Chapter 5 Oxidation Reduction Reactions

Transcription

1 Chapter 5 Oxidation Reduction Reactions Chemistry: The Molecular Nature of Matter, 7E Jespersen/Hyslop

2 Chapter in Context Define oxidation, reduction, oxidizing agents, reducing agents, and oxidation numbers Balance oxidation/reduction reactions Explore, at the molecular level, how acids react with metals Use the activity series to predict products of reactions Learn about the reaction of oxygen with organic compounds, metals, and nonmetals Perform calculations using the stoichiometry of oxidation/reduction reactions 2

3 Oxidation-Reduction Reactions Electron transfer reactions Electrons transferred from one substance to another Originally only combustion of fuels or reactions of metal with oxygen Important class of chemical reactions that occur in all areas of chemistry and biology Also called redox reactions to emphasize that reduction and oxidation must always occur together 3

4 Oxidation Reduction Reactions Involves 2 processes: Oxidation = Loss of electrons Na Na + + e Oxidation Half-Reaction Reduction = Gain of electrons Cl 2 + 2e 2Cl Reduction Half-Reaction Net reaction: 2Na + Cl 2 2Na + + 2Cl Oxidation and reduction always occur together Can't have one without the other 4

5 Oxidation Reduction Reaction Oxidizing Agent Substance that accepts electrons Accepts electrons from another substance Substance that is reduced Cl 2 + 2e 2Cl Reducing Agent Substance that donates electrons Releases electrons to another substance Substance that is oxidized Na Na + + e 5

6 Redox Reactions Very common Batteries car, flashlight, cell phone, computer Metabolism of food Combustion Chlorine Bleach Dilute NaOCl solution Cleans through redox reaction Oxidizing agent Destroys stains by oxidizing them 6

7 Redox Reactions e.g., Fireworks displays Net: 2Mg + O 2 2MgO Oxidation: Mg Mg e Loses electrons = oxidized Reducing agent Reduction: O 2 + 4e 2O Gains electrons = reduced Oxidizing agent 7

8 Guidelines For Redox Reactions Oxidation and reduction always occur simultaneously Total number of electrons lost by one substance equals total number of electrons gained by second substance For a redox reaction to occur, something must accept electrons that are lost by another substance 8

9 Oxidation Numbers Bookkeeping Method Way to keep track of electrons Not all redox reactions contain O 2 or ions Covalent molecules and ions often involved e.g., CH 4, SO 2, MnO 4, etc. Defined by set of rules How to divide up shared electrons in compounds with covalent bonds Often whole numbers but can be fractions Change in oxidation number of element during reaction indicates redox reaction has occurred 9

10 Hierarchy of Rules for Assigning Oxidation Numbers 1. Oxidation numbers must add up to charge on molecule, formula unit or ion 2. Atoms of free elements have oxidation numbers of zero 3. Metals in Groups 1A, 2A, and Al have +1, +2, and +3 oxidation numbers, respectively 4. H and F in compounds have +1 and 1 oxidation numbers, respectively 5. Oxygen has 2 oxidation number 6. Group 7A elements have 1 oxidation number 10

11 Hierarchy of Rules for Assigning Oxidation Numbers 7. Group 6A elements have 2 oxidation number 8. Group 5A elements have 3 oxidation number 9. When there is a conflict between two of these rules or ambiguity in assigning an oxidation number, apply rule with lower oxidation number and ignore conflicting rule Oxidation State Used interchangeably with oxidation number Indicates charge on monatomic ions Iron(III) means +3 oxidation state of Fe or Fe 3+ 11

12 Ex. 1 Assigning Oxidation Number 1. Li 2 O Li (2 atoms) (+1) = +2 (Rule 3) O (1 atom) ( 2) = 2 (Rule 5) sum = 0 (Rule 1) +2 2 = 0 so the charges are balanced to zero 2. CO 2 C (1 atom) (x) = x O (2 atoms) ( 2) = 4 (Rule 5) sum = 0 (Rule 1) x 4 = 0 or x = +4 C is in +4 oxidation state 12

