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Chapter 19

Chapter 19. Oxidation-Reduction Reactions. Section 1: Oxidation and Reduction. Standard 3.g.: Students know how to identify reactions that involve oxidation and reduction and how to balance oxidation-reduction reactions. Objective:

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Chapter 19

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  1. Chapter 19 Oxidation-Reduction Reactions

  2. Section 1: Oxidation and Reduction • Standard 3.g.: • Students know how to identify reactions that involve oxidation and reduction and how to balance oxidation-reduction reactions. • Objective: • We will assign oxidation numbers to reactant and product species and explain what an oxidation-reduction reaction is.

  3. Oxidation States • The oxidation number assigned to an element in a molecule is based on the distribution of electrons in that molecule. • Rules for Assigning Oxidation Numbers are in Table 1 on page 631.

  4. Rules • The oxidation number of any pure element is 0. • The oxidation number of a monatomic ion equals that charge on the ion. • The more electronegative element in a binary compound is assigned the number equal to the charge it would have if it were an ion. • The oxidation number of fluorine in a compound is always -1.

  5. Oxygen has an oxidation number of -2 unless it is combined with F, in which it is +1 or +2, or it is in a peroxide, in which it is -1. • Hydrogen is +1, unless combined with a metal, then it is -1. • In compounds, Group 1 is +1, Group 2 is +2, and Aluminum is +3. • The sum of the oxidation numbers of all atoms in a neutral compound is 0. • The sum of the oxidation numbers in a polyatomic ion equals the charge of the ion.

  6. Oxidation States of Chromium • Orange – K2Cr2O7 Potassium Dichromate • Yellow – Na2CrO4 Sodium Chromate • Green – CrCl3 Chromium(III) Chloride • Violet – Cr(NO3)3 Chromium(III) Nitrate

  7. Oxidation • The processes in which the atoms or ions of an element experience an increase in oxidation state. 2Na(s) + Cl2(g)  2NaCl(s) • The formation of sodium ion illustrates an oxidation process because each sodium atom loses an electron to become a sodium ion. Na  Na+ + e- • Sodium is considered oxidized because it increased in oxidation number.

  8. Reduction • The processes in which the atoms or ions of an element experience a decrease in oxidation state. • The formation of a chloride ion illustrates a reduction process because each chlorine atom gains an electron to become a chloride ion. Cl2 + 2e-  2Cl- • Chlorine is considered reduced because it decreased in oxidation number.

  9. Oxidation and Reduction as a Process • Any chemical process in which elements undergo changes in oxidation number is an oxidation-reduction reaction or redox reaction for short. • The part of the reaction involving oxidation or reduction alone can be written as a half-reaction.

  10. “Leo the Lion says Ger” • LEO = Loss of Electrons Oxidation • GER = Gain of Electrons Reduction

  11. Examples Identify the following as Oxidation or Reduction: • K  K+ + e- • S + 2e-  S2- • Mg  Mg2+ + 2e- • 2F-  F2 + 2e- • O2 + 4e-  2O2- • Mn2+  MnO4- + 5e-

  12. Examples Identify the following as Oxidation or Reduction: • K  K+ + e- Oxidation • S + 2e-  S2- • Mg  Mg2+ + 2e- • 2F-  F2 + 2e- • O2 + 4e-  2O2- • Mn2+  MnO4- + 5e-

  13. Examples Identify the following as Oxidation or Reduction: • K  K+ + e- Oxidation • S + 2e-  S2- Reduction • Mg  Mg2+ + 2e- • 2F-  F2 + 2e- • O2 + 4e-  2O2- • Mn2+  MnO4- + 5e-

  14. Examples Identify the following as Oxidation or Reduction: • K  K+ + e- Oxidation • S + 2e-  S2- Reduction • Mg  Mg2+ + 2e- Oxidation • 2F-  F2 + 2e- • O2 + 4e-  2O2- • Mn2+  MnO4- + 5e-

  15. Examples Identify the following as Oxidation or Reduction: • K  K+ + e- Oxidation • S + 2e-  S2- Reduction • Mg  Mg2+ + 2e- Oxidation • 2F-  F2 + 2e- Oxidation • O2 + 4e-  2O2- • Mn2+  MnO4- + 5e-

  16. Examples Identify the following as Oxidation or Reduction: • K  K+ + e- Oxidation • S + 2e-  S2- Reduction • Mg  Mg2+ + 2e- Oxidation • 2F-  F2 + 2e- Oxidation • O2 + 4e-  2O2- Reduction • Mn2+  MnO4- + 5e-

