2. Chapter 17 Oxidation and Reduction

3. Notes One Unit Six Redox Oxidation Numbers Identifying what is oxidized and what is reduced

4. 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) Cl 2 + 2 e -1 ------> 2 Cl -1

5. Redox: Reduction occurs when an atom gains one or more electrons. Ex: O + 2e -1  O 2- 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(NO 3 )  2 Ag + Cu(NO 3 ) 2.

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

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

8. 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).

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

11. Redox Burning: C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O+ Heat Rusting Iron: 4Fe + 3O 2  2Fe 2 O 3 + Heat Oxidation - Loss of e -1. Na(s)  Na +1 +1e -1 Reduction - Gain of e -1. Cl 2 + 2e -1  2Cl -1 Number line (Oxidation…Left or right?)

12. Oxidation Numbers Rules for Assigning Oxidation States The oxidation state of an atom in an uncombined element is 0. The oxidation state of a monatomic ion is the same as its charge. Oxygen is assigned an oxidation state of –2 in most of its covalent compounds. Important exception: peroxides (compounds containing the O2 2- group), in which each oxygen is assigned an oxidation state of –1) In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1 For a compound, sum total of Oxidation Numbers is zero. For an ionic species (like a polyatomic ion), the sum of the oxidation states must equal the overall charge on that ion.

13. Key Elements (99%) H +1 H -1 (99%)O -2 O -1 (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 (NO 3 -1 ) ion is always +5 (SO 4 -2 ) ion is always +6

14. Finding Oxidation Numbers +1-2 H2OH2O 2(+1)+1(-2)=Zero The sum of the oxidation numbers must be equal to _____ for a compound. Find Ox#’s for H 2 O? zero 2(H)+1(O)=Zero +1 -2 H 3 PO 4 Find Ox#’s for H 3 PO 4 ? 3(H)+4(O)=Zero +5 1(P)+ 3(+1)+4(-2)=Zero1(+5)+

15. Finding Oxidation #’s for Compounds +1-2 +1+5-2 H 3 PO 4 H2OH2O HNO 3 +1+5-2 H 2 SO 4 +1-2 +6 Hg 2 SO 4 +6+1-2 Na 2 Cr 2 O 7 +1 +6 -2 H 2 CO 3 +1 -2 +4 (NH 4 ) 2 CO 3 -3+1+4-2 Ca 3 (AsO 4 ) 2 +2+5 -2 Fe 2 (SO 4 ) 3 +6 +3 -2 Ba(ClO 4 ) 2 +2 +7 -2 Al 2 (CO 3 ) 3 +3+4 -2

16. Identifying OX, RD, SI Species Ca 0 + 2 H +1 Cl -1  Ca +2 Cl -1 2 + H 2 0 Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca 0  Ca +2, so Ca 0 is the species that is oxidized. Reduction = gain of electrons. The species becomes more negative in charge. For example, H +1  H 0, 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.

17. Oxidizing Agent and Reducing Agent: 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”.

18. Notes Two Unit Six Activity Series of Metals Balancing by Redox Electrolysis Lab A-Electrolysis

19. 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 H 2 from a solution by the reactions: Mg(s) + 2 H + (aq) → H 2 (g) + Mg 2+ (aq) Zn(s) + 2 H + (aq) → H 2 (g) + Zn 2+ (aq) Both metals react with the hydrogen ions, but magnesium metal can also displace zinc ions in solution by the reaction: Mg(s) + Zn 2+ → Zn(s) + Mg 2+metals

20. 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.

21. 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. 6 e-

22. Cu +2 (aq) + Mg (s)  Cu (s) + Mg +2 (aq) Zn +2 (aq) + Ag (s)  No RXN

23. Balancing By Redox Example One H 2 O + P 4 + H 2 SO 4  H 3 PO 4 + H 2 S #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1-20+1-2+6+1 -2 +5+1-2 5e -1 lost 8e -1 Gained X8 X5 285512 Multiply by 1. 12H 2 O +2P 4 +5H 2 SO 4  8H 3 PO 4 +5H 2 S

24. Balancing By Redox Example Two K 3 PO 4 + Cl 2  P 4 + K 2 O+ KClO 2 #1. Find the oxidation #’s. #2. ID the element (i) oxidized and (ii) reduced. #3. Find # of electrons lost or gained. #4. Cross-multiply. #5. Balance using whole # ratios. #6. Find whole number coefficients. +1-2 0+1-2+5+1-2 +3 0 5e -1 Gained 3e -1 Lost X3 X5 5/223/453 Multiply by 4. 12K 3 PO 4 +10Cl 2  3P 4 +8K 2 O+20KClO 2

25. Applications of Oxidation-Reduction Reactions

26. Batteries

27. Alkaline Batteries

28. Hydrogen Fuel Cells

29. Corrosion and…

30. …Corrosion Prevention

32. How to use the ½ Reaction Resource: Notice they are all written as Reductions (gaining of electrons) 1.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

33. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 Reducing Agent Stronger Reducing Agent Loses e - Oxidizing Agent Weaker Oxidizing Agent Gains e - StrongerWeaker

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

35. Electrolysis Lab- Demo KI (aq) What is available to react? K +1 I -1 H2OH2O Anode Reaction lowest reaction on right. Cathode Reaction highest reaction on left.

36. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 K +1 I -1 H2OH2O K + +e -  K(s) -2.92 An: lowest on right. Cath: highest on left. 2I -  I 2 (s)+2e - We see brown:I 2 (s) We see pink. We saw bubbles KI(aq)  

37. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 Na + +e -  Na(s) -2.71 An: lowest(right) Cath: highest(left) H 2 O  1/2O 2 (g)+2H + (pH=7)+2e - We saw bubbles! We saw copper on the pencil tip! Cu +2 SO 4 -2 H2OH2O CuSO 4 (aq)  

38. NO 3 - +2H + +e -  NO 2 (g)+H 2 O Fe 3+ +e -  Fe 2+ I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) Zn 2+ +2e -  Zn(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.77 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.76 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 Na + +e -  Na(s) -2.71 An: lowest(right) Cath: highest(left) H 2 O  1/2O 2 (g)+2H + (pH=7)+2e - We saw bubbles. We saw pink. We saw bubbles Na +1 SO 4 -2 H2OH2O Na 2 SO 4 (aq)  

39. Electrolysis lab A

40. Notes Three Unit Six Electrolysis Lab Results Concept Check Assignment Quiz-Balancing/Electrolysis

41. Na +1 +e -  Na(s) NO 3 - +2H + +e -  NO 2 (g)+H 2 O I 2 (s)+2e -  2I - Cu + +e -  Cu(s) Cu 2+ +2e -  Cu(s) SO 4 2- +4H + +2e -  SO 2 (g) Sn 4+ +2e -  Sn(s) 2H + +2e -  H 2 (g) Pb 2+ +2e -  Pb(s) Sn 2+ +2e -  Sn(s) Ni 2+ +2e -  Ni(s) Co 2+ +2e -  Co(s) 2H + (pH=7)+2e -  H 2 (g) Fe 2+ +2e -  Fe(s) Cr 3+ +3e -  Cr(s) 2H 2 O+2e -  2OH - +H 2 (g) +0.78 +0.53 +0.52 +0.34 +0.17 +0.15 0.00 -0.13 -0.14 -0.25 -0.28 -0.41 -0.44 -0.74 -0.84 Hg 2+ +2e -  Hg(l) +0.78 Ag + +e -  Ag(l) +0.80 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +0.82 NO 3 - +4H + +3e -  NO(g)+2H 2 O +0.96 Br 2 (l)+2e -  2Br - +1.06 1/2O 2 (g)+2H + (pH=7)+2e -  H 2 O +1.23 Mg +2 +e -  Mg(s)-2.37 An: lowest(right) Cath: highest(left) Mg(s)  2e -1 +Mg +2 Cu +2 SO 4 -2 H2OH2O CuSO 4 (aq)   Cl 2 (g)2e -  2Cl -1 +1.36 +2.37 Na +1 SO 4 -2 H2OH2O Na 2 SO 4 (aq) H +1 Cl -1 H2OH2O HCl(aq) Mg(s) Cu(s) -2.71 Cu +2 +Mg(s)  Cu(s)+ Mg +2 +2.71

42. Lab C Voltaic Cell Mg(s)  2e -1 +Mg +2 Mg(s) Cu(s) Cathode Anode Cu 2+ +2e -  Cu(s) e -1 SO 4 -2 H +1 Na +1 Spectator Ions SO 4 -2 Cu +2 H +1 Cl -1 Na +1 SO 4 -2 rxn Current Flow + -

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

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

45. Faraday’s Law: Lab B Fe(s) NO 3 -1 e -1 Anode? Cathode? Lose Mass? Gain Mass? Fe +3 3e -1 Fe +3 NO 3 -1 F = 96500 C/mole - Amp x second = C

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

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

48. Salt Bridge Cu +2 SO 4 -2 Cr +3 SO 4 -2 Cr +3 SO 4 -2 Cathode Anode Cu +2 + 2e -1  Cu Cr  Cr +3 +3e -1 Overall rxn: 2Cr+ x3 x2 3 6 3 2 6 2 3Cu +2  2Cr +3 + 3Cu +0.34 +0.74 emf=1.08volts 1.08 Cr +3 3e -1 SO 4 -2 Na +1 Cu(s)Cr(s) Salt Bridge e -1 ? rxn:

49. Ion-Electron Method for Balancing

50. UO 2 +2 + I 2  U +4 + IO 3 -1 (Acid) UO 2 +2 I 2  U +4 + H 2 O #2. Bal Non-O Elem. #3. + H 2 O. #4. + H +1. #5. + e -1 to bal +/-. 2 + H +1 4+ e -1 2 #1. Separate Half-rxn.  IO 3 -1 2 + H 2 O 6 + H +1 12 11 1 + e -1 10 5X 1X UO 2 +2 I 2  U +4 + H 2 O10 + H +1 20+ e -1 10  IO 3 -1 2 + H 2 O 6 + H +1 12 5 5 1 + e -1 10 84 UO 2 +2 5 + U +4 5 + H 2 O4+ H +1 8 + I 2 1  IO 3 -1 2

51. Ion-Electron Method for Balancing IO 3 -1 + Ti +3  I 2 + TiO 2 +1 (Acid) IO 3 -1 Ti +3  I 2 + H 2 O #2. Bal Non-O Elem. #3. + H 2 O. #4. + H +1. #5. + e -1 to bal +/-. 6 + H +1 12+ e -1 10 #1. Separate Half-rxn.  TiO 2 +1 1 + H 2 O 2 + H +1 4 1 2 1 + e -1 2 1X 5X IO 3 -1 Ti +3  I 2 + H 2 O6 + H +1 12+ e -1 10  TiO 2 +1 5 + H 2 O 10 + H +1 20 1 2 5 + e -1 10 8 4 IO 3 -1 2 + I 2 1 + H +1 8+ H 2 O4 + Ti +3 5  TiO 2 +1 5