13 Learning Check Assign oxidation numbers to all atoms: Example 1: ClO 4 O (4 atoms) ( 2) = 8 Cl (1 atom) ( 1) = 1 (molecular ion) sum 1 (violates Rule 1) Rule 5 for oxygen comes before Rule 6 for halogens O (4 atoms) ( 2) = 8 Cl (1 atom) (x) = x sum = 1 (Rule 1) 8 + x = 1 or x = 8 1 So x = +7; Cl is oxidation state +7 13

14 Learning Check Assign oxidation states to all atoms: MgCr 2 O 7 Mg =+2; O = 2; and Cr = x (unknown) [+2] + [2x] + [7 ( 2)] = 0 2x 12 = 0 x = +3 Cr is oxidation number of +3 KMnO 4 K =+1; O = 2; so Mn = x [+1] + [x] + [4 ( 2)] = 0 x 7 = 0 x = +7 Mn is oxidation number of +7 14

15 Redefine Oxidation-Reduction in Terms of Oxidation Number A redox reaction occurs when there is a change in oxidation number Oxidation Increase in oxidation number Electron loss Reduction Decrease in oxidation number Electron gain 15

16 Using Oxidation Numbers to Recognize Redox Reactions Sometimes literal electron transfer: decrease reduction increase oxidation Cu 2+ + Zn Zn 2+ + Cu Cu: oxidation number decreases by 2 reduction Zn: oxidation number increases by 2 oxidation 16

17 Using Oxidation Numbers to Recognize Redox Reactions Reduction and oxidation can be deduced from changes in oxidation numbers O: decrease reduction C: increase oxidation CH 4 + 2O 2 CO 2 + 2H 2 O O: oxidation number decreases by 2 reduction C: oxidation number increases by 8 oxidation 17

18 Ion Electron Method Way to balance redox equations Must balance both mass and charge Write skeleton equation Only ions and molecules involved in reaction Break into two half-reactions Oxidation Reduction Balance each half-reaction separately Recombine to get balanced net ionic equation 18

19 Balancing Redox Reactions Some redox reactions are simple: e.g., Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) Break into half-reactions Zn(s) Zn 2+ (aq) + 2e oxidation Reducing agent Cu 2+ (aq) + 2e Cu(s) reduction Oxidizing agent 19

20 Balancing Redox Reactions Zn(s) Zn 2+ (aq) + 2e Cu 2+ (aq) + 2e Cu(s) Each half-reaction is balanced for atoms oxidation reduction Same number of atoms of each type on each side Each half-reaction is balanced for charge Same sum of charges on each side Add both equations algebraically, canceling electrons NEVER have electrons in net ionic equation Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) 20

21 Balancing Redox Equations in Aqueous Solutions Many redox reactions in aqueous solution involve H 2 O and H + or OH Balancing the equation cannot be done by inspection Need method to balance equation correctly Start with acidic solution then work to basic conditions 21

22 Ion-Electron Method Acidic Solution 1. Divide equation into two half-reactions 2. Balance atoms other than H and O 3. Balance O by adding H 2 O to side that needs O 4. Balance H by adding H + to side that needs H 5. Balance net charge by adding e 6. Make electron gain equal electron loss; then add half-reactions 7. Cancel electrons and anything that is the same on both sides 22

23 Redox in Aqueous Solution Example 2: Mix solutions of K 2 Cr 2 O 7 and FeSO 4 Dichromate ion, Cr 2 O 7, oxidizes Fe 2+ to Fe 3+ Cr 2 O 7 is reduced to form Cr 3+ Acidity of mixture decreases as H + reacts with oxygen to form water Skeletal Eqn. Cr 2 O 7 + Fe 2+ Cr 3+ + Fe 3+ 23

24 Ex. 2 Ion Electron Method Balance in Acidic Solution Cr 2 O 7 + Fe 2+ Cr 3+ + Fe Break into half-reactions Cr 2 O 7 Cr 3+ Fe 2+ Fe Balance atoms other than H and O Cr 2 O 7 2Cr 3+ Put in 2 coefficient to balance Cr Fe 2+ Fe 3+ Fe already balanced 24

