Solutions

LabsDry Lab 4 Oxidation-Reduction Equations

#9 A Volumetric Analysis#15 Bleach Analysis

Chemical equationsChapters 8-11

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Will the substances mix?

• NaNO3 and H2O

– Miscible-ionic + polar

• C6H14 and H2O

– Immiscible-nonpolar + polar

• I2 and C6H14

– Miscible-nonpolar + nonpolar

• I2 and H2O

– immiscible-nonpolar + polar

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bent molecule104.5o

Aqueous solutions-solvent is water

•Polar molecule-unequal charge distribution gives water ability to dissolve many compounds

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–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

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Solutionhomogeneous 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

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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

SolvationSolvation of ionic compounds

• Polarity of water

• Ions surrounded by water molecules prevent recombining of ions (solvatedsolvated)

• Ions/shells free to move around, so dispersed uniformly throughout solution

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Electrolyte Conductivity Degree Examples of Dissociation

Strong high total strong acids (HCl, HNO3) many ionic salts (NaCl,

Sr(NO3)3-ions)

strong bases (NaOH,

Ba(OH)2, other IA/IIA hydroxides)

Weak low to partial weak organic acids

moderate (tap water, HCO2H- formic acid, C2H3O2)

weak bases (NH3)

Non none close to zero sugar, sugar solution, AgCl, Fe2O3 (neutral)

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Complete following dissociation equations: (all in water)

• CaCl2(s)Ca2+(aq) + 2Cl-(aq)• Fe(NO3)3(s) • KBr(s) • (NH4)2Cr2O7(s) Strong, Weak, or Nonelectrolytes• HClO4

• C6H12

• LiOH• NH3

• CaCl2• HC2H3O2

MolarityM = moles of solute Liter of solution

Moles = millimoles = micromoles liter milliliter microliter

BUTMoles does not equal millimoles or moles liter liter microliter

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• 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 = 0.061 mol NaCl

58.44 g NaCl 

• Convert 25.0 mL to 0.0250 L and substitute these two quantities into the defining equation for molarity.

• Molarity = 0.0611 mol NaCl = 2.44 M NaCl l0.0250 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. 

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• How many mL of solution are necessary if we are to have a 2.48 M NaOH solution that contains 31.52 g of the dissolved solid?

31.52 g NaOH 1 mol NaOH = 0.788 mol NaOH

40.00 g NaOH

2.48 M NaOH = 0.788 mol NaOH = .318 L = 318.mL NaOH

x L

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• Determine the molarity of Cl- ion in a solution prepared by dissolving 9.82 g of CuCl2 in enough water to make 600 mL of solution.

• 9.82 g CuCl2 1 mol CuCl2 = 0.073 mol CuCl2 134.45 g CuCl2• M = 0.073 mol CuCl2 = 0.1217 M CuCl2 .600 L • Ion = 2 mol Cl-

solute 1 mol CuCl2• 0.1217 M CuCl2 x 2 mol Cl- = 0.243 M

L solution 1 mol CuCl2

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• Calculate the mass of NaCl needed to prepare 175 mL of a 0.500 M NaCl solution.

0.500 M = x mol

.175 L

x mol = 0.0875 mol NaCl0.0875 mol NaCl 58.44 g NaCl = 5.11 g NaCl

1 mol NaCl

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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

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Common Terms of Solution Concentration

StockStock - routinely used solutions prepared in concentrated form

ConcentratedConcentrated - relatively large ratio of solute to solvent (5.0 M NaCl)

DiluteDilute - relatively small ratio of solute to solvent (0.01 M NaCl)

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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

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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.

• 0.25 M Ca(OCl2)– Ca(OCl2)(s) Ca2+(aq) + 2OCl-(aq) – Molarity of Ca2+ = molarity of Ca(OCl2) = 0.250 M– Molarity of 2OCl- = 2 x molarity of Ca(OCl2) =

0.50 M

• 2 M CrCl3– CrCl3 Cr3+(aq) + 2Cl-(aq)

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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.

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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= 0.050 mol KCl in 1

L soln

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DilutionDilution

• 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 molesi = molesf of MiVi = MfVf

– i/f stand for initial/final solution, respectively• Note that Vf is always larger than Vi

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Parts per million (mg/L)

• An aqueous solution with a total volume of 750 mL contains 14.38 mg of Cu2+. What is the concentration of Cu2+ in parts per million?

