Chapter 4

The Major Classes of Chemical ReactionsLearning Goals: Chapter 4

4.6 Elements in Redox Reactions

4.1 The Role of Water as a Solvent: Terms

4.2 Writing Equations for Aqueous Ionic Reactions

4.3 Precipitation Reactions

4.4 Acid-Base Reactions

4.5 Oxidation-Reduction (Redox) Reactions

4.7 Reversible Reactions: An Introduction to Chemical Equilibrium

Water is a polar molecule due to oxygen’s ability to attract electrons (called electronegativity) more so than hydrogen.

Oxygen atom “sucks” electron density from hydrogen atoms leading to a “polar molecule”.

e-e- It’s just symbolic language

HH

O

HH

O

The dissolution of an ionic compound in water

Ionic compounds dissolve when ion-water forces are larger than ion-ion forces holding the ionic crystal together.

“Likes dissolve likes”

Ionic substances conduct electricity, covalent substances do not.

Strong Electrolyte – 100% dissociation into ions NaCl (s) Na+ (aq) + Cl- (aq)H2O

Weak Electrolyte – not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

CH3OH Non-Electrolyte – no dissociation

An electrolyte is a substance that when dissolved in water results in a solution that can conduct electricity.

Strong Electrolyte100% Dissociation

Weak ElectrolyteLittle dissociation

Non-electrolyteNo Dissociation

A non-electrolyte is a substance that when dissolved, results in a solution that does not conduct electricity.

Solubility (S) is the maximum amount of a solute that can dissolve in a fixed quantity of a solvent at a specified temperature. (Units of g solute/100 g water)

Examples: Sucrose (sugar) - 203 g per 100 g H2O

NaCl - 39.12 g per 100 g H2O (very soluble)

AgCl - 0.0021 g per 100 g H2O (insoluble)

Ionic compounds dissociate into ions thus its chemical formula tells us the number of moles of different ions in solution.

How many moles of each ion are in the following solutions?

(a) 5.0 mol of ammonium sulfate dissolved in water

(b) 78.5 g of cesium bromide dissolved in water

(c) 7.42 x 1022 formula units of copper(II) nitrate dissolved in water

(e) 35 mL of 0.84 M glucose (C6H12O6)

Covalent compounds do not dissociate!

78.5 g CsBr mol CsBr

212.8 g CsBr = 0.369 mol CsBr

= 0.369 mol Cs+

= 0.369 mol Br-

(b) CsBr(s) Cs+(aq) + Br-(aq)

7.42x1022 formula units Cu(NO3)2

mol Cu(NO3)2 6.022x1023 formula units

= 0.123 mol Cu(NO3)2

= 0.123 mol Cu2+

= 0.246 mol NO3-

(c) Cu(NO3)2(s) Cu2+(aq) + 2NO3-(aq)

35 mL ZnCl2 1L

103mL = 2.9x110-2 mol ZnCl2

(d) ZnCl2(aq) Zn2+(aq) + 2Cl-(aq) 0.84 mol ZnCl2

L

= 2.9x110-2 mol Zn2+ = 5.8x10-2 mol Cl-

H2O

H2O

H2O

(a) (NH4)2SO4(s) 2NH4+(aq) + SO4

2-(aq)

5.0 mol (NH4)2SO42 mol NH4

+

1 mol (NH4)2SO4

= 10. mol NH4+

5.0 mol SO42-

H2O

What does a 3.5 M FeCl3 mean? What does 3.5 M FeCl3 mean?

(1) a homogeneous solution of 3.5 moles of dry 100% pure FeCl3 dissolved in 1.00 Liter total solution volume (not 1 L of liquid!).

(3) Note: It does not mean 3.5 moles of FeCl3 is dissolved in 1.00 liter of water!

(4) [Fe3+] = 3.5M and [Cl-] = 3 x 3.5 M

(5) It can be used as a conversion factor

3.5 moles FeCl31 Liter solution

3.5 M FeCl3 =

1. Precipitation Reaction-– an insoluble solid is formed from specific cation-anion combinations.

2. Acid-Base Reaction-– a protons donor substance reacts with a hydroxide donor substance forming a salt and water.

3. Oxidation-Reduction Reaction-electron donor substances react with react with substances that accept electrons.

There are 3-classes of chemical reactions that occur in aqueous solution.

Oxidation

Which is the

Loss of e-

Precipitation Acid-BaseNeutralization

OxidationReduction

Cations Anions H+ ions OH- ions

Combine to Form

Insoluble Precipitate

Predicted by

SolubilityRules

Involves Substances

Combine to Form

Salt and H2O

Reduction

Gain of e-

Which is called

Oxidizing Agent

Reducing Agent

Involves Substances Involves Substances

In a precipitation reactions a metal cation and non-metal anion combine & form an insoluble solid that “precipitates”.

AgNO3(aq) + Na2CrO4(aq) => Ag2CrO4 (s) + NaNO3(aq)

Spectator ions: any species that remains soluble.

Precipitation occurs when ionic attractive forces overcome water’s tendency to hydrate and dissolve.

How Do We Predict If A Precipitate Rx Will Occur?

We memorize solubility rules and apply the rule “if it can happen, it will happen”.

