CHEM 101 – CHAPTER 4
THREE MAJOR CLASSES OF CHEMICAL REACTIONS
The Role of Water as a Solvent
The Polar Nature of Water
Water’s great solvent power arises from…
 the uneven distribution of electron charge
o electrons are held closer to the oxygen atom
 its bent molecular shape
o it is not a linear molecule, it is bent at an angle of
104.5o
 it being a polar molecule
o the region near the oxygen atom is partially negative
and the region between the hydrogen atoms is partially
positive
Ionic Compounds in Water
How Ionic Compounds Dissolve:
Replacement of Charged
Attractions
 In an ionic solid, oppositely
charged ions are held together by
electrostatic attractions (positive
attracts negative)
 This diagram from chapter 2
reminds us of the structure of an
ionic compound…
 Water separates the ions by replacing these attractions with
stronger ones between several water molecules and each
ion
 Dissolution occurs because the attractions between each
type of ion and several water molecules outweigh the
attractions between the ions.
 Gradually, all the ions separate (dissociate) and become
solvated
o Surrounded closely by solvent molecules and then
move randomly in the solution
 Important understanding…
o Not all ionic compounds dissolve in water (table 4.1
on page 143)
 For these compounds (AgCl for example) the
attraction between the ions is greater than the
attraction between the ions and water
How Ionic Solutions Behave: Electrolytes and Electrical
Current
 When an ionic compound dissolves, the solution’s
electrical conductivity, the flow of electric current,
increases dramatically.
o Current flow implies the movement of charged
particles
 When the ionic compound dissolves, the separate
solvated ions moved toward the electrode of
opposite charge.
 Electrolyte – a substance that conducts a current when
dissolved in water
o Soluble ionic compounds are strong electrolytes
because they dissociate completely and conduct a
large current
Calculating the Amount (mol) of Ions in Solution
 The formula of the soluble ionic compound tells the
number of moles of the component ions in solution.
 Example…
o KBr (s) (in water)  K+ (aq) + Br – (aq)
 1 mol of KBr dissociates into 2 mol of ions
 1 mol of K+ ions and 1 mol of Br - ions
Sample Problem 4.1 p. 138
Using Molecular Scenes to Depict an Ionic Compound in
Aqueous Solution
Refer to the diagram on page 138 on the text…
The problem states that the four beakers contain aqueous
solutions of the strong electrolyte potassium sulfate.
 Potassium sulfate… potassium (K+) sulfate (SO42-)
o K2SO4
 K2SO4 (s) (in water)  2K+ (aq) + SO42- (aq)
(a) Beaker C best represents K2SO4 because there are 2
positive ions (1+) for 1 negative ion (2-)
(b) There are 9 particles in the diagram, so there is a total of
0.9 mol of particles
0.9 mol 6.022 x 1023 particles = 5.420 x 1023 particles
1 mol
Follow-Up Problem 4.1 p. 138
Refer to the diagram in the text…
The diagram implies 3 (2+) ions and 6 (1- ) ions
 The substance has to be BaCl2
o Ba2+ and Cl –
 9 x 0.05 mol = 0.45 total mol
9 particles 0.05 mol
particles
1 particle
1 mol
BaCl2
3 mol
particles
208.2 g = 31.2 g BaCl
BaCl2
1 mol
BaCl2
Sample Problem 4.2 p. 138
Determining Amount (mol) of Ions in Solution
(a)
(NH4)2SO4 (s) (in water)  2NH4+ (aq) + SO42- (aq)
5.0 mol (NH4)2SO4 2 mol NH4+
= 10. mol NH4+
1 mol (NH4)2SO4
5.0 mol (NH4)2SO4 1 mol SO42= 5.0 mol SO421 mol (NH4)2SO4
(b)
CsBr (s) (in water)  Cs+ (aq) + Br – (aq)
78.5 g CsBr 1 mol CsBr 1 mol Cs+
= 0.369 mol Cs+
212.8 g CsBr 1 mol CsBr
78.5 g CsBr 1 mol CsBr 1 mol Br - = 0.369 mol Br 212.8 g CsBr 1 mol CsBr
(c)
Cu(NO3)2 (s) (in water)  Cu2+ (aq) + 2NO37.42 x 1022
1 mol
1 mol Cu2+ = 0.123 mol Cu2+
formula units Cu(NO3)2
Cu(NO3)2
6.022 x 1023 1 mol
formula units Cu(NO3)2
Cu(NO3)2
7.42 x 1022
1 mol
2 mol NO3- = 0.246 mol NO3formula units Cu(NO3)2
Cu(NO3)2
6.022 x 1023 1 mol
formula units Cu(NO3)2
Cu(NO3)2
(d)
1 mol
Zn2+
1 mol
ZnCl2
= 2.9 x 10-2 mol Zn2+
35 mL 1 L
ZnCl2 ZnCl2
1000 mL
ZnCl2
0.84 mol
ZnCl2
1L
ZnCl2
35 mL 1 L
ZnCl2 ZnCl2
1000 mL
ZnCl2
0.84 mol 2 mol Cl- = 5.8 x 10-2 mol ClZnCl2
1L
1 mol
ZnCl2
ZnCl2
Follow-Up Problem 4.2 p. 139
(a)
KClO4 (s) (in water)  K+ (aq) + ClO4 – (aq)
2 mol KClO4 1 mol K+
= 2 mol K+
1 mol KClO4
2 mol KClO4 1 mol ClO4 = 2 mol ClO4
1 mol KClO4
(b)
Mg(C2H3O2)2 (s) (in water)  Mg2+ (aq) + 2C2H3O2354 g
1 mol
1 mol Mg2+
= 2.49 mol
Mg(C2H3O2)2 Mg(C2H3O2)2
Mg2+
142.40 g
1 mol
Mg(C2H3O2)2 Mg(C2H3O2)2
354 g
1 mol
2 mol C2H3O2- = 4.97 mol
Mg(C2H3O2)2 Mg(C2H3O2)2
C2H3O2142.40 g
1 mol
Mg(C2H3O2)2 Mg(C2H3O2)2
(c)
(NH4)2CrO4 (s) (in water)  2NH4+ (aq) + CrO421.88x1024
formula units
(NH4)2CrO4
1 mol
(NH4)2CrO4
2 mol NH4+ = 6.24 mol NH4+
6.022x1023
1 mol
formula units (NH4)2CrO4
(NH4)2CrO4
1.88x1024
formula units
(NH4)2CrO4
1 mol
(NH4)2CrO4
1 mol CrO4 - = 3.12 mol CrO4 -
6.022x1023
1 mol
formula units (NH4)2CrO4
(NH4)2CrO4
(d)
NaHSO4 (s) (in water)  Na+ (aq) + HSO4- (aq)
1.32 L
0.55 mol
1 mol Na+ = 0.73 mol Na+
NaHSO4 NaHSO4
1 L NaHSO4 1 mol
NaHSO4
1.32 L
0.55 mol
1 mol HSO4 - = 0.73 mol HSO4 NaHSO4 NaHSO4
1 L NaHSO4 1 mol
NaHSO4
Covalent Compounds in Water
 Water dissolves many covalent (molecular) compounds
also…
o Table sugar C12H22O11
o Ethanol CH3CH2OH (beverage alcohol)
o Automobile antifreeze HOCH2CH2OH (ethylene
glycol)
 These contain their own polar bonds, which interact with
the bonds of water
 Important difference from ionic compounds… most soluble
covalent substances do not separate (dissociate) into ions.
o HOCH2CH2OH (l) (in water)  HOCH2CH2OH (aq)
o Do not conduct electric current
 Nonelectrolytes – do not conduct electric current
Writing Equations for Aqueous Ionic Reactions
Chemists use three types of equations to represent aqueous
ionic reaction.