  17. Examples Identify the following as Oxidation or Reduction: • K  K+ + e- Oxidation • S + 2e-  S2- Reduction • Mg  Mg2+ + 2e- Oxidation • 2F-  F2 + 2e- Oxidation • O2 + 4e-  2O2- Reduction • Mn2+  MnO4- + 5e- Oxidation

  18. Redox Reactions • In order for a reaction to be a redox reaction, the atoms in the reaction must change oxidation state. If they don’t, it is not a redox reaction. 0 0 1+ 1- • Redox: 2Na + Cl2 2NaCl 4+ 2- 1+ 2- 1+ 4+ 2- • Not Redox: SO2 + H2O  H2SO3

  19. Homework Chapter 19.1 pg 635 #2-4 and Oxidation Worksheet

  20. Section 2: Balancing Redox Equations • Standard 3.g.: • Students know how to identify reactions that involve oxidation and reduction and how to balance oxidation-reduction reactions. • Objective: • We will balance redox equations by using the half-reaction method

  21. Balancing Redox Equations • In a normal equation, when we balance, we are only conserving mass, we are not looking at charge. • In a Redox Equation, when properly balanced, both mass and charge are conserved.

  22. Half-Reaction Method Consists of 7 steps: • Write the formula equation, if not given, and then write the net ionic equation. • Assign oxidation numbers and delete substances containing only elements that do not change oxidation state. • Write the half-reaction for oxidation, balance atoms, and balance charges.

  23. Write the half-reaction for reduction, balance atoms, and balance charges. • Conserve charge by adjusting the coefficients in front of electrons so that the number lost equals the number gained. • Combine the half-reactions and cancel out anything common to both sides of the equation. • Combine ions to form compounds shown in the original formula equation. Check all ions are balanced.

  24. When doing ionic equations, do not break up the covalent compounds • Covalent compound consists of 2 or more non-metals!!!!

  25. Example • Copper reacts with hot, concentrated sulfuric acid to form copper(II) sulfate, sulfur dioxide, and water. Write and balance the equation for this reaction. Step 1: Write Formula and Ionic Equations Formula: Cu + H2SO4 CuSO4 + SO2+ H2O Ionic: Cu + 2H+ + SO42-  Cu2+ + SO42- + SO2 + H2O

  26. Step 2: Assign Oxidations Numbers and Delete 0 1+ 6+ 2- 2+ 6+ 2- 4+ 2- 1+ 2- Cu + 2H+ + SO42-  Cu2+ + SO42- + SO2 + H2O Delete things that don’t change oxidation state. Cu + 2H+ + SO42-  Cu2+ + SO42- + SO2 + H2O Cu + SO42-  Cu2+ + SO2 Step 3: Write Oxidation Half-Reaction Cu  Cu2+ Atoms are balanced, look at charges 0  2+ Cu  Cu2+ + 2e-

  27. Step 4: Write Reduction Half-Reaction SO42- SO2 Need to balance Oxygen, so add H2O as needed SO42-  SO2+ 2H2O Need to balance H now, add H+ as needed SO42- + 4H+  SO2+ 2H2O Now balance charges, 6+  4+ on Sulfur SO42- + 4H+ + 2e-  SO2+ 2H2O

  28. Step 5: Conserve charge by adjusting the coefficients in front of electrons so that the number lost equals the number gained. Electron charge are equal so SKIP! Step 6: Combine Half-Reactions and Cancel Cu  Cu2+ + 2e- SO42- + 4H+ + 2e-  SO2+ 2H2O Cu + SO42- + 4H+ + 2e-  Cu2+ + 2e- + SO2+ 2H2O

  29. Step 7: Combine ions to form compounds Cu + SO42- + 4H+  Cu2+ + SO2+ 2H2O Need a SO42- to bond with Cu2+, must add to both sides, which uses up the 2 extra H+ Cu + 2H2 SO4  Cu SO4 + SO2+ 2H2O Check all elements are balanced and then YOU ARE DONE!!!

  30. If a redox reaction is happening with a base rather than an acid, sometimes you will need to add H2O and OH- rather than H2O and H+

  31. Let’s do another one: K2Cr2O7 + HCl + C2H5OH  CrCl3 + CO2 + KCl + H2O

  32. Homework Chapter 19.2 pg 641 #1-3

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