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Ch. 6 Oxidation-Reduction Reactions

Ch. 6 Oxidation-Reduction Reactions. Index. 5.1. Oxidation-reduction reactions involve electron transfer   5.2. The ion-electron method creates balanced net ionic equations for redox reactions   5.3. Metals are oxidized when they react with acids  

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Ch. 6 Oxidation-Reduction Reactions

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  1. Ch. 6 Oxidation-Reduction Reactions

  2. Index 5.1. Oxidation-reduction reactions involve electron transfer   5.2. The ion-electron method creates balanced net ionic equations for redox reactions   5.3. Metals are oxidized when they react with acids   5.4. A more active metal will displace a less active one from its compounds   5.5. Molecular oxygen is a powerful oxidizing agent   5.6. Redox reactions follow the same stoichiometric principles as other reactions 5.1. Oxidation-reduction reactions involve electron transfer

  3. Involve two processes: Oxidation – the loss of electrons, and Reduction – the gain of electrons Ca(s) + 2 H2O(l)→ Ca(OH)2(s) + H2(g) Oxidizer = Oxidizing agent = received the electrons and is reduced during the reaction Reducer = Reducing agent = donated the electrons and is oxidized during the reaction Oxidation –Reduction Reactions 5.1. Oxidation-reduction reactions involve electron transfer

  4. Guidelines For Redox Reactions • Oxidation and reduction always occur together • Total number of e- lost by one substance is the same as the total number of e- gained by the other • For a redox reaction to occur, something must accept the e- that are lost by another substance 5.1. Oxidation-reduction reactions involve electron transfer

  5. Oxidation Numbers • Oxidation number: • of any free element is zero • of any simple, monoatomic ion is equal to the charge on the ion • Fluorine in compounds is –1 • Group 1A in compounds is +1 • Oxygen in compounds is -2 • the sum of all oxidation numbers of the atoms in a molecule or polyatomic ion must equal the charge on the particle 5.1. Oxidation-reduction reactions involve electron transfer

  6. If there is a conflict between two rules apply the rule listed first Note that fractional values of oxidation numbers are allowed Special Rules 5.1. Oxidation-reduction reactions involve electron transfer

  7. Assign Oxidation States To All Atoms: • Fe2O3 • O: -2; Fe: +3 • Na2CO3 • Na: +1; O: -2; C: +4 • V(OH)3 • O: -2; H: +1; V: +3 • K2Cr2O7 • K: +1; O: -2; Cr: +6 5.1. Oxidation-reduction reactions involve electron transfer

  8. Your Turn! What is the oxidation number of F atoms in F2O? • -1 • -2 • -3 • -4 • none of these 5.1. Oxidation-reduction reactions involve electron transfer

  9. Your Turn! What is the oxidation number of P in H3PO3? • +1 • +2 • +3 • +4 • none of these 5.1. Oxidation-reduction reactions involve electron transfer

  10. Your Turn! What is the oxidation number of O atoms inH2O2? • -1 • -2 • -3 • -4 • none of these 5.1. Oxidation-reduction reactions involve electron transfer

