Chapter 4 Types of Chemical Reactions and Solution Chemistry.

Chapter 4Chapter 4

Types of Chemical Reactions Types of Chemical Reactions and Solution Chemistryand Solution Chemistry

TopicsTopics

Water as a solventWater as a solvent Electrolytes and nonelectrolytesElectrolytes and nonelectrolytes Calculations involving molarity of Calculations involving molarity of

solutionssolutions Precipitation reactionsPrecipitation reactions Acid base reactionsAcid base reactions Oxidation reduction reactionsOxidation reduction reactions

The Water Molecule, PolarityThe Water Molecule, Polarity

δ-

δ+ δ+

H H

O

δ- means a partial negative charge

δ+ means a partial positive charge

Thus, water has a partial negative end (0xygen) and a partial positive end (Hydrogen) – and it is called “polar” because of the unequal charge distribution

4.14.1 Water, the common solventWater, the common solvent

bond angle of water = 105bond angle of water = 105oo

Ions have charges and attract the opposite Ions have charges and attract the opposite charges on the water molecules. charges on the water molecules.

The process of breaking the ions of salts The process of breaking the ions of salts apart is called hydration apart is called hydration

Dissolving ionic salts in waterDissolving ionic salts in water and Hydrationand Hydration

NH4NO3(s) NH4+(aq) + NO3

-

(aq)

H2O (l)

Designates hydration of ions

How How Ionic solidsIonic solids dissolve in water dissolve in water

H HOH

H OH

HO

H HO

HHO

HH

O

HH

OH

H

O

HH

O

These ions have been surrounded by water, and are now dissolved!

These ions have been pulled away from the main crystal structure by water’s polarity.

HydrationHydration

Solubility in water and Solubility in water and Aqueous SolutionsAqueous Solutions

Water dissolves Water dissolves ionicionic compounds compounds (NaCl) and (NaCl) and polar covalentpolar covalent molecules molecules (ethanol C(ethanol C22HH55OH)OH)

The rule is: “The rule is: “like dissolves likelike dissolves like”” Polar dissolves polar.Polar dissolves polar. Nonpolar dissolves nonpolar.Nonpolar dissolves nonpolar. Oil is nonpolar.Oil is nonpolar.

– Oil and water don’t mix.Oil and water don’t mix. Many Salts are ionic- sea waterMany Salts are ionic- sea water

The Solution ProcessThe Solution Process Called Called “solvation”.“solvation”. Water breaks the + and - charged pieces Water breaks the + and - charged pieces

apart and surrounds them.apart and surrounds them. Solubility in water depends on the Solubility in water depends on the relative relative

attractions of ions for each other and attractions of ions for each other and attraction of ions for water moleculesattraction of ions for water molecules

In some ionic compounds, the attraction In some ionic compounds, the attraction between ionsbetween ions is greater than the attraction is greater than the attraction exerted by water (slightly soluble slats)exerted by water (slightly soluble slats)– Barium sulfateBarium sulfate and and calcium carbonatecalcium carbonate

Solids will dissolve if the attractive Solids will dissolve if the attractive force of the water molecules is force of the water molecules is stronger than the attractive force of the stronger than the attractive force of the crystal.crystal.

If not, the solids are insoluble.If not, the solids are insoluble. Water doesn’t dissolve nonpolar Water doesn’t dissolve nonpolar

molecules (like oil) because the water molecules (like oil) because the water molecules can’t hold onto them.molecules can’t hold onto them.

How does ethanol dissolve in waterHow does ethanol dissolve in water??