• 14.38 mg Cu2+ = 19.2 ppm Cu2+

0.750 L soln

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Molarity to PPM

• A solution is 3 x 10-7 M in manganese(VII) ion. What is the Mn7+ 3 x 10-7 M concentration in ppm?

3 x 10-7 mol Mn7+ 54.94 g Mn7+ 1000 mg Mn7+ = 0.0164 = 0.02 ppm Mn7+

L soln 1 mol Mn7+ 1 g Mn7+

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Homework:

Read 4.1-44, pp. 133-148

Q pp. 180-182, #11 c/d/i, 12, 15c, 16, 20, 23a, 28

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Solubility RulesSolubility 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

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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• For polyatomic ions, only subscripts afterafter

parenthesesparentheses become coefficients• Whether or not ionic compound dissolves to

appreciable extent in water depends on which cations/anions make up compound

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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

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(slightly soluble)

CrO42-

NO3-, ClO-, ClO4

-, HCO3-,

NH4+/ Group IA None

All metallic oxides (O-2) NH4+, Group IA metals

All metallic hydroxides (OH-) NH4+, Group IA/IIA from calcium down.

Important-NaOH/KOH Ba(OH)2, Sr(OH)2, and Ca(OH)2

marginally soluble

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Types of Solution ReactionsTypes of Solution Reactions Precipitation reactionsPrecipitation reactions

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Acid-base reactions (Neutralization)Acid-base reactions (Neutralization)NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

Oxidation-reduction reactionsOxidation-reduction reactionsFe2O3(s) + Al(s) Fe(l) + Al2O3(s)

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Precipitation ReactionsPrecipitation Reactions• 2 soluble substances combined

– AX(aq) + BZ(aq) AZ +BX– Determine, using solubility rules,

whether either will form solid (precipitateprecipitate: insoluble solid that settles from solution)

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Chemical equations for precipitation reactions can be written in several ways:

• Molecular equationMolecular equation: formulas of compounds are written as usual chemical formulas

– Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2HNO3(aq)

• Precipitation reacPrecipitation reaction: more accurately represented by ionic equation which shows compounds as being dissociated in solution

– Pb2+(aq) + 2NO3(aq) + 2H+(aq) + 2Cl–(aq) PbCl2(s) + 2H+(aq) + 2NO3 (aq)

• H+/NO3 ions not involved in formation of precipitate– Spectator ionsSpectator ions (identical species on both sides of equation can

be omitted from equation because ions not involved in reaction)– Net ionic equationNet ionic equation shows only species that actually undergo a

chemical change• Pb2+(aq) + 2Cl–(aq) PbCl2(s)

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Homework:

Read 4.5-4.6, pp. 148-top 156

Q pg. 182, #30, 34 a/c, 36 b/d, 38

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• What mass of Fe(OH)3 is produced when 35. mL of a 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M KOH solution?

0.250 M 0.180 M x g

Fe(NO3)3(aq) + 3KOH(aq) Fe(OH)3(aq) + 3KNO3(aq)

• 0.250 M Fe(NO3)3 = x mol = 0.00875 mol/1

.035 L• 0.180 M KOH = x mol = 0.00990 mol/3 = .00330 mol

.055 L limiting reactant• 0.00990 mol KOH 1 mol Fe(OH)3 106.88 g Fe(OH)3= .35 g Fe(OH)3

3 mol KOH 1 mol Fe(OH)3

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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, SO4

2-. Barium nitrate is added, which causes the sulfate to precipitate out as BaSO4. The original sample had a mass of 3.187 g. The dried BaSO4 has a mass of 2.005 g. What is the percent of sulfur in the original ore? (present in the 2.005 g)

2.005 g BaSO4 1 mol BaSO4 1 mol S 32.06 g S = 0.275 g S

233.36 g BaSO4 1 mol BaSO4 1 mol S

% S = 0.275 g S x 100% = 8.64% S in the ore

3.187 g in ore

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Homework:

Read 4.7, pp. 156-158

Q pg. 182, #40, 42, 44

Acid-Base TheoriesAcid-Base Theories

Attempt to explain what happens to molecules of acidic/basic substances in solution that

gives rise to their characteristic properties.