Example… the reaction of aqueous solutions of silver nitrate
and silver chromate
 Molecular equation – shows all the reactants and products
as if they were intact undissociated compounds.
2AgNO3 (aq) + Na2CrO4 (aq)  Ag2CrO4 (s) + 2NaNO3 (aq)
 Total ionic equation – shows all the soluble ionic
substances dissociated into ions
2Ag+ (aq) + 2NO3 – (aq) + 2Na+ (aq) + CrO42- (aq) 
AgCrO4 (s) + 2Na+ (aq) + 2NO3 – (aq)
o Spectator ions – they are not involved in the actual
chemical reaction…
2Ag+ (aq) + 2NO3 – (aq) + 2Na+ (aq) + CrO42- (aq) 
AgCrO4 (s) + 2Na+ (aq) + 2NO3 – (aq)
 Net ionic equation – eliminates the spectator ions and
shows only the actual chemical reaction
2Ag+ (aq)+ CrO42- (aq)  AgCrO4 (s)
Three types of chemical reactions…
1. Precipitation
2. Acid-base
3. Oxidation-reduction
Precipitation Reactions
The Key Event: Formation of a Solid from Dissolved Ions
 Precipitation reaction – two soluble ionic compounds
react to form an insoluble product.
o Precipitate – the insoluble product formed in a
precipitation reaction
 Why does a precipitate form?
o The electrostatic attraction between the ions outweighs
the tendency of the ions to remain solvated and move
throughout the solution.
 The key event…formation of an insoluble product through
the net removal of ions from solution.
 Example…
Molecular Equation
CaCl2 (aq) + 2NaF (aq)  CaF2 (s) + 2NaCl (aq)
Total Ionic Equation
Ca2+ (aq) + 2Cl - (aq) + 2Na+ (aq) + 2F - (aq) 
CaF2 (s) + 2Na+ (aq) + 2Cl- (aq)
Net Ionic Equation
Ca2+ (aq) + 2F - (aq)  CaF2 (s)
Predicting Whether a Precipitate Will Form
Example… sodium iodide + potassium nitrate
 When dissolved in water, the solutions that form from these
two compounds contain these ions…
o NaI (s) (in water)  Na+ (aq) + I – (aq)
o KNO3 (s) (in water)  K+ (aq) + NO3 – (aq)
 Possible precipitates…
o NaNO3
 This is a soluble compound… (table 4.1, rule #1
soluble ionic compounds)
o KI (s)
 This is a soluble compound… (table 4.1, rule #1
soluble ionic compounds)
 This reaction would not form a precipitate
Example… Pb(NO3)2 + NaI
 When dissolved in water, the solutions that form from these
two compounds contain these ions…
o NaI (s) (in water)  Na+ (aq) + I – (aq)
o Pb(NO3)2 (s) (in water)  Pb2+ (aq) + 2NO3 – (aq)
 Possible precipitates…
o NaNO3
 This is a soluble compound… (table 4.1, rule #1
soluble ionic compounds
o PbI2
 This is an insoluble compound… (table 4.1, rule
#3 soluble ionic compounds)
Molecular Equation
2NaI (aq) + Pb(NO3)2 (aq)  PbI2 (s) + 2NaNO3 (aq)
Total Ionic Equation
2Na+ (aq) + 2I - (aq) + Pb2+ (aq) + 2NO3 - (aq) 
PbI2 (s) + 2Na+ (aq) + 2NO3 - (aq)
Net Ionic Equation
2I - (aq) + Pb2+ (aq)  PbI2 (s)
Metathesis reactions – a reaction in which atoms or ions of two
compounds exchange bonding partners (also called doubledisplacement reactions)
Examples…
2NaI (aq) + Pb(NO3)2 (aq)  PbI2 (s) + 2NaNO3 (aq)
AgNO3 (aq) + KBr (aq)  AgBr (s) + KNO3 (aq)
Sample Problem 4.3 p. 143
Predicting Whether a Precipitation Reaction Occurs; Writing
Ionic Equztions
(a)
KF (s) (in water)  K+ (aq) + F – (aq)
Sr(NO3)2 (s) (in water)  Sr2+ (aq) + 2NO3 – (aq)
K+ (aq) + NO3 – (aq)  KNO3 (aq) [Soluble]
Sr2+ (aq) + F – (aq)  SrF2 (s) [Insoluble]
Molecular Equation
2KF (aq) + Sr(NO3)2 (aq)  SrF2 (s) + 2KNO3 (aq)
Total Ionic Equation
2K+ (aq) + 2F - (aq) + Sr2+ (aq) + 2NO3 (aq) 
SrF2 (s) + 2K+ (aq) + 2NO3 - (aq)
Net Ionic Equation
2F - (aq) + Sr2+ (aq)  SrF2 (s)
(b)
NH4ClO4 (s) (in water)  NH4+ (aq) + ClO4 – (aq)
NaBr (s) (in water)  Na+ (aq) + Br – (aq)
NH4+ (aq) + Br – (aq)  NH4Br (aq) [Soluble]
Na+ (aq) + ClO4 – (aq)  NaClO4 (aq) [Soluble]
NO REACTION…
Follow-Up Problem 4.3 p. 144
(a)
FeCl3 (s) (in water)  Fe3+ (aq) + 3Cl – (aq)
Cs3PO4 (s) (in water)  3Cs+ (aq) + PO43- (aq)
Fe3+ (aq) + PO43- (aq)  FePO4 (s) [Insoluble]
Cs+ (aq) + Cl – (aq)  CsCl (aq) [Soluble]
Balanced molecular
FeCl3 (aq) + Cs3PO4 (aq)  FePO4 (s) + 3CsCl (aq)
Total ionic
Fe3+ (aq) + 3Cl - (aq) + 3Cs+ + PO43- (aq) 
FePO4 (s) + 3Cs+ + 3Cl - (aq)
Net ionic
Fe3+ (aq) + PO43- (aq)  FePO4 (s)
(b)
NaOH(s) (in water)  Na+ (aq) + OH – (aq)
Cd(NO3)2 (s) (in water)  Cd2+ (aq) + 2NO3 – (aq)
Na+ (aq) + NO3 – (aq)  NaNO3 (aq) [soluble]
Cd2+ (aq) + OH – (aq)  Cd(OH)2 (s) [insoluble]
Balanced molecular
2NaOH (aq) + Cd(NO3)2 (aq)  Cd(OH)2 (s) + 2NaNO3 (aq)
Total ionic
2Na+ + 2OH (aq) + Cd2+ + 2NO3 - (aq) 
Cd(OH)2 (s) + 2Na+ (aq) + 2NO3 - (aq)
Net ionic
2OH (aq) + Cd2+  Cd(OH)2 (s)
(c)
MgBr2 (s) (in water)  Mg2+ (aq) + 2Br – (aq)
KC2H3O2 (s) (in water)  K+ (aq) + C2H3O2 – (aq)
Mg2+ (aq) + C2H3O2 – (aq)  Mg(C2H3O2)2 (aq) [Soluble]
K+ (aq) + Br – (aq)  KBr (aq) [Soluble]
NO REACTION
(d)
AgNO3 (s) (in water)  Ag+ (aq) + NO3 – (aq)
BaCl2 (s) (in water)  Ba2+ (aq) + 2Cl – (aq)
Ag+ (aq) + Cl – (aq)  AgCl (s) [insoluble]
Ba2+ (aq) + NO3 – (aq)  Ba(NO3)2 (aq) [soluble]
Balanced molecular
2AgNO3 (aq) + BaCl2 (aq)  2AgCl (s) + Ba(NO3)2 (aq)
Total ionic
2Ag+ (aq) + 2NO3 - (aq) + Ba2+ (aq) + 2Cl - (aq) 
2AgCl (s) + Ba2+ (aq) + 2NO3 - (aq)
Net ionic
Ag+ (aq) + Cl - (aq)  AgCl (s)
Sample Problem 4.4 p. 144
Using Molecules Depictions in Precipitation Reactions
(a)
Could be Na2SO4 or Ag2SO4
because the picture shows twice
as many 1+ ions as 2- (KCl would
need to be 1+ ions and 1- ions,
MgBr would need to be 2+ and 1ions)… Can only be Na2SO4
because Ag2SO4 is insoluble.