Ethanol Molecule Contains a Polar O-H Bond Similar to Those in the Water Molecule

The polar water molecule interacts strongly The polar water molecule interacts strongly with the polar-O-H bond in ethanolwith the polar-O-H bond in ethanol

A solution is a homogenous mixturehomogenous mixture of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)The solvent is the substance present in the larger amount

Solution Solvent Solute(s)

Soft drink (l)

Air (g)

Soft Solder (s)

H2O

N2

Pb

Sugar, CO2

O2, Ar, CH4

Sn

4.2 The nature of aqueous 4.2 The nature of aqueous solutions: strong and weak solutions: strong and weak

electrolyteselectrolytes

Electrolytes and NonelectrolytesElectrolytes and Nonelectrolytes ElectrolytesElectrolytes-- compounds that conduct an compounds that conduct an

electric current in electric current in aqueous solutionaqueous solution, or in , or in the the molten statemolten state

– all ionic compounds are electrolytes all ionic compounds are electrolytes because they dissociate into ions (they because they dissociate into ions (they are also called “are also called “saltssalts”)”)

barium sulfate- will conduct when barium sulfate- will conduct when moltenmolten, , but is but is insolubleinsoluble in water!in water!

NonelectrolytesNonelectrolytes- Do not conduct - Do not conduct an electric current an electric current – Most are Most are molecularmolecular materials, materials,

because they because they do not have ionsdo not have ions

Not all electrolytes conduct to the Not all electrolytes conduct to the same degreesame degree– there are there are strongstrong electrolytes, and electrolytes, and

weakweak electrolyteselectrolytes

– Conductivity depends on: Conductivity depends on: degree of degree of dissociation or ionizationdissociation or ionization

______________ ______________ ______________

Nonelectrolyte Weak electrolyte Strong electrolyte

Ethanol and table sugar

Acetic acid ammonia

Sodium chlorideHydrochloric acid

Dissociation of acids and bases:Dissociation of acids and bases:Strong and weak acids and basesStrong and weak acids and bases

Acids- form HAcids- form H++ ions when dissolved in water ions when dissolved in water

(According to (According to ArrheniusArrhenius)) Strong acids dissociate completely into HStrong acids dissociate completely into H++ and and

anionsanions

Strong acids- HStrong acids- H22SOSO44 HNO HNO33 HCl HBr HI HCl HBr HI

HClOHClO44

Bases - form OHBases - form OH-- ions when dissolved in water ions when dissolved in water Strong bases- KOH, NaOHStrong bases- KOH, NaOH

Weak acids- dissociate partiallyWeak acids- dissociate partially Acetic acid: HCAcetic acid: HC22HH33OO22 has 1% has 1%

dissociation in aqueous solutionsdissociation in aqueous solutions The most common weak base is The most common weak base is

ammonia, NHammonia, NH33

HCl H+ + Cl-

HNO3 H+ + NO3-

HC2H3O2(aq) H+ + C2H3O2-

Strong electrolyte, strong acid


Weak electrolyte, weak acid

H2SO4 H+ + HSO4-

HSO4- H+ + SO4

2-



H3PO4 H+ + H2PO4-

H2PO4- H+ + HPO4

2-

HPO42- H+ + PO4

3-




H2O

H2O

NaOH(s) NaNaOH(s) Na++(aq) + OH(aq) + OH--(aq)(aq)

NHNH33(aq) + H(aq) + H22O NHO NH44++(aq) + OH(aq) + OH--(aq)(aq)

H2O

4.34.3 The composition of solutionsThe composition of solutionsMolarity (M)Molarity (M)

Molarity: A concentration that expresses Molarity: A concentration that expresses the moles of solute in 1 L of solutionthe moles of solute in 1 L of solution

Molarity (MMolarity (M)) = = moles ofmoles of solute solute

litersliters solution solution

Molarity CalculationMolarity Calculation

If 4.0 g NaOH are used to make 500. mL of NaOH If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?solution, what is the molarity (M) of the solution?

4.0 g NaOH x 4.0 g NaOH x 1 mole NaOH1 mole NaOH = 0.10 mole NaOH= 0.10 mole NaOH

40.0 g NaOH 40.0 g NaOH

500. mL x 500. mL x 1 L _ 1 L _ = 0.500 L= 0.500 L

1000 mL1000 mL

0.10 mole NaOH0.10 mole NaOH = = 0.20 mole NaOH0.20 mole NaOH

0.500 L0.500 L 1 L1 L

= 0.20 M NaOH= 0.20 M NaOH

Calculating MolarityCalculating Molarity

1500 mL x 1500 mL x 1 L 1 L = 1.5 L = 1.5 L

1000 mL1000 mL

1.5 L x 1.5 L x 0.10 mole HCl0.10 mole HCl = 0.15 mole HCl = 0.15 mole HCl

1 L 1 L

An acid solution is a 0.10 M HCl. How An acid solution is a 0.10 M HCl. How many moles of HCl are in 1500 mL of many moles of HCl are in 1500 mL of this acid solution?this acid solution?

