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Oxidation/Reduction Reactions

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 .

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Oxidation/Reduction Reactions

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  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. • 2Zn(s) + O2(g)  2ZnO(s) is an example of a redox reaction, watch the following video to see what happens to the Zn metal and the oxygen gas. • 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. • In the previous example Zn loses two electrons, Zn(s)  Zn2+ + 2e- • Reduction • Early chemists thought of reduction as the loss of oxygen from a compound. • Now we refer to reduction as the gain of electrons. • In the previous example O must gain the electrons that Zn loses, O + 2e- O2-

  3. Redox reactions without oxygen: • Mg(s) + S(s)  MgS(s) • This reaction can be broken down into two steps: • Mg  Mg2+ + 2e-(loss of electrons: oxidation) • S + 2e-  S2- (gain of electrons: reduction) • H2(g) + Cl2(g)  2HCl(g) • Hydrogen and chlorine undergo the following changes to form a hydrogen chloride molecule: • H  H+ + e- (loss of electrons: oxidation) • Cl + e-  Cl- (gain of electrons: reduction)

  4. “LEO the lion says GER” Elements can be oxidized or reduced depending on the reaction. Observe Iron in the following reactions: • Oxidation • 4Fe(s) + 3O2(g)  2Fe2O3(s) - combustion or rusting • Fe  Fe3+ + 3e- - Loss of Electrons - Oxidation • Reduction • 2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g)-loss of oxygen • Fe3+ + 3e-  Fe -Gain of Electrons - Reduction

  5. Oxidation and Reduction must occur simultaneously: • If one element is oxidized then another must be reduced! • Example: 2AgNO3(aq) + Cu(s)  2Ag(s) + Cu(NO3)2(aq) • Net Ionic: 2Ag+ + Cu  2Ag + Cu2+ • Nitrates ions are spectator ions! • Half Reactions: • 2Ag+ + 2e- 2Ag Reduction • Cu  Cu2+ + 2e- Oxidation • These reactions occur simultaneously and the electrons lost must equal the electrons gained. • http://www.youtube.com/watch?v=a6RR4kPsnlE&feature=related

  6. When silver tarnishes it forms a black silver sulfide compound according to the following reaction: • 2Ag(s) + S(s)  Ag 2 S(s) • Observe the following video and write the oxidation and reduction half reactions: http://www.youtube.com/watch?v=a6RR4kPsnlE&feature=related • Ag  Ag+ + 1e- (oxidation) • S + 2e- S2- (reduction)

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

  8. Oxidation Numbers: • In order to recognize which reactant is oxidized and which is reduced it is important to be able to follow the movement of the electrons. • Oxidation numbers allow us to do this. An oxidation number is the real or apparent charge of an ion or an atom when we consider the bonds to be ionic. • Now learn the rules so that you are able to assign the oxidation number to any element!!!

  9. Oxidation Numbers: • Elements always have an oxidation number of zero. • K -K is 0 • N2 -N is 0 • Oxygen is always -2 unless it is in a peroxide and then it is -1. • HNO3 - O is -2 • H2O2- O is -1 • The charge on a monatomic ion is the oxidation number. • Fe3+ is +3 • Br1- is -1 • Hydrogen is always +1 unless it is in a metal hydride and then it is -1. • HNO3 - H is +1 • CaH2 - H is -1

  10. Determining Oxidation Numbers • K2CrO4 • O is -2 • K is +1 • Cr must be +6 • Cr2O3 • 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. • NO31- • O is -2 • N is +5

  11. Examples: • H is +1 because the molecule is neutral, S must be -2 • O is -2, there are two O in the molecule therefore S must be +4 • O is -2, there are three O in the ion therefore S must be +4 to give an overall charge of 2- • Determine the oxidation number of sulfur in the following molecules: • H2S • SO2 • SO32-

  12. Use of Oxidation Numbers: • In a redox reaction, one element loses electrons, while another gains them. Oxidation numbers allow you to keep track of these electron transfers and also allow you to determine whether a reaction is a redox reaction or not! • The following three reactions take place when coal with impurities of sulfur is burned. The result is the production of acid rain! • For each of these reactions determine the oxidation numbers of each element and then determine if the reaction is a redox reaction or not.

  13. 1. • S + O2 SO2 • ___ ____ ____ ,_____ • Redox? _______ • 2. • 2SO2 + O2 2SO3 • ___ ,____ ____ ____ ,____ • Redox? ________

  14. 3. • SO3 + H2O  H2SO4 • __, __ __ ,__ __, __, __ • Redox? _____ • 1. 0 0 +4 ,-2 Sulfur is oxidized, oxygen reducedREDOX • 2. +4, -2 0 +6, -2 Sulfur is oxidized, oxygen is reduced  REDOX. • 3. +6, -2 +1 , -2 +1, +6, -2 No change in the oxidation numbers of the elements therefore reaction number three is not a REDOX reaction.

  15. Reducing – Oxidizing Agents • Mg(s) + S(s)  MgS(s) • 0 0 +2,-2 (oxidation numbers) • Reducing Agent • The substance that loses electrons. • Mg is oxidized in the previous reaction but it is the REDUCING AGENT. • Oxidizing Agent • The substance that gains electrons. • S is reduced in the previous reaction but it is the OXIDIZING AGENT.