25 Ex. 2 Ion-Electron Method in Acid 3. Balance O by adding H 2 O to the side that needs O Cr 2 O 7 2Cr H 2 O Left side has seven O atoms Right side has none Add seven H 2 O to right side Fe 2+ Fe 3+ No O to balance 25

26 Ex. 2 Ion-Electron Method in Acid 4. Balance H by adding H + to side that needs H 14H + + Cr 2 O 7 2Cr H 2 O Right side has fourteen H atoms Left side has none Add fourteen H + to left side Fe 2+ Fe 3+ No H to balance 26

27 Ex. 2 Ion-Electron Method in Acid 5. Balance net charge by adding electrons. 6e + 14H + + Cr 2 O 7 2Cr H 2 O Net Charge = 14(+1) +( 2) = 12 Net Charge = 2(+3)+7(0) = 6 6 electrons must be added to reactant side Fe 2+ Fe 3+ + e 1 electron must be added to product side Now both half-reactions balanced for mass and charge 27

28 Ex. 2 Ion-Electron Method in Acid 6. Make electron gain equal electron loss; then add half-reactions 6e + 14H + + Cr 2 O 7 2Cr H 2 O 6 [ Fe 2+ Fe 3+ + e ] 6e + 6Fe H + + Cr 2 O 7 6Fe Cr H 2 O + 6e 7. Cancel anything that's the same on both sides 6Fe H + + Cr 2 O 7 6Fe Cr H 2 O 28

29 Ion-Electron in Basic Solution The simplest way to balance an equation in basic solution Use steps 1 7 above, then 8. Add the same number of OH to both sides of the equation as there are H + 9. Combine H + and OH to form H 2 O 10. Cancel any H 2 O that you can from both sides 29

30 Ex. 2 Ion-Electron Method in Base Returning to our example of Cr 2 O 7 and Fe Add to both sides of equation the same number of OH as there are H +. 6Fe H + 6Fe3+ + 2Cr 3+ + Cr 2 O OH + 7H 2 O + 14 OH 9. Combine H + and OH to form H 2 O. 7 6Fe H 2 O 6Fe + Cr 2 O Cr H 2 O + 14OH 10. Cancel any H 2 O that you can 6Fe H 2 O 6Fe + Cr 2 O Cr OH 30

31 Example 3: Ion-Electron Method Balance the following equation in basic solution: MnO 4 + HSO 3 Mn 2+ + SO 4 1. Break it into half-reactions MnO 4 Mn 2+ HSO 3 SO 4 2. Balance atoms other than H and O MnO 4 Mn 2+ Balanced for Mn HSO 3 SO 4 Balanced for S 31

32 Example 3: Ion-Electron Method 3. Add H 2 O to balance O MnO 4 Mn H 2 O H 2 O + HSO 3 SO 4 4. Add H + to balance H 8H + + MnO 4 Mn H 2 O H 2 O + HSO 3 SO 4 + 3H + 32

33 Example 3: Ion-Electron Method 5. Balance net charge by adding electrons. 5e + 8H + + MnO 4 Mn H 2 O 8 (+1) + ( 1) = = +2 Add 5 e to reactant side H 2 O + HSO 3 SO 4 + 3H ( 1) = (+1) = +1 Add 2 electrons to product side + 2e 33

34 Example 3: Ion-Electron Method 6. Make electron gain equal electron loss 2 [ 5e + 8H + + MnO 4 Mn H 2 O H 2 O + HSO 3 SO 2 5 [ 4 + 3H + + 2e Must multiply Mn half-reaction by 2 Must multiply S half-reaction by 5 Now have 10 electrons on each side ] ] 34

35 Example 3: Ion-Electron Method 6. Then add the two half-reactions 10e + 16H + + 2MnO 4 2Mn H 2 O 5H 2 O + 5HSO 3 5SO H e 10e H + + 2MnO 4 + 5H 2 O + 5HSO 3 7. Cancel anything that is the same on both sides. H + + 2MnO 4 + 5HSO 3 2Mn H 2 O + 5SO 4 Balanced in acid 3 2Mn H 2 O + 5SO H e 35