How many grams of KCl are present in How many grams of KCl are present in

2.5 L of 0.50 M KCl?2.5 L of 0.50 M KCl?

2.5 L x 2.5 L x 0.50 mole0.50 mole x x 74.6 g KCl 74.6 g KCl = 93 g = 93 g

KClKCl

1 L 1 mole KCl1 L 1 mole KCl

How many milliliters How many milliliters of an acid solution, which of an acid solution, which is 0.10 M HCl, contain 0.15 mole HCl?is 0.10 M HCl, contain 0.15 mole HCl?

0.15 mole HCl x0.15 mole HCl x 1 L soln 1 L soln x x 1000 mL1000 mL

0.10 mole HCl0.10 mole HCl 1 L 1 L

= 1500 mL HCl= 1500 mL HCl

How many grams of NaOH are required to prepare 400.How many grams of NaOH are required to prepare 400.mL of 3.0 mL of 3.0 MM NaOH solution? NaOH solution?

400. mL x 400. mL x 1 L 1 L = 0.400 L = 0.400 L 1000 mL1000 mL

0.400 L x 0.400 L x 3.0 mole NaOH 3.0 mole NaOH x x 40.0 g NaOH40.0 g NaOH 1 L1 L 1 mole NaOH 1 mole NaOH

= = 48 g NaOH48 g NaOH

1.0mg NaCl = 1.2X10-4 L1L

0.14mol NaClx1 mol NaCl

58.5g NaClx

1 g

1000 mgx

A sample of 0.14 M NaCl. What volume of sample contains 1.0 mg NaCl?

# mol = M X V (L) M

molLV

#)(

DilutionDilution

Adding more solvent to a known solution.Adding more solvent to a known solution. The moles of solute stay the same.The moles of solute stay the same. #moles = M x volume (L)#moles = M x volume (L)

# moles before dilution (1) = # moles after dilution# moles before dilution (1) = # moles after dilution

MM1 1 VV11 = M = M2 2 VV22

Stock solutionStock solution is a solution of known concentration is a solution of known concentration

used to make more dilute solutionsused to make more dilute solutions

Preparing a less concentrated solution from a more concentrated solution by dilution

Dilution

Add Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=

M1V1 = M2V2

M1 = 4.00 M2 = 0.200 V2 = 60.0 m L V1 = ?mL

V1 =M2V2

M1

= =

How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3?

Example

0.200 M X 60.0 mL

4.00 M3.00 mL

Dilution process

pipits

Wash bottle

Funnel

Volumetricflask

4.44.4 Types of Chemical ReactionsTypes of Chemical Reactions

Precipitation reactionsPrecipitation reactions Acid-base reactionsAcid-base reactions Oxidation-reduction reactionsOxidation-reduction reactions

4.5 Precipitation Reactions4.5 Precipitation Reactions

When aqueous solutions of ionic When aqueous solutions of ionic compounds are mixed together a solid compounds are mixed together a solid forms. forms.

A solid that forms from mixed solutions A solid that forms from mixed solutions is called is called precipitateprecipitate

If the substance is not part of the If the substance is not part of the solution, it is a precipitate solution, it is a precipitate

Precipitation ReactionsPrecipitation Reactions

________________________

________________________

________________________

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3

-

Na+ and NO3- are SpectatorSpectator ions

PbI2

Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

precipitate

Pb2+ + 2I- PbI2 (s)

Anions and cations switch partnersAnions and cations switch partners

Solubility rules for common ionic compounds in water at 250 C

Soluble CompoundsSoluble Compounds ExceptionsExceptions

Compounds containing alkali Compounds containing alkali metal ions and NHmetal ions and NH44

