Tiêu đề 6_Redox reactions.docx ✅

PRE-MEDICAL H 36 PHYSICAL CHEMISTRY Serial No. MODULE-1 Page No. 1. Some basic concepts of Chemistry 2. Atomic structure 3. Chemical Equilibrium 4. Ionic Equilibrium 5. Thermodynamics & Chemical Energetics 6. Redox reactions 7. Behaviour of Gases 8. Hints and Solutions
PRE-MEDICAL H 36 REDOX REACTION S.No. CONTENTS Page No. 1. Introduction 2. Oxidation number 3. Applications of oxidation number 4. Oxidation and reduction 5. Types of redox reactions 6. Balancing of redox reactions 7. Law of equivalence 8. Exercise-I (Conceptual Questions) 9. Exercise-II (Previous Years Questions) 10. Exercise-III (Analytical Questions) 11. Exercise-IV (Assertion & Reason)
PRE-MEDICAL H 36 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 same- P > 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 N 3 H  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. H 2 S  O.S of S = –2 H 2 SO 3  O.S of S = +4 H 2 SO 4  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
PRE-MEDICAL H 36 (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. H 2 , O 2 , N 2 , P 4 , S 8 = zero (v) Oxidation state of an element in any of its allotropic form is zero. Ex. C Diamond , C Graphite , S Monoclinic , S Rhombic = 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, NH 3 , H 2 O. (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 CaH 2 MgH 2     +1 –1 +1 –1 +2 –1 +2 –1 (xi) Oxidation state of oxygen in most of its compounds is –2except in- (a) Peroxides 22O  Oxidation state (O) = –1 Ex. H 2 O 2 , BaO 2 (b) Super oxides 12O Oxidation state (O) = –1/2 Ex. KO 2  –1/3 (d) OF 2 (Oxygen difluoride) Ex. F – O – F  +2 (e) O 2 F 2 (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. H 2 SO 4 If O.S. of S is x then 2(+1) + x + 4(–2) = 0 x – 6 = 0 = +6 Ex. H 2 SO 3 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.