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Solutions Labs Dry Lab 4 Oxidation-Reduction Equations #9 A Volumetric Analysis #15 Bleach Analysis Chemical equations Chapters 8-11.

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Presentation on theme: "Solutions Labs Dry Lab 4 Oxidation-Reduction Equations #9 A Volumetric Analysis #15 Bleach Analysis Chemical equations Chapters 8-11."— Presentation transcript:

1 Solutions Labs Dry Lab 4 Oxidation-Reduction Equations #9 A Volumetric Analysis #15 Bleach Analysis Chemical equations Chapters 8-11

2 1/21/20142 Will the substances mix? NaNO 3 and H 2 O –Miscible-ionic + polar C 6 H 14 and H 2 O –Immiscible-nonpolar + polar I 2 and C 6 H 14 –Miscible-nonpolar + nonpolar I 2 and H 2 O –immiscible-nonpolar + polar

3 1/21/20143 bent molecule o Aqueous solutions- solvent is water Polar molecule-unequal charge distribution gives water ability to dissolve many compounds

4 1/21/ –Strong forces in positive and negative ions of solid are replaced by strong water-ion interactions Hydration –Positive ends of water molecules are attracted to negatively charged anions –Negative ends of water molecules are attracted to positively charged cations

5 1/21/20145 Solution homogeneous mixture of two or more substances Solute- –If it and solvent in same phase, one in lesser amount –If it and solvent in different phases, one that changes phase –One dissolved Solvent- –If it and solute in same phase, one in greater amount –If it and solute in different phases, one retaining phase –One doing dissolving

6 1/21/20146 Electrolytes and Nonelectrolytes Many aqueous solutions of ionic compounds conduct electricity whereas water itself essentially does not conduct-electrolyte solutions Dissociation-ionic substances break apart completely into ions when dissolved in water Ionization-some molecular compounds (nonionic) break apart into ions when dissolved in water

7 Solvation Solvation of ionic compounds Polarity of water solvatedIons surrounded by water molecules prevent recombining of ions (solvated) Ions/shells free to move around, so dispersed uniformly throughout solution 1/21/20147

8 8 Electrolyte Conductivity Degree Examples of Dissociation Stronghightotal strong acids (HCl, HNO 3 ) many ionic salts (NaCl, Sr(NO 3 ) 3 -ions) strong bases (NaOH, Ba(OH) 2, other IA/IIA hydroxides) Weak low to partial weak organic acids moderate (tap water, HCO 2 H- formic acid, C 2 H 3 O 2 ) weak bases (NH 3 ) Nonnoneclose to zero sugar, sugar solution, AgCl, Fe 2 O 3 (neutral)

9 1/21/20149 Complete following dissociation equations: (all in water) CaCl 2 (s) Ca 2+ (aq) + 2Cl - (aq) Fe(NO 3 ) 3 (s) KBr(s) (NH 4 ) 2 Cr 2 O 7 (s) Strong, Weak, or Nonelectrolytes HClO 4 C 6 H 12 LiOH NH 3 CaCl 2 HC 2 H 3 O 2

10 Molarity M = moles of solute Liter of solution Moles = millimoles = micromoles liter milliliter microliter BUT Moles does not equal millimoles or moles liter liter microliter

11 1/21/ What is the molarity of a solution in which 3.57 g of sodium chloride, NaCl, is dissolved in enough water to make 25.0 ml of solution? Convert from mass of NaCl to moles of NaCl. –3.57 g NaCl x 1 mol NaCl = mol NaCl g NaCl Convert 25.0 mL to L and substitute these two quantities into the defining equation for molarity. Molarity = mol NaCl = 2.44 M NaCl l L solution We read this as 2.44 molar NaCl. It is important to understand that molarity means moles of solute per liter of solution, and not per liter of solvent.