++

NONO33--, HCO, HCO33

--, ClO, ClO33--

ClCl--, Br, Br--, I, I-- Halides of AgHalides of Ag++, Hg, Hg222+2+, Pb, Pb2+2+

SOSO442-2- Sulfates of AgSulfates of Ag++, Ca, Ca2+2+, Sr, Sr2+2+, ,

BaBa2+2+, Hg, Hg2+2+, Pb, Pb2+2+

Slightly soluble CompoundsSlightly soluble Compounds ExceptionsExceptions

COCO332-2-, PO, PO44

3-3-, CrO, CrO442-2-, S, S2-2- Compounds containing alkali Compounds containing alkali

metal ions and NHmetal ions and NH44++

OHOH--

Compounds containing alkali Compounds containing alkali metal ions and Bametal ions and Ba2+2+, Ca, Ca2+2+, , MgMg2+2+ are marginally soluble are marginally soluble

Solubility Rules Solubility Rules Predicting reaction’s productPredicting reaction’s product

All nitrates are solubleAll nitrates are soluble

Alkali metals ions and NHAlkali metals ions and NH44++ ions are ions are

solublesoluble

Halides are soluble except AgHalides are soluble except Ag++, Pb, Pb+2+2, , and Hgand Hg22

+2+2

Most sulfates are soluble, except PbMost sulfates are soluble, except Pb+2+2, , BaBa+2+2, Hg, Hg+2+2,and Ca,and Ca+2+2

Solubility RulesSolubility Rules

Most hydroxides are slightly soluble Most hydroxides are slightly soluble (insoluble) except NaOH and KOH(insoluble) except NaOH and KOH

Sulfides, carbonates, chromates, and Sulfides, carbonates, chromates, and phosphates are insolublephosphates are insoluble

Lower number rules supersede so NaLower number rules supersede so Na22S S

is solubleis soluble

4.6 Describing reactions in solution4.6 Describing reactions in solution

Types of equations used to represent Types of equations used to represent chemical reactions: chemical reactions:

– Formula equationFormula equation

– Complete ionic equationComplete ionic equation

– Net ionic equationNet ionic equation

Writing Net Ionic EquationsWriting Net Ionic Equations

1. Write the balanced formula equation.

2. Write the net ionic equation showing the strong electrolytes

3. Determine precipitate from solubility rules

4. Cancel the spectator ions on both sides of the ionic equation

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

Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

What mass of solid is formed when What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M is mixed with 100.00 mL of 0.100 M sodium hydroxide?sodium hydroxide?

Ba(Cl)Ba(Cl)22 +2NaOH Ba(OH) +2NaOH Ba(OH)2 2 (s)(s) + 2NaCl + 2NaCl BaBa2+2+ + 2Cl + 2Cl-- + 2Na + 2Na++ + 2OH + 2OH-- Ba(OH) Ba(OH)2 2 (s)(s) + 2Na + 2Na++ +2Cl +2Cl--

4.7 Stoichiometry of precipitation 4.7 Stoichiometry of precipitation reactionsreactions

ExampleExample Calculate mass of solid NaCl that must be to 1.50 L Calculate mass of solid NaCl that must be to 1.50 L

0f 0.100M AgNO0f 0.100M AgNO33 solution to precipitate all of the solution to precipitate all of the

AgAg++ ions in the form of AgCl ions in the form of AgCl

AgAg++(aq) + Cl(aq) + Cl-- (aq) AgCl(s) (aq) AgCl(s)

#moles Ag#moles Ag++ = #moles AgNO = #moles AgNO33

= 1.50L X = 0.15 mol Ag= 1.50L X = 0.15 mol Ag+ +

1 L

0.100 mol Ag+

#mole Cl#mole Cl- - required =#mole NaCl = #moles Agrequired =#mole NaCl = #moles Ag++

(1:1 mole ratio)(1:1 mole ratio)

