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Chemistry 30 – Unit 2 Electrochemical Changes

To accompany Inquiry into Chemistry. PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca. Chemistry 30 – Unit 2 Electrochemical Changes. Section 12.1 Characterizing Oxidation Reduction. Chapter 12, Section 12.1.

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Chemistry 30 – Unit 2 Electrochemical Changes

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  1. To accompany Inquiry into Chemistry PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca Chemistry 30 – Unit 2 Electrochemical Changes

  2. Section 12.1 Characterizing Oxidation Reduction

  3. Chapter 12, Section 12.1 • Oxidation: Historically, the reaction of a substance with O2 • Today, oxidation: loss of electrons • Reduction: Historically, the reaction of a metal ore to produce a smaller mass of pure metal • Today, reduction: gain of electrons

  4. gain reduction oxidation lose electrons electrons reduction gaining is oxidation losing is Memory Tools: Chapter 12, Section 12.1 LEO the Lion says GER Oil Rig

  5. Chapter 12, Section 12.1 • Oxidation and reduction cannot happen individually – if one substance loses electrons (oxidation), another substance must gain electrons (reduction) • Oxidation/reduction reactions called “redox reactions” • Redox Song

  6. Zn(s) + Cu2+(aq) + 2 NO3ˉ(aq) Zn2+(aq) + 2 NO3ˉ(aq) + Cu(s) and finally reduced to a net-ionic equation: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Chapter 12, Section 12.1 Zn(s) + Cu(NO3)2(aq) Zn(NO3)2(aq) + Cu(s) • Consider the example: This equation can be rewritten as a total ionic equation: spectators

  7. electrons lost oxidation electrons gained reduction causes reduction called “reducing agent” causes oxidation called “oxidizing agent” Chapter 12, Section 12.1 • Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Note: the oxidizing agent gets reduced; the reducing agent gets oxidized!!!

  8. Chapter 12, Section 12.1 • The net-ionic equation: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) can be written as two half-reactions, Cu2+(aq) + 2 e–Cu(s) reduction Zn(s) Zn2+(aq) + 2 e–oxidation

  9. Chapter 12, Section 12.1 • Half-Reactions WS • Discuss questions 3 and 4, page 437 • Spontaneity of redox reactions – Nelson Lab or Investigation 12.A, page 438

  10. Chapter 12, Section 12.1 • Example: Cu(s), Pb(s), Ag(s) and Zn(s) are reacted with solutions of Cu(NO3)2(aq), Pb(NO3)2(aq), AgNO3(aq), and Zn(NO3)2(aq) • The experimental results are summarized in the chart:

  11. strongest oxidizing agent AgNO3(aq) Zn(NO3)2(aq) Cu(NO3)2(aq) Pb(NO3)2(aq) oxidizing agents reducing agents strongest reducing agent Chapter 12, Section 12.1 Cu2+(aq) Zn2+(aq) Ag+(aq) Pb2+(aq)  spontaneous;  non-spontaneous

  12. Chapter 12, Section 12.1 • Generalizations to this point: • Oxidizing agents: metal ions, e.g. Ag+(aq), and non-metal elements, e.g. Cl2(g) • Reducing agents: non-metal ions, e.g. Br –(aq), and metal elements, e.g. Fe(s)

  13. “SOA” “SRA” Chapter 12, Section 12.1 • Results of experiment can be summarized in a table of redox half-reactions: Ag+(aq) + e– Ag(s) Cu2+(aq) + 2 e– Cu(s) Pb2+(aq) + 2 e–Pb(s) Zn2+(aq) + 2 e– Zn(s) Where are the weakest OA and RA? Examine data chart from previous slide. How can chart of half-reactions be used to predict whether or not a reaction is spontaneous?