12 1/21/ How many mL of solution are necessary if we are to have a 2.48 M NaOH solution that contains g of the dissolved solid? g NaOH 1 mol NaOH = mol NaOH g NaOH 2.48 M NaOH = mol NaOH =.318 L = 318.mL NaOH x L

13 1/21/ Determine the molarity of Cl - ion in a solution prepared by dissolving 9.82 g of CuCl 2 in enough water to make 600 mL of solution g CuCl 2 1 mol CuCl 2 = mol CuCl g CuCl 2 M = mol CuCl 2 = M CuCl L Ion = 2 mol Cl - solute 1 mol CuCl M CuCl 2 x 2 mol Cl - = M L solution 1 mol CuCl 2

14 1/21/ Calculate the mass of NaCl needed to prepare 175 mL of a M NaCl solution M = x mol.175 L x mol = mol NaCl mol NaCl g NaCl = 5.11 g NaCl 1 mol NaCl

15 1/21/ Solution Concentration Amounts of chemical present in solutions-expressed as concentration (amount of solute dissolved in given amount of solvent) Most common unit of concentration used for aqueous solutions-molarity –# moles of solute per liter of solution

16 1/21/ Common Terms of Solution Concentration Stock Stock - routinely used solutions prepared in concentrated form Concentrated Concentrated - relatively large ratio of solute to solvent (5.0 M NaCl) Dilute Dilute - relatively small ratio of solute to solvent (0.01 M NaCl)

17 1/21/ To calculate the concentration of each type of ion in a solution Dissociate solid into its cations/anions This gives how many moles of ions are formed for each mole of solid Multiply number of ions by given molarity

18 1/21/ Calculate the molarity of all the ions in each of the following solutions: Always write out the dissociation equation, so you have ion-to-solute mole ratio M Ca(OCl 2 ) –Ca(OCl 2 )(s) Ca 2+ (aq) + 2OCl - (aq) –Molarity of Ca 2+ = molarity of Ca(OCl 2 ) = M –Molarity of 2OCl - = 2 x molarity of Ca(OCl 2 ) = 0.50 M 2 M CrCl 3 –CrCl 3 Cr 3+ (aq) + 2Cl - (aq)

19 1/21/ Standard solution-solution whose concentration is accurately known a. Put a weighed amount of a substance (solute) into a volumetric flask and add a small quantity of water. b. Dissolve the solid in the water by gently swirling the flask (with stopper in place). c. Add more water (with gentle swirling) until the level of the solution just reaches the mark etched on the neck of the flask. Then mix solution thoroughly by inverting the flask several times.

20 1/21/ Example To prepare a 0.50 molar (0.50 M) solution of KCl, we first measure out 0.50 mole of solid KCl, that is 37.3 g When water is added up to this mark, the flask will contain 1.0 L of the KCl solution, and will contain 0.50 mole of KCl The number of moles of KCl in given amounts of the above solution is easy to find For instance, 0.10 L of 0.50 M KCl will contain 0.10 L soln x 0.50 mol KCl= mol KCl in 1 L soln

21 1/21/ Dilution Used to prepare less concentrated solution from more concentrated solution Adding more water to given amount of solution does not change number of moles of solute present in solution Moles of solute before dilution = moles of solute after dilution Since moles of solute equals solution volume (V) times molarity (M) we have moles i = moles f of M i V i = M f V f –i/f stand for initial/final solution, respectively Note that V f is always larger than V i

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24 1/21/ Parts per million (mg/L) An aqueous solution with a total volume of 750 mL contains mg of Cu 2+. What is the concentration of Cu 2+ in parts per million? mg Cu 2+ = 19.2 ppm Cu L soln

25 1/21/ Molarity to PPM A solution is 3 x M in manganese(VII) ion. What is the Mn 7+ 3 x M concentration in ppm? 3 x mol Mn g Mn mg Mn 7+ = = 0.02 ppm Mn 7+ L soln 1 mol Mn 7+ 1 g Mn 7+

26 1/21/ Homework: Read , pp Q pp , #11 c/d/i, 12, 15c, 16, 20, 23a, 28

27 1/21/ Solubility Rules Solubility of solute: amount dissolved in given quantity of solvent at given temperature Major ideas: –Many salts dissociate into ions in aqueous solution –If solid forms from combination of selected ions in solution Solid must contain anion part and cation part Net charge on solid must be zero –Simple solubility rules used to predict products of reactions in aqueous solutions

28 1/21/ Any ionic compound can be broken apart into its cations and anion Compounds containing 3/more different atoms break apart into appropriate polyatomic ion(s) Charges of all ions must add up to zero Subscripts on monatomic ions become coefficients for ions after parenthesesFor polyatomic ions, only subscripts after parentheses become coefficients Whether or not ionic compound dissolves to appreciable extent in water depends on which cations/anions make up compound