Solubility Rules For Ionic Compounds Silberberg

1. Salts of Group 1A and ammonium ion (NH4+)

2. NO3-, CH3COO- or C2H3O2

- , ClO4-

3. Cl-, Br- I- except those of Ag+, Pb2+, Cu+, and Hg22+.

1. All OH, except those of Group 1A and the larger members of Group 2A (beginning with Ca2+).

2. CO32- and PO4

3-) are insoluble, except those of Group 1A(1) and NH4

+.

3. All S2- except those of Group 1A, Group 2A and NH4+.

Soluble Ionic Compounds

Insoluble Ionic Compounds

The “driving force” of a precipitation reaction is the act of PRECIPITATION.

If it can happen, it will.

1. Note the ions present in the reactants.2. Consider the possible cation-anion combinations.3. Refer to the table of solubility rules and decide whether any of the ion combinations is insoluble. 4. If a combination is insoluble, that RXN will occur. 5. Write the molecular, ionic and net ionic equation for the reaction.

How To Predict Whether A Precipitation Reaction

Look at ions that are possible from the reactants, then use “Solubility Rules” !

Example: Pb(NO3)2(aq) + NaI(aq) ==> ?

Example: The reaction of the salts: Pb(NO3)2(aq) + NaI(aq) ==> PbI2(s)

1. Recognize => IONIC => Solubility Rules Apply

2. Consult the Solubility table to see if PbI2 or NaNO3 are insoluble.

3. PbI2 is insoluble so that reaction will occur.

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

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

- PbI2 + 2Na+ + 2NO3-

4. Now write 3-equations 1)molecular, 2) ionic and 3) net equation as:

There are 3-types of equations that are written for precipitation, acid-base and redox reactions1. The molecular equation

shows all reactants and products as undissociated compounds.

2. The total ionic equationshows all of the soluble ionic substances dissociated into ions.

3. The net ionic equationeliminates the spectator ions and shows the actual chemical change taking place.

Pb2+ + 2I- PbI2 (s)

Na+ and NO3- are spectator ions---they don’t do much but sit there!

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

-

precipitate

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

Putting It All Together: Predicting Precipitation

1. Know ionic nomenclature so you can write the correct ionic formula of reactants and products.

2. Write the molecular equation by writing the chemical formula for reactants and products.

3. Refer to the table of solubility rules and decide whether any of the ion combinations is insoluble. 4. If a candidate is insoluble, that reaction will occur.

5. Remove the spectator ions and write the net ionic equation that summarized the reaction.

0. Know how to read a Table of Solubility Rules

3. Break the compounds into their ions and write the ionic equation for the reaction.

Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions.

Learning Check: Precipitation Reactions

(b) ammonium perchlorate(aq) + sodium bromide(aq)

(a) sodium sulfate(aq) + strontium nitrate(aq)

(c) silver nitrate(aq) + sodium chromate(aq)

Solubility Rules For Ionic Compounds in Water

Molecular Equation Na2SO4(aq) + Sr(NO3)2 (aq) 2NaNO3(aq) + SrSO4(s)

2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3

-(aq) 2Na+(aq) +2NO3-(aq)+ SrSO4(s)

SO42-(aq)+ Sr2+(aq) SrSO4(s)

Ionic Equation

Net Ionic Equation

3. Cancel the spectator ions to obtain the net ionic equation

1. Convert names formulas, write a balanced equation showing intact compounds (not dissociated).

2. Dissociate the formula equation to an ionic equation showing cations and anions with charge and phase.

NH4ClO4(aq) + NaBr (aq) NH4Br (aq) + NaClO4(aq)

Examing the solubility table shows that none of the possible combinations of ions would result in an insoluble precipitate (NH4ClO4 is soluble as is NH4Br, NaBr and NaClO4)

In this case all ions are spectator ions.

NH4+ + ClO4+(aq) + Na+ + Br- (aq) NH4+ + Br -(aq) +

Na+ + ClO4 -(aq)

Molecular Equation

Ionic Equation

Does a precipitate form when silver nitrate is mixed with a solution of sodium chromate? Write the molecular, ionic and net ionic equations for the reaction.

1. Write the balanced molecular equation.

2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.

3. Cancel the spectator ions on both sides of the ionic equation to obtain the net ionic equation

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

2. Acid-Base Reactions

The effects of acid rain on a statue of George Washington taken in 1935 (left) and 2001 (right) marble.

1. Precipitation Reaction-– an insoluble solid is formed from specific cation-anion combinations.

2. Acid-Base Reaction-– a protons donor substance reacts with a hydroxide donor substance forming a salt and water.

3. Oxidation-Reduction Reaction-electron donor substances react with react with substances that accept electrons.

There are 3-classes of chemical reactions that occur in aqueous solution.

