H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 1 Chapter 19 Redox Equilibrium I: Redox Reactions 19.1Redox Reactions 19.2Balancing.

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H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 1 Chapter 19 Redox Equilibrium I: Redox Reactions 19.1Redox Reactions 19.2Balancing Redox Equations

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Redox Reactions (SB p.172) Redox Reactions OxidationReduction Addition of oxygenRemoval of oxygen Removal of hydrogenAddition of hydrogen Loss of electronGain of electron Increase in O.N.Decrease in O.N. (An 'imaginary charge')

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 3 Rules for Assigning Oxidation Number 19.1 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.

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 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 Redox Reactions (SB p.176)

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Redox Reactions (SB p.172) Definition of oxidizing and reducing agents Oxidizing agentReducing agent 1. 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.

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Balancing Redox Equations (SB p.180) Half Equation Method Combine the two half equations and eliminate electrons: For (1) x 2 =(3) and (2) x 5 =(4) (3) + (4) : 2MnO 4- (aq) + 10I - (aq) + 16H + (aq) 2Mn 2+ (aq) + 5I 2 (aq) + 8H 2 O(l) 4 Balance the no. of charges: (with a charge of –2 on both sides) 2I - (aq) I 2 (aq) + 2e - …(2) Balance the no. of charges: (with a charge of +2 on both sides) MnO 4 - (aq) + 8H + (aq) + e - Mn 2+ (aq) + 4H 2 O(l)…(1) 3 Balance the no. of I atoms: 2I - (aq) I 2 (aq) Balance the no. of O atoms by adding H 2 O molecules: MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l) Balance the no. of H atoms by adding H + (aq) ions: MnO 4 - (aq) + 8H + (aq) Mn 2+ (aq) + 4H 2 O(l) 2 I - (aq) I 2 (aq)MnO 4 - (aq) Mn 2+ (aq)1 OxidationReductionStep

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Balancing Redox Equations (SB p.182) Oxidation Number Method 3. BrO 3 - (aq) +6 I - (aq) Br - (aq) + 3 I 2 (aq) 4. BrO 3 - (aq) +6 I - (aq) Br - (aq) + 3 I 2 (aq) + 3H 2 O(l) 5. BrO 3 - (aq) +6 I - (aq) + 6H + Br - (aq) + 3 I 2 (aq) + 3H 2 O(l) 1.BrO 3 - (aq) + I - (aq) Br - (aq) + I 2 (aq) O.N. increased by 1 2. BrO 3 - (aq) + I - (aq) Br - (aq) + I 2 (aq) O.N. decreased by 6 Example

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 8 The END