= 0.15 mol NaCl X 0.15 mol NaCl X 1mol NaCl

58.45 g NaCl

= 8.77 g NaCl

4.8 Acid-Base Reactions4.8 Acid-Base Reactions

For the purpose of this chapter: For the purpose of this chapter:

An acid is a proton donor An acid is a proton donor a a base is a proton acceptorbase is a proton acceptor usually usually

(donates OH(donates OH- - ions to the solution)ions to the solution)

acidbase acid base

Describing acid-base reactionsDescribing acid-base reactions

HCl (aq) + NaOH (aq) NaCl (aq) + H2O

H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O

H+(aq) + OH- (aq) H2O (l)

HC2H3O2(aq)+NaOH (aq) NaC2H3O2(aq) +H2O

HC2H3O2(aq) +Na+ + OH- C2H3O2- +Na+ +H2O

HC2H3O2(aq) + OH- C2H3O2- +H2O

Weak acid

Strong acid

When acid and bases with equal When acid and bases with equal amounts of hydrogen ion Hamounts of hydrogen ion H++ and and hydroxide ions OHhydroxide ions OH-- are mixed, the are mixed, the resulting solution is neutral. resulting solution is neutral. This reaction is called This reaction is called Neutralization

reaction• HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

• H2SO4(aq) + 2KOH (aq) K2SO4(aq) + 2H2O(l)• Notice salt (NaCl) and water are the products • Salt =Salt = ionic compound whose cation comes from

a base and anion from an acid.• Neutralization between acid and metal

hydroxide produces water and a salt.

Neutralization Reactions and SaltsNeutralization Reactions and Salts

ExampleExample

How many mL of 2.00 M HHow many mL of 2.00 M H22SOSO44 are required to are required to

neutralize 50.0 mL of 1.00 M KOH?neutralize 50.0 mL of 1.00 M KOH?

HH22SOSO44 + 2KOH + 2KOH KK22SOSO44 + 2H + 2H22OO

0.0500 L x 0.0500 L x 1.00 mole KOH1.00 mole KOH x x 1 mole H1 mole H22SOSO44 x x

1 L 2 mole KOH1 L 2 mole KOH

1 L 1 L x x 1000 mL 1000 mL = 12.5 mL = 12.5 mL

2 mole 2 mole HH22SOSO44 1 L 1 L

AnotherAnother solutionsolution HH22SOSO44 + 2KOH + 2KOH KK22SOSO44 + 2H + 2H22OO

#mole KOH = (M) KOH X (V) KOH#mole KOH = (M) KOH X (V) KOH

#mole KOH = 0.05 L X1M = 0.05 mol KOH#mole KOH = 0.05 L X1M = 0.05 mol KOH

#mole H#mole H22SOSO44 == ## mole KOH Xmole KOH X2mol KOH

1 mol H2SO4

= 0.025 mol H2SO4

#mole H2SO4 = M H2SO4 X vol (L) H2SO4

0.025 mol H2SO4 = 2.00 M X VV = 0.0125 L = 12.5 mL

Acid - Base TitrationsAcid - Base Titrations

Often called a Often called a neutralization titrationneutralization titration Because the Because the acid neutralizes the base acid neutralizes the base

Often called Often called volumetric analysisvolumetric analysis since titration is since titration is made to determine concentrationsmade to determine concentrations

It involves delivery of a measured volume of It involves delivery of a measured volume of solution of known concentration solution of known concentration (titrant) (titrant)

Titrant is added to the unknown Titrant is added to the unknown (analyte)(analyte)

until the until the equivalence (stoichiometric) pointequivalence (stoichiometric) point is is reached where enough titrant has been added to reached where enough titrant has been added to neutralize it. neutralize it.

