1. Chapter 4: The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Precipitation Reactions and Acid-Base Reactions 4.3 Oxidation - Reduction (Redox) Reactions 4.4 Counting Reactants and Products in Precipitation, Acid-Base, and Redox Processes 4.5 Reversible Reactions: An Introduction to Chemical Equilibrium C142B Lecture Notes Campbell / Callis

2. The Role of Water as a Solvent: The solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong Electrolytes. NaCl(s) + H 2 O(l) Na + (aq) + Cl - (aq) When sodium chloride dissolves into water the ions become solvated, and are surrounded by water molecules. These ions are called “aqueous” and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

3. Fig 4.1 (P135) Electrical Conductivity of Ionic Solutions

4. Fig. 4.2

5. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I Problem: How many moles of each ion are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 10 21 formula units of iron (III) chloride dissolved in water d) 75.0 mL of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water a) Na 2 CO 3 (s) 2 Na + (aq) + CO 3 -2 (aq) moles of Na + = 4.0 moles Na 2 CO 3 x = 8.0 moles Na + and 4.0 moles of CO 3 -2 are present H2OH2O 2 mol Na + 1 mol Na 2 CO 3

6. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II b) RbF(s) Rb + (aq) + F - (aq) H2OH2O moles of RbF = c) FeCl 3 (s) Fe +3 (aq) + 3 Cl - (aq) H2OH2O moles of FeCl 3 = 9.32 x 10 21 formula units x moles of Cl - = moles Fe +3 =

7. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II b) RbF(s) Rb + (aq) + F - (aq) H2OH2O moles of RbF = 46.5 g RbF x = 0.445 moles RbF 1 mol RbF 104.47 g RbF thus, 0.445 mol Rb + and 0.445 mol F - are present c) FeCl 3 (s) Fe +3 (aq) + 3 Cl - (aq) H2OH2O moles of FeCl 3 = 9.32 x 10 21 formula units 1 mol FeCl 3 6.022 x 10 23 formula units FeCl 3 x = 0.0155 mol FeCl 3 moles of Cl - = 0.0155 mol FeCl 3 x = 0.0465 mol Cl - 3 mol Cl - 1 mol FeCl 3 and 0.0155 mol Fe +3 are also present.

8. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III d) ScBr 3 (s) Sc +3 (aq) + 3 Br - (aq) H2OH2O Converting from volume to moles: Moles of ScBr 3 = 75.0 mL x x 1 L 10 3 mL Moles of Br - = e) (NH 4 ) 2 SO 4 (s) 2 NH 4 + (aq) + SO 4 - 2 (aq) H2OH2O Moles of NH 4 + = 7.8 moles (NH 4 ) 2 SO 4 x = 15.6 mol NH 4 + 2 mol NH 4 + 1 mol(NH 4 ) 2 SO 4 and 7.8 mol SO 4 - 2 are also present.

9. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III d) ScBr 3 (s) Sc +3 (aq) + 3 Br - (aq) H2OH2O Converting from volume to moles: Moles of ScBr 3 = 75.0 mL x x = 0.042 mol ScBr 3 1 L 10 3 mL 0.56 mol ScBr 3 1 L Moles of Br - = 0.042 mol ScBr 3 x = 0.126 mol Br - 3 mol Br - 1 mol ScBr 3 0.042 mol Sc +3 are also present e) (NH 4 ) 2 SO 4 (s) 2 NH 4 + (aq) + SO 4 - 2 (aq) H2OH2O Moles of NH 4 + = 7.8 moles (NH 4 ) 2 SO 4 x = 15.6 mol NH 4 + 2 mol NH 4 + 1 mol(NH 4 ) 2 SO 4 and 7.8 mol SO 4 - 2 are also present.

10. Fig. 4.3

11. The Solubility of Ionic Compounds in Water The solubility of Ionic Compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compounds may be several orders of magnitude less than ones that are called “soluble” in water, for example: Solubility of NaCl in water at 20 o C = 365 g/L Solubility of MgCl 2 in water at 20 o C = 542.5 g/L Solubility of AlCl 3 in water at 20 o C = 699 g/L Solubility of PbCl 2 in water at 20 o C = 9.9 g/L Solubility of AgCl in water at 20 o C = 0.009 g/L Solubility of CuCl in water at 20 o C = 0.0062 g/L