Oxidation

Which is the

Loss of e-

Precipitation Acid-BaseNeutralization

OxidationReduction

Cations Anions H+ ions OH- ions

Combine to Form

Insoluble Precipitate

Predicted by

SolubilityRules

Involves Substances

Combine to Form

Salt and H2O

Reduction

Gain of e-

Which is called

Oxidizing Agent

Reducing Agent

Involves Substances Involves Substances

Chemistry uses 3-definitions of acids and bases: 1) Arrehenius 2) Bronsted-Lowry 3) Lewis

Name Acid Definition Base Definition

Arrhenius Substance that increases H+

Substances that increase OH-

Brønsted-Lowry

Substances that donate H+

Substances that accept H+

Lewis Electron-pair acceptor

Electron-pair donar

Chem 7 Chem 11

Acid: substance that has a covalent H atom in its formula, and releases a proton H+ when dissolved in water.

Base: a substance that contains OH in its formula, and releases hydroxide ions (OH-) when dissolved in water.

MOH(s) ==> OH- + M+ H2O

HCl(aq), HBr(aq), HI (aq), HNO3(aq), H2SO4(aq), HClO4(aq)HCl(g), HBr(g), HI (g), HNO3(l), H2SO4(l), HClO4(l)

NaOH(s), KOH (s), LiOH (s), Mg(OH)2(s), Ca(OH)2(s)NaOH(aq), KOH (aq), LiOH (aq), Mg(OH)2(aq), Ca(OH)2(aq)

HA(g) ==> H+(aq) + A-(aq)H2O

Chemists symbolize the reactions of acids with water two confusing ways. Don’t let it bug you.

HA(g) ===> H+ + A-H2O

HCl(g) ===> H+ + Cl-H2O

HNO3(l) ===> H+ + NO3-H2O

HCl(g) + H2O => H3O+ + Cl-

HNO3(g) + H2O => H3O+ + NO3-

HA(g) + H2O => H3O+ + A-

H3O+ RepresentationH+ representation

Generalized Acid HAGeneralized Acid HA

Acids and bases are classified as either strong or weak. We use arrows to symbolize the difference.

KOH(aq) ==> OH-(aq) + K+ (aq)

H2SO4(aq) ==> 2H+(aq) + SO42-(aq)Strong Acid

Strong Base

100% ionized = strong electrolyte = ==> arrow

NH4OH(aq) <==> OH-(aq) + NH4+ (aq)

HNO2(aq) <==> H+(aq) + NO2-(aq)Weak Acid

Weak Base

<20% ionized = weak electrolyte = <==> arrow

Strong acids and strong bases completely dissociate or completely ionize in water:

Before dissociation

Afterdissociation

HA HA A-H3O+

H3O+

A-

HA(g) + H2O(l) H3O+(aq) + A-(aq) HA H+ + A-

Strong acids and strong bases dissociate completely, conduct electricity and are strong electrolytes.

Strong StrongHCl(aq) H+(aq) + Cl-(aq)

HNO3(aq) H+ (aq) + NO3-(aq)

H2SO4(aq) H+ (aq) + HSO4-(aq)

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

Ca(OH)2(aq) Ca2+(aq) + 2OH-(aq)

Uni-directional arrow used to heavily product-favored

Group I and II Hydroxides

Strong Bases

sodium hydroxide: NaOH

calcium hydroxide Ca(OH)2

potassium hydroxide: KOH

strontium hydroxide Sr(OH)2

barium hydroxide Ba(OH)2

lithium hydroxide: LiOH

Strong Acids

hydrochloric acid HClhydrobromic acid HBrhydroiodic acid HInitric acid HNO3sulfuric acid H2SO4perchloric acid HClO4

H-X acidsOxide containing acids

We must memorize common strong acids and strong bases. All are strong electrolytes that dissociate completely in solution.

Remember anything not strong is weak!

Acid Anion Anion NameHF F- fluoride ionCH3COOH CH3COO- acetate ionHCN CN- cyanide ionHNO2 NO2

- nitrite ion H2CO3 CO3

2- carbonate ionH2SO3 SO3

2- sulfite ion

H3PO4 PO43- phosphate ion

(COOH)2 (COO)22- oxalate ion

Common Weak Acids and Their Anions

Weak acids and weak bases dissociate only to a slight extent in water Ka << 1.

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Before dissociation

Afterdissociation

HA HA A-H3O+

HAHA

HA

HA

HA

HA

reactant-favored

HA H+ + A-

Weak acids and weak bases do not dissociate or ionize to a large extent in solution, and so are weak electrolytes (small value of Ka).

CH3CO2H(aq) <=> CH3CO2- (aq) + H+

H2SO3 (aq) <=> HSO3- (aq) + H+(aq)

All not strong is weak!

HNO2 (aq) <=> NO2- (aq) + H+(aq)

Acids can have one, two or three acidic protons depending on their structure.

Monoprotic acids--only one H+ availableHCl H+ + Cl- Strong electrolyte, strong acid

HNO3 H+ + NO3- Strong electrolyte, strong acid

CH3COOH H+ + CH3COO- Weak electrolyte, weak acid

Diprotic acids--two acidic H+ available for reactionH2SO4 H+ + HSO4

-

HSO4- H+ + SO4

2-

Strong electrolyte, strong acid

Weak electrolyte, weak acid

Triprotic acids--three acidic H+

H3PO4 H+ + H2PO4-

H2PO4- H+ + HPO4

2-

HPO42- H+ + PO4

3-

Weak electrolyte, weak acid

Acids and bases have distinct properties.Acids:

• Acrid sour taste• React with metals (Group I,II) to yield H2 gas• Changes plant dye litmus from blue to red• React with carbonates and bicarbonates to

produce CO2 gas

Base

Acid

Bases: • Bitter taste• Slippery feel• Changes plant dye litmus from red to blue• React and neutralizes the effects of acids

200 Million MT H2SO4

50 Million MT NaOH/yr3 million containers

Common Acids

Common Bases

, soft drinks

Most anti-perspirants, watertreatment plants, paper

Acids and bases are everywhere. Determining the Molarity of H+ Ions in Aqueous Solutions of AcidsNitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M nitric acid?

