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Unit 7: Notes One

Unit 7: Notes One. Redox Oxidation Numbers Identifying what is oxidized and what is reduced Activity Series of Metals Chapter 17 Oxidation and Reduction. Oxidation Reduction Chemisty: Redox Chemistry. Oxidation and Reduction reactions always take place simultaneously.

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Unit 7: Notes One

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  1. Unit 7: Notes One Redox Oxidation Numbers Identifying what is oxidized and what is reduced Activity Series of Metals Chapter 17 Oxidation and Reduction

  2. Oxidation Reduction Chemisty: Redox Chemistry Oxidation and Reduction reactions always take place simultaneously. Loss of electrons – oxidation (Increase in Oxidation Number) Ex:Na ------> Na+1 + e-1 Gain of electrons - reduction ( Decrease in Oxidation Number) Cl2 + 2 e-1 ------> 2 Cl-1

  3. Redox: Reduction occurs when an atom gains one or more electrons. Ex: O + 2e-1 O2- Oxidation Foccurs when an atom or ion loses one or more electrons. Ex: Fe  Fe+3 + 3e-1 LEO goes GER Copper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO3)  2 Ag + Cu(NO3)2.

  4. Redox reactions involve electron transfer: • Lose e - =Oxidation • Cu (s) + 2 Ag+1 (aq) Cu +2 (aq) + 2 Ag(s) • Gain e - =Reduction

  5. Loses electrons Loses hydrogen Gains oxygen Oxidation occurs when a molecule does any of the following: If a molecule undergoes oxidation, it has been oxidized and it is the reducing agent (aka reductant).

  6. Reduction occurs when a molecule does any of the following: Gains electrons Gains hydrogen Loses oxygen If a molecule undergoes reduction, it has been reduced and it is the oxidizing agent (aka oxidant).

  7. zinc is being oxidized while the copper is being reduced. Why?

  8. Redox Burning: C6H12O6 + 6O2 6CO2 + 6H2O+ Heat Rusting Iron: 4Fe + 3O22Fe2O3 + Heat Oxidation - Loss of e-1. Na(s)Na+1 +1e-1 Reduction - Gain of e-1. Cl2+ 2e-1 2Cl-1 Number line (Oxidation…Left or right?)

  9. Oxidation #’s: Key Elements • (99%) H+1 H-1 (Hydrides, only with Metals) • (99%)O-2 O-1 (Peroxides) • (Always) Li+1, Na+1, K+1, Rb+1, Cs+1, Fr+1 • (Always) Be+2, Mg+2, Ca+2, Ba+2, Sr+2, Ra+2 • (Always) Al+3 • (with only a metal) F-1, Cl-1, Br-1, I-1 • (NO3-1) ion is always +5 • (SO4-2) ion is always +6

  10. Finding Oxidation Numbers The sum of the oxidation numbers must be equal to _____ for a compound. zero Find Ox#’s for H2O? +1 -2 H2O 2 (H)+ 1 (O) = Zero 2 (+1)+ 1 (-2) = Zero Find Ox#’s for H3PO4? -2 +1 +5 H3PO4 3 (H)+ 1 (P)+ 4 (O) = Zero 3 (+1)+ 1 (+5)+ 4 (-2) = Zero

  11. Identifying OX, RD, SI Species • Ca0 + 2 H+1Cl-1Ca+2Cl-12 + H20 • Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca0Ca+2, so Ca0 is the species that is oxidized. • Reduction = gain of electrons. The species becomes more negative in charge. For example, H+1H0, so the H+1 is the species that is reduced. • Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1 Cl-1, so the Cl-1 is the spectator ion.

  12. Oxidizing Agents and Reducing Agents: Oxidizing agent gets reduced itself and reducing agent gets oxidized itself, so a strong oxidizing agent should have a great tendency to accept e and a strong reducing agent should be willing to lose e easily. What are strong oxidizing agents- metals or non metals? Why? Spectator Ions are ions that do NOT change their oxidation number from the reactant side of a RXN to the product side of a RXN. They are just “hanging out”.