12. The Solubility of Covalent Compounds in Water The covalent compounds that are very soluble in water are the ones with -OH group in them and are called “Polar” and can have strong polar (electrostatic)interactions with water. Examples are compound such as table sugar, sucrose (C 12 H 22 O 11 ); beverage alcohol, ethanol (C 2 H 5 -OH); and ethylene glycol (C 2 H 6 O 2 ) in antifreeze. CH H H O H Methanol = Methyl Alcohol Other covalent compounds that do not contain a polar center, or the -OH group are considered “Non-Polar”, and have little or no interactions with water molecules. Examples are the hydrocarbons in Gasoline and Oil. This leads to the obvious problems in Oil spills, where the oil will not mix with the water and forms a layer on the surface! Octane = C 8 H 18 and / or Benzene = C 6 H 6

13. Acids - A group of Covalent molecules which loose Hydrogen ions to water molecules in solution HI(g) + H 2 O(l) H + (aq) + I - (aq) When gaseous hydrogen Iodide dissolves in water, the attraction of the oxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the Iodide ion for the hydrogen atom, and it is lost to the water molecule to form an Hydronium ion and an Iodide ion in solution. We can write the Hydrogen atom in solution as either H + (aq) or as H 3 O + (aq) they mean the same thing in solution. The presence of a Hydrogen atom that is easily lost in solution is an “Acid” and is called an “acidic” solution. The water (H 2 O) could also be written above the arrow indicating that the solvent was water in which the HI was dissolved. HI(g) + H 2 O(l) H 3 O + (aq) + I - (aq) HI(g) H + (aq) + I - (aq) H 2 O(l)

14. Fig. 4.4

15. Strong Acids and the Molarity of H + Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of pure H 2 SO 4 into sufficient water to produce 2.30 Liters of sulfuric acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H + are released for every mole of acid: H 2 SO 4 (l) + 2 H 2 O(l) 2 H 3 O + (aq) + SO 4 - 2 (aq) Molarity of H 3 O + (or simply H + ) = Moles H 2 SO 4 = = 1.58 moles H 2 SO 4 155 g H 2 SO 4 1 mole H 2 SO 4 98.09 g H 2 SO 4 x Molarity of SO 4 - 2 =

16. Strong Acids and the Molarity of H + Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of pure H 2 SO 4 into sufficient water to produce 2.30 Liters of sulfuric acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H + are released for every mole of acid: H 2 SO 4 (l) + 2 H 2 O(l) 2 H 3 O + (aq) + SO 4 - 2 (aq) Molarity of H + = 2 x 0.687 mol H + / 2.30 liters = 0.597 Molar in H + Moles H 2 SO 4 = = 1.58 moles H 2 SO 4 155 g H 2 SO 4 1 mole H 2 SO 4 98.09 g H 2 SO 4 x 1.58 mol SO 4 -2 2.30 L solution Molarity of SO 4 - 2 = = 0.687 Molar in SO 4 - 2

17. Fig. 4.5

18. Precipitation Reactions: A solid product is formed When ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together. Pb(NO 3 ) (aq) + NaI(aq) Pb +2 (aq) + 2 NO 3 - (aq) + Na + (aq) + I - (aq) When we add these two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have Lead Nitrate and Sodium Iodide, or Lead Iodide and Sodium Nitrate formed, to determine which will happen we must look at the Solubility Rules (P 141) to determine what could form. Rule 3a in Table 4.1 indicates that Lead Iodide will be insoluble, so a precipitate will form! The remainder (sodium sulfate) is soluble (Rule 4a). Pb(NO 3 ) 2 (aq) + 2 NaI(aq) PbI 2 (s) + 2 NaNO 3 (aq)

19. Precipitation Reactions: Will a Precipitate form? If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate? KCl(aq) + NH 4 NO 3 (aq) = K + (aq) + Cl - (aq) + NH 4 + (aq) + NO 3 - (aq) By exchanging cations and anions we see that we could have Potassium Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility rules, Table 4.1: If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate, will we get a precipitate? Table 4.1 shows that: Therefore we get: Na 2 SO 4 (aq) + Ba(NO 3 ) 2 (aq) BaSO 4 ( ) + 2 NaNO 3 ( )

20. Precipitation Reactions: Will a Precipitate form? If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate? KCl(aq) + NH 4 NO 3 (aq) = K + (aq) + Cl - (aq) + NH 4 + (aq) + NO 3 - (aq) By exchanging cations and anions we see that we could have Potassium Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility rules, Table 4.1 shows all possible products as soluble, so there is no net reaction! KCl(aq) + NH 4 NO 3 (aq) = No. Reaction! If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate, will we get a precipitate? Table 4.1 shows that Barium Sulfate is insoluble, therefore we will get a precipitate! Na 2 SO 4 (aq) + Ba(NO 3 ) 2 (aq) BaSO 4 (s) + 2 NaNO 3 (aq)

