4) a[+4, +2] b[+3, +1] 6. The oxidation state of oxygen atom in potassium superoxide is :(1) Zero (2) –1/2 (3) –1 6.4 (4) –2 APPLICATIONS OF OXIDATION NUMBER : (A) To compare the strength of acid and base :Strength of acid  Oxidation Number 1 Strength of base  Oxidation number Example : order of acidic strength in HClO, HClO2, HClO3, HClO4 will be. Solution : Oxidation number of chlorine HClO (Hypo chlorous acid) +1 HClO2 (Chlorous acid) +3 HClO3 (Chloric acid) +5 HClO4 (Perchloric acid) +7  Strength of acid  Oxidation Number So the order will be :HClO4 > HClO3 > HClO2 > HClO (B) To determine the oxidising and reducing nature of the substances :Oxidising agents are the substances which accept electrons in a chemical reaction i.e., electron acceptors are oxidizing agent. Reducing agents are the substances which donate electrons in a chemical reaction i.e., electron donors are reducing agent. Highest O.S. +4 +5 +5 +6 +7 +6 +7 +8 +8 +2 +1 C N P S Cl Cr Mn Os Ru O H Elements 0 0 –2 –1 Lowest O.S. –4 –3 –3 –2 –1 0 0 (a) Ex. (b) Ex. If effective element in a compound is present in maximum oxidation state then the compound acts as oxidizing agent. KMnO4 K2Cr2O7 H2SO4 SO3 H3PO4 HNO3 HClO4            +7 +6 +6 +6 +5 +5 +7 If effective element in a compound is present in minimum oxidation state then the compound acts as reducing agent. PH3 NH3 CH4       –3 –3 –4 (c) Ex. (C) If effective element in a compound is present in intermediate oxidation state then the compound can act as well as reducing agent. HNO2 H3PO3 SO2 H2O2        +3 +3 +4 –1 To calculate the equivalent weight of compounds : The equivalent weight of an oxidising agent or reducing agent is that weight which accepts one mole electrons in a chemical reaction or looses one mole electron in chemical reaction. Molecular weight (a) Equivalent weight of oxidant = No.of electrons gained by one mole Example : In acidic medium 6e– + Cr2O72 + 14H+  2Cr3+ + 7H2O Here atoms which undergoes reduction is Cr. Its O.S. is decreasing from +6 to +3 Molecular weight of K 2Cr2O 7 M Equivalent weight of K2Cr2O7 = = 3 2 6 2 Note :- [6 in denominator indicates that 6 electron were gained by Cr2O7 as it is clear from the given balanced equation] Molecular weight (b) Equivalent weight of a reducant = No.of electrons lost by one mole In acidic medium, C 2 O 42   2CO2 + 2e– M 2 (c) In different conditions a compound may have different equivalent wts. Because, it depends upon the number of electron gained or lost by that compound in that reaction. Example : (i) MnO 4  Mn+2 (acidic medium) (+7) (+2) M 158 Here 5 electrons are taken so equivalent weight = = = 31.6 5 5 (ii) MnO 4  MnO2 (neutral medium) or (Weak alkaline medium) (+7) (+4) M 158 Here, only 3 electrons are gained, so eq. wt. = = = 52.7 3 3 Note : When only alkaline medium is given consider it as weak alkaline medium. (iii) MnO 4  MnO 4 2 (strong alkaline medium) (+7) (+6) M Here, only one electron is gained, so eq. wt. = = 158 1 Note : KMnO4 acts as an oxidant in every medium although with different strength which follows the order – Acidic medium > neutral medium > alkaline medium Here, Total electrons lost = 2 So, equivalent weight = (D) While, K2Cr2O7 acts as an oxidant only in acidic medium as follows Cr2O72  2Cr3+ (2×6)  (2×3) M 294 Here, 6 electrons are gained so eq. wt. = = = 49 6 6 To determine the possible molecular formula of compound : Since the