  13. Activity Series of Metals (in your packet) The activity series of metals is an empirical tool used to predict products in displacement reactions and reactivity of metals with water and acids in replacement reactions and ore extraction. It can be used to predict the products in similar reactions involving a different metal.The activity series is a chart of metals listed in order of declining relative reactivity. The top metals are more reactive than the metals on the bottom. For example, both magnesium and zinc can react with hydrogen ions to displace H2 from a solution by the reactions:Mg(s) + 2 H+(aq) → H2(g) + Mg2+(aq)Zn(s) + 2 H+(aq) → H2(g) + Zn2+(aq)Both metals react with the hydrogen ions, but magnesium metal can also displace zinc ions in solution by the reaction:Mg(s) + Zn2+ → Zn(s) + Mg2+

  14. If Metal is higher On chart than Ion RXN will happen

  15. Reduction: Cu+2(aq) + 2 e-  Cu(s) The Cation becomes a solid metal (the + charge GAINS ELECTRONS to become a zero charge. Oxidation: Cu(s) Cu+2(aq) + 2 e- The metal becomes a cation (the zero charge metal LOSES ELECTRONS To become a + charge.

  16. Redox: Oxidation Reduction Reaction 3Cu+2 (aq) + 2Fe (s) 3 Cu (s) + 2Fe+3 (aq) The paired reduction and oxidation Electrons transfer from the metal to the cation if the metal Is above (ie higher) on the Activity Series Chart in your packet. This reaction will occur because Fe (metal) is higher than Cu so the reaction will occur. 6 e-

  17. Cu+2 (aq) + Mg (s) Cu (s) + Mg+2 (aq) Mg is higher than Cu on the activity series Zn+2 (aq) + Ag (s) No RXN Because Ag is lower than Zn on The activity series

  18. Notes Two Unit Seven • Activity Series of Metals • How to use ½ RXN sheet • Electrolysis • Lab A-Electrolysis

  19. Activity Series of Metals ELA701 • Three chemical RXNS to make metals: • Heating only: 2PbO(s)  2Pb(s) + 1 O2(g) • Heating with Coke (Carbon) 2MnO3(s) + 3C(s)  4Mn(s) + 3CO2(g) • Electricity through molten oxide 2Li2O(s)  4Li(s) + 1O2(g)

  20. How to use the ½ Reaction Resource: • Notice they are all written as Reductions (gaining of electrons) • Find the highest one on the left hand side and write it in the forward direction. • 2. Find the lowest one on the right and write it backwards as an oxidation ,along with changing the sign of the voltage. • 3. Make sure to balance electrons lost and gained (you DO NOT MULTIPLY the voltages !!!) • 4. So you will have two ½ cell reactions and you can cancel electrons and write the WHOLE CELL RXN. • Let’s Practice????? X

  21. Stronger 1/2O2(g)+2H+(pH=7)+2e-H2O Weaker +1.23 Br2(l)+2e-2Br- +1.06 Oxidizing Agent Reducing Agent NO3-+4H++3e-NO(g)+2H2O +0.96 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Ag++e- Ag(l) +0.80 Hg2++2e- Hg(l) +0.78 NO3-+2H++e-NO2(g)+H2O +0.78 Fe3++e-Fe2+ +0.77 Gains e- I2(s)+2e-2I- +0.53 Loses e- +0.52 Cu++e-Cu(s) Cu2++2e-Cu(s) +0.34 SO42-+4H++2e-SO2(g) +0.17 Sn4++2e-Sn(s) +0.15 2H++2e-H2(g) 0.00 Pb2++2e-Pb(s) -0.13 -0.14 Sn2++2e-Sn(s) Ni2++2e-Ni(s) -0.25 -0.28 Co2++2e-Co(s) -0.41 2H+(pH=7)+2e-H2(g) -0.44 Fe2++2e-Fe(s) Weaker Oxidizing Agent Stronger Reducing Agent Cr3++3e-Cr(s) -0.74 -0.76 Zn2++2e- Zn(s) 2H2O+2e-2OH-+H2(g) -0.84