36 Ex. 3 Ion-Electron Method in Base 8. Add same number of OH to both sides of equation as there are H + H + + 2MnO 4 2Mn H 2 O + + OH + 5HSO 3 5SO 4 + OH 9. Combine H + and OH to form H 2 O H 2 O + 2MnO 4 + 5HSO Cancel any H 2 O that you can 2MnO 4 + 5HSO 3 2Mn H 2 O + OH + 5SO 4 2 2Mn H 2 O + 5SO OH 36

37 Learning Check Balance each equation in the solution indicated by the skeletal reaction using the ion electron method. MnO 4 + C 2 O 4 MnO 2 + CO 3 + H + in acid 2MnO 4 + 3C 2 O 4 + 2H 2 O 2MnO 2 + 4H + + 6CO 3 ClO + VO 3 ClO 3 + V(OH) 3 + OH in base ClO + 4H 2 O + 2VO 3 ClO 3 + 2V(OH) 3 + 2OH 37

38 Acids as Oxidizing Agents Metals often react with acid Form metal ions and Molecular hydrogen gas Molecular equation Zn(s) + 2HCl(aq) H 2 (g) + ZnCl 2 (aq) Net ionic equation Zn(s) + 2H + (aq) H 2 (g) + Zn 2+ (aq) Zn is oxidized H + is reduced H + is the oxidizing reagent Zn is the reducing reagent 38

39 Oxidation of Metals by Acids Ease of oxidation process depends on metal Metals that react with HCl or H 2 SO 4 Easily oxidized by H + More active than hydrogen (H 2 ) e.g., Mg, Zn, alkali metals Mg(s) + 2H + (aq) Mg 2+ (aq) + H 2 (g) 2Na(s) + 2H + (aq) 2Na + (aq) + H 2 (g) Metals that don t react with HCl or H 2 SO 4 Not oxidized by H + Less active than H 2 e.g., Cu, Pt 39

40 Anion Determines Oxidizing Power Acids are divided into two classes: 1. Nonoxidizing Acids Anion is weaker oxidizing agent than H 3 O + Only redox reaction is 2H + + 2e H 2 or 2H 3 O + + 2e H 2 + 2H 2 O HCl(aq), HBr(aq), HI(aq) H 3 PO 4 (aq) Cold, dilute H 2 SO 4 (aq) Most organic acids (e.g., HC 2 H 3 O 2 ) 40

41 2. Oxidizing Acids Anion is stronger oxidizing agent than H 3 O + Used to react metals that are less active than H 2 No H 2 gas formed HNO 3 (aq) Concentrated Dilute Very dilute, with strong reducing agent H 2 SO 4 (aq) Hot, conc d, with strong reducing agent Hot, concentrated 41

42 Nitrate Ion as Oxidizing Agent A. Concentrated HNO 3 NO 3 more powerful oxidizing agent than H + NO 2 is product Partial reduction of N (+5 to +4) NO 3 (aq) + 2H + (aq) + e NO 2 (g) + H 2 O e.g., oxidation reduction Cu(s) + 2NO 3 (aq) + 4H + (aq) Cu 2+ (aq) + 2NO 2 (g) + 2H 2 O Reducing agent Oxidizing agent 42

43 Nitrate Ion as Oxidizing Agent B. Dilute HNO 3 NO 3 is more powerful oxidizing agent than H + NO is product Partial reduction of N (+5 to +2) NO 3 (aq) + 4H + (aq) + 3e NO(g) + 2H 2 O Used to react metals that are less active than H 2 e.g., Reaction of copper with dilute nitric acid 3Cu(s) + 8HNO 3 (dil, aq) 3Cu(NO 3 ) 2 (aq) + 2NO(g) + 4H 2 O 43

44 Reactions of Sulfuric Acid A. Hot, concentrated H 2 SO 4 Becomes potent oxidizer SO 2 is product Partial reduction of S (+6 to +4) SO 4 + 4H + + 2e SO 2 (g) + 2H 2 O e.g., Cu + 2H 2 SO 4 (hot, conc.) CuSO 4 + SO 2 + 2H 2 O B. Hot, concentrated with strong reducing agent H 2 S is product Complete reduction of S (+6 to 2) SO H + + 8e H 2 S(g) + 4H 2 O e.g., 4Zn + 5H 2 SO 4 (hot, conc.) 4ZnSO 4 + H 2 S + 4H 2 O 44

