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Chapter 19. Redox Equilibrium I: Redox Reactions. 19.1 Redox Reactions 19.2 Balancing Redox Equations. 19.1 Redox Reactions (SB p.172). Redox Reactions. (An 'imaginary charge'). 19.1 Redox Reactions (SB p.175). Rules for Assigning Oxidation Number.
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Chapter 19 Redox Equilibrium I: Redox Reactions 19.1 Redox Reactions 19.2 Balancing Redox Equations
19.1 Redox Reactions (SB p.172) Redox Reactions (An 'imaginary charge')
19.1 Redox Reactions (SB p.175) Rules for Assigning Oxidation Number • The oxidation number of an element is 0. • For a simple ionic compound, the oxidation number of a constituent element is the same as the charge on the ion. • The oxidation number of some elements in their compounds are always the same. • In an uncharged compound, the sum of oxidation numbers of all constituent atoms is 0.
19.1 Redox Reactions (SB p.176) • 5. In polyatomic ion, the sum of oxidation numbers of all constituent atoms is equal to the charge of the ion. • 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. • The oxidation number of an element can be different in different compounds.
19.1 Redox Reactions (SB p.172) Definition of oxidizing and reducing agents
19.2 Balancing Redox Equations (SB p.180) Step Reduction Oxidation 1 MnO4-(aq) Mn2+(aq) I -(aq) I2(aq) 2 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) Balance the no. of I atoms: 2I-(aq) I2(aq) 3 Balance the no. of charges: (with a charge of +2 on both sides) MnO4-(aq) + 8H+(aq) + e- Mn2+(aq) + 4H2O(l)…(1) Balance the no. of charges: (with a charge of –2 on both sides) 2I-(aq) I2(aq) + 2e-…(2) 4 Combine 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) Half Equation Method
19.2 Balancing Redox Equations (SB p.182) • BrO3-(aq) + I-(aq) Br-(aq) + I2(aq) • +5 -1 -1 0 O.N. increased by 1 2. BrO3-(aq) + I-(aq) Br-(aq) + I2(aq) +5 -1 -1 0 O.N. decreased by 6 3. 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) Oxidation Number Method Example