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Chapter 20.1

Chapter 20.1. Redox Reactions Oxidation & Reduction . Do Mg burning!. Review of Types of Reactions. *Synthesis/Combination **ALWAYS REDOX A + B  AB * SOMETIMES *Decomposition AB  A + B **Single-replacement AX + B  A + BX Double-replacement AX + BY  AY + BX **Combustion

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Chapter 20.1

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  1. Chapter 20.1 Redox Reactions Oxidation & Reduction Do Mg burning!

  2. Review of Types of Reactions • *Synthesis/Combination **ALWAYS REDOX • A + B  AB * SOMETIMES • *Decomposition • AB  A + B • **Single-replacement • AX + B  A + BX • Double-replacement • AX + BY  AY + BX • **Combustion • Makes heat/light

  3. REDOX! • Are there changes in the “charges” from one side of the equation to the other? • Then it can be called a REDOX reaction!

  4. Review: Oxidation Numbers (Charges) • Get out Ion Sheet & Activity Series • Go to PAGE11 in Workbook—Chapter 3 • Go over RULES for giving “oxidation numbers” • Assign “oxidation numbers” to EACH element in an equation.

  5. OXIDATION RULES • The oxidation number of any element in it’s free state (uncombined with other elements) is zero! • The ox# of alkali metal is always 1+ and alkaline earth metals is 2+. • Metallic halides are always 1-. • Oxygen is almost always 2-. Hydrogen is 1+ or 1-. • Oxygen family is always 2- and Nitrogen family is 3-. • All others are assigned to make the charges in the molecule = ZERO! Let’s do #4 and #5 from book on back of worksheet. REDOX Worksheet—put #’s in.

  6. Why do the charges “change”? • Well, it’s the electrons of course!!!! When is it not? Silver Nitrate + Potassium Iodide

  7. Reduction-Oxidation Reactions A reaction in which electrons are transferred from one species to another. • Oxidation means the loss of electrons (LEO) (Used to refer ONLY to reactions when a substance combine with oxygen—burning) • Reduction means the gain of electrons (GER) Let’s do an example…..,but first…….

  8. LEO the lion says GER

  9. LET’S TAKE A CLOSER LOOK… Equation: 2Na(s) + Cl2(g)  2NaCl(s) Net Ionic: 2Na(s) + Cl2(g)  2Na+ + 2Cl-

  10. When magnesium reacts with oxygen, each magnesium atom transfers two electrons to each oxygen atom.

  11. A REDOX EXAMPLEWhat kind of reaction is this? Equation: 2Mg(s) + O2(g)  2MgO(s) Net Ionic: 2Mg(s) + O2(g)  2Mg2+ + 2O2- 2+ 2- 0 0 0 0

  12. Changes in Oxidation Number Equation: 2K + Br2 2KBr N. I. Equation: 2K + Br2  2K+ + 2Br- • K is OXIZED from 0 to +1 state --- LOSS e- • Br is REDUCED From 0 to -1 state ---GAIN e- LEO the Lion says GER Loss of electrons = Oxidation Gain of electrons = Reduction

  13. WHAT CHARGE DOES AN ELECTRON HAVE? Equation: 2KBr(aq) + Cl2(aq)  2KCl(aq) + Br2(aq) Net Ionic Equation: 2Br-(aq) + Cl2(aq)  2Cl-(aq) + Br2(aq) Gained an e- Lost an e- e- e- 0 0

  14. Gaining and Losing ElectronsSome Confusing Vocabulary • If an electron is lost, that element is oxidized (LEO) • If an electron is gained, the element is reduced (GER) Equation: 2KBr(aq) + Cl2(aq)  2KCl(aq) + Br2(aq) Net Ionic Equation: 2Br-(aq) + Cl2(aq)  2Cl-(aq) + Br2(aq) - + - 0 + 0 0 0 REDUCED? What’s OXIZED &