sum of oxidation number of all the atoms present in a compound is zero, so the validity of the formula can be confirmed. GOLDEN KEY POINTS SOME OXIDIZING AGENTS / REDUCING AGENTS WITH EQUIVALENT WEIGHT Species MnO 4 (O.A.) Changed to Mn 2 in acidic medium Reaction MnO 4 + 8H+ + 5e–  Mn2+ + 4H2O Electrons exchanged or change in O.N. 5 MnO 4 + 3e– + 2H2O  MnO2 + 4OH– 3 MnO 4 + e–  MnO 24  1 Cr 3 CrO 72  + 14H+ + 2e–  2Cr3+ + 7H2O 6 Mn 2  MnO2 + 4H+ + 2e–  Mn2+ + 2H2O 2 Cl– Cl2 + 2e–  2Cl– 2 Cu+ Cu2+ + e–  Cu+ 1 S2O32  (R.A.) S4O 62  2 S2O32   S4O 62  + 2e– H2O2(O.A.) H2O H2O2 + 2H+ + 2e–  2H2O 2 H2O2(R.A.) O2 2 Fe2+ (R.A.) Fe3+ H2O2  O2 + 2H+ + 2e– (O.N. of oxygen in H2O2 is –1 per atom) 1 Fe2+  Fe3+ + e– MnO 4 (O.A.) MnO 2 in neutral medium MnO 4 (O.A.) MnO24  in basic medium Cr2O72 (O.A.) MnO2(O.A.) in acidic medium in acidic medium Cl2 (O.A.) in bleaching powder CuSO 4 (O.A.) in iodometric titration 2 (for two molecules) Illustrations Illustration 7. To find the n-factor in the following chemical changes.  H  Mn2+ (i) KMnO4  Power by: VISIONet Info Solution Pvt. Ltd Website : www.edubull.com H2 O (ii) KMnO4   Mn4+ Mob no. : +91-9350679141 Eq. wt. E= M 5 E= M 3 M 1 M E= 6 E= E= M 2 M 2 M E= 1 2M E= =M 2 M E= 2 M E= 2 E= E= M 1  H OH (concentrated basic medium)  Cr3+  Mn6+ (iv) K2Cr2O7  (iii) KMnO4  (v) C 2O 42   CO2 (vi) FeSO4  Fe2O3 (vii) Fe2O3  FeSO4 Solution (i) In this reaction, KMnO4 which is an oxidizing agent, itself gets reduced to Mn2+ under acidic conditions. n = |1 × (+7) – 1 × (+2) | = 5 (ii) In this reaction, KMnO4 gets reduced to Mn4+ under neutral or slightly (weakly) basic conditions. n = |1 × (+7) – 1 × (+4) | = 3 (iii) In this reaction, KMnO4 gets reduced to Mn6+ under basic conditions. n = |1 × (+7) – 1 × (+6) | = 1 (iv) In this reaction, K2Cr2O7 which acts as an oxidizing agent reduced to Cr3+ under acidic conditions. (It does not react under basic conditions.) n = |2 × (+6) – 2 × (+3) | = 6 (v) In this reaction C 2O 42  (oxalate ion) gets oxidized to CO2 when it is reacted with an oxidizing agent. n = |2 × (+3) – 2 × (+4) | = 2 (vi) In this reaction, ferrous ions get oxidized to ferric ions. n = |2 × (+2) – 1 × (+3) | = 1 (vii) In this reaction, ferric ions are getting reduced to ferrous ions. n = |2 × (+3) – 2 × (+2) | = 2 Illustration 8. Suppose that there are three atoms A, B, C and their oxidation number are 6, –1, –2, respectively. Then the molecular formula of compound will be. Solution Since, the charge on a free compound is zero. So +6 = (–1×4) + (–2) +6 = –6 Or +6 = (–1×2) + (–2×2) = –2 + (–4) = –6 So molecular formula, AB4C or AB2C2 1. BEGINNER’S BOX-2 Molecular weight of KMnO4 in acidic medium and neutral medium will be respectively (1) 7 × equivalent weight and 2 × equivalent weight (2) 5 × equivalent weight and 3 × equivalent weight (3) 4 × equivalent weight and 5 × equivalent weight (4) 2 × equivalent weight and 4 × equivalent weight 2. In acidic medium, equivalent weight of K2Cr2O7 (Molecular weight = M) is (1) M/3 (2) M/4 (3) M/6 (4) M/2 6.5 OXIDATION AND REDUCTION There are two concepts of oxidation and reduction. (A) Classical / old concept :OXIDATION REDUCTION Addition of O2 Addition of H2 (1) (2) (3) (4) 2Mg + O2  2MgO C + O2  CO2 Removal of H2 H2S + Cl2  2HCl + S (oxidation of H2S) 4HI + O2  2I2 + 2H2O (oxidation of HI) Addition of electronegative element Fe + S  FeS (oxidation of Fe) SnCl2 + Cl2  