1. Oxidation/Reduction Reactions REDOX REACTONS! All chemical reactions fall into two categories those that are redox and those that are not redox! Redox reactions involve the transfer of electrons. http://www.youtube.com/watch?v=e6Xxz-VBE6s

2. What is oxidation and reduction? Oxidation Early chemists thought of oxidation as the combining of an element with oxygen, as in combustion. Now we refer to oxidation as the loss of electrons. Reduction Early chemists thought of reduction as the loss of oxygen from a compound. Now we refer to reduction as the gain of electrons.

3. “LEO the lion says GER” Oxidation 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) - combustion or rusting Fe  Fe 3+ + 3e - - Loss of Electrons - Oxidation Reduction 2Fe 2 O 3 (s) + 3C(s)  4Fe(s) + 3CO 2 (g)-loss of oxygen Fe 3+ + 3e -  Fe-Gain of Electrons - Reduction

4. Oxidation and Reduction must occur simultaneously: Example:2AgNO 3 (aq) + Cu(s)  2Ag(s) + Cu(NO 3 ) 2 (aq) Net Ionic:2Ag + + Cu  2Ag + Cu 2+ Nitrates ions are spectator ions! Half Reactions: 2Ag + + 2e -  2AgReduction Cu  Cu 2+ + 2e - Oxidation These reactions occur simultaneously and the electrons lost must equal the electrons gained.http://www.youtube.com/watch?v=a6RR4kPsnlE&feature=relat edhttp://www.youtube.com/watch?v=a6RR4kPsnlE&feature=relat ed

5. Reducing – Oxidizing Agents Reducing Agent The substance that loses electrons. Cu is oxidized in the previous reaction but it is the REDUCING AGENT. Oxidizing Agent The substance that gains electrons. Ag + is reduced in the previous reaction but it is the OXIDIZING AGENT.

6. Chapter 20.1- Read pages 631  638 Complete question #1, #2 page 634 Complete question #3  #8 page 638

7. Oxidation Numbers: Elements always have an oxidation number of zero. K -K is 0 N 2 -N is 0 Oxygen is always -2 unless it is in a peroxide and then it is -1. HNO 3 - O is -2 H 2 O 2 - O is -1 The charge on a monatomic ion is the oxidation number. Fe 3+ is +3 Br 1- is -1 Hydrogen is always +1 unless it is in a metal hydride and then it is -1. HNO 3 - H is +1 CaH 2 - H is -1

8. Determining Oxidation Numbers K 2 CrO 4 O is -2 K is +1 Cr must be +6 Cr 2 O 3 O is -2 Cr must be +3 Note that elements can have more than one oxidation number. The sum of oxidation numbers in a neutral compound must be zero. The sum of oxidation numbers in a polyatomic ions must equal the charge on the ion. NO 3 1- O is -2 N is +5

9. Chapter 20.2-Read pages 639  643 Page 641 #9 & # 10 http://www.youtube.com/watch ?v=EHe8-AFMsMA http://www.youtube.com/watch ?v=EHe8-AFMsMA Page 643 #13  #16

10. Balancing Redox Reactions Watch the following videos to help you balance redox equations: http://www.youtube.com/watch?v=KIyGr- 1snMY&feature=relmfu http://www.youtube.com/watch?v=KIyGr- 1snMY&feature=relmfu http://www.youtube.com/watch?v=-B3RWeC_7oI http://www.youtube.com/watch?v=TBmwhTzc41o &feature=fvwrel http://www.youtube.com/watch?v=TBmwhTzc41o &feature=fvwrel

11. Balancing Redox Reactions: Example: Sn+ Ag +  Sn 2+ + Ag 1.Assign oxidation numbers. 0 +1 +2 0 Sn+ Ag +  Sn 2+ + Ag

12. 2. Oxidation occurs when the oxidation number increases. Reduction occurs when the oxidation number decreases. Write the two half reactions. Oxidation: Sn  Sn 2+ Reduction: Ag +  Ag 3. Use electrons to balance the charges in the half reactions. In oxidation the electrons appear on the right. In reduction the electrons appear on the left.

13. Oxidation: Sn  Sn 2+ + 2e - Reduction: Ag + + 1e -  Ag 4. If the number of electrons transferred is not equal multiple by a whole number so that the number of electrons lost equals the number gained. Oxidation: Sn  Sn 2+ + 2e - Reduction: (Ag + + 1e -  Ag) x2

14. 5. Add the half reactions: Oxidation: Sn  Sn 2+ + 2e - Reduction: 2Ag + + 2e -  2Ag Net Balanced Redox Reaction: Sn + 2 Ag +  Sn 2+ + 2Ag

15. Redox reactions are usually too complex to use trial and error method. Balance the following example that is in an acidic solution (assume the presence of H 2 O and H + ): HNO 3 + Fe 2+  Fe 3+ + NO 2 1.Assign oxidation numbers. +1+5-2 +2 +3 +4 -2 HNO 3 + Fe 2+  Fe 3+ + NO 2

16. 2.Write the half reactions: Oxidation: Fe 2+  Fe 3+ Reduction: HNO 3  NO 2 3. Balance the half reactions. Use water to balance the oxygen (a), then hydrogen ions to balance the hydrogen(b), then electrons to balance the charges(c). a) Oxidation: Fe 2+  Fe 3+ Reduction: HNO 3  NO 2 + H 2 O

17. b) Oxidation: Fe 2+  Fe 3+ Reduction: HNO 3 + H +  NO 2 + H 2 O c) Oxidation: Fe 2+  Fe 3+ + 1e - Reduction: HNO 3 + H + + 1e -  NO 2 + H 2 O

18. 4.Multiply the oxidation and reduction equations by whole numbers so that the number of electrons transferred is equal. Oxidation: Fe 2+  Fe 3+ + 1e - Reduction: HNO 3 + H + + 1e -  NO 2 + H 2 O 5. Add the reactions so that electrons cancel. Net:Fe 2+ + HNO 3 + H +  Fe 3+ + NO 2 + H 2 O 6. Check the equation is balance by charge and atom.

19. Chapter 20.3- Read pages 645  654 Page 647 #17 & #18 Page 649 #19 & #20 Page 652 #21 Page 654 #22,23,24 & 25 CHAPTER 20 Review the study guide on page 656 Complete the odd numbered questions on page 658 Complete the Standardized Test Prep