45 Redox Reactions of Metals Acids reacting with metal Special case of more general phenomena Single Replacement Reaction Reaction where one element replaces another A + BC AC + B 1. Metal A can replace metal B If A is more active metal, or 2. Nonmetal A can replace nonmetal C If A is more active than C 45

46 Single Replacement Reaction Left = Zn(s) + CuSO 4 (aq) Center = Cu 2+ (aq) reduced to Cu(s) Zn(s) oxidized to Zn 2+ (aq) Right = Cu(s) plated out on Zn bar Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) 46

47 Single Replacement Reaction Zn 2+ ions take place of Cu 2+ ions in solution Cu atoms take place of Zn atoms in solid Cu 2+ oxidizes Zn 0 to Zn 2+ Zn 0 reduces Cu 2+ to Cu 0 More active Zn 0 replaces less active Cu 2+ Zn 0 is easier to oxidize! 47

48 Activity Series of Metals Cu less active, can't replace Zn 2+ Can't reduce Zn 2+ Cu(s) + Zn 2+ (aq) No reaction General phenomenon Metal that is more easily oxidized will displace one that is less easily oxidized from its compounds Activity Series Metals at bottom more easily oxidized (more active) than those at top This means that given metal ion will be displaced from its compounds by any metal below it in table 48

49 Activity Series of Metals 49

50 How Activity Series Generated 2H + (aq) + Sr(s) Sr 2+ (aq) + H 2 (g) H + oxidizes Sr 0 to Sr 2+ Sr 0 reduces H + to H 2 More active Sr 0 replaces less active H + Sr 0 is easier to oxidize! H 2 (g) + Sr 2+ (aq) NO REACTION! Why? H 2 less active, can't replace Sr 2+ Can't reduce Sr 2+ 50

51 Learning Check: Metal Activity Using the following observations, rank these metals from most reactive to least reactive: Cu(s) + HCl(aq) no reaction Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Mg(s) + ZnCl 2 (aq) MgCl 2 (aq) + Zn(s) Mg > Zn > H > Cu 51

52 Reactivity Varies by Metal Metals at very bottom of table Very strong reducing agents Very easily oxidized Na down to Cs Alkali and alkaline earth metals React with H 2 O as well as H + 2Na(s) + 2H 2 O H 2 (g) + 2NaOH(aq) 52

53 Reactivity Varies by Metal Ag slight reaction (top of activity series) not pictured 2HCl(aq) + Ag(s) 2AgCl(aq) + H 2 (g) Fe very little reactivity (near top of activity series) 2HCl(aq) + Fe(s) FeCl 2 (aq) + H 2 (g) Zn somewhat reactive (middle of activity series) 2HCl(aq) + Zn(s) ZnCl 2 (aq) + H 2 (g) Mg very reactive (bottom of activity series) 2HCl(aq) + Mg(s) MgCl 2 (aq) + H 2 (g) 53

54 Using Activity Series to Predict Reactions If M is below H Can displace H from solutions containing H + 2H + (aq) H 2 (g) If M is above H Doesn't react with nonoxidizing acids HCl, H 3 PO 4, etc. In general Metal below replaces ion above 54

55 Uses of Activity Series Predictive tool for determining outcome of single replacement reactions Given a metal (M ) and the ion of a different metal (M ' n+ ) Look at chart and draw arrow from M to M ' n+ Arrow that points up from bottom left to top right will occur Arrow that points down from top left to bottom right will NOT occur 55

56 Learning Check 2Au 3+ (aq) + 3Ca(s) Au(s) + Ca 2+ (aq) 2Au(s) + 3Ca 2+ (aq) rxn occurs NO reaction Sn(s) + Na + (aq) NO reaction Mn(s) + Co 2+ (aq) Cu(s) + H + (aq) Co(s) + Mn 2+ (aq) rxn occurs NO reaction 56