SnCl4 (oxidation of SnCl2) Removal of electropositive element 2NaI + H2O2  2NaOH + I2 (oxidation of NaI) N2 + Cl2  2NH3 H2 + Cl2  HCl Removal of O2 CuO + C Cu + CO (reduction of CuO) H2O + C  CO + H2 (reduction of H2O) Addition of electropositive element CuCl2 + Cu  Cu2Cl2 (reduction of CuCl2) HgCl2 + Hg  Hg2Cl2 (reduction of HgCl2) Removal of electronegative element 2FeCl3 + H2  2FeCl2 + 2HCl (reduction of FeCl3) (B) Electronic / Modern Concept (1) De-electronation (2) Oxidation process are those process in which one or more e–s are lost by an atom, ion or molecule. (3) Example :(a) Zn  Zn+2 + 2e– M  Mn+ + ne– (b) Sn+2  Sn+4 + (4–2)e– M n1  M  n 2 (n2 – n1)e– (c) Cl–  Cl + e– A–n  A + ne– (d) MnO 4 2  MnO 4 + (2 – 1)e– A  n1  A  n 2 (n1 – n2)e– 6.6 Electroation Reduction process are those process in which one or more e–s are gained by an atom, ion or molecule. (a) Cu+2 2e–  Cu Mn+ + ne– M (b) Fe+3 + (3–2)e–  Fe+2 M x1 (x1 – x2)e–  M  x 2 (c) O + 2e–  O2– A + xe  A–x (d) [Fe(CN)4]3– (4 – 3)e–  [Fe(CN)4]–4 A  n1 (n1 – n2)e–  – A  n 2 TYPES OF REDOX REACTIONS (A) Intermolecular redox reaction :- When oxidation and reduction takes place separately in different compounds, then the reaction is called intermolecular redox reaction. SnCl2 + 2FeCl3  SnCl4 2FeCl2 Sn+2  Sn+4 (Oxidation) Fe+3  Fe+2 (Reduction) (B) Inteamolecular redox reaction :- During the chemical reactions place in single compound then there action is called intermolecular redox reaction. 2KClO3  2KCl + 3O2 Cl+5 O–2 Cl–1 O0 Reduction Oxidation (C) Disproportion reaction :- when reduction an oxidation take place in the same element of the same compound then the reaction is called disproportionation reaction. H2O2  H2O + 1/2O2 O–1 O–2 O0 Reduction Oxidation (D) Comproportionation reaction : Reverse of disproportionation reaction known as comproportionation reaction Ex. ClO– + Cl– Cl2 + OH– BEGINNER’S BOX-3 1. 2. 3. Oxidation is defined as(1) Gain of electrons (3) Loss of electron (2) Decreases in positive valency (4) Addition of electropositive element Reduction is defined as (1) Increases in positive valency (3) Loss of protons (2) Gain of electrons (4) Decreases in negative valency In the reaction MnO 4 + SO32 + H+  SO 42 + Mn2+ + H2O (1) MnO 4 and H+ both are reduced (2) MnO 4 is reduced and H+ is oxidised (3) MnO 4 is reduced and SO32 is oxidised (4) MnO 4 is oxidised and SO32 is reduced 4. The charge on cobalt in [Co(CN)6]–3 is – (1) –6 (2) –3 (3) +3 (4) +6 5. Which of the following halogen always show only one oxidating state its compounds? (1) Cl (2) F (3) Br (4) I 6. Which of the following reaction do not involve oxidation-reduction ? (1) 2Rb +2H2O  2RbOH + H2 (2) 2CuI2  2CuI + I2 (3) NH4Cl + NaOH  NaCl + NH3 + H2O (4) 3Mg + N2  Mg3N2 7. The fast reaction between water and sodium is the example of (1) Oxidation (2) Reduction (3) Intermolecular redox (4) Intramolecular redox 8. Choose the redox reaction from the following :(1) Cu + 2H2SO4  CuSO4 + SO2 + 2H2O (2) BaCl2 + H2SO4  BaSO4 + 2HCl (3) 2NaOH + H2SO4  Na2SO4 + 2H2O (4) KNO3 + H2SO4  2HNO3 + K2SO4 9. Which of the following not a redox reaction ? (1) MnO 4  MnO2 + O2 (2) Cl2 + H2O  HCl + HClO (3) 2CrO 4 2 + 2H+  2CrO 72  + H2O (4) MnO 4 + 8H+ + 5Ag  Mn+2 + 4H2O + 5Ag+ 10. In the reaction 6Li + N2 2LI3N (1) Li undergoes reduction (3) N undergoes oxidation (2) Li undergoes oxidation (4) Li is oxidation 11. H2O2 + H2O2 2H2O + O2 is an example of dispropotionation because :(1) Oxidation number of oxygen only decreases (2) Oxidation number of oxygen only increases (3) Oxidation number of oxygen decreases as well as increase (4) Oxidation number of oxygen neither decreases nor increases 6.7 BALANCING OF REDOX REACTION :(A) Oxidation number change method (B) Ion electron method. (A) Oxidation number change method :This method was given by Jonson. In a balanced redox reaction, total increase in oxidation number must be equal to total decreases in oxidation number. This equivalence provides the basis for balancing redox reactions. The general procedure involves the following steps :(i) Select the atom in oxidising agent whose oxidation number decreases and indicate the gain of electrons. (ii) Select the atom in reducing agent whose oxidation number increases and indicate the loss of electrons. (iii) Now cross multiply i.e. multiply oxidizing agent by the number of loss of electrons and reducing agent by number of gain of electrons. (iv) Balance the number of atoms on both sides whose oxidation numbers change in the reaction. (v) In order to balance oxygen atom, add H2O molecules to the side deficient in oxygen. (vi) Then balance the number of H atoms by adding H+ ions to the side deficient in hydrogen. Illustrations Illustration 9. Balance the following reaction by the oxidation number method :Cu + HNO3  Cu(NO3)2 + NO2 + H2O Solution Write the oxidation number of all the atoms. 0 +1+5–2 +2+5–2 +4–2 +1–2 Cu + HNO3  Cu(NO3)2 + NO2 + H2O There is change in oxidation number of Cu and N. 0 +2+5–2 Cu  Cu(NO3)2 ……….(1) (Oxidation no. is increased by 2) +5 +4 HNO3  NO2 ……….(2) (Oxidation no. is decreased by 1) To make increase and decrease equal, eq. (2) is multiplied by 2. Cu + 2HNO3  Cu(NO3)2 + 2NO2 + H2O Balancing nitrates ions, hydrogen and oxygen, the following equation is obtained. Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O This is the balanced equation. Illustration 10. Balance the following reaction by the oxidation number method :MnO 4 + Fe+2  Mn+2 + Fe+3 Solution Write the oxidation number of all the atoms. +7 –2 MnO 4 + Fe+2  Mn+2 + Fe+3 Change in oxidation number has occurred in Mn and Fe. +7 MnO 4  Mn+2 ………..(1) (Decrement in oxidation no. by 5) +2 +3 Fe Fe ………..(2) (Increment in oxidation no. by 1) To make increase and decrease equal, eq. (2) is multiplied by 5. MnO 4 + 5Fe+2  Mn+2 + 5Fe+3 To balance oxygen, 4H2O are added to R.H.S. and to balance hydrogen, 8H+ are added to L.H.S MnO 4 + 5Fe+2 + 8H+  Mn+2 + 5Fe+3 + 4H2O This is the balanced equation. (B) Ion-Electron method :This method was given by Jette and La Mev in 1972. The following steps are followed while balancing redox reaction (equations) by this method. (i) Write the equation in ionic from. (ii) Split the redox equation into two half reactions, one representing oxidation and the other representing reduction. (iii) Balance these half reactions separately and then add by multiplying with suitable coefficients so that the electrons are cancelled. Balancing is done using following substeps. (a) Balance all other atoms except H and O. (b) Then balance oxygen atoms by adding H2O molecules to the side deficient in oxygen. The number of H2O molecules added is equal to the deficiency of oxygen atoms. (c) Balance hydrogen atoms by adding H+ ions equal to the deficiency in the side which is deficient