(b)
Could be Ba(NO3)2 or CaF2
because the picture shows twice
as many 1- ions as 2+ ions
(NH4NO3 would need to be 1+
and 1- ions and MgSO4 would need to be 2+ and 2- ions)…
Can only be Ba(NO3)2 because CaF2 is insoluble.
(c)
Balanced molecular equation
Na2SO4 (aq) + Ba(NO3)2 (aq)  BaSO4 (s) + 2NaNO3 (aq)
 Precipitate is barium sulfate
Total ionic equation
2Na+ (aq) + SO42- (aq) + Ba2+ (aq) + 2NO3- (aq)
 BaSO4 (s) + 2Na+ (aq) + 2NO3 - (aq)
 Spectator ions… Na+ (aq) and NO3 – (aq)
Net ionic equation
SO42- (aq) + Ba2+ (aq)  BaSO4 (s)
(d)
Step 1 – find the limiting reactant
4 Ba2+
0.010 mol 1 mol BaSO4
= 0.040 mol BaSO4
particles Ba2+ ions
1 Ba2+
1 mol Ba2+ ions
particle
5 SO42particles
0.010 mol 1 mol BaSO4
SO42- ions
1 SO421 mol SO42particle
ions
= 0.050 mol BaSO4
The limiting reactant is Ba2+
Step 2 – Calculate the mass (g) of product (BaSO4)
0.040 mol BaSO4 233.4 g BaSO4 = 9.3 g BaSO4
1 mol BaSO4
Follow-Up Problem 4.4 p. 145
(a)
Could be Zn(NO3)2 or PbCl2 because it has 2+ ions and 1- ions,
but it can only be Zn(NO3)2 because PbCl2 is insoluble.
(b)
Could be Cd(OH)2 or Ba(OH)2 because it has 2+ ions and 1ions, but it can only be Ba(OH)2 because Cd(OH)2 is insoluble.
(c)
Balanced Molecular
Zn(NO3)2 (aq) + Ba(OH)2 (aq)  Zn(OH)2 (s) + Ba(NO3)2 (aq)
 The precipitate is zinc hydroxide
 The spectator ions are Ba2+ (aq) and NO3 – (aq)
Total Ionic
Zn2+ (aq) + 2NO3 - (aq) + Ba2+ (aq) + 2OH - (aq)
 Zn(OH)2 (s) + Ba2+ (aq) + 2NO3 - (aq)
Net Ionic
Zn2+ (aq) + 2OH - (aq)  Zn(OH)2 (s)
(d)
Step 1 – Determine the limiting reactant
4 Zn2+
particles
6 OH particles
0.050 mol
Zn2+ ions
1 Zn2+
particle
1 mol Zn(OH)2 = 0.200 mol Zn(OH)2
1 mol Zn2+ ion
0.050 mol 1 mol Zn(OH)2 = 0.150 mol Zn(OH)2
OH - ions
1 OH 2 mol OH - ion
particle
The limiting reactant is OH –
Step 2 – Calculate the mass of Zn(OH)2 produced
0.150 mol
Zn(OH)2
99.43 g Zn(OH)2 = 15 g Zn(OH)2
1 mol Zn(OH)2
Acid-Base Reactions
Reactions that involve water as reactant or product, in addition
to its common role as solvent
 Acid-base reactions (also called neutralization reactions)
occurs when an acid reacts with a base
 Definitions of acids and bases have changed over the years,
but we will use definitions that apply to substances found
commonly in the lab (If you take CHEM 102, you will see
another definition in chapter 18)
o Acid
 Substance that produces H+ when dissolved in
water
 HX (in water) H+ (aq) + X – (aq)
o Base
 Substance that produce OH – when dissolved in
water
 MOH (in water)  M+ (aq) + OH – (aq)
Acids and the Solvated Proton
Acidic solutions arise when certain covalent H-containing
molecules dissociated into ions in water
 HBr for example…
o Contain a polar bond (H – Br) where the Br end is
partially negative and the H end is partially positive
 When hydrogen bromide gas dissolves in water,
the hydrogen and bromine atoms are attracted to
the polar ends of the water molecules and it
breaks the bonds that hold the hydrogen and
bromine atoms together leaving a solvated H+ ion
and a solvated Br – ion
H+ ions attract the
negative end of the
water molecule to form
H3O+ (we call a
hydronium ion)
**As a general notation
for these various
species, we write H+ (aq), but later in the chapter and in much of
the text, we use H3O+ (aq)
Acids and Bases as Electrolytes
Acids and bases are categorized in terms of their “strength.”
The stronger the acid or base, the more it dissociates into ions in
water.
 Strong acids and bases are strong electrolytes and conduct a
large current
o Strong acids
 Binary acids with Group 7A (17) (HCl, HBr, HI)
 Oxoacids that have 2 or more oxygens than
hydrogens in their formula (HNO3, H2SO4,
HClO4)
o Strong Bases
 Group 1A (1) hydroxides [LiOH, NaOH, KOH,
RbOH, CsOH]
 Group 2A (2) hydroxides [Ca(OH)2, Sr(OH)2,
Ba(OH)2]
 Weak acids and bases are weak electrolytes and conduct
only a small current.
o Weak acids
 HF, H3PO4, CH3COOH or HC2H3O2
o Weak bases
 NH3
Because a strong acid (or strong base) dissociates completely,
we can determine the molarity of H+ (or OH -) and the amount
(mol) or number of each ion in solution…
Sample Problem 4.5 p. 147
Determining the Number of H+ (or OH-] Ions in Solution
25.3 mL 1 L
1.4 mol HNO3 = 0.035 mol HNO3
HNO3
1000 mL 1 L
0.035 mol 1 mol H+ ions 6.022x1023
= 2.1x1022 H+ ions
HNO3
ions
1 mol HNO3 1 mol H+ ions
Follow-Up Problem 4.5 p. 147
451 mL KOH 1 L
1.20 mol KOH = 5.4x10-3 mol KOH
1000 mL 1 L
Structural Features of Acids and Bases
 Acids have one or more H atoms as part of their structure,
which are either completely released (strong) or partially
released (weak) as protons in water.