29 1/21/2014 Soluble-good deal of solid visibly dissolves when added to water Slightly soluble-only small amount of solid dissolves (Ionic compound can have low solubility and still be strong electrolyte) Insoluble-no solid dissolves (relative term- does not mean that no solute dissolves) 29

30 1/21/ (slightly soluble) CrO 4 2- NO 3 -, ClO -, ClO 4 -, HCO 3 -, NH 4 + / Group IA None All metallic oxides (O -2 ) NH 4 +, Group IA metals All metallic hydroxides (OH-) NH 4 +, Group IA/IIA from calcium down. Important-NaOH/KOH Ba(OH) 2, Sr(OH) 2, and Ca(OH) 2 marginally soluble

31 1/21/ Types of Solution Reactions 4 Precipitation reactions AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) 4 Acid-base reactions (Neutralization) NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) 4 Oxidation-reduction reactions Fe 2 O 3 (s) + Al(s) Fe(l) + Al 2 O 3 (s)

32 1/21/ Precipitation Reactions 2 soluble substances combined –AX(aq) + BZ(aq) AZ +BX precipitate –Determine, using solubility rules, whether either will form solid (precipitate: insoluble solid that settles from solution)

33 1/21/ Chemical equations for precipitation reactions can be written in several ways: Molecular equationMolecular equation: formulas of compounds are written as usual chemical formulas –Pb(NO 3 ) 2 (aq) + 2HCl(aq) PbCl 2 (s) + 2HNO 3 (aq) Precipitation reacPrecipitation reaction: more accurately represented by ionic equation which shows compounds as being dissociated in solution –Pb 2+ (aq) + 2NO 3 (aq) + 2H + (aq) + 2Cl – (aq) PbCl 2 (s) + 2H + (aq) + 2NO 3 (aq) H + /NO 3 ions not involved in formation of precipitate –Spectator ions –Spectator ions (identical species on both sides of equation can be omitted from equation because ions not involved in reaction) –Net ionic equation –Net ionic equation shows only species that actually undergo a chemical change Pb 2+ (aq) + 2Cl – (aq) PbCl 2 (s)

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36 1/21/ Homework: Read , pp. 148-top 156 Q pg. 182, #30, 34 a/c, 36 b/d, 38

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38 1/21/ What mass of Fe(OH) 3 is produced when 35. mL of a M Fe(NO 3 ) 3 solution is mixed with 55 mL of a M KOH solution? M M x g Fe(NO 3 ) 3 (aq) + 3KOH(aq) Fe(OH) 3 (aq) + 3KNO 3 (aq) M Fe(NO 3 ) 3 = x mol = mol/1.035 L M KOH = x mol = mol/3 = mol.055 Llimiting reactant mol KOH 1 mol Fe(OH) g Fe(OH) 3 =.35 g Fe(OH) 3 3 mol KOH 1 mol Fe(OH) 3

39 1/21/ Gravimetric Analysis- measurement of weight An ore sample is to be analyzed for sulfur. As part of the procedure, the ore is dissolved and the sulfur is converted to sulfate ion, SO Barium nitrate is added, which causes the sulfate to precipitate out as BaSO 4. The original sample had a mass of g. The dried BaSO 4 has a mass of g. What is the percent of sulfur in the original ore? (present in the g) g BaSO 4 1 mol BaSO 4 1 mol S g S = g S g BaSO 4 1 mol BaSO 4 1 mol S % S = g S x 100% = 8.64% S in the ore g in ore

40 1/21/ Homework: Read 4.7, pp Q pg. 182, #40, 42, 44

41 Acid-Base Theories Attempt to explain what happens to molecules of acidic/basic substances in solution that gives rise to their characteristic properties.