in hydrogen atoms. (d) Balance the charge by adding electrons tot the die which is rich in +ve charge. i.e. deficient in electrons. Number of electrons added is equal to the deficiency. (e) Multiply the half equations with suitable coefficients to equalize the number of electrons. (iv) Add these half equations to get an equation which is balanced with respect to charge and atoms. (v) If the medium of reaction is basic, OH– ions are added to both sides of balanced equation , which is equal to number of H+ ions in Balanced Equation. Illustrations Illustration 11. Balance the following reaction by ion-electron method in acidic medium :Cr2O72 + C 2O 42   Cr3+ + CO2 Solution Cr2O72 + C 2O 42   Cr3+ + CO2 (a) Write both the half reaction. (Reduction half reaction) Cr2O72  Cr3+ (b) C 2 O 42   CO2 (Oxidation half reaction) Atoms other than H and O are balanced Cr2O72  2Cr3+ (c) C 2 O 42   2CO2 Balance O-atoms by the addition of H2O to another side Cr2O72  2Cr3+ + 7H2O (d) C 2 O 42   2CO2 Balance H-atoms by the addition of H+ to another side Cr2O72 14 H+  2Cr3+ + 7H2O (e) C 2 O 42   2CO2 Now, balance the charge by the addition of electron (e–) ………(1) Cr2O72 14H+ + 6e–  2Cr3+ + 7H2O (f) C 2 O 42   2CO2 + 2e– ………(2) Multiply equations by a constant to get the same number of electrons on both side. In the above case second equation is multiplied by 3 and then added to first equation Cr2O72 + 14H+ + 6e– 2Cr3+ + 7H2O 6CO2 + 6e– 3C 2O 24  Cr2O72 3C 2O 24  + + 2Cr3+ + 6CO2 + 7H2O Illustration 12. Balance the following reaction by ion-electron method : OH  I– + CrO 24  Cr(OH)3 + IO3  Solution  OH  I– + CrO 24  Cr(OH)3 + IO3  (a) Separate the two half reactions. Cr(OH)3  CrO 24  (Oxidation half reaction) (b) IO3  I– (Reduction of half reaction) Balance O-atoms by adding H2O. H2O + Cr(OH)3  CrO 24  (c) IO3  I– + 3H2O Balance H-atoms by adding H+ to side having deficiency and add equal no of OH– ions to the side ( medium is known) H2O + Cr(OH)3  CrO 24  + 5H+ 5OH– + H2O + Cr(OH)3  CrO 24  + 5H+ + 5OH– or 5OH– + Cr(OH)3  CrO 24  + 4H2O IO3 + 6H+  I– + 3H2O IO3 + 6H+ + 6OH–  I– + 3H2O + 6OH– (d) IO3 + 3H2O  I– + 6OH– or Balance the charges by adding electrons 5OH– + Cr(OH)3  CrO 24  + 4H2O + 3e– (e) IO3 + 3H2O + 6e–  I– + 6OH– Multiply first equation by 2 and add to second to give 100 H– + 2Cr(OH)3 2CrO 24  + 8H2O + 6e– IO3 + 3H2O + 6e– 4OH– + 2Cr(OH)3 + IO3 6.8  I– + 6OH– 5H2O + 2CrO 24  + I– – LAW OF EQUIVALENCE The law states that one equivalent of an element combine with one equivalent of the other, and in a chemical reaction equivalent and mill equivalent of reactants react in equivalent to give same no. of equivalent or milli equivalents of products separately. According : (i) aA + bB  mM + nN m. eq of A = m. eq of B = m. eq of M = m. eq of N (ii) In a compound MxNy m. eq of MxNy = m. eq of M = m. eq of N GOLDEN KEY POINTS FOR ACID-BASE (NEUTRALIZATION REACTION) OR REDOX REACTION : N1V1 = N2V2 is always true. But M1V1 = M2V2 (may or may not be true) But M1×n1×V1 = M2×n2×V2 (always true when n terms represent n-factor) Illustrations Illustration 13. Calculate the normality of a solution containing 15.8 g of KMnO4 in 50 mL acidic solution . W  1000 Solution Normality (N) = E  VmL molar mass of KMnO 4 158 Where W = 15.8 g, V = 50 mL E= = = 31.6 5 Valence factor So, N = 10 Illustration 14. Calculate the normality of a solution containing 50 mL of 5 M solution K2Cr2O7 in acidic medium. Solution Normality (N) = Molarity × Valence factor = 5 × 6 = 30 N Illustration 15. Find the number of moles of KMnO4 needed to oxidize one mole Cu2S in acidic medium . The reaction is KMnO4 + Cu2S  Mn2+ + Cu2+ + SO2 Solution From law of equivalence Equivalents of Cu2S = equivalents of KMnO4 Moles of Cu2S × v.f. = moles of KMnO4 × v.f. 1×8 = n2×5 8 n2 = = 1.6 5 Illustration 16. The number of moles of oxalate ions oxidized by one mole of MnO 4 ion in acidic medium. 