  14. Chapter 12, Section 12.1 • Spontaneity Generalization: OA spontaneous RA RA non-spontaneous OA

  15. Chapter 12, Section 12.1 Take the chart of half-reactions with Zn, Cu, Pb, and Ag elements and ions and combine with the two data charts on the next page to produce a single chart of half-reactions formatted in the standard way

  16. Chapter 12, Section 12.1

  17. Chapter 12, Section 12.1 Ag+(aq) + e– Ag(s) Cu2+(aq) + 2 e– Cu(s) Pb2+(aq) + 2 e–Pb(s) Zn2+(aq) + 2 e– Zn(s) Chart 2: Pb2+(aq) + 2 e– Pb(s) Cd2+(aq) + 2 e– Cd(s) Zn2+(aq) + 2 e– Zn(s) V2+(aq) + 2 e– V(s) Chart 1: Cd2+(aq) + 2 e–Cd(s) V2+(aq) + 2 e–V(s) Be2+(aq) + 2 e–Be(s) Ra2+(aq) + 2 e–Ra(s)

  18. Chapter 12, Section 12.1 Ag+(aq) + e– Ag(s) Cu2+(aq) + 2 e– Cu(s) Pb2+(aq) + 2 e–Pb(s) Cd2+(aq) + 2 e– Cd(s) Zn2+(aq) + 2 e– Zn(s) V2+(aq) + 2 e– V(s) Be2+(aq) + 2 e– Be(s) Ra2+(aq) + 2 e– Ra(s)

  19. Chart on page 7 of Data Booklet has been prepared in this manner You can use the chart and the spontaneity generalization developed earlier to predict spontaneity of redox reactions Chapter 12, Section 12.1

  20. Chapter 12, Section 12.1 purple box bottom of page • Examples: • #5a, 6a, and Review 6a page 440

  21. Must be done in this order! Chapter 12, Section 12.2 • Writing balanced half-reactions in acidic or neutral media: • Step 1: balance elements other than O or H by inspection • Step 2: balance O’s using H2O(l)’s • Step 3: balance H’s using H+(aq)’s • Step 4: balance total charge using electrons (e–)

  22. + 2 e– Step 4 2 + 2 e– Step 1 Step 4 Chapter 12, Section 12.2 • Examples: • Zn2+(aq) Zn(s) • Cl–(aq)Cl2(g) neutral media

  23. + 14 H+(aq) + 7 H2O(l) + 6 e– 2 Step 3 Step 1 Step 2 Step 4 Chapter 12, Section 12.2 • Example: acidic medium • Cr2O72-(aq) Cr3+(aq) You’ll find this half-reaction on page 7 of your data booklet

  24. Chapter 12, Section 12.2 • You are not required to write half-reactions in basic media, though you should be able to use given half-reactions if they are in basic media

  25. Chapter 12, Section 12.2 • Example: Practice Problem 1a, page 448 • Both of the required half-reactions are in your data booklet • Half-reactions, those you write yourself, and those from your data chart can be used to balance redox reaction equations MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l) (Ag(s) Ag+(aq) + e–) MnO4–(aq) + 8 H+(aq) + 5 Ag(s) Mn2+(aq) + 4 H2O(l) + 5 Ag+(aq) 5 x OA RA

  26. 4 + 2 e– 2 x ( + 2 H+(aq) + e– ) + H2O(l) Chapter 12, Section 12.2 • Practice Problem 1b, page 448 • Neither of these half-reactions are in your data booklet you must write each yourself Hg(l) + Cl–(aq) HgCl42–(s) NO3–(aq) NO2(g) Hg(l) + 4 Cl–(aq) + 2 NO3–(aq) + 4 H+(aq) HgCl42–(s) + 2 NO2(g) + 2 H2O(l)

  27. 1c AsH3(s) + 4 H2O(l) H3AsO4(aq) + 8 H+(aq) + 8 e– 4 x (Zn2+(aq) + 2 e–Zn(s)) AsH3(s) + 4 H2O(l) + 4 Zn2+(aq) 4 Zn(s) + H3AsO4(aq) + 8 H+(aq) Data Booklet (basic) 1d I2(s) + 6 H2O(l) 2 IO3–(aq) + 12 H+(aq) + 10 e– 5 x (OCl–(aq) + H2O(l) + 2 e– Cl–(aq) + 2 OH−(aq)) I2(s) + 11 H2O(l) + 5 OCl–(aq) 2 IO3–(aq) + 12 H+(aq) + 5 Cl–(aq) + 10 OH−(aq) Chapter 12, Section 12.2 • Do Practice Problems 1c, 1d page 448