  11. Determining if a Redox reaction is occurring • If the oxidation numbers change, a redox reaction is occurring • Increasing oxidation # denotes oxidation • Decreasing oxidation # denotes reduction • If the oxidation numbers remain static, no redox is occurring 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  12. Balancing Redox Reactions: Ion-Electron Method • Identify the half-reactions • Balance each atom in the half reaction, saving H and O for last • Balance O by adding 1 water molecule for each needed O • Balance H by adding 1 H+ ion for each needed H 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  13. Balancing Redox Reactions: (Cont.) • Balance charges by adding electrons to the more positive side • Find the least common multiple of electrons for the two half- reactions. Multiply each reaction by the factor needed to achieve the LCM of electrons • Add the half reactions, canceling like substances that appear on both sides 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  14. Balancing Basic Reactions • The simplest way to balance reactions in basic solution is to first balance them as if they were in acidic solution, then “convert” to basic solution • Additional Steps for Basic Solutions 8) To both sides of the equation, add the same number of OH- ions as there are H+. 9) Combine H+ and OH- to form H2O 10) Cancel H2O molecules that are on both sides of the reaction. 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  15. Balance, Using the Ion-Electron Method • MnO4- + C2O42-→ MnO2 + CO32- (acidic) • MnO4- + 4H+ +3e-→ MnO2 + 2H2O • C2O42- + 2H2O → 2CO32- + 4H+ +2e- • 2MnO4- + 8H+ +6e-+ 3C2O42- + 6H2O → 2MnO2 + 4H2O + 6CO32- + 12H+ +6e- • 2MnO4- + 3C2O42- + 2H2O → 2MnO2 + 6CO32- + 4H+ • ( )×2 • ( )×3 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  16. Balance, Using the Ion-Electron Method • Mn(s) + Cu2+(aq)→ Cu(s) + MnO2(s) (basic) • Mn + 2H2O → MnO2 + 4H+ + 4e- • Cu2+(aq) + 2e-→ Cu(s) • Mn + 2H2O + 2Cu2+(aq) + 4e-→ MnO2 + 4H+ + 4e-+ 2Cu(s) • Mn + 2H2O + 2Cu2+(aq)→ MnO2 + 4H+ + 2Cu(s) • Mn + 2H2O + 2Cu2+(aq) +4OH-(aq)→ MnO2 + 4H+ + 4OH-(aq)+ 2Cu(s) • Mn + 2Cu2+(aq) +4OH-(aq)→ MnO2 + 2H2O (aq)+ 2Cu(s) • ( )×1 • ( )×2 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  17. Balance, Using the Ion-Electron Method • H2SO3(aq) + H2CO3(aq) →H2C2O4(aq) + SO42- (aq) • H2SO3(aq) + H2O → SO42- (aq)+ 4H+(aq) + 2e- • 2 H2CO3(aq) + 2 H+ + 2e- →H2C2O4(aq) + 2H2O • H2SO3(aq) + 2 H2CO3(aq) →H2C2O4(aq) + SO42- (aq) + H2O + 2 H+ 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  18. Balance, Using the Ion-Electron Method • ClO- + VO3-(aq) →ClO3-(aq)+ V(OH)3(s) (basic) • ClO- + 2H2O →ClO3-(aq) + 4H+ + 4e- • 3H+ + 2e- + VO3- (aq) → V(OH)3(s) • ClO- + 2H2O + 2H+ + 2VO3- (aq) → ClO3-(aq) + 2V(OH)3(s) • ClO- + 4H2O + 2VO3- (aq) → ClO3-(aq) + 2V(OH)3(s) + 2OH- 5.2 The ion–electron method creates balanced net ionic equations for redox reactions

  19. A + BC → AC + B Metal A can replace metal B if it is a more active metal, or Nonmetal A can replace nonmetal C if it is more active than C. An activity series arranges metals according to their ease of oxidation (Table 5.2) Single Displacement (Replacement) Reaction 5.4 A more active metal will displace a less active one from its compounds

  20. Activity Series 5.4 A more active metal will displace a less active one from its compounds

  21. Interactions between different metals: Galvanic Cells 5.1. Oxidation-reduction reactions involve electron transfer

  22. Interactions between different metals: Galvanic Corrosion 5.1. Oxidation-reduction reactions involve electron transfer

  23. Interactions between different metals: Sacrificial Anodes 5.1. Oxidation-reduction reactions involve electron transfer

  24. Your Turn! Which of the following will react with Cu(s)? • HNO3(conc) • ZnCl2(aq) • AgNO3(aq) • None of these 5.4 A more active metal will displace a less active one from its compounds

  25. Oxygen Reacts With Many Substances The products depend, in part, on how much oxygen is available • Combustion of hydrocarbons • O2 plentiful; CH4 + 2O2→ CO2 + 2H2O • O2 limited: 2CH4 + 3O2 → 2 CO + 4H2O • O2 scant: CH4 + O2 → C + 2H2O • Organic compounds containing O also produce carbon dioxide and water • C2H5OH+ 3O2→ 2CO2 + 3H2O • Organic compounds containing S produce sulfur dioxide • 2C2H5SH + 9O2→4CO2 + 6H2O + 2 SO2 5.5 Molecular oxygen is a powerful oxidizing agent

  26. Other Oxygen Reactions • Organic compounds containing S produce sulfur dioxide • 2C2H5SH + 9O2→4CO2 + 6H2O + 2 SO2 • Many metals corrodeor tarnish when exposed to oxygen • 4Fe + 3O2→2Fe2O3 • 4Ag + O2 →2Ag2O • Most nonmetals react with oxygen directly • Plentiful: C + O2→CO2 • Limited: 2C + O2 →CO 5.5 Molecular oxygen is a powerful oxidizing agent

  27. Your Turn! What is the coefficient on O2 when octane, C8H18 is combusted with scant oxygen? • 1 • 2 • 3 • 4 • none of these 2 C8H18 + 9 O2→16 C + 18 H2O 5.5 Molecular oxygen is a powerful oxidizing agent