5 2 3 5 (1) (2) (3) (4) 2 5 5 3 2  Solution Equivalent of C 2O 4 = equivalents of MnO 4 5 x (mole) × 2 = 1 × 5 ; x = 2 Illustration 17. What volume of 6 M HCl and 2 M HCl should be mixed to get two litre of 3 M HCl? Solution Let, the volume of 6 M HCl required to obtain 2 L of 3M HCl = x L  Volume of 2 M HCl required = (2 – x)L M1V1 + M2V2 = M3V3 6M HCl 2M HCl 3M HCl 6 × (x) + 2 × (2 – x) = 3 × 2  6x + 4 – 6x = 6  4x = 2  x = 0.5 L Hence, volume of 6 M HCl required = 0.5 L Volume of 2 M HCl required = 1.5 L Illustration 18. In a reaction vessel, 1.184 g of NaOH is required to be added for completing the reaction. How many milli litre of 0.15 M NaOH should be added for this requirement ? Solution Amount of NaOH present in 1000 mL of 0.15 M NaOH = 0.15×40 = 6 g 6  1 ml of this solution contain NaOH = ×10–3 g 1000 1  1.184 g of NaOH will be present in = ×1.184 = 197.33 mL 6  103 Illustration 19. What weight of Na2CO3 of 85% purity would be required to prepare 45.6 mL of 0.253 NH2SO4? Solution Meq. of Na2CO3 = Meq. of H2SO4 = 45.6×0.235 WNa 2 CO3 WNa 2 CO3  ×1000 = 45.6 × 0.235  ×1000 = 45.6 × 0.235 E Na 2 CO3 106 / 2    WNa2 CO3 = 0.5679 g For 85 g of pure Na2CO3, weighed sample = 100 g  For 0.5679 g of pure Na2CO3, weighed sample = 100 ×0.5679 = 0.6681 g 85 Illustration 20. The number of moles of KMnO4 that will be required to react with 2 mol of ferrous oxalate is 6 2 4 (1) (2) (3) (4) 1 5 5 5 Solution Mn7+ + 5e–  Mn2+] × 3  Fe2   Fe3  e  C2O42   2CO2  2e   3 moles of KMnO4 = 5 moles of FeC2O4 6  2 mol of ferrous oxalate  mole of KMnO4 5 Hence, (A) is the correct answer. Illustration 21. What volume of 6 M HNO3 is needed to oxidize 8 g of Fe2+ to Fe3+, HNO3 gets converted to NO? (1) 8 mL (2) 7.936 mL (3) 32 mL (4) 64 mL 2+ Solution Meq. of HNO3 = Meq. of Fe 8 or 6 × 3 × V = × 1000 56 V = 7.936 mL n-factor = 3 +5 2+  ( NO3  NO) Hence, (B) is the correct answer. Illustration 22. Which of the following is/are correct ? (1) g mole wt. = mol. wt. in g = wt. of 6.02×1023 molecules (2) mole = NA molecule = 6.02×1023 molecules (3) mole = g molecules (4) None of the above Solution Ans. (A), (B) and (C) 1. BEGINNER’S BOX-4 In the half reaction : 2ClO  Cl2 (1) 5 electrons are gained (2) 5 electrons are liberated (3) 10 electrons are gained (4) 10 electrons are liberated  3 2. The number of electrons required to balance the following equation:NO3 + 4H– + e–  2H2O + NO are(A) 5 (B) 4 (C) 3 (D) 2 3. Which of the following equation is balanced one- Edubull (1) 5BiO 3 + 22H+ + Mn2+  5Bi3+ + 7H2O + MnO 4 (2) 5BiO 3 + 14H+ + 2Mn2+  5Bi3+ + 7H2O + 2MnO4 (3) 2BiO 3 + 4H+ + Mn2+  2Bi3+ + 2H2O + MnO 4 (4) 6BiO 3 + 12H+ + 3Mn2+  6Bi3+ + 6H2O + 3MnO 4 ANSWER KEY 1. (2) 2. (3) 3. BEGINNER'S BOX-1 (1) 4. (2) 5. (4) 6. (2) (2) 6. (3) BEGINNER'S BOX-2 1. (2) 2. (3) 1. 8. (3) (1) 2. 9. (2) (3) 3. 10. BEGINNER'S BOX-3 (3) 4. (3) 5. (2) 11. (3) 1. (3) 2. (3) 3. BEGINNER'S BOX-4 (2) 7. (3)