21. Fig. 4.6

23. Predicting Whether a Precipitation Reaction Occurs (see Table 4.1); Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Ca(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (s) + NaNO 3 (aq) Ca 2+ (aq) + 2 NO 3 - (aq) +2 Na + (aq) + SO 4 -2 (aq) CaSO 4 (s) + 2Na + (aq) + 2 NO 3 - (aq) Molecular Equation Total Ionic Equation Net Ionic Equation Ca 2+ (aq) + SO 4 -2 (aq) CaSO 4 (s) Spectator Ions are Na + and NO 3 - b) Ammonium Sulfate and Magnesium Chloride are added together. In exchanging ions, no precipitates will be formed, accd. To Table 4.1, so there will be no reactions occurring! All ions are spectator ions!

24. Acid - Base Reactions : Neutralization Rxns. An Acid is a substance that produces H + (H 3 O + ) ions when dissolved in water. A Base is a substance that produces OH - ions when dissolved in water. Acids and Bases are electrolytes, and their strength is categorized in terms of their degree of dissociation in water to make hydronium or hydroxide ions, resp.. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids and bases dissociate only to a tiny extent (<<100%) and are weak electrolytes. The generalized reaction between an Acid and a Base is: HX(aq) + MOH(aq) MX(aq) + H 2 O(l) Acid + Base = Salt(aq) + Water

25. Fig. 4.7

26. Table 4.2 (P 143) Selected Acids and Bases Acids Bases Strong (100% to H + ) Strong (100% to OH - ) Hydrochloric, HCl Sodium hydroxide, NaOH Hydrobromic, HBr Potassium hydroxide, KOH Hydroiodoic, HI Calcium hydroxide, Ca(OH) 2 Nitric acid, HNO 3 Strontium hydroxide, Sr(OH) 2 Sulfuric acid, H 2 SO 4 Barium hydroxide, Ba(OH) 2 Perchloric acid, HClO 4 Weak (tiny % to H + ) Weak (tiny % yield of OH - ) Hydrofluoric, HF Ammonia, NH 3 Phosphoric acid, H 3 PO 4 Acidic acid, CH 3 COOH (or HC 2 H 3 O 2 )

27. Writing Balanced Equations for Neutralization Reactions - I Problem: Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) Calcium Hydroxide(aq) and HydroIodic acid(aq) b) Lithium Hydroxide(aq) and Nitric acid(aq) c) Barium Hydroxide(aq) and Sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution: a) Ca(OH) 2 (aq) + 2HI(aq) CaI 2 (aq) + 2H 2 O(l) Ca 2+ (aq) + 2 OH - (aq) + 2 H + (aq) + 2 I - (aq) Ca 2+ (aq) + 2 I - (aq) + 2 H 2 O(l) 2 OH - (aq) + 2 H + (aq) 2 H 2 O(l)

28. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO 3 (aq) c) Ba(OH) 2 (aq) + H 2 SO 4 (aq)

29. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO 3 (aq) LiNO 3 (aq) + H 2 O(l) Li + (aq) + OH - (aq) + H + (aq) + NO 3 - (aq) Li + (aq) + NO 3 - (aq) + H 2 O(l) OH - (aq) + H + (aq) H 2 O(l) c) Ba(OH) 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2 H 2 O(l) Ba 2+ (aq) + 2 OH - (aq) + 2 H + (aq) + SO 4 2 - (aq) BaSO 4 (s) + 2 H 2 O(l)

30. Fig. 4.8

31. Finding the Concentration of Acid from an Acid - Base Titration Volume (L) of base (difference in buret readings) needed to titrate Moles of base needed to titrate Moles of acid which were titrated M (mol/L) of original acid soln. M (mol/L) of base molar ratio volume (L) of acid

32. Potassium Hydrogenphthalate KHC 8 H 4 O 4 C O OK+K+ C O H O K+K+ O C O O C O H+H+ = “KHP” = a common acid used to titrate bases (M = 204.2 g/mol)