  28. Chapter 12, Section 12.2 • Half-reactions WS • Do question 14 a,b,c (acidic conditions) on page 475 (yes I do mean page 475) • Time to work

  29. Chapter 12, Section 12.2 • 14 a • (both half-reactions are in the Data Booklet) MnO2(s) + 4 H+(aq) + 2 e–Mn2+(aq) + 2 H2O(l) 2 Cl–(aq) Cl2(g) + 2 e– MnO2(s) + 4 H+(aq) + 2 Cl–(aq) Mn2+(aq) + 2 H2O(l) + Cl2(g)

  30. 2 x(NO(g) + 3 H+(aq) + 3 e– NH2OH(aq)) Chapter 12, Section 12.2 3 x(Sn(s) Sn2+(aq) + 2 e–) • 14 b 2 NO(g) + 6 H+(aq) + 3 Sn(s) 3 Sn2+(aq) + 2 NH2OH(aq)

  31. 3 x (Cd2+(aq) + 2 e– Cd(s)) Chapter 12, Section 12.2 2 x (V2+(aq) + 3 H2O(l) VO3–(aq) + 6 H+(aq) + 3 e–) • 14 c 3 Cd2+(aq) + 2 V2+(aq) + 6 H2O(l) 3 Cd(s) + 2 VO3–(aq) + 12 H+(aq)

  32. Chapter 12, Section 12.2 • Read Reducing Iron (to make steel), pages 452-4

  33. Chapter 12, Section 12.3 • Assigning oxidation numbers is a method of electron bookkeeping – they make it possible to see electron gain and loss without writing half-reactions • Oxidation Number Rules: • Pure elements have an oxidation number of 0. e.g. in Cl2 each Cl atom has an oxidation number of 0; in Fe each Fe atom has an oxidation number of 0.

  34. Chapter 12, Section 12.3 • The oxidation number of hydrogen in compounds is +1. This is true in molecular compounds and acids. Exception: in metal hydrides, e.g. LiH, H is -1 • The oxidation number of oxygen in compounds is -2. Exceptions: peroxides, e.g. H2O2, where oxygen is -1 and the compound OF2, where oxygen is +2

  35. Chapter 12, Section 12.3 • In monatomic ions, e.g. Cl– and Fe3+, the oxidation number equals the ion charge • Oxidation numbers of atoms in compounds add to zero. Oxidation numbers of atoms in polyatomic ions add to the ion charge Note oxidation numbers can be fractions Do Oxidation State WS

  36. 0 +7 -2 +1 +2 +2 +1-2 5 Fe(s) + 2 MnO4–(aq) + 16 H+(aq) 5 Fe2+(aq) + 2 Mn2+(aq) + 8 H2O(l) Chapter 12, Section 12.3 (oxidation numbers in red) • Uses for oxidation numbers: • Determining whether a substance is oxidized or reduced – e.g. in the reaction: Fe changes from 0 to +2; it is oxidized Mn in MnO4–(aq) changes from +7 to +2 in Mn2+; it is reduced an increase in oxidation number is oxidation; a decrease in oxidation number is reduction

  37. Chapter 12, Section 12.3 • Determining the OA and RA in a redox reaction. The OA is reduced (has a decrease in oxidation number), the RA is oxidized (has an increase in oxidation number) • Determining whether or not a reaction is redox. If it is, one* substance will increase in oxidation number (oxidation) and one* will decrease (reduction). If no change in oxidation numbers, the reaction is not redox.

  38. +1 -1 +2 -2 +1 +3 -2+1 Redox +3-1 +1 -2 +1 +3-2 +1-1 Not Redox -3 +1 0 +4 -2 +1 -2 Redox +4-2 +1 -2 +1+5-2 +2-2 Redox Chapter 12, Section 12.3 H2O2(aq) + 2 Fe(OH)2(s) 2 Fe(OH)3(s) • Do Practice Problem 11, page 461 PCl3(l) + 3 H2O(l) H3PO3(aq) + 3 HCl(aq) 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l) 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)

  39. +7 -2 +1 +2 +1 -2 MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l) Chapter 12, Section 12.3 4. Determining the number of electrons transferred in a redox reaction or half-reaction. e.g. Note that Mn changes from +7 to +2, a change of 5 electrons. The number of electrons transferred can be determined without a half-reaction!