Titration A solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point– the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

Endpoint –the point at which the color of indicator changes

TitrationTitration Equivalence point is marked by using an Equivalence point is marked by using an

indicatorindicator Where the indicator changes color is the Where the indicator changes color is the

endpointendpoint End point does not match always the End point does not match always the

equivalence point.equivalence point. A successful titration requires:A successful titration requires:

– A rapid known exact reactionA rapid known exact reaction– Endpoint is very close to the equivalence pointEndpoint is very close to the equivalence point– Accurate Measurement of volume of titrant Accurate Measurement of volume of titrant

Accurate determination of a solution Accurate determination of a solution

concnetration is called concnetration is called

TitrationTitration

A 50.00 mL sample of aqueous Ca(OH)A 50.00 mL sample of aqueous Ca(OH)22

requires 34.66 mL of 0.0980 M Nitric acid requires 34.66 mL of 0.0980 M Nitric acid

for neutralization. What is [Ca(OH)for neutralization. What is [Ca(OH)22 ]? ]?

# of H# of H++ x M x MA A x Vx VA A = = # of OH# of OH-- x M x MB B x Vx VBB

ExampleExample

In the titration of KHP (molar mass = In the titration of KHP (molar mass = 204.22g/mol), 34.67 mL of NaOH solution 204.22g/mol), 34.67 mL of NaOH solution was required to react with 0.1082 g KHP. was required to react with 0.1082 g KHP. Calculate the molarity of NaOH solution. Calculate the molarity of NaOH solution.

SolutionSolution

KHPmol

NaOHmol1

KHPg22.204

KHPmol1

L1067.34

mol10298.53

4

NaOH(aq) + KHP(aq) → NaKP(aq) + H2O(l)

There are 5.298X10-4 mol of sodium hydroxide in 34.67 mL of solution. #mole NaOH = MNaOH X VNaOH

= 1.528 M NaOH

0.1082 g KHP× # mol NaOH =

= 5.298X10-4 mol NaOH

NaOHV

molNaOH# =MNaOH =

)(2)()(211

2

00

sClNagClsNa

Each sodium atom loses one electron:

Each chlorine atom gains one electron:

eNaNa10

10 CleCl

4.9 Oxidation - Reduction (Redox) 4.9 Oxidation - Reduction (Redox) ReactionsReactions

Definitions

eNaNa10

Loss of Electrons = Oxidation

Sodium is oxidized

Gain of Electrons = Reduction10

CleCl Chlorine is reduced

Increase of the oxidation state

Decrease of the oxidation state

Definitions - The substance that loses electrons is called reducing agent (Electron donor)(Electron donor) - The substance that gains electrons is called oxidizing agent (Electron acceptor)

Mg(s) + S(s) → MgS(s)Mg is oxidized – loses e-

S is reduced – gains e-

Mg is the reducing agent

S is the oxidizing agent

0 0+2 -2

Not All Reactions are Redox Reactions

- Reactions in which there has been no change in oxidation number are not redox reactions.

Examples:

)()()()( 3

2511111

3

251

aqONNasClAgaqClNaaqONAg

)()()()(22

2

1

4

26

2

1

4

26

2

1121

lOHaqOSNaaqOSHaqHONa

Assigning Oxidation States

• An “oxidation state” (oxidation number) is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction.

• Generally, a bonded atom’s oxidation number is the charge it would have if the electrons in the bond were assigned to the atom of the more electronegative element

Rules for Assigning Oxidation States

1)The oxidation StateState of any uncombined element is zero.

2)The oxidation StateState of a monatomic ion equals its charge.

11

2

00

22

ClNaClNa


3) The oxidation statestate of oxygen in compounds is -2, except in peroxides, such as H2O2 where it is -1.

4) The oxidation statestate of hydrogen in compounds is +1, except in metal hydrides, like NaH, where it is -1.

2

2

1

OH

Rules for Assigning Oxidation states

5) The sum of the oxidation states of the atoms in the compound must equal 0.

2

2

1

OH2(+1) + (-2) = 0 H O

2

122

)(

HOCa(+2) + 2(-2) + 2(+1) = 0 Ca O H


6) The sum of the oxidation StateStates in the formula of a polyatomic ion is equal to its ionic charge.

3

2?

ONX + 3(-2) = -1N O

24

2?

OS

X = +5 X = +6

X + 4(-2) = -2S O

NaIO3

Na =? O = ?

3x-2 + 1 + ? = 0

I =

IF7

F = ?