57 Reaction of Nonmetals with O 2 Many nonmetals react directly with O 2 to form nonmetal oxides Sulfur reacts with O 2 Forms SO 2 S(s) + O 2 (g) 2SO 2 (g) Nitrogen reacts with O 2 Forms various oxides NO, NO 2, N 2 O, N 2 O 3, N 2 O 4, and N 2 O 5 Dinitrogen oxide, N 2 O Laughing gas used by dentists Propellant in canned whipped cream 57

58 Oxygen as an Oxidizing Agent Oxygen reacts with many substances Combustion Rapid reaction of substance with oxygen that gives off both heat and light Hydrocarbons are important fuels Products depend on how much O 2 is available 1. Complete Combustion O 2 plentiful CO 2 and H 2 O products e.g., CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O 2C 8 H 18 (g) + 25O 2 (g) 16CO 2 (g) + 18H 2 O 58

59 Oxidation of Organic Compounds 2. Incomplete Combustion Not enough O 2 a. Limited O 2 supply CO(g) is the carbon product 2CH 4 (g) + 3O 2 (g) 2CO(g) + 4H 2 O b. Very limited O 2 C(s) is the carbon product e.g., CH 4 (g) + O 2 (g) C(s) + 2H 2 O Gives tiny black particles Soot lamp black Component of air pollution 59

60 Oxidation of Organic Compounds 3. Combustion of Organics containing O Still produce CO 2 and H 2 O Need less added O 2 C 12 H 22 O 11 (s) + 12O 2 (g) 12CO 2 (g) + 11H 2 O 4. Combustion of Organics containing S Produce SO 2 as product 2C 4 H 9 SH + 15O 2 (g) 8CO 2 (g) + 10H 2 O + 2SO 2 (g) SO 2 turns into acid rain when mixed with water SO 2 oxidized to SO 3 SO 3 reacts with H 2 O to form H 2 SO 4 60

61 Non-combustion reactions of O 2 Reactions with O 2 are not always combustion Example: luminol reaction Fe in blood catalyzes 2H 2 O 2 2H 2 O + O 2 O 2 reacts with luminol (C 8 H 5 N 3 O 2 ) C 8 H 5 N 3 O 2 + O 2 C 8 H 5 NO 4 + N 2 Energy from the reaction released as light So the appearance of light suggests blood is present 61

62 Reaction of Metals with O 2 Corrosion Direct reaction of metals with O 2 Many metals corrode or tarnish when exposed to O 2 e.g., 2Mg(s)+ O 2 (g) 2MgO(s) 4Al(s) + 3O 2 (g) 4Fe(s) + 3O 2 (g) 4Ag(s) + O 2 (g) 2Al 2 O 3 (s) 2Fe 2 O 3 (s) 2Ag 2 O(s) 62

63 Learning Check: Complete Following Reactions Aluminum metal and oxygen gas forms aluminum oxide solid 4Al(s) + 3O 2 (g) 2Al 2 O 3 (s) Solid sulfur (S 8 ) burns in oxygen gas to make gaseous sulfur trioxide S 8 (s) + 12O 2 (g) 8SO 3 (g) Copper metal is heated in oxygen to form black copper(ii) oxide solid 2Cu(s) + O 2 (g) 2CuO(s) 63

64 Stoichiometry in Redox Reactions Like any other stoichiometry problem Balance redox reaction Use stoichiometric coefficients to relate mole of one substance to moles of another Types of problems Start with mass or volume of one reactant and find mass or volume of product Perform titrations Have limiting reactant calculations Calculate percentage yields 64

65 Stoichiometry in Redox Reactions Example: How many grams of Na 2 SO 3 (126.1 g/mol) are needed to completely react with 12.4 g of K 2 Cr 2 O 7 (294.2 g/mol)? First need balanced redox equation 8H + (aq) + Cr 2 O 7 (aq) + 3SO 3 (aq) 3SO 4 (aq) + 2Cr 3+ (aq) + 4H 2 O Then do calculations 1. g K 2 Cr 2 O 7 moles K 2 Cr 2 O 7 moles Cr 2 O 7 (aq) 2. moles Cr 2 O 7 (aq) moles 3SO 4 (aq) 3. moles SO 3 (aq) moles Na 2 SO 3 g Na 2 SO 3 65