42 1/21/ Acids and Bases acidsSolutions of acids –Sour taste –Change blue litmus to red (abr) –Dissolve certain metals –React w/carbonates to produce carbon dioxide gas basesSolutions of bases –Bitter taste –Feel slippery –Change red litmus to blue (brb) Properties of acid neutralized by addition of base, and vice versa

43 1/21/ Arrhenius Svante Arrhenius proposed this acid-base theory AcidAcid: substance that produces hydrogen ions (H +, protons) in aqueous solution –HNO 3 (aq) H + (aq) + NO 3 – (aq) BaseBase: substance that produces hydroxide ions (OH – ) in aqueous solution –Ca(OH) 2 (aq) Ca 2+ (aq) + 2OH – (aq) All are electrolytes undergoing dissociation in aqueous solution Neutralization is combination of H + /OH – to form water –H + (aq) + OH – (aq) H 2 O(l)

44 1/21/ Strong acid-completely dissociates into its ions-strong electrolytes –HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, HClO 4 Strong base-soluble ionic compounds containing hydroxide ion (OH-)-strong electrolytes –Group IA (Li, Na, K, Rb, Cs)/heavy Group IIA (Ca, Sr, Ba) metal hydroxides, NH 2 + Weak acid-dissociates (ionizes) only to a slight extent in aqueous solutions-weak electrolytes –Acetic acid, HF Weak base-very few ions are formed in aqueous solutions-weak electrolyte –NH 3

45 1/21/ Acid/base both completely ionized in solution, ionic equation: 2H + (aq) + 2Cl – (aq) + Mg 2+ (aq) + 2OH – (aq) 2H 2 O(l) + Mg 2+ (aq) + 2Cl – (aq) –Notice that Mg 2+ ion and Cl – ion are spectator ions. Net ionic equation is: –2H + (aq) + 2OH – (aq) 2H 2 O(l) –or H + (aq) + OH – (aq) H 2 O(l) (coefficients can be cancelled) Neutralization equation for any strong acid and strong base is: –H + (aq) + OH – (aq) H 2 O(l) If we had carried out this reaction with two moles of HCl for each mole of base, there would be no left over acid or base, and the solution would be described as neutral.

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47 1/21/ Brønsted-Lowry Theory More general theory of acid-base reactions than Arrhenius –Acid –Acid: substance that donates proton –Base –Base: substance that accepts proton

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50 1/21/ All Arrhenius acids are also Brønsted acids All Arrhenius bases are also Brønsted bases

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52 Acid-Base reactions that form gases 2HCl + Na 2 S H 2 S + 2NaCl –2H + + S 2- H 2 S (net ionic equation) Carbonates/bicarbonates + acids –First form carbonic acid which is unstable –If carbonic acid is present in solution in sufficient concentrations, it decomposes to form CO 2 gas and water 1/21/201452

53 Writing Equations for Acid-Base Neutralization Reactions Neutralization reaction: Neutralization reaction: between acid/base that produces water/salt

54 1/21/ In neutralization reaction, H + from acid/ OH - from base combine to form water Metal ion from base/nonmetal ion from acid form salt Neutralization reaction of hydrochloric acid and magnesium hydroxide 2HCl(aq) +Mg(OH) 2 (aq) 2H 2 O(l) +MgCl 2 (aq) Acid base watersalt

55 1/21/ Neutralization List species present in solution before reaction Write balanced net ionic reaction Find out number of moles of acid we need to neutralize Using stoichiometry of reaction, find out required moles of base to neutralize acid (limiting reactant if necessary) Determine volume of OH- needed to give that many moles Make sure moles of acid equals moles of base

56 1/21/ How many mL of a M NaOH solution is needed to just neutralize mL of a M HCl solution? H +, Cl-, Na +, OH -, H 2 O H + (aq) + OH - (aq) H 2 O( l ) M 1 V 1 = M 2 V 2 Since # H + = OH -, we multiply both sides by 1 (van Hoft factor) (0.800 M)(x mL) = (0.600 m)(40.00 mL) = mL

57 1/21/ What volume of a M HCl solution is needed to neutralize 25.0 mL of a M NaOH? 1)List species and decide what reaction will occur H+ Cl-Na+ OH- Na+(aq) + Cl-(aq) NaCl(s)-soluble-spectator ions H+(aq) + OH-(aq) H 2 O(l) 2)Write balanced net ionic equation H+(aq) + OH-(aq) H 2 O(l) 3)Determine limiting reactant Problem requires addition of just enough H+ ions to react exactly with OH- ions present, so not concerned with determining limiting reactant 4)Calculate moles of reactant needed –M 1 V 1 = M 2 V 2 –(0.100 M)(x mL) = (0.350 M)(25.0 mL) –87.5 mL