  40. +6 -2 +1 +3 +1-2 Cr2O72-(aq) + 14 H+(aq) + 6 e– 2 Cr3+(aq) + 7 H2O(l) Chapter 12, Section 12.3 • Another example: • Note that Cr changes from +6 to +3, a change of 3, yet the half-reaction has 6 e–. Why??? • There are 2 Cr’s, each changing by 3 2 x 3 = 6

  41. +7 -2 +1 +2 +1 -2 MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l) Chapter 12, Section 12.3 • Determining which substance in a half-reaction is really functioning as the OA (or RA) e.g.: Note that the Mn in the MnO4−(aq) is reduced MnO4−(aq) is the OA H+(aq) is a very necessary spectator 6. Balancing equations by the oxidation number method

  42. +1-2+1 0 +1 -1 +1 +1-2 +1 -2 2 NaOH(aq) + Cl2(g) NaCl(aq) + NaClO(aq) + H2O(l) Chapter 12, Section 12.2 • Disproportionation Reaction – a redox reaction where the same element, is both oxidized and reduced (acts as both OA and RA) • Example: • Note that Cl in Cl2 goes from 0 to -1, and from 0 to +1 in NaClO(aq)

  43. +4-2 +1 -2 +1+5-2 +2-2 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) Chapter 12, Section 12.3 • Another example: • N is oxidized from +4 to +5 and reduced from +4 to +2

  44. Chapter 12, Section 12.3 • Redox reactions are balanced by making electrons gained and lost equal • This can be done using half-reactions or by observing change in oxidation number • Try balancing the following equation by inspection: S8(s) + KMnO4(aq) + HCl(aq) SO2(g) + MnCl2(s) + KCl(aq) + H2O(l)

  45. 32 5 Insert coefficients in equation to make these numbers equal Chapter 12, Section 12.3 0 +1+7 -2 +1 -1 +4 -2 +2 -1 +1 -1 +1 -2 • Try it using change in oxidation number: S8(s) + KMnO4(aq) + HCl(aq) SO2(g) + MnCl2(s) + KCl(aq) + H2O(l) 32 32 48 96 40 5e–/Mn 5e– /KMnO4 4e–/S 32e– /S8 RA OA Usually the other coefficients will be easily found

  46. +1 -2 +1 -1 0 +1 -2 Chapter 12, Section 12.3 16 8 H2S(g) + H2O2(aq) S8(s) + H2O(l) 8 1e−/O 2e− /H2O2 2e−/S 2e− /H2S RA OA Because of the 8 in S8, it is necessary to put 8 in front of H2S and because of 2e−/2e− ratio, 8 must also go in front of the H2O2 Do WS 56 B,C

  47. Chapter 12, Section 12.4 Buret containing solution titrant • Redox Titrations Statement: (applies to all titrations: “titration of with ” sample titrant Know this! Erlenmeyer flask containing solution sample

  48. Chapter 12, Section 12.4 • Oxidizing and reducing agents can be titrated to find concentrations (just like acids and bases) • KMnO4(aq) is used as the titrant in many redox titrations – acidifiedMnO4–(aq) is a very strong OA in the half-reaction: MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l) dark purple colourless

  49. Chapter 12, Section 12.4 • As long as some RA is left in the sample solution, the purple colour of the MnO4–(aq) disappears. Once the RA is used up the purple colour remains. Ideally the endpoint is a very light purple

  50. 2 x (MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l)) Chapter 12, Section 12.4 5 x (H2O2(l) O2(g) + 2 H+(aq) +2 e–) • Example Practice Problem 19, page 469 2 MnO4–(aq) + 16 H+(aq) + 5 H2O2(l) 5 O2(g) + 10 H+(aq) + 2 Mn2+(aq) + 8 H2O(l) 6 2 MnO4–(aq) + 6 H+(aq) + 5 H2O2(l) 5 O2(g) + 2 Mn2+(aq) + 8 H2O(l) n1 0.02045 mol/L 38.95 mL n2 m=?

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