  16. Lead contamination in drinking water can result in brain and nervous system damage. If tap water is slightly acidic it can dissolve the lead solder in pipes by the following reaction: • Pb(s) + 2H+(aq)  Pb2+(aq) + H2(g) • In this case the hydrogen ions from the acidic tap water oxidize the lead, while the lead reduces the hydrogen ions.

  17. Reactivity of Metals • Metals (M) vary in their reactivity with acids (HA). • M(s) + 2HA(aq)  H2(g) + MA(aq) • The metal is oxidized, the hydrogen is reduced, the metal dissolves and hydrogen gas is produced. • The metal is considered to be the reducing agent.

  18. Complex ions or neutral compounds may also be oxidizing and reducing agents. Consider the reaction that occurs when nitrogen dioxide gas bubbles in a solution containing silver ions. • Ag+(aq) + NO2(g) + H2O(l)  Ag(s) + NO3-(aq) +2H+(aq) • +1 +4,-2 +1, -2 0 +5, -2 +1 • Silver is reduced and nitrogen in the nitrogen dioxide is oxidized. • The oxidation numbers of oxygen and hydrogen do not change. • NO2 is the reducing agent. Silver ions are the oxidizing agent.

  19. Use the oxidation numbers to identify the oxidizing and reducing agents in the following reactions: • A) CH4 + 2 O2 CO2 + 2H2O • B) Pb + H2SO4  H2 + PbSO4 • C) 2Mg + CO2  C + 2MgO • Oxidizing Agent Reducing Agent • A) O2 CH4 • B) H2SO4Pb • C) CO2 Mg

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

  21. Balancing Redox Equations: • Many redox equations can be balanced through trial and error! You have been doing this for a least two years without even knowing it! : ) • Do the following: • ___Al + ____HCl ____ AlCl3 + ___ H2 • 2 6 2 3 • Redox reactions occurring in aqueous solutions are often far more complicated than this!

  22. Steps for Balancing: • Step 1: Identify and write the two half-reactions. • Step 2: Balance the elements and charges for each half-reaction: • 2 a) Balance all the elements except hydrogen and oxygen. • 2b) Balance the oxygen atoms by adding H2O to the appropriate side. • 2c) Balance the hydrogen by adding H+ to the appropriate side. • 2d) Balance the charges by adding electrons to the appropriate side. Step 3: Multiply one or both of the half-reactions by a whole number so that the number of electrons gained and lost is equal. Step 4: Combine the half-reactions. Eliminate anything common to the product and reactant sides. Step 5: Check that all of the elements are balanced and that the total charge on each side is the same.

  23. Example: • Balance: MnO4- + Fe2+  Mn2+ + Fe3+ • Step 1: • Reduction: MnO4-  Mn2+ Oxidation: Fe2+  Fe3+ +7, -2 + 2 +2 +3 • Step 2: • b) MnO4-  Mn2+ + 4H2O • c) MnO4-+ 8H+ Mn2+ + 4H2O • d) MnO4-+ 8H+ + 5e- Mn2+ + 4H2O

  24. Oxidation: • Fe2+  Fe3+ +1e- • Step 3: • MnO4-+ 8H+ + 5e- Mn2+ + 4H2O • (Fe2+  Fe3+ +1e-) x5

  25. Step 4: • Reduction: MnO4-+ 8H+ + 5e- Mn2+ + 4H2O • Oxidation: 5Fe2+ 5 Fe3+ +5e- • Net Reaction: MnO4-+ 8H+ +5Fe2+ Mn2+ + 4H2O +5 Fe3+ • Step 5: • Notice that the final redox equation is balance by atom and by charge.

  26. 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=-B3RWeC_7oI • http://www.youtube.com/watch?v=TBmwhTzc41o&feature=fvwrel

  27. Balancing Redox Reactions: • Example: • Sn + Ag+ Sn2+ + Ag • 1. Assign oxidation numbers. • 0 +1 +2 0 • Sn + Ag+ Sn2+ + Ag

  28. 2. Oxidation occurs when the oxidation number increases. Reduction occurs when the oxidation number decreases. Write the two half reactions. • Oxidation: Sn Sn2+ • 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.

  29. Oxidation: Sn Sn2+ + 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 Sn2+ + 2e- • Reduction: (Ag+ + 1e- Ag) x2

  30. 5. Add the half reactions: • Oxidation: Sn Sn2+ + 2e- • Reduction: 2Ag+ + 2e-  2Ag • Net Balanced Redox Reaction: • Sn + 2 Ag+ Sn2+ + 2Ag

  31. 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 H2O and H+): • HNO3 + Fe2+ Fe3+ + NO2 • 1. Assign oxidation numbers. • +1+5-2 +2 +3 +4 -2 • HNO3 + Fe2+ Fe3+ + NO2

  32. 2. Write the half reactions: • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3 NO2 • 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+ Fe3+ • Reduction: HNO3 NO2 + H2O

  33. b) • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3 + H+  NO2 + H2O • c) • Oxidation: Fe 2+ Fe3+ + 1e- • Reduction: HNO3 + H+ + 1e- NO2 + H2O

  34. 4. Multiply the oxidation and reduction equations by whole numbers so that the number of electrons transferred is equal. • Oxidation: Fe 2+ Fe3+ + 1e- • Reduction: HNO3 + H+ + 1e- NO2 + H2O • 5. Add the reactions so that electrons cancel. • Net: Fe 2++ HNO3 + H+  Fe3+ + NO2 + H2O • 6. Check the equation is balance by charge and atom.

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

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