58 28.0 mL of M HNO 3 and 53.0 mL of M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H + or OH - ions in excess after the reaction goes to completion? H +, NO 3 -, K +, OH - for KNO 3 which is soluble 28.0 mL HNO 3 1L mol H + = 7.00 x mol H mL L HNO mL KOH 1L mol OH - = 1.70 x mol OH mL L KOH Limiting reactant is H + so 7.00 x mol H + will react with = 7.00 x mol OH - to form 7.00 x mol H 2 O x mol OH x mol OH - leaves 1.00 x mol OH - Volume of combined solution is sum of individual volumes 28.0 mL mL = 81.0 mL = 8.1 x L Molarity of OH - in excess is Mol OH x mol OH - = M OH - L solution 8.1 x L 1/21/201458

59 1/21/ Acid-Base Titrations analyte titrantProcedure for determining concentration of unknown acid (or base) solution (analyte) using known (standardized) concentration (titrant) of base (or acid) solution In titration of acid solution of unknown concentration, a known volume of a standardized base solution is added to acid until acid is just neutralized equivalence or stoichiometric pointPoint at which exactly enough base has been added to neutralize acid-equivalence or stoichiometric point indicatorPoint often marked by indicator, a substance that changes color at (or very near) equivalence point endpointPoint where indicator actually changes-endpoint

60 1/21/ Goal is to choose indicator so endpoint (where indicator changes color) occurs exactly at equivalence point (where just enough titrant has been added to react with all the analyte) Because concentration (M) and required volume (V) of base are known, number of moles of base needed to neutralize acid can be calculated Use mole ratios from balanced neutralization reaction M 1 V 1 =M 2 V 2

61 1/21/ Example What is the volume of 0.05-molar HCl that is required to neutralize 50 mL of a 0.10-molar Mg(OH) 2 solution? –Every mole of Mg(OH) 2 dissociates to produce 2 moles of OH- ions, so a 0.10 M Mg(OH) 2 solution will be a 0.20 M OH- solution –Solution will be neutralized when # moles of H+ ions added is equal to # mole of OH- originally in the solution –Moles = molarity x volume –Moles of OH- = (0.20 M)(50 mL) = 10 millimoles = # moles H+ added –Volume = moles/molarity –Volume of HCl = 10 millimoles/0.05 M = 200 mL Another way to look at it: –M 1 V 1 =M 2 V 2 –(0.05 M)(V) = (0.10 M)(50 mL)(2 moles)

62 1/21/ You want to determine the molar mass of an acid. The acid contains one acidic hydrogen per molecule. You weigh out a g sample of the pure acid and dissolve it, along with 3 drops of phenolphthalein indicator, in distilled water. You titrate the sample with M NaOH. The pink endpoint is reached after addition of mL of the base. Calculate the molar mass of the acid M NaOH = x mol NaOH = mol NaOH L 1 :1 ratio of OH- to H mol H + = g acid= g/mol molar mass acid

63 1/21/ Homework: Read 4.8, pp Q pg. 183, #46 a/c, 48 b/c, 50 a/b, 52, 54

64 Redox Reactions The transfer of electrons

65 1/21/ LEO goes GER In redox reactions electrons are transferred from one atom to another –Oxidation Chemical change in which atom, ion, or molecule loses electrons –Reduction Chemical change in which atom, ion, or molecule gains electrons Oxidation/reduction occur simultaneously

66 1/21/ Substance being oxidized is called reducing agent because it donates electrons which cause reduction in another substance. Substance being reduced is called oxidizing agent because it accepts electrons from another substance. In the reaction of metallic sodium with chlorine gas, an electron is transferred from the sodium atom to a chlorine atom. 2Na + Cl 2 2Na+ + 2Cl– Sodium is oxidized and chlorine is reduced. In sodium-chlorine reaction: Sodium- reducing agent Chlorine- oxidizing agent

67 1/21/ Separated into two half-reactions Oxidation half-reaction Reduction half-reaction For sodium-chlorine reaction they are oxidation 2Na 2Na + + 2e – reduction 2e – + Cl 2 2Cl – Summation of two half-reactions yields overall redox reaction shown above

