ch19---redox-equilibrium-I.ppt

5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 21Chapter 19Redox Equilibrium I:Redox Reactions19.1 Redox Reactions 19.2 Balancing Redox Equations5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 2219.1 Redox Reactions (SB p.172)Redox ReactionsOxidationReductionAddition of oxygenRemoval of oxygenRemoval of hydrogenAddition of hydrogenLoss of electronGain of electronIncrease in O.N.Decrease in O.N.(An imaginary charge)5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 23Rules for Assigning Oxidation Number19.1 Redox Reactions (SB p.175)1.The oxidation number of an element is 0.2.For a simple ionic compound,the oxidation number of a constituent element is the same as the charge on the ion.3.The oxidation number of some elements in their compounds are always the same.4.In an uncharged compound,the sum of oxidation numbers of all constituent atoms is 0.5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 245.In polyatomic ion,the sum of oxidation numbers of all constituent atoms is equal to the charge of the ion.6.The oxidation number of a constituent element in a compound can be obtained by arithmetic calculation.This is done by first assigning reasonable oxidation numbers to the other elements.7.The oxidation number of an element can be different in different compounds.19.1 Redox Reactions (SB p.176)5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 2519.1 Redox Reactions (SB p.172)Definition of oxidizing and reducing agentsOxidizing agentReducing agent1.Undergoes reduction.Undergoes oxidation.2.Oxidizes a reducing agent.Reduces an oxidizing agent.3.Gain of electron(s).Loss of electron(s).4.Decrease in oxidation number.Increase in oxidation number.5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 2619.2 Balancing Redox Equations (SB p.180)Half Equation MethodCombine the two half equations and eliminate electrons:For(1)x 2=(3)and(2)x 5=(4)(3)+(4):2MnO4-(aq)+10I-(aq)+16H+(aq)2Mn2+(aq)+5I2(aq)+8H2O(l)4Balance the no.of charges:(with a charge of 2 on both sides)2I-(aq)I2(aq)+2e-(2)Balance the no.of charges:(with a charge of+2 on both sides)MnO4-(aq)+8H+(aq)+e-Mn2+(aq)+4H2O(l)(1)3Balance the no.of I atoms:2I-(aq)I2(aq)Balance the no.of O atoms by adding H2O molecules:MnO4-(aq)Mn2+(aq)+4H2O(l)Balance the no.of H atoms by adding H+(aq)ions:MnO4-(aq)+8H+(aq)Mn2+(aq)+4H2O(l)2I-(aq)I2(aq)MnO4-(aq)Mn2+(aq)1OxidationReductionStep5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 2719.2 Balancing Redox Equations (SB p.182)Oxidation Number Method3.BrO3-(aq)+6 I-(aq)Br-(aq)+3 I2(aq)4.BrO3-(aq)+6 I-(aq)Br-(aq)+3 I2(aq)+3H2O(l)5.BrO3-(aq)+6 I-(aq)+6H+Br-(aq)+3 I2(aq)+3H2O(l)1.BrO3-(aq)+I-(aq)Br-(aq)+I2(aq)2.+5 -1 -1 0 O.N.increased by 12.BrO3-(aq)+I-(aq)Br-(aq)+I2(aq)+5 -1 -1 0 O.N.decreased by 6Example5/21/20245/21/20245/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 28