What is the H+ molarity of 0.70 M H2SO4?

Of H+ .466 M H3PO4?

Of 2.5 M NaOH ?

Determining the Molarity of H+ Ions in Aqueous Solutions of Strong Acids or BasesNitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M nitric acid?

What is the molarity of H+ in a 0.70 M H2SO4?

One mole of H+(aq) is released per mole of nitric acid (HNO3)

HNO3(l) H+(aq) + NO3-(aq)

1.4M HNO3(aq) is 1.4M H+(aq).

H2O

What is the molarity of H+ in a 0.466 M H3PO4?

NaOH (aq) + HCl (aq) ==> H2O + Na+ + Cl-

base + acid ==> Water + salt

A salt’s cation comes from a parent base and anion comes from a parent acid!

Acids react with bases in a chemical reaction called a “neutralization reaction” forming a salt and water.

NaOH (aq) + HCl (aq) ==> H2O + Na+ + Cl-

Chemists use general symbols to represent the “neutralization reaction” to form a salt and water.

MOH (aq) + HX (aq) ==> H2O + M+ + X-

Alkali Metal Cation

Halide Anion

base + acid ==> Water + salt1. The molecular equation

shows all reactants and products as undissociated compounds.

2. The total ionic equationShows all soluble ionic substances dissociated into ions.

3. The net ionic equationEliminate spectator ions and show actual chemical change

Ca2+ and SO42- are spectator ions---they just watch!

Learning check: write the molecular, ionic and net equation for the neutralization between calcium hydroxide and sulfuric acid.

Ca(OH)2 (aq) + H2SO4 (aq) ===> 2H2O + CaSO4

Ca2+ + 2OH- + 2H+ + SO42- ===> 2H2O + Ca2+ + SO42-

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

Learning check: Write molecular, ionic and net equations for the following acid base neutralization reactions.

strontium hydroxide(aq) + perchloric acid(aq)

barium hydroxide(aq) + sulfuric acid(aq)

Acetic acid(aq) + potassium

Nitric acid(aq) + barium hydroxide(aq)

Ba2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq)

2H2O(l)+Ba2+(aq)+SO42-(aq)

Writing Ionic Equations for Acid-Base Reactions

strontium hydroxide(aq) + perchloric acid(aq)

barium hydroxide(aq) + sulfuric acid(aq)(b) Ba(OH)2(aq) + H2SO4(aq) 2H2O(l) + BaSO4(aq)

(b) Sr2+(aq)+2OH-(aq)+ 2H+(aq)+2ClO4-(aq)

2H2O(l)+Sr2+(aq)+2ClO4-(aq)

(a) Sr(OH)2(aq)+2HClO4(aq) 2H2O(l)+Sr(ClO4)2(aq)

(c) 2OH-(aq)+ 2H+(aq) 2H2O(l)

(c) 2OH-(aq)+ 2H+(aq) 2H2O(l)

Weak acids dissociate to a very small extent and this fact is reflected in their equations as well using a double-arrow (<==>).

Molecular equationNaOH(aq) + CH3COOH(aq) CH3COONa(aq) + H2O

Total ionic equation Na+(aq)+ OH-(aq) + CH3COOH(aq)

CH3COO-(aq) + Na+(aq) + H2O(l)

Net ionic equation OH-(aq) + CH3COOH(aq) CH3COO-(aq) + H2O(l)

Weak acid is not dissociated!

1. Name and characterize each as a base, acid or salt?

HF(g), HI(aq), LiOH(aq), Mg(OH)2, Na2SO4 CH3COONH4

2. Name and classify the following as strong, weak acid or base?

HClO4, Sr(OH)2, HClO2, NH3(g), H3PO4(aq), H2SO4(aq), HNO3(aq)

3. Write the M/I/NI equation for the reactions of a) hydrochloric acid with calcium hydroxide and b) phosphoric acid with sodium hydroxide

4. What acids and bases were reacted to form the following salts? Show using balanced equations.

1) NaNO2 2) CaSO4 3) Mg(PO4)2

Learning Check: Acids and Bases

3. What is the correct formula of the salt formed in the neutralization reaction of hydrochloric acid with calcium hydroxide?

4. What is the chemical formula of the salt produced by the

neutralization of sodium hydroxide with sulfurous acid?