68 1/21/ Oxidation number Oxidation state Used to help keep account of electrons in reactions Designates positive or negative character –Positive oxidation #: atom lost electrons compared to uncombined atom –Negative oxidation #: atom gained electrons Oxidation numbers are useful in writing formulas, in recognizing redox reactions, and in balancing redox reactions –Element oxidized in reaction whenever its oxidation number increases as result of reaction –When oxidation number of element decreases in reaction, it is reduced

69 1/21/ (+1 when it is in a covalent compound, even if it is listed 2 nd )

70 1/21/ Assign oxidation states for: CaF 2 C 2 H 6 H 2 O I Cl 5 KMnO 4 SO 4 2-

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72 Activity Series Different metals vary in ease of oxidation Activity series gives order of decreasing ease of oxidation Predicts outcome of reactions between metals and either metal salts or acids –Any metal on list can be oxidized by ions of elements below it 1/21/201472

73 Balancing Redox Reactions Acid solutions Basic solutions

74 1/21/ Oxidation number change method Redox equation balanced by comparing increases and decreases in oxidation # Tip-off – If you are asked to balance an equation and if you are not told whether the reaction is a redox reaction or not, you can use the following procedure.

75 1/21/ Step 1: Try to balance the atoms in the equation by inspection. If you successfully balance the atoms, go to Step 2. If you are unable to balance the atoms, go to Step 3. Step 2: Check to be sure that the net charge is the same on both sides of the equation. If it is, you can assume that the equation is correctly balanced. If the charge is not balanced, go to Step #3. Step 3: If you have trouble balancing the atoms and the charge by inspection, determine the oxidation numbers for the atoms in the formula, and go to Step 4. Step 4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced. Step 5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number (number of electrons lost = number of electrons gained). Step 6: Add coefficients to the formulas so as to obtain the correct ratio of the atoms whose oxidation numbers are changing. Step 7: Balance the rest of the equation by inspection.

76 1/21/ NO 2 (g) + H 2 (g) NH 3 (g) + H 2 O(l) The atoms in this equation can be balanced by inspection. (You might first place a 2 in front of the H 2 O to balance the Os, then 7/2 in front of the H 2 to balance the Hs, and then multiply all the coefficients by 2 to get rid of the fraction.) 2NO 2 (g) + 7H 2 (g) 2NH 3 (g) + 4H 2 O(l) We therefore proceed to Step #2. For the reaction between NO 2 and H 2, the net charge on both sides of the equation in Step #1 is zero. Because the charge and the atoms are balanced, the equation is correctly balanced.

77 1/21/ HNO 3 (aq) + H 3 AsO 3 (aq) NO(g) + H 3 AsO 4 (aq) + H 2 O(l) Step #1: Try to balance the atoms by inspection. –The H and O atoms are difficult to balance in this equation. Step #3: Is the reaction redox? –The N atoms change from +5 to +2, so they are reduced. –The As atoms, which change from +3 to +5, are oxidized. Step #4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced. –As +3 to +5 Net Change = +2 –N +5 to +2 Net Change = -3 Step #5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number. –As atoms would yield a net increase in oxidation number of +6. (Six electrons would be lost by three arsenic atoms.) –2 N atoms would yield a net decrease of -6. (2 N gain 6 e-). –Thus the ratio of As atoms to N atoms is 3:2. Step #6: To get the ratio identified in Step 5, add coefficients to the formulas which contain the elements whose oxidation number is changing. –2HNO 3 (aq) + 3H 3 AsO 3 (aq) NO(g) + H 3 AsO 4 (aq) + H 2 O(l) Step #7: Balance the rest of the equation by inspection. –2HNO 3 (aq) + 3H 3 AsO 3 (aq) 2NO(g) + 3H 3 AsO 4 (aq) + H 2 O(l)