33. Finding the Concentration of Base from an Acid - Base Titration - I Problem: A titration is performed between Sodium Hydroxide and KHP (204.2 g/mol) to standardize the base solution, by placing 50.00 mg of solid Potassium Hydrogenphthalate in a flask with a few drops of an indicator. A buret is filled with the base, and the initial buret reading is 0.55 mL; at the end of the titration the buret reading is 33.87 mL. What is the concentration of the base? Plan: Use the molar mass of KHP to calculate the number of moles of the acid, from the balanced chemical equation, the reaction is equal molar, so we know the moles of base, and from the difference in the buret readings, we can calculate the molarity of the base. Solution: The rxn is: KHP(aq) + OH - (aq) --> KP - (aq) + H 2 O or HKC 8 H 4 O 4 (aq) + OH - (aq) KC 8 H 4 O 4 - (aq) + H 2 O(aq)

34. Finding the Concentration of Base from an Acid - Base Titration - II moles KHP = 50.00 mg KHP Volume of base = Final buret reading - Initial buret reading = one mole of H + : one mole of OH - ; therefore ____________ moles of KHP will titrate _____________ moles of NaOH. molarity of NaOH = = moles L molarity of base =

35. Finding the Concentration of Base from an Acid - Base Titration - II moles KHP = x = 0.00024486 mol KHP 50.00 mg KHP 204.2 g KHP 1 mol KHP 1.00 g 1000 mg Volume of base = Final buret reading - Initial buret reading = 33.87 mL - 0.55 mL = 33.32 mL of base one mole of H + : one mole of OH - ; therefore 0.00024486 moles of KHP will titrate 0.00024486 moles of NaOH. molarity of NaOH = = 0.07348679 moles per liter 0.00024486 moles 0.03332 L molarity of base = 0.07349 M

36. Fig. 4.9

37. Fig. 4.10

38. Fig. 4.11

40. Highest and Lowest oxidation numbers of Chemically reactive main-group Elements 1 +1 H non-metals metalloids metals F Cl Br I At ONCB 1A +1 2A +2 2 3 4 5 6 7 LiBe 3A +3 +4 -4 +4 4A5A6A7A +5+6+7 -3-2 S Se Te Po P As Sb Bi Si Ge Sn Pb Al Ga In Tl NaMg KCa RbSr CsBa RaFr Period Fig. 4.12

41. Fig. 4.13

42. Rn Xe Kr Ar NeFONCBBeLi HeH Period 1 2 3 4 5 6 IA IIAIIIAIVAVAVIAVIIA VIIIA Common Ox #s: Main Group Elements Na K Rb Cs Mg Ca Sr Ba Al Ga In Tl Si Ge Sn Pb P As Sb Bi S Se Te Po Cl Br I At +1 -1 +1 +2 +3 +3, +2 +1 +3,+1 +4,+2 -1,-4 +4,-4 +4,+2 -4 +4,+2, -4 +4,+2 +7,+5 +3,+1 +7,+5 +3,+1 +7,+5 +3,+1 +7,+5 +3,+1 +2 +6,+4 +2 -1,-2 +6,+4 +2,-2 +6,+4 -2 +6,+4 -2 +6,+4 +2,-2 +3 +5,+3 -3 +5,+3 -3 +5,+3 -3 all from +5 -3

43. Transition Metals Possible Oxidation States IIIBIVBVBVIBVIIBIBIIB VIIIB ScTiVCrMnFeCoNiCuZn YZrNbMoTcRuRhPdAgCd HgAuPtIrOsReWTaHfLa +3 +4,+3 +2 +4,+3 +2 +2,+1 +1 +3,+1 +2 +4,+2 +3,+2 +7,+6 +4,+3 +2 +6,+3 +2 +5,+4 +3+2 +5,+4 +2 +4,+3 +1 +8,+5 +4,+3 +7,+5 +4 +6,+5 +4,+3 +5,+4 +3 +6,+5 +4 +7,+5 +4 +8,+6 +4,+3 +2

44. Determining the Oxidation Number of an Element in a Compound Problem: Determine the oxidation number (Ox. No.) of each element in the following compounds. a) Iron III Chloride b) Nitrogen Dioxide c) Sulfuric acid Plan: We apply the rules in Table 4.3, always making sure that the Ox. No. values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge. Solution: a) FeCl 3 This compound is composed of monoatomic ions. The Ox. No. of Cl - is -1, for a total of -3. Therefore the Fe must be +3. b) NO 2 c) H 2 SO 4