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Acids and BasesAcids and Bases

• Solutions of acidsacids– Sour taste– Change blue litmus to red (abr)– Dissolve certain metals– React w/carbonates to produce carbon dioxide gas

• Solutions of basesbases – Bitter taste– Feel slippery– Change red litmus to blue (brb)

• Properties of acid neutralized by addition of base, and vice versa

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Svante ArrheniusArrhenius proposed this acid-base theory

• AcidAcid: substance that produces hydrogen ions (H+, protons) in aqueous solution

– HNO3(aq) H+(aq) + NO3–(aq)

• BaseBase: substance that produces hydroxide ions (OH–) in aqueous solution

– Ca(OH)2(aq) Ca2+(aq) + 2OH–(aq)

• All are electrolytes undergoing dissociation in aqueous solution

• Neutralization is combination of H+/OH– to form water

– H+(aq) + OH–(aq) H2O(l)

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• Strong acid-completely dissociates into its ions-strong electrolytes

– HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4

• 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, NH2

+

• 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

– NH3

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• Acid/base both completely ionized in solution, ionic equation:

2H+(aq) + 2Cl–(aq) + Mg2+(aq) + 2OH–(aq) 2H2O(l) + Mg2+(aq) + 2Cl–(aq)

– Notice that Mg2+ ion and Cl– ion are spectator ions.

• Net ionic equation is:– 2H+(aq) + 2OH–(aq) 2H2O(l)

– or H+(aq) + OH–(aq) H2O(l) (coefficients can be cancelled)

• Neutralization equation for any strong acid and strong base is:

– H+(aq) + OH–(aq) H2O(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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BrBrøønsted-Lowry Theorynsted-Lowry Theory

• More general theory of acid-base reactions than Arrhenius

– AcidAcid: substance that donates proton– BaseBase: substance that accepts proton

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• All Arrhenius acids are also Brønsted acids

• All Arrhenius bases are also Brønsted bases

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Acid-Base reactions that form gasesAcid-Base reactions that form gases

• 2HCl + Na2S H2S + 2NaCl

– 2H+ + S2- H2S (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 CO2 gas and water

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Writing Equations for Acid-Base Neutralization Reactions

Neutralization reaction: Neutralization reaction: between acid/base that produces water/salt

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• 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) 2H2O(l) +MgCl2(aq)

Acid base water salt

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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

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• How many mL of a 0.800 M NaOH solution is needed to just neutralize 40.00 mL of a 0.600 M HCl solution?

• H+, Cl-, Na+, OH-, H2O

• H+ (aq) + OH- (aq) H2O(l)

• M1V1 = M2V2

• Since # H+ = OH-, we multiply both sides by 1 (van Hoft factor)

(0.800 M)(x mL) = (0.600 m)(40.00 mL)

= 30.00 mL

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What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of a 0.350 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) H2O(l)

2) Write balanced net ionic equation• H+(aq) + OH-(aq) H2O(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)4) Calculate moles of reactant needed Calculate moles of reactant needed – MM11VV11 = M = M22VV22

– (0.100 M)(x mL) = (0.350 M)(25.0 mL)(0.100 M)(x mL) = (0.350 M)(25.0 mL)– 87.5 mL87.5 mL

• 28.0 mL of 0.250 M HNO3 and 53.0 mL of 0.320 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+, NO3-, K+, OH- for KNO3 which is soluble

28.0 mL HNO3 1L 0.250 mol H+ = 7.00 x 10-3 mol H+

1000 mL L HNO3

53.0 mL KOH 1L 0.320 mol OH- = 1.70 x 10-2 mol OH-

1000 mL L KOH

• Limiting reactant is H+ so 7.00 x 10-3 mol H+ will react with = 7.00 x 10-3 mol OH- to form 7.00 x 10-3 mol H2O.

• 1.70 x 10-2 mol OH- - 7.00 x 10-3 mol OH- leaves 1.00 x 10-2 mol OH-

• Volume of combined solution is sum of individual volumes

28.0 mL + 53.0 mL = 81.0 mL = 8.1 x 10-2 L

• Molarity of OH- in excess is

Mol OH- 1.00 x 10-2 mol OH- = 0.123 M OH-

L solution 8.1 x 10-2 L04/10/23 58

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Acid-Base TitrationsAcid-Base Titrations• Procedure for determining concentration of unknown

acid (or base) solution (analyteanalyte) using known (standardized) concentration (titranttitrant) 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

• Point at which exactly enough base has been added to neutralize acid-equivalenceequivalence or or stoichiometric pointstoichiometric point

• Point often marked by indicatorindicator, a substance that changes color at (or very near) equivalence point

• Point where indicator actually changes-endpointendpoint

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• 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

• M1V1=M2V2

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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:– M1V1=M2V2

– (0.05 M)(V) = (0.10 M)(50 mL)(2 moles)

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• You want to determine the molar mass of an acid. The acid contains one acidic hydrogen per molecule. You weigh out a 2.879 g sample of the pure acid and dissolve it, along with 3 drops of phenolphthalein indicator, in distilled water. You titrate the sample with 0.1704 M NaOH. The pink endpoint is reached after addition of 42.55 mL of the base. Calculate the molar mass of the acid.