 Strong bases have either OH- (NaOH) or O2- (K2O) as part
of their structure. The oxide ion is not stable in water and
reacts to form OH – ion
o K2O (s) + H2O (l)  2K+ (aq) + OH – (aq)
The Key Event: Formation of H2O from H+ and OH –
Example…
Molecular Equation
2HCl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) + 2H2O (l)
Total Ionic Equation
2H+ (aq) + 2Cl- (aq) + Ba2+ (aq) + 2OH - (aq)
 Ba2+ (aq) + 2Cl - (aq) + 2H2O (l)
Net ionic
2H+ + 2OH - (aq)  2H2O (l)
H+ + OH - (aq)  H2O (l)
The key event in aqueous reactions between a strong acid and
a strong base is that an H+ ion from the acid and an OH- from
the base form a water molecule.
An acid and a base form a salt solution and water
 In the above example the salt is BaCl2 (aq)
 The cation of the salt comes from the base and the anion
comes from the acid
 HX (aq) + MOH (aq)  MX (aq) + H2O (l)
 Metathesis reactions (double replacement)
o Precipitation reactions
o Acid-base reactions
Another example…
3HCl (aq) + Al(OH)3 (s)  AlCl3 (aq) + 3H2O (l)
Sample Problem 4.6 p. 148
Writing Ionic Equations for Acid-Base Reactions
(a)
Balanced Molecular
HCl (aq) + KOH (aq)  KCl (aq) + H2O (l)
Total Ionic
H+ (aq) + Cl - (aq) + K+ (aq) + OH - (aq)
 K+ (aq) + Cl - (aq) + H2O (l)
 Spectator ions are Cl – (aq) and K+ (aq)
Net Ionic
H+ + OH - (aq)  H2O (l)
(b)
Balanced Molecular
Sr(OH)2 (aq) + 2HClO4 (aq)  Sr(ClO4)2 (aq) + 2H2O (l)
Total Ionic
Sr2+ (aq) + 2OH - (aq) + 2H+ (aq) + 2ClO4- (aq)
 Sr2+ (aq) + 2ClO4 - (aq) + 2H2O (l)
 Spectator ions are Sr2+ (aq) and ClO4 – (aq)
Net Ionic
2OH - (aq) + 2H+ (aq)  2H2O (l)
OH - (aq) + H+ (aq)  H2O (l)
(c)
Balanced Molecular
Ba(OH)2 (aq) + H2SO4 (aq)  BaSO4 (s) + 2H2O (l)
Total Ionic
Ba2+ (aq) + 2OH - (aq) + 2H+ (aq) + SO42- (aq)
 BaSO4 (s) + 2H2O (l)
 This is a precipitation reaction as well as a
neutralization reaction… there are NO spectator ions
Net Ionic
Ba2+ (aq) + 2OH - (aq) + 2H+ (aq) + SO42- (aq)
 BaSO4 (s) + 2H2O (l)
Follow-Up Problem 4.6 p. 149
Balanced Molecular
Ca(OH)2 (aq) + 2HNO3 (aq)  Ca(NO3)2 (aq) + 2H2O (l)
Total Ionic
Ca2+(aq) + 2OH - (aq) + 2H+(aq) + 2NO3 - (aq)
 Ca2+(aq) + 2NO3 - (aq) + 2H2O (l)
Net Ionic
2OH - (aq) + 2H+(aq)  2H2O (l)
OH - (aq) + H+(aq)  H2O (l)
Proton Transfer in Acid-Base Reactions
3 types of reactions…
Reaction Between Strong Acid and Strong Base
HCl (g) + H2O (l)  H3O+ (aq) + Cl – (aq)
H3O+ (aq) + Cl – (aq) + Na+ (aq) + OH H2O (l) + Cl- (aq) + Na+ (aq) + HOH (l)
Net Ionic
H3O+ (aq) + OH-  H2O (l) + HOH (l)
H3O+ (aq) + OH-  2H2O (l)
Key points…
 An acid-base reaction is a proton-transfer process
 Bronsted & Lowry definitions…
o An acid is a molecule (or ion) that donates a proton
o A base is a molecule (or ion) that accepts a proton
 In an aqueous reaction between strong acid and strong base,
H3O+ ion acts as the acid and donates a proton to OH –
ion which acts as the base and accepts it.
Gas-Forming Reactions: Acids with Carbonates (or Sulfites)
When a carbonate, such as K2CO3 is treated with an acid, such
as HCl, one of the products is carbon dioxide…
2HCl (aq) + K2CO3 (aq)  2KCl (aq) + [H2CO3 (aq)]
[H2CO3 (aq)]  H2O (l) + CO2 (g)
___________________________________________________
2HCl (aq) + K2CO3 (aq)  2KCl (aq) + H2O (l) + CO2 (g)
Total ionic equation
2H3O+ (aq) + Cl - (aq) + 2K+ (aq) + CO32- (aq)
 2K+ (aq) + Cl - (aq) + CO2 (g) + H2O (l) + 2H2O (l)
Net Ionic equation
2H3O+ (aq) + CO32- (aq)  CO2 (g) + H2O (l) + 2H2O (l)
2H3O+ (aq) + CO32- (aq)  CO2 (g) + 3H2O (l)
 2H+ transferred to the oxygen in CO32o H3O+ acts as the acid (donates proton… H+)
o CO32- acts as the base (accepts proton… H+)
Key point…
 This is an acid-base reaction
Ionic sulfites (SO32-) react in a similar manner to form H2O (l)
and SO2 (g)…
2H3O+ (aq) + SO32- (aq)  [H2SO3 (aq)] + 2H2O (l)
2H3O+ (aq) + SO32- (aq)  SO2 (g) + H2O (l) + 2H2O (l)
2H3O+ (aq) + SO32- (aq)  SO2 (g) + 3H2O (l)
Reactions of Weak Acids and Bases
Molecular Equation
CH3COOH (aq) +NaOH (aq)  CH3COONa (aq) + H2O (l)
Total ionic equation
CH3COOH (aq) + Na+ (aq) + OH - (aq)
 CH3COO – (aq) + Na+ (aq) + H2O (l)
Net ionic equation
CH3COOH (aq) + OH - (aq)  CH3COO – (aq) + H2O (l)
Reaction of vinegar and baking soda (sodium hydrogen
carbonate)
Sample Problem 4.7 p. 151
Writing Proton-Transfer Equations for Acid-Base Reactions
(a)
HI (g) + H2O (l)  H3O+ (aq) + I – (aq)
Total ionic (balanced)
2H3O+ (aq) + 2I – (aq) + Ca2+ (aq) + 2OH – (aq)
 2I – (aq) + Ca2+ (aq) + 4H2O (l)
 Protons transferred from H3O to OH –
 Ca2+ and I – are the spectator ions
 Salt formed is calcium iodide (CaI2)
Net ionic equation
2H3O+ (aq) + 2OH – (aq)  4H2O (l)
H3O+ (aq) + OH – (aq)  H2O (l)
(b)
Total ionic (balanced)
K+ (aq) + OH – (aq) + CH3CH3COOH (aq)
 K+ (aq) + H2O (l) + CH3CH3COO – (aq)
 Proton transferred from CH3CH3COOH to OH –
 K+ is the only spectator ion
Net ionic equation
OH – (aq) + CH3CH3COOH (aq) H2O (l) + CH3CH3COO – (aq)
Follow-Up Problem 4.7 p. 151
HBr (s) + H2O (l)  H3O+ (aq) + Br – (aq)
Total ionic equation
2H3O+ (aq) + 2Br – (aq) + Ca2+ (aq) + 2HCO3- (aq)
 2CO2 (g) + 4H2O (l) + 2Br – (aq) + Ca2+ (aq)
 The spectator ions are Ca2+ and Br  The salt is calcium bromide (CaBr2)
 Protons are transferred from H3O+ to HCO3 –
Net ionic equation
2H3O+ (aq) + 2HCO3- (aq)  2CO2 (g) + 4H2O (l)
H3O+ (aq) + HCO3- (aq)  CO2 (g) + 2H2O (l)
Quantifying Acid-Base Reactions by Titration
Titration – a process by which a known concentration of one
solution is used to determine the unknown concentration of
another.