  22. e-1 Electrolysis Electrolysis -electric current produced chemical reaction Cathode -reduction -oxidation Anode Which is the…  anode Cathode=? Anode=? Electron flow? Mass Gain=? Cathode(spoon) Cathode Mass Loss=? Copper

  23. Electrolysis Lab- Demo KI (aq) What is available to react? K+1 H2O I-1 Cathode Reaction highest reaction on left. AnodeReaction lowest reaction on right. 

  24. KI(aq) 1/2O2(g)+2H+(pH=7)+2e-H2O +1.23 Br2(l)+2e-2Br- +1.06 NO3-+4H++3e-NO(g)+2H2O +0.96 K+1 I-1 H2O 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Ag++e-Ag(l) +0.80 Cath: highest on left. Hg2++2e-Hg(l) +0.78  NO3-+2H++e-NO2(g)+H2O +0.78 Fe3++e-Fe2+ +0.77 I2(s)+2e-2I- +0.53 We see pink. We saw bubbles +0.52 Cu++e-Cu(s) An: lowest on right. Cu2++2e-Cu(s) +0.34  2I-I2(s)+2e- SO42-+4H++2e-SO2(g) +0.17 Sn4++2e-Sn(s) +0.15 2H++2e-H2(g) 0.00 We see brown:I2(s) Pb2++2e-Pb(s) -0.13 -0.14 Sn2++2e-Sn(s) Ni2++2e-Ni(s) -0.25 Co2++2e-Co(s) -0.28 -0.41 2H+(pH=7)+2e-H2(g) -0.44 Fe2++2e-Fe(s) Cr3++3e-Cr(s) -0.74 -0.76 Zn2++2e-Zn(s) 2H2O+2e-2OH-+H2(g) -0.84 K++e-K(s) -2.92

  25. 1/2O2(g)+2H+(pH=7)+2e-H2O +1.23 CuSO4(aq) Br2(l)+2e-2Br- +1.06 NO3-+4H++3e-NO(g)+2H2O +0.96 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Cu+2 H2O SO4-2 Ag++e-Ag(l) +0.80 Hg2++2e-Hg(l) +0.78 Cath: highest(left) NO3-+2H++e-NO2(g)+H2O +0.78 Fe3++e-Fe2+ +0.77  I2(s)+2e-2I- +0.53 +0.52 Cu++e-Cu(s) We saw copper on the pencil tip! Cu2++2e-Cu(s) +0.34 An: lowest(right) SO42-+4H++2e-SO2(g) +0.17  Sn4++2e-Sn(s) +0.15 H2O1/2O2(g)+2H+(pH=7)+2e- 2H++2e-H2(g) 0.00 Pb2++2e-Pb(s) -0.13 We saw bubbles! -0.14 Sn2++2e-Sn(s) Ni2++2e-Ni(s) -0.25 Co2++2e-Co(s) -0.28 -0.41 2H+(pH=7)+2e-H2(g) -0.44 Fe2++2e-Fe(s) Cr3++3e-Cr(s) -0.74 -0.76 Zn2++2e-Zn(s) 2H2O+2e-2OH-+H2(g) -0.84 Na++e-Na(s) -2.71