• 0.1704 M NaOH = x mol NaOH = 0.00725 mol NaOH

0.04255 L• 1 :1 ratio of OH- to H+

• 0.00725 mol H+ = 2.879 g acid = 397.1 g/mol

molar mass acid

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Homework:

Read 4.8, pp. 158-164

Q pg. 183, #46 a/c, 48 b/c, 50 a/b, 52, 54

Redox Reactions

The transfer of electrons

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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

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•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 + Cl2 2Na+ + 2Cl–Sodium is oxidized and chlorine is reduced.

In sodium-chlorine reaction:

Sodium-reducing agentChlorine-oxidizing agent

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• Separated into two half-reactions• Oxidation half-reaction• Reduction half-reaction

• For sodium-chlorine reaction they are• oxidation 2Na 2Na+ + 2e–

• reduction 2e– + Cl2 2Cl–

• Summation of two half-reactions yields overall redox reaction shown above

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Oxidation numberOxidation numberOxidation stateOxidation 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

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

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Assign oxidation states for:

• CaF2

• C2H6

• H2O

•ICl5• KMnO4

• SO42-

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Activity SeriesActivity 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

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Balancing Redox Reactions

Acid solutions

Basic solutions

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Oxidation number change Oxidation number change methodmethod

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. 

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• 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.

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NO2(g) + H2(g)      NH3(g) + H2O(l)  

• The atoms in this equation can be balanced by inspection. (You might first place a 2 in front of the H2O to balance the O’s, then 7/2

in front of the H2 to balance the H’s, and then

multiply all the coefficients by 2 to get rid of the fraction.)

•    2NO2(g) + 7H2(g)    2NH3(g) + 4H2O(l)

• We therefore proceed to Step #2. For the reaction between NO2 and H2, 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.

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HNO3(aq) + H3AsO3(aq)    NO(g) + H3AsO4(aq) + H2O(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.– 2HNO3(aq) + 3H3AsO3(aq)   NO(g) + H3AsO4(aq) + H2O(l)

• Step #7:  Balance the rest of the equation by inspection.– 2HNO3(aq) + 3H3AsO3(aq)   2NO(g) + 3H3AsO4(aq) + H2O(l)

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Cu(s) + HNO3(aq)    Cu(NO3)2(aq) + NO(g) + H2O(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) + HNO3(aq)    Cu(NO3)2(aq) + 2NO(g) + H2O(l)

• Step #7:  Balance the rest of the equation by inspection.– 3Cu(s) + 8HNO3(aq) 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

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Some Examples:

• Al(s)  +  MnO2(s)   Al2O3(s)  +  Mn(s)

• SO2(g) + HNO2(aq)  H2SO4(aq)  + NO(g)

• HNO3(aq) + H2S(aq)  NO(g) + S(s) + H2O(l)

• Al(s) + H2SO4(aq)  Al2(SO4)3(aq) + H2(g)

Redox Reactions

Half-reaction in acidic solutions

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Cr2O72-(aq) + HNO2(aq)  Cr3+(aq) + NO3

-(aq) (acidic)

• Step #1:   Write the skeletons of the oxidation and reduction half-reactions. – Cr2O7

2-    Cr3+ – HNO2       NO3

-

• Step #2:    Balance all elements other than H and O. – Cr2O7

2-     2Cr3+ (need 2 in front of chromium)– HNO2      NO3

- • Step #3:    Balance the oxygen atoms by adding H2O

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 H2O molecules.• Cr2O7

2-      2Cr3+  +  7H2O – The second half-reaction needs one more oxygen

atom on the left, so we add one H2O molecule. • HNO2    +  H2O     NO3

- 

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• Step #4:    Balance H atoms by adding H+ ions on the side of the arrow where H atoms are needed.– 1st half-reaction needs 14 H atoms on left to balance 14

hydrogen atoms in 7 H2O molecules-add 14 H+ ions to left.  • Cr2O7

2-  +  14H+      2Cr3+  +  7H2O – 2nd half-reaction needs 3 H atoms on right to balance 3 hydrogen

atoms on left-add 3 H+ ions to the right.  • HNO2    +  H2O      NO3

-  +  3H+ • Step #5:    Balance the charge by adding electrons.