 In a typical acid-base titration, a standardized solution of
base, whose concentration is known, is added to a solution
of acid whose concentration is unknown. (or vise-versa)
 Before titration – an acid-base indicator is added
(phenolphthalein is clear in acids and pink in bases)
 Equivalence point – when the amount (mol) of H+ ions in
the original volume of acid has reacted with the same
amount (mol) of OH – ions from the buret.
 End point – occurs when a tiny excess of OH – ions change
the indicator from clear to pink permanently
o The tiny excess is considered insignificant when doing
the calculations
Sample Problem 4.8 p. 152
Finding the Concentration of an Acid from a Titration
Step 1 – Writing the balanced equation
NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
Step 2 – Calculate the volume of NaOH added
33.87 mL – 0.55 mL = 33.32 mL NaOH added
Step 3 – Convert volume of NaOH to liters
33.32 mL 1 L
1000 mL
= 0.03332 L NaOH
Step 4 – Calculate mol of NaOH added
0.03332 L NaOH 0.1524 mol NaOH = 5.078x10-3 mol NaOH
1 L NaOH
Step 5 – Calculate mol of HCl
5.078x10-3 mol NaOH 1 mol HCl
= 5.078x10-3 mol HCl
1 mol NaOH
Step 6 – Calculate molarity of HCl
Molarity = mol HCl / volume (L) of HCl
5.078x10-3 mol HCl 1000 mL
50.00 mL
1L
= 0.1016 M HCl
Follow-Up Problem 4.8 p. 153
Step 1 – Write a balanced equation
2HCl (aq) + Ba(OH)2  BaCl2 (aq) + 2H2O (l)
Step 2 – Calculate mol HCl in 50.00 mL
50.00 mL HCl 1 L
0.1016 mol
HCl
1000 mL 1 L
= 5.080x10-3 mol
HCl
Step 3 – Calculate mol Ba(OH)2 needed
5.080x10-3 mol HCl 1 mol Ba(OH)2 = 2.54x10-3 mol Ba(OH)2
2 mol HCL
Step 4 – Calculate volume of Ba(OH)2 needed
2.52x10-3 mol
Ba(OH)2
1L
= 0.01966 L Ba(OH)2
0.1292 mol
Ba(OH)2
IF you wanted to do this in ONE equation…
50.00
1 L HCL 0.1016
mL HCl
mol HCl
1000 mL 1 L HCl
HCl
1 mol
Ba(OH)2
2 mol
HCl
1L
= 0.01966
Ba(OH)2 L Ba(OH)2
0.1292
mol
Ba(OH)2
Oxidation-Reduction (Redox) Reactions
The key chemical event in an oxidation-reduction (or redox)
reaction is the net movement of electrons from one reactant to
another.
 The movement occurs from the reactant (or atom in the
reactant) with less attraction for electrons to the reactant (or
atom) with more attraction for electrons.
 Occurs in both the formation of both ionic and covalent
compounds
o Transfer of electrons in ionic compounds
 2Mg (s) + O2 (g)  2MgO (s)
 Mg (s) loses 2e- when it forms a Mg2+ ion
 O2 (g) gains 2e – when it forms a O2- ion
 Shift of electrons in covalent compounds
o H2 (g) + Cl2 (g) 2HCl (g)
 Electrons are held more closely by the chlorine
atom than the hydrogen atom (so the hydrogen
end is more positive and the chlorine end is
more negative)
Some Essential Redox Terminology
 Oxidation – the loss of electrons
o Mg  Mg2+ + 2e- (Mg is oxidized)
 Reduction – the gain of electrons
o ½ O2 + 2e-  O2- (Oxygen is reduced)
 Oxidizing agent – the species doing the oxidizing
(causing the electron loss)
o O2 oxidizes Mg, so O2 is the oxidizing agent (it
takes the electrons from Mg)
 Reducing agent – the species doing the reducing
(causing the gain of electrons)
o Mg reduces O2, so Mg is the reducing agent (it
provides the electrons for oxygen)
Using Oxidation Numbers to Monitor Electron Charge
Chemists have devised a “bookkeeping” system to monitor
which atom loses electron charge and which atom gains it…
 Each atom is assigned an oxidation number, or oxidation
state
o The charge the atom would have if electrons where
transferred completely, not shared
 Each element in a binary ionic compound has a full charge
because the atom transferred its electron(s)
o The atoms oxidation number equals its ionic charge
 In NaCl, the oxidation number for Na is +1 and
the oxidation number for Cl is -1
 To determine the oxidation numbers of atoms in a covalent
compound (or a polyatomic ion), we use a set of rules…
(see the chart below)
 An oxidation number (O.N.) has the sign before the
number (example… +2), whereas an ionic charge has the
sign after the number (example… 2+)
 Oxidation rules must be memorized for the exam…
Sample Problem 4.9 p. 155
Determining the Oxidation Number of Each Element in a
Compound (or Ion)
(a)
Zinc chloride is ionic… its formula is ZnCl2… the charge of the
zinc ion is Zn2+ so its oxidation number is +2 for a total of +2
charge… the charge of each chloride ion is Cl- so its oxidation
number is -1 and since there are 2 Cl in the formula there is a
total of -2 charge… (The sum of the oxidation numbers equal
zero)
(b)
Sulfur trioxide is covalent… its formula is SO3… since sulfur
does not have a rule; we use the rule for oxygen to determine its
oxidation number… Rule #5 states that oxygen is -1 in
peroxides and -2 in all other compounds except with fluorine.
Each oxygen is -2 for a total of -6. Since there is one sulfur
atom, its oxidation number must be -6 so that the oxidation
numbers add up to zero.
(c)
Nitric acid is covalent… its formula is HNO3… we have two
rules… #3 states that hydrogen is +1 with nonmetals, so
hydrogen is +1. Rule #5 states that oxygen is -1 in peroxides
and -2 in all other compounds except with fluorine, so each
oxygen is -2 and -6 total. Nitrogen does not have a rule… if the
oxidation numbers must add up to zero, nitrogen must be +5




1 hydrogen at +1
1 nitrogen at +5
3 oxygens at -2 each totaling -6
Oxidation numbers add up to zero
(d)
Dichromate ion is a polyatomic ion (use the rules)… its formula
is Cr2O72-… Rule #5 states that oxygen is -1 in peroxides and -2
in all other compounds except with fluorine… so each oxygen is
-2 and -14 total. In this case (polyatomic ion), the oxidation
numbers must add up to the charge of the polyatomic ion.