  26. 1/2O2(g)+2H+(pH=7)+2e-H2O +1.23 Na2SO4(aq) Br2(l)+2e-2Br- +1.06 NO3-+4H++3e-NO(g)+2H2O +0.96 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Na+1 H2O SO4-2 Ag++e-Ag(l) +0.80 Hg2++2e-Hg(l) +0.78 Cath: highest(left) NO3-+2H++e-NO2(g)+H2O +0.78  Fe3++e-Fe2+ +0.77 I2(s)+2e-2I- +0.53 +0.52 Cu++e-Cu(s) We saw pink. We saw bubbles Cu2++2e-Cu(s) +0.34 An: lowest(right) SO42-+4H++2e-SO2(g) +0.17  Sn4++2e-Sn(s) +0.15 H2O1/2O2(g)+2H+(pH=7)+2e- 2H++2e-H2(g) 0.00 Pb2++2e-Pb(s) -0.13 We saw bubbles. -0.14 Sn2++2e-Sn(s) Ni2++2e-Ni(s) -0.25 Co2++2e-Co(s) -0.28 -0.41 2H+(pH=7)+2e-H2(g) -0.44 Fe2++2e-Fe(s) Cr3++3e-Cr(s) -0.74 -0.76 Zn2++2e-Zn(s) 2H2O+2e-2OH-+H2(g) -0.84 Na++e-Na(s) -2.71

  27. Electrolysis lab A

  28. Notes #3 • Balancing by Oxidation-Reduction

  29. Finding Oxidation #’s for Compounds +4 +1 -2 +1 -2 H2O H2CO3 +1 +5 -2 -3 +1 +4 -2 H3PO4 (NH4)2CO3 -2 +2 +5 +1 +5 -2 Ca3(AsO4)2 HNO3 +1 -2 +3 +6 +6 -2 H2SO4 Fe2(SO4)3 +1 +6 -2 +2 -2 +7 Hg2SO4 Ba(ClO4)2 +3 +4 -2 +6 +1 -2 Na2Cr2O7 Al2(CO3)3

  30. Balancing By Redox Example One #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. H2O + P4+ H2SO4 H3PO4+ H2S #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1 -2 0 +1 +6 -2 +1 +5 +1 -2 -2 5e-1 X8 lost 12 2 5 8 5 8e-1 X5 Gained Multiply by 1. 12H2O + 2P4+ 5H2SO4 8H3PO4+ 5H2S

  31. Balancing By Redox Example Two #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. K3PO4 + Cl2 P4+ K2O+ KClO2 #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1 +5 -2 0 0 +1 -2 +1 -2 +3 5e-1 X3 Gained 3 5/2 3/4 2 5 3e-1 X5 Lost Multiply by 4. 12K3PO4 + 10Cl2  3P4+ 8K2O+ 20KClO2

  32. Applications of Oxidation-Reduction Reactions

  33. Batteries

  34. Alkaline Batteries

  35. Hydrogen Fuel Cells

  36. Corrosion and…

  37. …Corrosion Prevention

  38. Notes Three Unit Seven • Electrolysis Lab Results • Concept Check Assignment • Quiz-Balancing/Electrolysis

  39. CuSO4(aq) Cl2(g)2e-2Cl-1 +1.36 1/2O2(g)+2H+(pH=7)+2e-H2O +1.23 Br2(l)+2e-2Br- +1.06 H2O Cu+2 SO4-2 NO3-+4H++3e-NO(g)+2H2O +0.96 Na2SO4(aq) 1/2O2(g)+2H+(pH=7)+2e-H2O +0.82 Ag++e-Ag(l) +0.80 Hg2++2e-Hg(l) +0.78 Na+1 H2O SO4-2 NO3-+2H++e-NO2(g)+H2O +0.78 I2(s)+2e-2I- +0.53 HCl(aq) +0.52 Cu++e-Cu(s) Cu2++2e-Cu(s) +0.34 SO42-+4H++2e-SO2(g) +0.17 H+1 H2O Cl-1 Sn4++2e-Sn(s) +0.15 2H++2e-H2(g) 0.00 Mg(s) Cu(s) Pb2++2e-Pb(s) -0.13 Cath: highest(left) -0.14 Sn2++2e-Sn(s) Ni2++2e-Ni(s) -0.25  Co2++2e-Co(s) -0.28 -0.41 2H+(pH=7)+2e-H2(g) An: lowest(right) -0.44 Fe2++2e-Fe(s) Cr3++3e-Cr(s) -0.74  Mg(s)2e-1+Mg+2 +2.37 2H2O+2e-2OH-+H2(g) -0.84 Mg+2+e-Mg(s) -2.37 Cu+2+ Mg(s) Cu(s)+ +2.71 Mg+2 -2.71 Na+1+e-Na(s)