1. Identify the following as a strong or weak acid or base a salt. If a salt what is the parent acid and base?

HF, HI, LiOH, Ca(OH)2, Na2SO4 CH3COO-, NH4+

2. Classify the following as strong, weak acid or base?

HClO4, Sr(OH)2, HClO2, NH3, H3PO4, H2SO4

Learning check: Name the type of reaction and write M/I/NI ionic equations for the following:

1. hydrochloric acid and potassium hydroxide2. sodium carbonate and hydrochloric acid3. aluminum nitrate and sodium phosphate4. potassium chloride + iron(II) nitrate5. Iron sulfide and hydrochloric acid6. acetic acid and magnesium hydroxide7. sodium sulfite and hydrochloric acid

Reactions between acids and carbonate containing compounds yield gaseous CO2 and H2O + salt.

Molecular equation NaHCO3(aq) + CH3COOH(aq)

CH3COONa(aq) + CO2(g) + H2O(l)

Total ionic equation Na+(aq)+ HCO3

-(aq) + CH3COOH(aq)

CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)

Net ionic equation HCO3

-(aq) + CH3COOH(aq)

CH3COO-(aq) + CO2(g) + H2O(l)

Oxidation

Involves simultaneous

Which is the

Loss of e-

OxidationReduction

Reduction

Gain of e-

Which is called

Oxidizing Agent

Reducing Agent

WhichIncreases oxidation number

Decreases oxidation number

Metals tend to loose electrons(oxidized) forming cationsMetals are reducing agents

Non-metals tend to gain electrons (be reduced) forming anions which are oxidizing agents.

Redox reactions trends can be understood by applying the properties of elements in the periodic table.

Mg loses electron(s)Mg(s) is oxidizedMg is the reducing agentMg increases its oxidation number

Oxidizing PerspectiveO gains electron(s)O is reducedO is the oxidizing agentO decreases its oxidation number

Reduction Perspective

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

Reduced

We learn rules to figure out what is oxidized and what is reduced!

An oxidation-reduction reaction occurs when there is a net movement of electrons from one reactant to another. One reactant gives the other receives!

2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-)

O2 + 4e- 2O2- Reduction half-reaction (gain e-)

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

2Mg + O2 2MgO Net Equation

Ionic Equation

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

Reduced

Oxidation and reduction occur simultaneously.

1. The oxidation state of a free elements in atomic or molecular for are assigned an oxidation number of 0.

Na, Be, K, Pb, H2, O2, P4 = 0

2. The oxidation number of a an monoatomic ion is equal to the charge on the ion (Group Number Rule)

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. H2O (H = +1 and O = 2) Sum = 0

A set of rules called “assigning oxidation numbers” is used to identify what species is being oxidized and reduced in a chemical reaction.

5. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.

6. The oxidation number of oxygen is usually –2. In H2O2 and O2

2- it is –1 (exceptions).

HCO3-

O = -2 H = +1

3x(-2) + 1 + ? = -1C = +4

Oxidation numbers of all the elements in HCO3

- ?

7. When there is a conflict the lower numbered rules take priority over the higher numbered rules.

4. Fluorine has an oxidation state of –1 in its compounds

1. The oxidation state of any free element = 0

2. In monatomic ions, the oxidation number is equal to the charge on the ion (Use Group Number)

3. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.

4. Fluorine has an oxidation state of –1 in its compounds

5. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds (–1).

6. The oxidation number of oxygen is usually –2. In H2O2 and O2

2- it is –1 (exceptions).

7. When there is a conflict the lower numbered rules take priority over the higher numbered rules.

Quick Table Assigning Oxidation Numbers

HCO3-

O = -2 H = +1

3x(-2) + 1 + ? = -1

C = +4

IF7

F = -1

7x(-1) + ? = 0

I = +7

NaIO3

Na = +1 O = -2

3x(-2) + 1 + ? = 0

I = +5

K2Cr2O7

O = -2 K = +1

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

Cr = +6

Assign oxidation numbers to the following ions and the elements within a compound.

1.! Na2SO3

! 1.!Na +1! 2.!S +4! 3.!O -2

2.! PF3

! 1.!P +3! 2.!F -1

3.! CrO32-

! 1.!Cr +4! 2.!O -2

4.! Cr2O72-

! 1.!Cr +6! 2.!O -2

5.! (NH4)3PO4

! 1.!N -3! 2.!H +1! 3.!P +5! 4.!O -2

Assign oxidation numbers to the following ions and the elements within a compound.

Elements that make up compounds can take on a range of “oxidation states”.

Example: N

Chemists use oxidation rules to identify what is being oxidized and what is being reduced in a chemical reaction.

CH4 + 2O2 CO2 + 2H2O

2HgO 2Hg + O2

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

The reactions below occur as written. Use the rules for assigning oxidation numbers to the oxidizing and reducing agents in the reactions below? (increase in ON = reduced, decrease = oxidized)

-2+1

-2

+1-4 -4 -2

00+2

+2 +2 00

0

Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)

1. Assign oxidation numbers2. Identify the species being oxidized (reactant that increases in oxidation number) and the specie being reduced (reactant that decrease in oxidation number).3. The species that is oxidized is the reducing agent, the species that is reduced is called the oxidizing agent.