  28. Ore Analysis A 0.3000 g sample of tin ore (containing tin (II)) is titrated with 8.08 mL of 0.0500 M KMnO4 to oxidize the all the tin(II) to tin(IV) and reducing the MnO4- to MnO2 under acid conditions. What was the percentage tin in the original sample? 5.6 Redox reactions follow the same stoichiometric principles as other reactions

  29. Your Turn! A 25.0 g sample of granite contains a vein of copper. What is the % of Cu present if 25.00 mL of concentrated 15 M HNO3 are reacted? The Cu is converted to Cu2+ and the NO3- ions are converted to NO2. a) 12 b) 38 c) 48 d) 95 e) none of these 47.7 % 5.6 Redox reactions follow the same stoichiometric principles as other reactions

  30. Key Points About Redox Reactions • Oxidation (electron loss) always accompanies reduction (electron gain). • The oxidizing agent is reduced, and the reducing agent is oxidized. • The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent in the balanced equation.

  31. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Figure 21.1 A summary of redox terminology. OXIDATION One reactant loses electrons. Zn loses electrons. Reducing agent is oxidized. Zn is the reducing agent and becomes oxidized. Oxidation number increases. The oxidation number of Zn increases from 0 to +2. REDUCTION Other reactant gains electrons. Hydrogen ion gains electrons. Oxidizing agent is reduced. Hydrogen ion is the oxidizing agent and becomes reduced. Oxidation number decreases. The oxidation number of H decreases from +1 to 0.

  32. Half-Reaction Method for Balancing Redox Reactions • Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. • Each reaction is balanced for mass (atoms) and charge. • One or both are multiplied by some integer to make the number of electrons gained and lost equal. • The half-reactions are then recombined to give the balanced redox equation. • Advantages: • The separation of half-reactions reflects actual physical separations in electrochemical cells. • The half-reactions are easier to balance especially if they involve acid or base. • It is usually not necessary to assign oxidation numbers to those species not undergoing change.

  33. Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) Cr2O72- Cr3+ I-I2 Cr2O72- Cr3+ 6e- + 14H+(aq) + Cr2O72- Cr3+ 2 + 7H2O(l) Balancing Redox Reactions in Acidic Solution Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) 1. Divide the reaction into half-reactions - Determine the O.N.s for the species undergoing redox. +6 -1 +3 0 Cr is going from +6 to +3 I is going from -1 to 0 2. Balance atoms and charges in each half-reaction - 14H+(aq) + 2 + 7H2O(l) net: +6 Add 6e- to left. net: +12

  34. 2 + 2e- X 3 I-I2 I-I2 I-I2 6e- + 6e- + 14H+ + Cr2O72- Cr3+ 2 + 7H2O(l) 6 3 + 6e- 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) 14H+(aq) + Cr2O72- Cr3+ + 7H2O(l) Balancing Redox Reactions in Acidic Solution continued 2 2 + 2e- Cr(+6) is the oxidizing agent and I(-1)is the reducing agent. 3. Multiply each half-reaction by an integer, if necessary - 4. Add the half-reactions together - Do a final check on atoms and charges.

  35. 14H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH- 7H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 14OH- 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) Balancing Redox Reactions in Basic Solution Balance the reaction in acid and then add OH- so as to neutralize the H+ ions. + 14OH-(aq) + 14OH-(aq) Reconcile the number of water molecules. Do a final check on atoms and charges.

  36. Cr2O72- I- Figure 21.2 The redox reaction between dichromate ion and iodide ion. Cr3+ + I2

  37. PROBLEM: Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution: MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq) C2O42- CO32- MnO4- MnO2 MnO4- MnO2 C2O42- CO32- 2 MnO4- MnO2 C2O42-2CO32- Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method PLAN: Proceed in acidic solution and then neutralize with base. SOLUTION: +7 +4 +3 +4 4H+ + + 2H2O + 2H2O + 4H+ +3e- +2e-

  38. 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- C2O42- + 2H2O 2CO32- + 4H+ + 2e- C2O42- + 2H2O 2CO32- + 4H+ + 2e- 4H+ + MnO4- +3e- MnO2+ 2H2O 4H+ + MnO4- +3e- MnO2+ 2H2O 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O X 3 X 2 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 2MnO2(s) + 6CO32-(aq) + 4H+(aq) 2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 2MnO2(s) + 6CO32-(aq) + 2H2O(l) Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method continued: + 4OH- + 4OH-

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