  40. Lab C Voltaic Cell e-1 Anode rxn Mg(s) Mg(s)2e-1+Mg+2 + rxn Cathode Cu2++2e-Cu(s) H+1 Cu(s) Spectator Ions - Cl-1 Cu+2 SO4-2 SO4-2 H+1 Na+1 Cl-1 SO4-2 Current Flow Na+1

  41. Notes Four Unit Seven • Faraday’s Law Lab B • This as a chemical process that uses electricity to produce industrial quantities of specific chemicals.

  42. Application of Faraday’s law F = 96500 C/mol e-) A x s = C

  43. Faraday’s Law: Lab B Fe(s) Cathode? Anode? e-1 Fe(s) F = 96500 C/mole- Amp x second = C Lose Mass? Fe+3 Gain Mass? 3e-1 Fe+3 NO3-1 NO3-1 NO3-1

  44. Faraday’s Law Calculation One How many grams of Gold will be plated, using a current of 3.0 amps for 1.5 hours? 1. Balanced Equation Au+3+3e-1Au(s) 2. Calculate Coulombs. 60 min 60 Sec 3.0 ampx 1.5 Hour x x =16000C 1 min 1 hour 3. Calculate moles e-1. 1mole e-1 16000C x = 0.17 mol e-1 96500C 4. Calculate moles of substance. 1 m Au(s) 0.17 mol e-1x =0.056mol Au(s) 3mol e-1 5. Calculate grams. 197.0gAu 0.056mol Au(s)x = 11g Au 1mol Au

  45. Faraday’s Law Calculation Two How many grams of Silver will be plated, using a current of 2.0 amps for 45 minutes? 1. Balanced Equation Ag+1+1e-1Ag(s) 2. Calculate Coulombs. 60 Sec 2.0 ampx 45 Hour x =5400C 1 min 3. Calculate moles e-1. 1mole e-1 5400C x = 0.056 mol e-1 96500C 4. Calculate moles of substance. 1 m Ag(s) 0.056 mol e-1x =0.056mol Au(s) 1mol e-1 5. Calculate grams. 107.9gAg 0.05596mol Au(s)x = 6.0 g Ag 1mol Ag

  46. rxn: Cathode +0.34 Salt Bridge 6 e-1 ? 3 Cu+2 + 2e-1 Cu 3 x3 1.08 rxn: Anode +0.74 6 2 2 x2 Cr Cr+3+3e-1 Na+1 SO4-2 Overall rxn: Na+1 3Cu 2Cr+ 3Cu+2 2Cr+3+ emf= 1.08volts Cu(s) Cr(s) Salt Bridge SO4-2 Cu+2 3e-1 Cr+3 SO4-2 Cu+2 Cr+3 SO4-2 3e-1 Cu+2 SO4-2 Cu+2 Cr+3 SO4-2 SO4-2 Cu+2 SO4-2 Cr+3 Cu+2 SO4-2 SO4-2

  47. Ion-Electron Method for Balancing

  48. Ion-Electron Method for Balancing #1. Separate Half-rxn. #2. Bal Non-O Elem. #3. + H2O. #4. + H+1 . #5. + e-1 to bal +/-. UO2+2 + I2 U+4 + IO3-1 (Acid) 5X 1 UO2+2 1  U+4 + H2O 2 + H+1 4 + e-1 2 2 1 I2  IO3-1 + H2O + H+1 12 + e-1 6 10 1X 8 4 5 UO2+2  U+4 + H2O 10 + H+1 20 + e-1 5 10 2 1 I2  IO3-1 + H2O + H+1 12 + e-1 6 10 1 5 UO2+2 + I2 2 + U+4 5 + H+1 8  IO3-1 + H2O 4

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