78 1/21/ Cu(s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) Step #1: Try to balance the atoms by inspection. –The N atoms and the O atoms are difficult to balance by inspection. Step #3: Is the reaction redox? –The copper atoms are changing their oxidation number from 0 to +2, and some of the nitrogen atoms are changing from +5 to +2. Step #4: Determine the net increase in oxidation number for element oxidized and the net decrease in oxidation number for element reduced. –Cu 0 to +2 Net Change = +2 –Some N +5 to +2 Net Change = -3 Step #5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number. –We need three Cu atoms (net change of +6) for every 2 nitrogen atoms that change (net change of -6). –Because some of the nitrogen atoms are changing and some are not, we need to be careful to put the 2 in front of a formula in which all of the nitrogen atoms are changing or have changed. –The 3 for the copper atoms can be placed in front of the Cu(s). Step #6: To get the ratio identified in Step 5, add coefficients to the formulas which contain the elements whose oxidation number is changing. –3Cu(s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2NO(g) + H 2 O(l) Step #7: Balance the rest of the equation by inspection. –3Cu(s) + 8HNO 3 (aq) 3Cu(NO 3 ) 2 (aq) + 2NO(g) + 4H 2 O(l)

79 1/21/ Some Examples: Al(s) + MnO 2 (s) Al 2 O 3 (s) + Mn(s) SO 2 (g) + HNO 2 (aq) H 2 SO 4 (aq) + NO(g) HNO 3 (aq) + H 2 S(aq) NO(g) + S(s) + H 2 O(l) Al(s) + H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (aq) + H 2 (g)

80 Redox Reactions Half-reaction in acidic solutions

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82 1/21/ Cr 2 O 7 2- (aq) + HNO 2 (aq) Cr 3+ (aq) + NO 3 - (aq) (acidic) Step #1: Write the skeletons of the oxidation and reduction half-reactions. –Cr 2 O 7 2- Cr 3+ –HNO 2 NO 3 - Step #2: Balance all elements other than H and O. –Cr 2 O Cr 3+ (need 2 in front of chromium) –HNO 2 NO 3 - Step #3: Balance the oxygen atoms by adding H 2 O molecules on the side of the arrow where O atoms are needed. –The first half-reaction needs seven oxygen atoms on the right, so we add seven H 2 O molecules. Cr 2 O Cr H 2 O –The second half-reaction needs one more oxygen atom on the left, so we add one H 2 O molecule. HNO 2 + H 2 O NO 3 -

83 1/21/ Step #4: Balance H atoms by adding H + ions on the side of the arrow where H atoms are needed. –1 st half-reaction needs 14 H atoms on left to balance 14 hydrogen atoms in 7 H 2 O molecules-add 14 H + ions to left. Cr 2 O H + 2Cr H 2 O –2 nd half-reaction needs 3 H atoms on right to balance 3 hydrogen atoms on left-add 3 H + ions to the right. HNO 2 + H 2 O NO H + Step #5: Balance the charge by adding electrons. –Sum of charges on left side of chromium half-reaction is +12 (-2 for the Cr 2 O 7 2- plus +14 for the 14 H + ). –Sum of charges on right side of chromium half-reaction is +6 (for the 2 Cr 3+ ). –Add 6 electrons to left side-sum on each side becomes +6. 6e - + Cr 2 O H + 2Cr H 2 O –Sum on the left side of the nitrogen half reaction is zero. –Sum on the right side of the nitrogen half-reaction is +2 (-1 for the nitrate plus +3 for the 3 H + ). –Add 2 es to right side-sum on each side becomes zero. HNO 2 + H 2 O NO H + + 2e -

84 1/21/ Step #6: If # es lost in oxidation half #es in reduction half, multiply one or both reactions by # making # es gained = # lost. –For Cr half-reaction to gain 6 electrons, N half-reaction must lose 6 electrons-multiply coefficients in N half-reaction by 3. 6e - + Cr 2 O H + 2Cr H 2 O 3(HNO 2 + H 2 O NO H + + 2e - ) –3HNO 2 + 3H 2 O 3NO H + + 6e - Step #7: Add the 2 half-reactions. –3 H 2 O in 2 nd half-reaction cancel 3 of 7 H 2 O in 1 st half- reaction to yield 4 H 2 O on the right of the final equation. –9 H + on right of 2 nd half-reaction cancel 9 of 14 H + on left of 1 st half-reaction leaving 5 H + on left of final equation. Cr 2 O HNO 2 + 5H + 2Cr NO H 2 O Step #8: Check to make sure that the atoms and the charge balance. –The atoms in our example balance and the sum of the charges is +3 on each side, so our equation is correctly balanced. Cr 2 O 7 2- (aq) + 3HNO 2 (aq) + 5H + (aq) 2Cr 3+ (aq) + 3NO 3 - (aq) + 4H 2 O(l)