7x-1 + ? = 0

I =

K2Cr2O7

O = ? K = ?

7x-2 + 2x+1 + 2x? = 0

Cr =??

What is the oxidation states of all atoms in the following ?

4.10 Balancing oxidation-reduction Equations

Two systematic methods for balancing redox equations are available, and are based on the fact that the total electrons gained in reduction equals the total lost in oxidation.

The two methods:

1) Use oxidation state changes

2) Use half-reactions (the method to be used her)

1: write separate half-reaction equations

for oxidation and reduction

2. balance the atoms in the half reactions

3. add enough electrons to one side of each

half-reaction to balance the charges

4. multiply each half-reaction by a number to

make the electrons equal in both

5. add the balanced half-reactions to show an

overall equation and cancel identical

species

6. Check that elements and charges are balanced

Balancing redox equations using half-reactionsBalancing redox equations using half-reactions

Example

2Mg (s) + O2 (g) 2MgO (s)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-

2Mg + O2 2MgO

Balancing redox equationsBalancing redox equationsin an acidic solutionin an acidic solution

For reactions in acidic solution For reactions in acidic solution an 8 step procedure.an 8 step procedure.

Write separate half reactionsWrite separate half reactions For each half reaction balance all For each half reaction balance all

reactants except H and Oreactants except H and O Balance O using HBalance O using H22OO

Balance H using HBalance H using H++

Balance charge using eBalance charge using e-- Multiply equations to make Multiply equations to make

electrons equalelectrons equal Add equations and cancel Add equations and cancel

identical speciesidentical species Check that charges and elements Check that charges and elements

are balanced.are balanced.

Example: Balance the following equation

I- + OCI- I3- + Cl- (acidic solution)

Oxidation : I- I3-

3I- I3-

3I- I3- + 2e-

Reduction : OC1- C1-

OC1- C1- + H2O

OC1- C1- + H2O

2H+ + OC1- C1- + H2O

1+ -1

2H+ + OC1- + 2e- C1- + H2O (2)

3I- I3- + 2e- (1)

(1) + (2)

2H+ + OC1- + 3I- C1- + I3- + H2O

• Do everything you would with acid, but add one more step.

• Add enough OH- to both sides to neutralize the H+

• For each H atom that is lacking, add one molecule of H2O to the side that requires it. At the same time add one unit of OH- in the opposite side of the half-reaction

Balancing equations in basic solution

Example

First half reaction:MnO4

- MnO2

(a) MnO4- MnO2 + 2H2O

(b) 4 H2O + MnO4- MnO2 + 2H2O + 4 OH-

(c) 4 H2O + MnO4- + 3 e- MnO2 + 2H2O + 4 OH-

(d) 2 H2O + MnO4- + 3 e- MnO2 + 4 OH- (1)

Balance the following equation in basic solution

MnO4- + C2O4

2- MnO2 + CO32-

Second half reaction:

C2O42- CO3

2-

(a) C2O42- 2 CO3

2-

(b) 2 H2O + C2O42- 2 CO3

2-

(c) 4 OH- + 2 H2O + C2O42- 2 CO3

2- + 4 H2O

(d) 4 OH- + 2 H2O + C2O42- 2 CO3

2- + 4 H2O + 2 e-

(e) 4 OH- + C2O42- 2 CO3

2- + 2 H2O + 2 e- (2)

2 H2O + MnO4- + 3 e- MnO2 + 4 OH- (1)

We eliminate electrons by multiplying (1) by 2 and (2) by 3.

(2) X 3:

12 OH- + 3 C2O42- 6 CO3

2- + 6 H2O + 6 e- (3)

4 H2O + 2 MnO4- + 6 e- 2 MnO2 + 8 OH- (4)

(1) X 2:

(3) + (4) to eliminate the electrons and common species.

4 OH- + 3 C2O42- + 2 MnO4

- 6 CO32- + 2 H2O + 2 MnO2

Chapter 4 Types of Chemical Reactions and Solution Chemistry.

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Transcript of Chapter 4 Types of Chemical Reactions and Solution Chemistry.