Identify the oxidizing agent and reducing agent in each of the following:

Zn is oxidizedZn Zn2+ + 2e- Zn is the reducing agent

Cu2+ is reducedCu2+ + 2e- Cu Cu2+ is the oxidizing agent

Identify the oxidizing agent and reducing agent in each of the following:

Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)Oxidized

Reduced

0 0-2 -2+2 +2 +6+6

Copper wire reacts with silver nitrate to form silver metal and soluble cupric nitrate. What is the oxidizing agent and the reducing agent in this reaction?

Copper wire reacts with silver nitrate to form silver metal and soluble cupric nitrate. What is the oxidizing agent and the reducing agent in this reaction? (Translate words to equations)

Ag+ is reduced Ag+ is the oxidizing agent

Cu is oxidized Cu is the reducing agent

Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)0 0+2+1

OxidizedReduced

Identify the oxidizing agent and reducing agent in each of the following:

(a) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)

(b) PbO(s) + CO(g) Pb(s) + CO2(g)

(c) 2H2(g) + O2(g) 2H2O(g)

1. Know the rules for assigning oxidation numbers

2. Assign oxidation numbers to all reactants and products

3. Increasing ON = Reactant Oxidized (reducing agent)

Decreasing ON = Reactant Reduced (oxidizing agent)

Identify the oxidizing agent and reducing agent in each of the following:

(c) 2H2(g) + O2(g) 2H2O(g)

(b) PbO(s) + CO(g) Pb(s) + CO2(g)

+2 -2 +2 -2 0 +4 -2

0 0 +1 -2

The O.N. of C increases; it is oxidized; CO is the reducing agent.The O.N. of Pb decreases; it is reduced; PbO is the oxidizing agent.

The O.N. of H increases; it is oxidized; it is the reducing agent.The O.N. of O decreases; it is reduced; it is the oxidizing agent.

1. When an element combines with O2 the element has been oxidized and O2 reduced.

2. When an element combines with a halogen the element is oxidized and the halogen reduced.

3. When a metal combines with something, the metal is been oxidized. Whatever the metal combined with, is reduced.

Using oxidation numbers Balancing Redox Reactions

• Few redox reactions are easy to balance by inspection.

• Most redox reactions involve H+ or OH- as sometimes they are reactants or products.

• We normally use H+ or OH- in excess and we use them with water to balance a redox reaction.

• We use the “Half-Reaction (Ion-Electron) Method”

Ion-Electron Method To Balance Redox Rxtns1. Divide the equation into two half-reactions-oxidation and reduction2. Balance all atoms other than H and O3. Balance O by adding H2O4. Balance H by adding H+

5. Balance net charge by adding e-

6. Balance e- gain to e- loss and add the half-reactions7. Add the half reactions and cancel common factors. 8. Add to both sides of the equation the same number of OH-

as there are H+

9. Combine H+ and OH- to form H2O and cancel common H2O10. Cancel common H2O

Balancing Redox Reactions In Acid

Balancing the Equation

SO32-(aq) + MnO4

-(aq) → SO42-(aq) + Mn2+(aq)

Sulfite ion + permanganate ion → sulfate ion + managenese

Note that the equation is not balanced in mass (atoms) or in charge!

Ion-Electron (Half-reaction) Method

SO32-(aq) + MnO4

-(aq) → SO42-(aq) + Mn2+(aq)

1. Determine the oxidation states of the various species.

4+ 6+7+ 2+

SO32-(aq) → SO4

2-(aq)

2. Identify the oxidizing and reducing agents. Write the half-reactions

MnO4-(aq) → Mn2+(aq)

3. Balance atoms other than H and O:

Already balanced for elements S and Mn

4+ 6+

7+ 2+

Example 5-64. Balance O by adding H2O to the side that needs O:

H2O(l) + SO32-(aq) → SO4

2-(aq)

MnO4-(aq) → Mn2+(aq) + 4 H2O(l)

5. Balance hydrogen by adding H+:

H2O(l) + SO32-(aq) → SO4

2-(aq) + 2 H+(aq)

8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)

Balancing Redox Equations

H2O(l) + SO32-(aq) → SO4

2-(aq) + 2 H+(aq) + 2 e-(aq)

5 e-(aq) + 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)

6. Add e- to balance charge. Charge should be equal on both sides.

7. Multiply the half-reactions to balance all e- in both equations:

5 H2O(l) + 5 SO32-(aq) → 5 SO4

2-(aq) + 10 e-(aq) + 10 H+(aq)

16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)

Balancing Redox Reactions8. Add both equations and simplify:

5 SO32-(aq) + 2 MnO4

-(aq) + 6H+(aq) →

5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l)

9. Check the result

10. If a basic solution is desired then add OH- to neutralize the H+ on both sides of the equation. Cancel common water.

5 SO32-(aq) + 2 MnO4

-(aq) + 6H+ + 6OH- (aq) →

5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l) + 6OH-

5 SO32-(aq) + 2 MnO4

-(aq) + 3H2O (aq) →

5 SO42-(aq) + 2 Mn2+(aq) + 6OH-

Balancing Redox Reactions

• Method used to balance redox reactions

Oxidation:

Reduction:

Overall:

Sn2+ → Sn4+(aq) + 2 e-

Fe3+(aq) + 1 e- → Fe2+

Fe3+(aq) + Sn2+ → Fe2+ + Sn4+3+ 2+ 2+ 4+

Sn2+ → Sn4+(aq) + 2 e-

2 Fe3+(aq) + 2 e- → 2Fe2+

Atoms and mass are balanced but charge is not. Let’s multiply by 2 to get charge to cancel

Fe3+(aq) + Sn2+ → Fe2+ + Sn4+

+ 4OH-

4H+ + MnO4- +3e- MnO2+ 2H2Ox2

C2O42- + 2H2O 2CO3

2- + 4H+ + 2e-x 3

8H+ + 2MnO4- + 6e- 2MnO2+ 4H2O

3C2O42- + 6H2O 6CO3

2- + 12H+ + 6e-

2MnO2-(aq) + 3C2O4

2-(aq) + 2H2O(l) 2MnO2(s) + 6CO32-(aq) + 4H+(aq)

+ 4OH-

2MnO2-(aq) + 3C2O4

2-(aq) + 4OH-(aq) 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

The redox reaction between dichromate and iodide anions

Cr2O72- I- Cr3+ + I2

Balancing redox reactions by the half-reaction methodPROBLEM: Permanganate ion is a strong oxidizing agent and its deep purple

color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate anion to form carbonate anion and solid manganese dioxide. Balance the skeleton ionic reaction that occurs between KMnO4 and Na2C2O4 in basic solution.

MnO4-(aq) + C2O4

2-(aq) MnO2(s) + CO32-(aq)

PLAN: Proceed in acidic solution and then neutralize with base.

SOLUTION:MnO4

- MnO2 C2O42- CO3

2-

MnO4- MnO2 C2O4

2- CO32-2

MnO4- MnO2 + 2H2O4H+ + C2O4

2- 2CO32- + 2H2O + 4H+

+7 +4 +3 +4

+ 3e-+ 2e-

R O

Balancing redox reactions in acidic solution

Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(aq)

1. Divide the equation into two half-reactions---an oxidation and a reduction use oxidation numbers if necessary.

2. Balance all atoms other than H and O

3. Balance O by adding H2O (count)

4. Balance H by adding H+

5. Balance net charge by adding e-

6. Balance e- gain to e- loss and add the half-reactions

7. Add the half reactions and cancel common factors.

8. For basic reactions neutralize the H+ with OH- but add it on both sides. Cancel common water.

Cr2O72-(aq) + 6I-(aq) 2Cr3+(aq) + 3I2(aq) + 7H2O(l)14H+(aq) +

Balancing redox reactions in basic solution

❐ Add OH- to neutralize the H+ ions.

14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l)

+ 14OH-(aq) + 14OH-(aq)

14H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH-

❐ Reconcile the number of water molecules.

Do a final check on atoms and charges.

7H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 14OH-

1. Write the unbalanced equation for the reaction ion ionic form.

Show a balanced equation for the oxidation of Fe2+ to Fe3+ by Cr2O7

2- (to Cr3+) in acid solution?

Fe2+ + Cr2O72- Fe3+ + Cr3+

2. Identify what is being oxidized and reduced using oxidation numbers. Separate the equation into two half-reactions.

Oxidation:

Cr2O72- Cr3+

+6 +3Reduction:

Fe2+ Fe3++2 +3

3. Balance the atoms other than O and H in each half-reaction.

Cr2O72- 2Cr3+

4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.

Cr2O72- 2Cr3+ + 7H2O

14H+ + Cr2O72- 2Cr3+ + 7H2O

5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.

Fe2+ Fe3+ + 1e-

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

6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.

6Fe2+ 6Fe3+ + 6e-

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

7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.

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

6Fe2+ 6Fe3+ + 6e-Oxidation:

Reduction:

14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

8. Verify that the number of atoms and the charges are balanced.

14x1 – 2 + 6x2 = 24 = 6x3 + 2x3

9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.

How Can We Predict Redox Reactions?Show the oxidation and reduction that occur and write an

overall ionic equation for the reaction of hydrochloric acid with iron to produce hydrogen and iron(II).

Fe(s) Fe2+(aq) + 2 e-

2H+ (s) + 2e- (g) H2

Fe(s) + 2H+(aq) Fe2+(aq) + H2(g)

1. Translate words to chemical formula HCl + Fe(s).2. What will be oxidized and reduced by looking at the activity series3. Write the half-reactions maintaining mass and charge balance4. Sum the two half-reactions

There are 5-Main Types of Redox Reactions1. Combination A + B => AB

i. Element + Element => Compoundii. Element + Compound => New Compoundiii. Compound + Compound => New Compound

2. Decomposition AB => A + B

3. Displacement AB + C => AC + B

4. Disproportionation--common element or element in compound is both oxidized and reduced

5. Combustion CH4 + O2 => CO2 + H2O

-produce fewer products than reactants.

-produce more products than reactants.

-methathesis (switching places)

1. Combination Reactions: A + B => C

S + O2 SO2

0 0 +4 -22. Non-Metal + Non-metal Binary Molec Cmpd

1. Metal + Non-metal Binary Ionic Compound2Na(s)+ Cl2 2NaCl

4. Compound + Compound New Compound

CaO(s)+ H2O Ca(OH)2

-2+2 -2+1 -2 +1+2

Combination reactions produce fewer products than reactants.