  15. Chapter 20.1 Redox Reactions Oxidation & Reduction

  16. LEO the lion says GER

  17. When magnesium reacts with oxygen, each magnesium atom transfers two electrons to each oxygen atom.

  18. Gaining and Losing ElectronsREVIEW: Some Confusing Vocabulary • If an electron is Lost, that element is Oxidized. • BUT THAT ELEMENT IS ALSO THE “REDUCING AGENT” • The part that is oxidized is the Reducing Agent • If an electron is Gained, the element is Reduced. • BUT THAT ELEMENT IS ALSO THE “OXIDIZING AGENT” • The part that is reducedis the Oxidizing Agent • They are opposites! Similar to conjugates

  19. REAL-LIFE EX OF REDOX • Hydrogen Peroxide on a wound • Silver Tarnishing • Chlorine bleach to whiten laundry • Photography

  20. ANOTHER REDOX EXAMPLEWhat kind of reaction is this? • Rusting is a Redox reaction! 4Fe (s) + 3O2 (g)  2Fe2O3 (s) As a reactant,IRONhas a zero oxidation state but as a product (in Fe2O3) iron has a 3+ oxidation state. Three electrons per atom had to be transferred (LOST) in order for this to happen. Note that OXYGENalso changed from a zero oxidation state to a 2- oxidation state. Oxygen needed toGain 2 electrons per atom. 3+ 2- 0 0

  21. EXAMPLE-Study Guide: Equation: 2Cd(s) + NiO(s)  CdO(s) + Ni(s) Net Ionic Equation: 2Cd(s) + NiO(s)  CdO(s) + Ni(s) 2- 2- 2+ 2+ 0 0 What’s OXIZED & REDUCED? Who’s the RA & OA?

  22. Chapter 20.2 Balancing Redox Reactions

  23. EXAMPLES • Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l) • See Handout • From Study Guide • Zn + MnO2 + H2O  Zn(OH)2 + Mn2O3 Do #12-14 in book with whiteboards—One on each Notecard— In Groups-1 person do it normal-then compare

  24. 20.2 Balancing Redox Equations • Oxidation-Number Method • STEPS to Balance • Write the Basic Equation. • Assign Oxi # to everything. • ID what was oxidized and reduced. • Write the equation for the oxidized substance and an equation for the reduced substance. • Indicate how many electrons were lost in oxidation and how many gained in reduction. • Use a least common multiple to get the e- equal. • Enter the coefficients back into the original equation. • Balance the rest of the elements. Give Handout with “Balancing Red-Ox Equations”

  25. Chapter 20.3 Half Reactions

  26. 20.3 Half Reactions • Half Reaction: One of the two parts to a REDOX reaction • Species: Any kind of chemical involved in a process (so we don’t have to specify what kind of particle)

  27. Balancing with Half-Reactions • See the steps on page 651 • Use half reactions and balance charges and atoms. • IF IN ACID: Add H20 & H+ where needed to balance. • IF IN BASE: Add H2O and OH- where needed to balance. • Add the two reactions back together If DONE, do #15-17 & #24-26

  28. REDOX REACTION EXAMPLE What “type” of reaction is this? A reaction occurs when a piece of zinc is dipped into a solution of copper (II) nitrate. Write the conventional (molecular), ionic and net ionic equations. Fig. 9-9, p. 245

  29. REDOX REACTION EXAMPLESingle-Replacement Redox Reaction • Conventional (Molecular) Equation Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s) • Ionic Equation Zn(s) + Cu+2(aq) + 2NO3-(aq) → Zn+2(aq) + 2NO3-(aq) + Cu(s) • Net Ionic Equation Zn(s) + Cu+2(aq) → Zn+2(aq)+ Cu(s) 0 0

  30. What if we did the opposite? • Cu(s) + Zn(NO3)2(aq)→ • NR!!! • Cu(s) + Zn(NO3)2(aq)→ NR

  31. Cu (s) + 2AgNO3 (aq)  2Ag (s) + Cu(NO3)2 (aq)

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