– Sum of charges on left side of chromium half-reaction is +12 (-2 for the Cr2O7

2- plus +14 for the 14 H+). – Sum of charges on right side of chromium half-reaction is +6 (for

the 2 Cr3+). – Add 6 electrons to left side-sum on each side becomes +6.

• 6e-  +  Cr2O72-  +  14H+    2Cr3+  +  7H2O

– 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 e’s to right side-sum on each side becomes zero.

• HNO2    +  H2O      NO3-  +  3H+  +  2e-

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• Step #6:    If # e’s lost in oxidation half ≠ #e’s in reduction half, multiply one or both reactions by # making # e’s 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-  +  Cr2O7

2-  +  14H+      2Cr3+  +  7H2O • 3(HNO2    +  H2O      NO3

-  +  3H+  +  2e-) – 3HNO2    +  3H2O      3NO3

-  +  9H+  +  6e- • Step #7:    Add the 2 half-reactions.

– 3 H2O in 2nd half-reaction cancel 3 of 7 H2O in 1st half-reaction to yield 4 H2O on the right of the final equation.

– 9 H+ on right of 2nd half-reaction cancel 9 of 14 H+ on left of 1st half-reaction leaving 5 H+ on left of final equation.

• Cr2O72-  +  3HNO2   +  5H+      2Cr3+  +  3NO3

-   +  4H2O • 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.

• Cr2O72-(aq) + 3HNO2(aq) +  5H+(aq)

2Cr3+(aq) + 3NO3- (aq) + 4H2O(l)

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Some Examples

• MnO4-(aq) + Br(aq) MnO2(s) + BrO3

-(aq)

• I2(s)  +  OCl-(aq)  IO3-(aq) + Cl-(aq)

• Cr2O72-(aq) + C2O4

2(aq)Cr3+(aq) + CO2(g)

• Mn(s) + HNO3(aq)  Mn2+(aq) + NO2(g)

Redox Reactions

Half-reaction in basic solutions

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• 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 1-10.

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Cr(OH)3(s) + ClO3-(aq)    CrO4

2-(aq) + Cl-(aq)   (basic)• Step #1:

– Cr(OH)3     CrO42-

– ClO3-        Cl-

• Step #2: (Not necessary for this example) • Step #3:

– Cr(OH)3  +  H2O      CrO42-

– ClO3-        Cl-  +  3H2O

• Step #4: – Cr(OH)3  +  H2O     CrO4

2-  +  5H+ – ClO3

-  +  6H+        Cl-  +  3H2O • Step #5:

– Cr(OH)3  +  H2O      CrO42-  +  5H+  +  3e-

– ClO3-  +  6H+  +  6e-      Cl-  +  3H2O

• Step #6:   – 2(Cr(OH)3  +  H2O      CrO4

2-  +  5H+  +  3e- ) • 2Cr(OH)3  +  2H2O      2CrO4

2-  +  10H+  +  6e- – ClO3

-  +  6H+  +  6e-        Cl-  +  3H2O • Step #7:

– 2Cr(OH)3(s) + ClO3-(aq)  2CrO4

2-(aq) + Cl-(aq) + H2O(l) + 4H+

(aq)

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• 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 + ClO3- + 4OH-

2CrO42- + Cl- + H2O + 4H+ + 4OH-

• Step #9: Combine the 4 H+ ions and the 4 OH- ions on the right of the equation to form 4 H2O.

– 2Cr(OH)3 + ClO3- + 4OH-

2CrO42- + Cl- + H2O + 4H2O

• Step #10:  Cancel or combine the H2O molecules.

– 2Cr(OH)3(s)  +  ClO3-(aq)  +  4OH-(aq)  

  2CrO42-(aq)  +  Cl-(aq)  +  5H2O(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.

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Examples

• CrO42-(aq) + S2-(aq) Cr(OH)3(s) + S(s)

• MnO4-(aq) + I-(aq)  MnO2(s) + IO3

-(aq)

• H2O2(aq) + ClO4-(aq)  O2(g) + ClO2

-(aq)

• S2-(aq) + I2(s)  SO42-(aq) + I-(aq)

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Homework:Read 4.9-4.10, pp. 164-179Q pp. 183-184, #58 a/b/c/df/h/I, 60 b/c/g, 62 a/c/e, 64 a/c, 66 a/cDo one additional exercise and one challenge problem.Submit quizzes by email to me:http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch04_ace1.xml

http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch04_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch04_ace3.xml