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86 1/21/ Some Examples MnO 4 - (aq) + Br(aq) MnO 2 (s) + BrO 3 - (aq) I 2 (s) + OCl - (aq) IO 3 - (aq) + Cl - (aq) Cr 2 O 7 2- (aq) + C 2 O 4 2 (aq) Cr 3+ (aq) + CO 2 (g) Mn(s) + HNO 3 (aq) Mn 2+ (aq) + NO 2 (g)

87 Redox Reactions Half-reaction in basic solutions

88 1/21/ Steps 1-7: Begin by balancing the equation as if it were in acid solution. If you have H+ ions in your equation at the end of these steps, proceed to Step #8. Otherwise, skip to Step#11. Step 8: Add enough OH- ions to each side to cancel the H+ ions. (Be sure to add the OH- ions to both sides to keep the charge and atoms balanced.) Step 9: Combine the H+ ions and OH- ions that are on the same side of the equation to form water. Step 10: Cancel or combine the H2O molecules. Step 11: Check to make sure that the atoms and the charge balance. If they do balance, you are done. If they do not balance, re-check your work in Steps

89 1/21/ Cr(OH) 3 (s) + ClO 3 - (aq) CrO 4 2- (aq) + Cl - (aq) (basic) Step #1: –Cr(OH) 3 CrO 4 2- –ClO 3 - Cl - Step #2: (Not necessary for this example) Step #3: –Cr(OH) 3 + H 2 O CrO 4 2- –ClO 3 - Cl - + 3H 2 O Step #4: –Cr(OH) 3 + H 2 O CrO H + –ClO H + Cl - + 3H 2 O Step #5: –Cr(OH) 3 + H 2 O CrO H + + 3e - –ClO H + + 6e - Cl - + 3H 2 O Step #6: –2(Cr(OH) 3 + H 2 O CrO H + + 3e - ) 2Cr(OH) 3 + 2H 2 O 2CrO H + + 6e - –ClO H + + 6e - Cl - + 3H 2 O Step #7: –2Cr(OH) 3 (s) + ClO 3 - (aq) 2CrO 4 2- (aq) + Cl - (aq) + H 2 O(l) + 4H + (aq)

90 1/21/ Step #8: Because there are 4 H + on the right side of our equation above, we add 4 OH - to each side of the equation. 2Cr(OH) 3 + ClO OH - 2CrO Cl - + H 2 O + 4H + + 4OH - Step #9: Combine the 4 H + ions and the 4 OH - ions on the right of the equation to form 4 H 2 O. –2Cr(OH) 3 + ClO OH - 2CrO Cl - + H 2 O + 4H 2 O Step #10: Cancel or combine the H 2 O molecules. –2Cr(OH) 3 (s) + ClO 3 - (aq) + 4OH - (aq) 2CrO 4 2- (aq) + Cl - (aq) + 5H 2 O(l) Step #11: The atoms in our equation balance, and the sum of the charges in each side is -5. Our equation is balanced correctly.

91 1/21/ Examples CrO 4 2- (aq) + S 2- (aq) Cr(OH) 3 (s) + S(s) MnO 4 - (aq) + I - (aq) MnO 2 (s) + IO 3 - (aq) H 2 O 2 (aq) + ClO 4 - (aq) O 2 (g) + ClO 2 - (aq) S 2- (aq) + I 2 (s) SO 4 2- (aq) + I - (aq)

92 1/21/ Homework: Read , pp Q pp , #58 a/b/c/df/h/I, 60 b/c/g, 62 a/c/e, 64 a/c, 66 a/c Do one additional exercise and one challenge problem. Submit quizzes by to me: mdahl/ace/launch_ace.html?folder_path=/chemistry/book_cont ent/ _zumdahl/ace&layer=act&src=ch04_ace1.xml mdahl/ace/launch_ace.html?folder_path=/chemistry/book_cont ent/ _zumdahl/ace&layer=act&src=ch04_ace2.xml mdahl/ace/launch_ace.html?folder_path=/chemistry/book_cont ent/ _zumdahl/ace&layer=act&src=ch04_ace3.xml


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