3. Compound + Element New Compound

PCl3+ Cl2 PCl5+5+3 0 -1

2. Decomposition Reactions

2H2O 2H2 + O2

2HgO 2Hg + O2

1. Compound Element A + Element B

2. Compound Compound A + Element B

--produces more products than reactants through the action of heat (thermal decomposition) or electricity (electrolytic decomposition).

2KClO3 2KCl + 3O2

+1 +5 -2 +1 -1 0heat

electricity

heat

3. Displacement Reactions

+4

0

A + BC AC + B

Metal Displacement

TiCl4(aq) + 2Mg(s) Ti + 2MgCl20 0 +2

Zn(s) + CuSO4 (aq) => ZnSO4 (aq) + Cu (s)

Ca(s) + 2H2O(aq) Ca(OH)2 + H2 H2 Displacement0 +1 +2 -2

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

+1 0-2→

A more “reactive metal” will displace a “less reactive metal” or H2 within their compounds to form the oxidized metal and the reduced form of the less active metal or H2.

0+2 +2

-1 -1

Methathesis =displacementdouble

Activity Series For MetalsPure metals higher in the series will be oxidized by a dissolved elemental ion below it in the series which is reduced.

H2 DISPLACEMENT Metals above H2 react with acids (H+), or H2O to displace H2(g). H+ reduced to H2.

Al(s) + H+ => Al3+ + H2(g)Ag(s) + H+ => No reaction

METAL DISPLACEMENTMore active metal + Salt of less active metal => Reduced Less active metal + Salt of more active metal.

Al(s) + Fe3+(aq) => Al3+ + Fe(s)Zn(s) + CuSO4 => ZnSO4 + Cu(s)

Solid Zn rod dunked into a solution of CuSO4.

Zn Rod

CuSO4

Metal Displacement (Activity Series)

CuSO4

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

A solid Zn rod dunked into a solution of copper sulfate (CuSO4)

The Zn is oxidized and the Cu2+ is reduced to solid Cu(s)

Use activity series: Zn metal is higher than Cu in the series.It will displace Cu2+

from solution.

1. Fe(s) + 3Cu2+ => 2Fe3+ + Cu(s)

2. Sn(s) + Ca2+ => Sn2+ + Ca(s)

3. Au(s) + 3AgNO3(aq) => Au(NO3)3

+ 3 Ag(s)

4. 2 Al(s) + Cr2O3(s) ! Al2O3(s) + 2 Cr(s)

5. Pt(s) + 4 HCl(aq) ! PtCl4(aq) + 2 H2(g)

Predict whether the following reactions will occur as written.

More active metal + Salt of less active metal => Reduced Less active metal + Salt of more active metal.

F2

Cl2

Br2

I2

Stre

ngth

as o

xidi

zing

age

nt

A more active halogen will displace a less active (heavier) halide from their binary salts.

-10 -1+1 0 +1Cl2 + 2NaI ===> I2 + 2NaClBr2 + 2NaI ===> I2 + 2NaBrI2(s) + 2F- ===> No reaction

Cl2 and Br2 both are higher and will displace I-

from its compounds, forming I2 and the chloride and bromide salt respectively. A less active halogen will not displace a more active halide from their salts.

Halogen + Salt of Less Active Halide gives Less Active Halide + salt of more active halide.

4. Disproportionation ReactionA single element in a compound is both oxidized and reduced (both oxidizing and reducing agent in one)

H2O2 + 2OH- 2H2O + O2 +1 +1 -2-1 +1-2 0

Cl2 + 2OH- ClO- + Cl- + H2O0 +1 -1+1-2 -2 -2+1

5. Combustion ReactionIn a combustion reaction a fuel or typically a hydrocarbon reacts with oxygen forming carbon dioxide and water. The oxygen is increases in O.N. and is therefore oxidized (reducing agent).

CH4 + 2O2 CO2 + 2H2O +4 +1-1 0

2C4H10 + 13O2 8CO2 + 10H2O

-2+4 -2

Identifying the Type of Redox ReactionClassify each of the following redox reactions as a combination, decomposition, or displacement reaction, write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents:

(a) magnesium(s) + nitrogen(g) magnesium nitride (aq)(b) hydrogen peroxide(l) water(l) + oxygen gas(c) aluminum(s) + lead(II) nitrate(aq) aluminum nitrate(aq) + lead(s)

Identifying the Type of Redox Reaction

(b)Decomposition H2O2(l) H2O(l) + O2(g)

+1 -1 +1 -2 0

1/2 or

(a) Combination0 0 +2 -3

Mg is the reducing agent; N2 is the oxidizing agent. 3Mg(s) + N2(g) Mg3N2 (aq)

2 H2O2(l) 2 H2O(l) + O2(g)H2O2 is the oxidizing and reducing agent.

(c)Displacement

Al(s) + Pb(NO3)2(aq) Al(NO3)3(aq) + Pb(s)

0 +2 +5 -2 +3 +5 -2 0

2Al(s) + 3Pb(NO3)2(aq) 2Al(NO3)3(aq) + 3Pb(s)

Pb(NO3)2 is the oxidizing and Al is the reducing agent.