66 Ex. Redox Stoichiometry (cont.) grams K 2 Cr 2 O 7 moles K 2 Cr 2 O 7 moles Cr 2 O 7 (aq) 12.4 g K 2 Cr 2 O 7 1 mol K mol Cr 2 O 7 moles Cr 2 O 7 (aq) moles 3SO 3 (aq) mol Cr 2 O moles SO 3 (aq) moles Na 2 SO 3 g Na 2 SO g K mol Cr 2 Cr 2 Cr 2 O 2 3 mol SO 7 O O mol Cr2O 1 mol K Cr O mol SO mol Na2SO 0.126mol SO3 1 mol SO g Na 1 mol Na 2 2 SO SO g Na 2 SO 3 66

67 Redox Titrations Equivalence point reached when the number of moles of oxidizing and reducing agents have been mixed in the correct stoichiometric ratio No simple indicators to detect endpoints Three very useful oxidizing agents that change color 1. KMnO 4 : Deep purple of MnO 4 fades to almost colorless Mn 2+ (very pale pink) 2. K 2 Cr 2 O 7 : Bright yellow orange of Cr 2 O 7 changes to pale blue green of Cr IO 3 : When reduced to I 2 (s) in presence of I, forms I 3 which forms dark blue complex with starch 67

68 Example: Redox Titration I reacts with IO 3 in acidic solution to form I 2 (s). If ml of M I is needed to titrate ml of a solution containing IO 3, what is the molarity of the solution? 1. Write Unbalanced Equation I (aq) + IO 3 (aq) I 2 (s) I (aq) is oxidized to I 2 IO 3 (aq) is reduced to I 2 68

69 Ex. Redox Titration (cont) 2. Balance Equation Note: the reaction occurs in acidic solution 5 [ 2I (aq) I 2 (s) + 2e ] 2IO 3 (aq) + 12H + (aq) + 10e I 2 (s) + 6H 2 O 10I (aq) + 2IO 3 (aq) + 12H + (aq) 6I 2 (s) + 6H 2 O Not done as not lowest whole number coefficients 5I (aq) + IO 3 (aq) + 6H + (aq) 3I 2 (s) + 3H 2 O 69

70 Example: Redox Titrations Calculate mmol of I titrated mmol I 12.34mL I 7.007mmol I 1 ml I Convert to mmol of IO 3 present 1 mmol IO mmol I 5 mmol I Convert to M of IO 3 solution 1.401mmol IO 25.00mL IO 3 3 = M IO mmol IO3 70

71 Example: Ore Analysis A g sample of tin ore was dissolved in acid solution converting all the tin to tin(ii). In a titration, 8.08 ml of M KMnO4 was required to oxidize the tin(ii) to tin(iv). What was the percentage tin in the original sample? 3Sn 2+ (aq) + 2MnO 4 (aq) + 8H + (aq) 3Sn 4+ (aq) + 2MnO 2 (s) + 4H 2 O M of KMnO 4 V = mol KMnO 4 mol KMnO 4 mol Sn/mol KMnO 4 = mol Sn 2+ mol Sn 2+ MM = g Sn 2+ in sample % Sn = g Sn/g sample 100% 71

72 Example: Ore Analysis (cont.) M of KMnO 4 ml KMnO 4 = mmol KMnO M KMnO ml = mmol KMnO4 mmol KMnO 4 mmol MnO 4 mmol Sn 2+ 1 mmol MnO 0.404mmol KMnO 4 1 mmol KMnO = mmol Sn 2+ Mol Sn 2+ g/mol = g Sn in original sample 2 1 mmol Sn 0.606mmol Sn 2 1 mmol Sn = g Sn %Sn = g Sn/ g sample 100% g Sn 100 = 24.0% Sn g ore mmol Sn 2 mmol MnO 118.7mg Sn 1 mmol Sn 4 1 g 1000mg 72