Since chromium does not have a rule, we determine its oxidation
number from the fact that the two chromiums must add up to
+12 so that we have a -2 total charge. So the oxidation number
for each of the chromium atoms is +6
 7 oxygen at -2 (-14 total)
 2 chromium at +6 (+12 total)
 The net charge is 2- which is the charge of the polyatomic
ion
Follow-Up Problem 4.9 p. 156
(a)
Scandium oxide is ionic (scandium is a metal and oxygen a
nonmetal)… the formula is Sc2O3… there is NOT a rule for
scandium, so we must start with oxygen. Rule #5 states that
oxygen is -1 in peroxides and -2 in all other compounds except
with fluorine… so each oxygen atom is -2 and oxygens total -6.
If the oxidation numbers need to total zero, each scandium must
be +3 and the two scandiums total +6.
 3 oxygen at -2 (total -6)
 2 scandium at +3 (total +6)
 Sum of the oxidation numbers is zero
(b)
Gallium chloride is ionic (gallium is a metal and chlorine a
nonmetal)… the formula is GaCl3… there is NOT a rule for
gallium, so we must start with chlorine. Rule #6 states for
elements in Group 7A (17), their oxidation number is -1 in
combination with metals, nonmetals (except F), and other
halogens lower in the group (Br & I). Each chlorine atom is -1
for a total of -3. Since the oxidation numbers must add up to
zero, the one gallium atom must be +3
 3 chlorine at -1 (total -3)
 1 gallium at +3 (total +3)
 Sum of the oxidation numbers is zero
(c)
The hydrogen phosphate ion is a polyatomic ion… the formula
is HPO42-… Rule #1 states that hydrogen is +1 with nonmetals.
There is one hydrogen atom for a total of +1. Rule #5 states that
oxygen is -1 in peroxides and -2 in all other compounds except
with fluorine, so oxygen is -2… There are 4 oxygens for a total
of -8. Because the net charge has to equal 2-, the one
phosphorus atom has to be +5.




1 hydrogen at +1 (total +1)
4 oxygen at -2 (total -8)
1 phosphorus at +5 (total +5)
Net charge is 2-
(d)
Iodine trifluoride is covalent (so it follows the rules)… its
formula is IF3… fluorine has its own rule. (Rule #4 states that
fluorine is -1 in all compounds) Each of the fluorine atoms is -1
for a total of -3. Iodine is lower in Group 7A (17) so it is not -1.
Since there is only one iodine atom, its oxidation number must
be +3 if the oxidation numbers must add up to zero.
 1 iodine at +3 (total +3)
 3 fluorines at -1 (total of -3)
 Sum of the oxidation numbers is zero
Sample Problem 4.10 p. 156
Identifying Redox Reactions
(a)
CaO (s) + CO2 (g)  CaCO3 (s)
Reactants
Ca +2
O -2
C +4
Products
Ca +2
O -2
C +4
This is NOT a redox reaction…
(b)
4KNO3 (s)  2K2O (s) + 2N2 (g) + 5O2 (g)
Reactants
K +1
N +5
O -2
Products
K +1
N=0
O=0
Nitrogen is reduced… it gained electrons (went from +5 to 0)
Oxygen is oxidized… it lost electrons (went from -2 to 0)
(c)
H2SO4 (aq) + 2NaOH (aq)  Na2SO4 (aq) + H2O (l)
Reactants
H +1
S +6
O -2
Na +1
Products
H +1
S +6
O -2
Na +1
This is NOT a redox reaction…
Follow-Up Problem 4.10 p. 156
(a)
NCl3 (l) + 3H2O (l)  NH3 (aq) + 3HOCl (aq)
Reactants
N +3
Cl -1
H +1
O -2
Products
N -3
H +1
O -2
Cl +1
Nitrogen is reduced… It gained electrons (went from +3 to -3)
Chlorine is oxidized… It lost electrons (went from -1 to +1)
(b)
AgNO3 (aq) + NH4I (aq)  AgI (s) + NH4NO3 (aq)
Reactants
Ag +1
N +5, -3
O -2
H +1
I -1
This is NOT a redox reaction
Products
Ag +1
N +5, -3
O -2
H +1
I -1
(c)
2H2S (g) + 3O2 (g)  2SO2 (g) + 2H2O (g)
Reactants
H +1
S -2
O=0
Products
H +1
S +4
O -2
Sulfur is oxidized… it lost electrons (went from -2 to +4)
Oxygen is reduced… it gained electrons (went from 0 to -2)
Using Oxidation Numbers to Identify Oxidizing and
Reducing Agents
 The reactant that contained the atom that was oxidized is
the reducing agent
 The reactant that contained the atom that was reduced is the
oxidizing agent
Sample Problem 4.11 p. 157
Identifying Oxidizing and Reducing Agents
(a)
2Al (s) + 3H2SO4 (aq)  Al2(SO4)3 (aq) + 3H2 (g)
Reactants
Al = 0
H +1
S +6
O -2
Products
Al +3
H=0
S +6
O -2
Aluminum was oxidized… it lost electrons (went from 0 to +3)
 Aluminum is the REDUCING AGENT
Hydrogen was reduced… it gained electrons (went from +1 to 0)
 H2SO4 is the OXIDIZING AGENT
(b)
PbO (s) + CO (g)  Pb (s) + CO2 (g)
Reactants
Pb +2
O -2
C +2
Products
Pb = 0
O -2
C +4
Lead was reduced… it gained electrons (it went from +2 to 0)
 PbO is the OXIDIZING AGENT
Carbon was oxidized… it lost electrons (it went from +2 to +4)
 CO is the REDUCING AGENT
(c)
2H2 (g) + O2 (g)  2H2O (g)
Reactants
H=0
O=0
Products
H +1
O -2
Hydrogen was oxidized… it lost electrons (went from 0 to +1)
 Hydrogen was the REDUCING AGENT
Oxygen was reduced… it gained electrons (went from 0 to -2)
 Oxygen was the OXIDIZING AGENT
Follow-Up Problem 4.11 p. 157
(a)
2Fe(s) + 3Cl2(g)  2FeCl3(s)
Reactants
Fe = 0
Cl = 0
Products
Fe +3
Cl -1
Iron was oxidized… it lost electrons (went from 0 to +3)
 Iron was the REDUCING AGENT
Chlorine was reduced… it gained electrons (went from 0 to -1)
 Chlorine was the OXIDIZING AGENT
(b)
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
Reactants
C -3
H +1
O=0
Products
C +4
H +1
O -2
Carbon was oxidized… it lost electrons (went from -3 to +4)
 C2H6 is the REDUCING AGENT
Oxygen was reduced… it gained electrons (went from 0 to -2)
 O2 is the OXIDIZING AGENT
(c)
5CO (g) + I2O5(s)  I2(s) + 5CO2 (g)
Reactants
C +2
O -2
I +5
Products
C +4
O -2
I=0
Carbon was oxidized… it lost electrons (went from +2 to +4)
 CO is the REDUCING AGENT
Iodine was reduced… it gained electrons (went from +5 to 0)
 I2O5 is the OXIDIZING AGENT
Balancing Redox Equations
Oxidation number method of balancing equations…
 Follow the five steps in Sample Problem 4.12
Sample Problem 4.12 p. 158
Balancing Redox Equations by the Oxidation Number Method