45. Determining the Oxidation Number of an Element in a Compound Problem: Determine the oxidation number (Ox. No.) of each element in the following compounds. a) Iron III Chloride b) Nitrogen Dioxide c) Sulfuric acid Plan: We apply the rules in Table 4.3, always making sure that the Ox. No. values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge. Solution: a) FeCl 3 This compound is composed of monoatomic ions. The Ox. No. of Cl - is -1, for a total of -3. Therefore the Fe must be +3. b) NO 2 The Ox. No. of oxygen is -2 for a total of -4. Since the Ox. No. in a compound must add up to zero, the Ox. No. of N is +4. c) H 2 SO 4 The Ox. No. of H is +1, so the SO 4 2- group must sum to -2. The Ox. No. of each O is -2 for a total of -8. So the Sulfur atom is +6.

46. Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the Rx: a) Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g) b) S 8 (s) + 12 O 2 (g) 8 SO 3 (g) c) NiO(s) + CO(g) Ni(s) + CO 2 (g) Plan: First we assign an oxidation number (O.N.) to each atom (or ion) based on the rules in Table 4.3. A reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction). A reactant is the oxidizing agent if it contains an atom that is reduced ( O.N. decreased). Solution: a) Assigning oxidation numbers: Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g) 0 +1 +2 0 HCl is the oxidizing agent, and Zn is the reducing agent!

47. Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: S 8 (s) + 12 O 2 (g) 8 SO 3 (g) ____ is the reducing agent and ____ is the oxidizing agent. c) Assigning oxidation numbers: NiO (s) + CO (g) Ni (s) + CO 2 (g) ___ is the reducing agent and ____ is the oxidizing agent.

48. Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: S 8 (s) + 12 O 2 (g) 8 SO 3 (g) 00 +6 -2 S 8 is the reducing agent and O 2 is the oxidizing agent c) Assigning oxidation numbers: S [0] S[+6] S is Oxidized O[0] O[-2] O is Reduced NiO (s) + CO (g) Ni (s) + CO 2 (g) +2 -2 +2 -2 0 +4 -2 Ni[+2] Ni[0] Ni is Reduced C[+2] C[+4] C is oxidized CO is the reducing agent and NiO is the oxidizing agent

49. Balancing REDOX Equations: The oxidation number method Step 1) Assign oxidation numbers to all elements in the equation. Step 2) From the changes in oxidation numbers, identify the oxidized and reduced species. Step 3) Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. Draw tie-lines between these atoms to show electron changes. Step 4) Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients. Step 5) Complete the balancing by inspection, adding states of matter.

50. REDOX Balancing using Ox. No. Method - I ___ H 2 (g) +___ O 2 (g) ___ H 2 O (g) 0 +1- 1 e - 0-2 +2 e - electrons lost must = electrons gained therefore multiply Hydrogen reaction by 2! and we are balanced! 22

51. REDOX Balancing Using Ox. No. Method - II Fe +2 (aq) + MnO 4 - (aq) + H 3 O + (aq) Fe +3 (aq) + Mn +2 (aq) + H 2 O (aq) -1e - +2+3 +7 +2+5 e - 5 Fe +2 (aq) + MnO 4 - (aq) +__ H 3 O + (aq) 5 Fe +3 (aq) + Mn +2 (aq) +__ H 2 O (aq) Multiply Fe +2 & Fe +3 by ___ to correct for the electrons gained by the Manganese. __Fe +2 (aq) + MnO 4 - (aq) + H 3 O + (aq) __ Fe +3 (aq) + Mn +2 (aq) + H 2 O (aq) Next balance the # O and H atoms:

52. REDOX Balancing Using Ox. No. Method - II Fe +2 (aq) + MnO 4 - (aq) + H 3 O + (aq) Fe +3 (aq) + Mn +2 (aq) + H 2 O (aq) -1e - +2+3 +7 +2+5 e - 5 Fe +2 (aq) + MnO 4 - (aq) +8 H 3 O + (aq) 5 Fe +3 (aq) + Mn +2 (aq) +12 H 2 O (aq) Multiply Fe +2 & Fe +3 by five to correct for the electrons gained by the Manganese. 5 Fe +2 (aq) + MnO 4 - (aq) + H 3 O + (aq) 5 Fe +3 (aq) + Mn +2 (aq) + H 2 O (aq) Next balance the # O and H atoms: Make 4 water molecules from the oxygens of the MnO 4 -, which requires 8 protons from the acid (H 3 O + ions). These 8 H 3 O + ions will also make 8 more water molecules, for a total of 4+8 = 12 water molecules.