(a)
Cu(s) + HNO3 (aq)  Cu(NO3)2(aq) + NO2(g) + H2O(l)
Step 1 – Assign oxidation numbers to all atoms
Reactants
Cu = 0
H +1
N +5
O -2
Products
Cu +2
H +1
N +4
O -2
Step 2 – Identify the oxidized and reduced reactants
Copper was oxidized…it lost electrons (went from 0 to +2)
Nitrogen was reduced…it gained electrons (went from +5 to +4)
Step 3 – Compute the number of electrons of electrons lost in
oxidation and electrons gained in reduction
Copper lost 2 electrons… Cu(s)  Cu+2 + 2eNitrogen gained 1 electron… N+5 + 1e N+4
Step 4 – Multiply the number of electrons by factor(s) that make
the electrons lost equal to the electrons gained as balancing
coefficients
Cu(s) + 2HNO3 (aq)  Cu(NO3)2(aq) + 2NO2(g) + H2O(l)
Step 5 – Complete the balancing by inspection
Cu(s) + 2HNO3 (aq)  Cu(NO3)2(aq) + 2NO2(g) + H2O(l)
Balance nitrogen…
Cu(s) + 4HNO3 (aq)  Cu(NO3)2(aq) + 2NO2(g) + H2O(l)
Balance hydrogen…
Cu(s) + 4HNO3 (aq)  Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
(b)
PbS(s) + O2(g)  PbO(s) + SO2(g)
Step 1 – Assign oxidation numbers to all atoms
Reactants
Pb +2
S -2
O=0
Products
Pb +2
S +4
O -2
Step 2 – Identify the oxidized and reduced reactants
Sulfur is oxidized…it lost electrons (went from -2 to +4)
Oxygen was reduced…it gained electrons (went from 0 to -2)
Step 3 – Compute the number of electrons of electrons lost in
oxidation and electrons gained in reduction
Sulfur lost 6 electrons… S-2  S+4 + 6eOxygen gained 2 electrons… O2 + 4e-  O2- + O2Step 4 – Multiply the number of electrons by factor(s) that make
the electrons lost equal to the electrons gained as balancing
coefficients
PbS(s) + 3/2 O2(g)  PbO(s) + SO2(g)
Multiply by 2 to get rid of the fraction in the coefficient…
2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g)
Follow-Up Problem 4.12 p. 159
K2Cr2O7(aq) + HI(aq)  KI(aq) + CrI3(aq) + I2(s) + H2O(l)
Step 1 – Assign oxidation numbers to all atoms
Reactants
K +1
Cr +6
O -2
H +1
I -1
Products
K +1
Cr +3
O -2
H +1
I -1, -1, 0
Step 2 – Identify the oxidized and reduced reactants
Chromium is reduced… it gained electrons (went from +6 to +3)
Iodine was oxidized… it lost electrons (went from -1 to 0)
Step 3 – Compute the number of electrons of electrons lost in
oxidation and electrons gained in reduction
Chromium gained 3 electrons… Cr+6 + 3e  Cr+3
Iodine lost 1 electron… I -  I + 1e
Step 4 – Multiply the number of electrons by factor(s) that make
the electrons lost equal to the electrons gained as balancing
coefficients
K2Cr2O7(aq) + 3HI(aq)  KI(aq) + CrI3(aq) + 3I2(s) + H2O(l)
Step 5 – Complete the balancing by inspection
K2Cr2O7(aq) + 14HI(aq)
 2KI(aq) + 2CrI3(aq) + 3I2(s) + 7H2O(l)
Quantifying Redox Reactions by Titration
In an acid-base titration, a known concentration of a base is used
to find and unknown concentration of an acid (or vice versa)…
 In a redox titration, a known concentration of oxidizing
agent is used to find an unknown concentration of reducing
agent (or vice versa)
o Measuring iron content in water
o Quantifying vitamin C in fruits and vegetables
In the example above…
 MnO4- (permanganate ion) is a common oxidizing agent
(found in the KMnO4 (aq) in this example… K+ ions and
MnO4- ions in solution)
o In the above example it is being used to determine the
C2O42- (oxalate ion)… (found in Na2C2O4 (aq) in this
example… Na+ ions and C2O42- ions in solution)
o When all the C2O42- has been oxidized, the next drop
of MnO4- will turn the solution light purple.
 This is the end point of the titration which we
assume is the equivalence point
o We can then calculate the C2O42- concentration from
the known volume of Na2C2O4 solution and the known
volume and concentration of KMnO4 solution.
Sample Problem 4.13 p. 160
Finding the Amount of Reducing Agent by Titration
2KMnO4(aq) + 5CaC2O4(aq) + 8H2SO4(aq) 
2MnSO4(aq) + K2SO4(aq) + 5CaSO4(aq) + 10CO2(g) + 8H2O(l)
Step 1 – calculate the amount (mol) of KMnO4 added
2.05 mL
1L
4.88x10-4 mol = 1.00x10-6 mol
KMnO4 soln
KMnO4
KMnO4
1000 mL 1 L soln
Step 2 – calculate amount (mol) of CaC2O4
1.00x10-6 mol
KMnO4
5 mol CaC2O4
= 2.50x10-6 mol CaC2O4
2 mol KMnO4
Step 3 – calculate amount (mol) of Ca2+ ion
2.50x10-6 mol
CaC2O4
1 mol Ca2+
= 2.50x10-6 mol Ca2+
1 mol CaC2O4
Follow-Up Problem 4.13 p. 160
(a)
Step 1 – calculate the amount (mol) of KMnO4 added
6.53 mL
1L
4.56x10-3
KMnO4 soln
mol KMnO4
1000 mL 1 L
= 2.97x10-5 mol
KMnO4
Step 2 – calculate amount (mol) of CaC2O4
2.97x10-5 mol
KMnO4
5 mol CaC2O4 = 7.425x10-5 mol CaC2O4
2 mol KMnO4
Step 3 – calculate amount (mol) of Ca2+ ion
= 7.425x10-5 mol
CaC2O4
1 mol Ca2+
= 7.425x10-5 mol Ca2+
1 mol CaC2O4
Step 4 – calculate the molarity of Ca2+ ion in milk
7.425x10-5 mol Ca2+ 1000 mL
2.50 mL
1L
= 2.97x10-2 M Ca2+
(b)
Step 5 – convert to grams/liter (g/L)
2.97x10-2 mol Ca2+
lL
40.08 g Ca2+ = 1.19 g Ca
1 mol Ca2+
1L
Elements in Redox Reactions
Combination Redox Reactions
In a combination redox reaction, two or more reactants, at least
one of which is an element, form a compound.
X+YZ
Combining Two Elements
1. Metal and nonmetal form an ionic compound (metal is the
reducing agent and the nonmetal is the oxidizing agent…
the metal is oxidized and the nonmetal is reduced)
2K(s) + Cl2(g)  2KCl (see picture below)
4Al(s) + 3O2(g)  2Al2O3(s)
2. Two nonmetals form a covalent compound…
N2(g) + 3H2(g)  2NH3(g)
P4(s) + 6Cl2(g)  4PCl3(l)
N2(g) + O2(g)  2NO(g)
Combining Compound and Element
2NO(g) + O2(g)  2NO2(g)
PCl3(l) + Cl2(g)  PCl5(s)
Decomposition Redox Reactions
In a decomposition redox reaction, a compound forms two or
more products, at least one of which is an element.
ZX+Y
In any decomposition reaction, the reactant absorbs enough
energy (heat or electricity) for one or more bonds to break.
Thermal Decomposition
2KClO3(s) + heat  2KCl(s) + 3O2(g) (Note how this is written
in the text and in the example below…)
Electrolytic Decomposition
In the process of electrolysis, a compound absorbs electrical
energy and decomposes into its elements…
2H2O(l) + electricity  2H2(g) + O2(g)
MgCl2(l) + electricity  Mg(l) + Cl2(g)
Displacement Redox Reactions and Activity Series
In any displacement reaction, the number of substances on the
two sides of the equation remains the same, but atoms (or ions)
exchange places.
 In double-displacement (metathesis) reactions
(precipitation and acid-base reactions), atoms (or ions) of
two compounds exchange places… these are NOT redox
reactions
o Example… AB + CD  AC + BD
 In single-displacement reactions, one of the substances is
an element; therefore all single displacement reactions are
redox reactions
o Example… X + YZ  XZ + Y
In the above example, lithium displaces hydrogen from water…
The Activity series of metals
Metals are ranked by their ability to displace H2 from various
sources and another metal from solution
 The most active metals displace H2 from water… Group
1A(1) metals and Ca, Sr, and Ba from Group 2A(2)
 Slightly less reactive metals displace H2 from steam… Al
and Zn are examples
 Still less reactive metals displace H2 from acids… Ni and
Sn are examples
 The least reactive metals cannot displace H2 from any
source… examples silver (Ag), gold (Au), and platinum
(Pt) do not reaction with water or acid
 An atom of one metal displaces the ion of another (see the
example below… shows Cu displacing Ag+ ions)
Figure 4.20 – The Activity Series of Metals
Activity Series of the Halogens
Reactivity of the elements decreases down Group 7A(17)
 F2 > Cl2 > Br2 > I2
 2Br –(aq) + Cl2(aq)  Br2(aq) + 2Cl –(aq)
 Bromine was oxidized… it lost electrons (it is the reducing
agent)
 Chlorine was reduced… it gained electrons (it is the
oxidizing agent)
o The halogen higher in the group is the stronger
oxidizer
Combustion Reactions
Combustion is the process of combining with oxygen, most
commonly with the release of heat and the production of light,
as in a flame.
 All of these reactions are redox reactions because
elemental oxygen is a reactant.
2CO(g) + O2(g)  2CO2(g)
Combustion of Butane
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)
Biological Respiration
C6H12O6(aq) + 6O2(g)  6CO2(g) + 6H2O(g) + energy
Sample Problem 4.14 p. 166
Identifying the Type of Redox Reaction
(a)
3Mg(s) + N2(g)  Mg3N2(s)
[Combination] X + Y  Z
 Magnesium is oxidized… it lost electrons (went from 0 to
+2)… so, magnesium is the reducing agent
 Nitrogen is reduced… it gained electrons (went from 0 to 3)… so, nitrogen is the oxidizing agent
(b)
2H2O2(l)  2H2O(l) + O2(g) [Decomposition]
ZX+Y
 Oxygen is both oxidized and reduced… oxygen is -1 in a
peroxide, so it goes from -1 to -2 (in water) and -1 to 0 (in
O2)
 H2O2 is both the reducing agent and oxidizing agent
(c)
2Al(s) + 3Pb(NO3)2(aq)  2Al(NO3)3(aq) + 3Pb(s)
[Displacement]
X + YZ  XZ + Y
Total ionic equation…
2Al(s)+ 3Pb2+(aq)+ 6NO3-(aq)2Al3+(aq) + 6NO3-(aq) + 3Pb(s)
Net ionic equation…
2Al(s) + 3Pb2+(aq)  2Al3+(aq) + 3Pb(s)
 Aluminum is oxidized… it lost electrons (went from 0 to
+3) so, aluminum is the reducing agent
 Lead is reduced… it gained electrons (went from +2 to 0)
so, Pb(NO3)2 is the oxidizing agent
Follow-Up Problem 4.14 p. 167
(a)
S8(s) + 16F2(g)  8SF4(g)
[Combination]
X+YZ
 Sulfur is oxidized… it lost electrons (went from 0 to +4)
so, S8 is the reducing agent
 Fluorine is reduced… it gained electrons (went from 0 to 1) so, F2 is the oxidizing agent
(b)
2CsI(aq) + Cl2(aq)  2CsCl(aq) + I2(aq)
[Displacement]
X + YZ  XZ + Y
Total ionic equation…
2Cs+(aq) + 2I –(aq) + Cl2(aq)  2Cs+(aq) + 2Cl –(aq) + I2(aq)
Net ionic equation…
2I –(aq) + Cl2(aq)  2Cl –(aq) + I2(aq)
 Iodine is oxidized… it lost electrons (it went from -1 to 0)
so, CsI is the reducing agent
 Chlorine is reduced…it gained electrons (it went from 0 to
-1) so, Cl2 is the oxidizing agent
(c)
3Ni(NO3)2(aq) + 2Cr(s)  3Ni(s) + 2Cr(NO3)3(aq)
Total ionic equation…
3Ni2+(aq)+ 6NO3-(aq)+ 2Cr(s)  3Ni(s)+ 2Cr3+(aq) + 6NO3(aq)
Net ionic equation…
3Ni2+(aq) + 2Cr(s) 3Ni(s) + 2Cr3+(aq)
 Nickel was reduced… it gained electrons (went from +2 to
0) so, Ni(NO3)2 is the oxidizing agent
 Chromium was oxidized… it lost electrons (went from 0 to
+3) so, Cr is the reducing agent
The Reversibility of Reactions and the Equilibrium State
Up to this point…
 We have viewed reactions as transformations of reactants
into products until they are complete (until the limiting
reactant is used up)
o Many reactions stop before this happens…
 The reason this happens is because there are two
opposing reactions taking place simultaneously
 The forward reaction… left to right
 The reverse reaction… right to left
 Both reactions occur at the same rate
 No further changes appear in the amounts
of reactants or products
o Dynamic equilibrium
 On the macroscopic level, the
reaction is static, but it is dynamic
on the molecular level
In principle, every reaction is reversible and will reach dynamic
equilibrium if all the reactants and products are present.
Example…
CaCO3(s) + heat  CaO(s) + CO2(g)
[breakdown]
CaO(s) + CO2(g)  CaCO3(s)
[formation]
In the top example, the reaction (breakdown) goes to completion
because the reverse reaction cannot take place without the
carbon dioxide.
In the bottom example, as CO2 starts to be produced
(breakdown) the reverse reaction (formation) starts to take place.
Eventually, the rate at which CaO(s) and CO2(g) is being
formed will be equal to the rate at which the CaCO3(s) is being
formed. The concentrations of CaCO3, CaO, and CO2 no
longer change.
Reaction reversibility applies to other systems as well…
1. Weak acids in water.
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO –(aq)
2. Weak bases in water.
NH3(aq) + H2O(aq)  NH4+(aq) + OH –(aq)
3. Gas-forming reactions.
2HCl(aq) + K2CO3(aq)   2KCl(aq) + H2O(l) + CO2(g)
4. Other acid-base and precipitation reactions.
 These processes seem to go to completion because the
ions become tied up in the products
o Water in acid-base reactions
o Insoluble solids in precipitation reactions
 Water and insoluble solids in fact do dissociate to a very
small extent and do reach equilibrium
o The final reaction mixture is almost all product…