Download Chapter 4 2013

The Major Classes of Chemical Reactions
Chapter 4
Learning Goals: Chapter 4
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
4.6 Elements in 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.
e-
It’s just symbolic
language
eO
H
H
O
H
H
Ionic compounds dissolve when ion-water forces are
larger than ion-ion forces holding the ionic crystal
together.
“Likes dissolve likes”
Oxygen atom
“sucks” electron
density from
hydrogen atoms
leading to a
“polar molecule”.
Ionic substances conduct electricity, covalent
substances do not.
Strong Electrolyte – 100% dissociation into ions
NaCl (s)
H 2O
Na+ (aq) + Cl- (aq)
Weak Electrolyte – not completely dissociated
CH3COOH
CH3COO- (aq) + H+ (aq)
Non-Electrolyte – no dissociation
The dissolution of an ionic compound in water
CH3OH
An electrolyte is a substance that when dissolved in
water results in a solution that can conduct electricity.
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)
Non-electrolyte
No Dissociation
Strong Electrolyte
100% Dissociation
Weak Electrolyte
Little dissociation
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
Covalent compounds do not dissociate!
(e) 35 mL of 0.84 M glucose (C6H12O6)
What does a 3.5 M FeCl3 mean?
H 2O
2NH4+(aq) + SO42-(aq)
(a) (NH4)2SO4(s)
5.0 mol (NH4)2SO4
2 mol NH4+
1 mol
(NH4)2SO4
= 10. mol NH4+
5.0 mol SO42-
H 2O
(b) CsBr(s)
Cs+(aq) + Br-(aq)
mol CsBr
78.5 g CsBr
212.8 g CsBr
(c) Cu(NO3)2(s)
H 2O
= 0.369 mol CsBr
= 0.369 mol Cs+
= 0.369 mol Br-
Cu2+(aq) + 2NO3-(aq)
= 0.123 mol Cu2+
7.42x1022
mol Cu(NO3)2
formula
= 0.123 mol Cu(NO3)2
units Cu(NO3)2 6.022x1023 formula units
= 0.246 mol NO
3
H 2O
(d) ZnCl2(aq)
Zn2+(aq) + 2Cl-(aq)
1L
0.84 mol ZnCl2
35 mL ZnCl2
= 2.9x110-2 mol ZnCl2
L
103mL
= 2.9x110-2 mol Zn2+
= 5.8x10-2 mol Cl-
What does 3.5 M FeCl3 mean?
3.5 M FeCl3 =
3.5 moles FeCl3
1 Liter solution
(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
-
There are 3-classes of chemical reactions that
occur in aqueous solution.
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.
Precipitation
Involves Substances
Cations
Anions
Involves Substances
Involves Substances
H+ ions
OH- ions
Combine to Form
Insoluble
Precipitate
Salt and H2O
Predicted by
In a precipitation reactions a metal cation and nonmetal anion combine & form an insoluble solid that
“precipitates”.
Oxidation
Reduction
Combine to Form
3. Oxidation-Reduction Reaction-electron donor
substances react with react with substances that accept
electrons.
Acid-Base
Neutralization
Solubility
Rules
Oxidation Reduction
Which is the
Loss of e-
Gain of e-
Which is called
Reducing
Agent
Oxidizing
Agent
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”.
AgNO3(aq) + Na2CrO4(aq) => Ag2CrO4 (s) + NaNO3(aq)
Precipitation occurs
when ionic attractive
forces overcome water’s
tendency to hydrate and
dissolve.
Spectator ions:
any species that
remains soluble.
Solubility Rules For Ionic Compounds Silberberg
The “driving force” of a precipitation reaction is
the act of PRECIPITATION.
Soluble Ionic Compounds
1. Salts of Group 1A and ammonium ion (NH4+)
2. NO3-, CH3COO- or C2H3O2- , ClO43. Cl-, Br- I- except those of Ag+, Pb2+, Cu+, and Hg22+.
Insoluble Ionic Compounds
1. All OH, except those of Group 1A and the larger
members of Group 2A (beginning with Ca2+).
2. CO32- and PO43-) are insoluble, except those of Group
1A(1) and NH4+.
3. All S2- except those of Group 1A, Group 2A and NH4+.
If it can happen, it will.
How To Predict Whether A Precipitation Reaction
Example: Pb(NO3)2(aq) + NaI(aq) ==> ?
2. Consult the Solubility table to see if PbI2 or
NaNO3 are insoluble.
3. PbI2 is insoluble so that reaction will occur.
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.
There are 3-types of equations that are written
for precipitation, acid-base and redox reactions
1. The molecular equation
shows all reactants and products as undissociated compounds.
precipitate
PbI2(s) + 2NaNO3(aq)
2. The total ionic equation
shows all of the soluble ionic substances dissociated into ions.
Pb2+ + 2NO3- + 2Na+ + 2I-
Pb(NO3)2(aq) + NaI(aq) ==> PbI2(s)
1. Recognize => IONIC => Solubility Rules Apply
Look at ions
that are possible
from the
reactants, then
use “Solubility
Rules” !
2NaI(aq) + Pb(NO3)2(aq)
Example: The reaction of the salts:
PbI2 (s) + 2Na+ + 2NO3-
Na+ and NO3- are spectator ions---they don’t do much but sit there!
3. The net ionic equation
eliminates the spectator ions and shows the actual chemical
change taking place.
Pb2+ + 2IPbI2 (s)
Learning Check: Precipitation Reactions
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.
(a) sodium sulfate(aq) + strontium nitrate(aq)
(b) ammonium perchlorate(aq) + sodium bromide(aq)
(c) silver nitrate(aq) + sodium chromate(aq)
4. Now write 3-equations 1)molecular, 2) ionic and 3) net equation as:
2NaI(aq) + Pb(NO3)2(aq)
2Na+ +
2I-
+
Pb2+ +
2I-(aq) + Pb2+(aq)
2NO3
PbI2(s) + 2NaNO3(aq)
-
PbI2 + 2Na+ + 2NO3PbI2(s)
Putting It All Together: Predicting Precipitation
0. Know how to read a Table of Solubility Rules
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. Break the compounds into their ions and write the
ionic equation for the reaction.
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.
Solubility Rules For Ionic Compounds in Water
1. Convert names formulas, write a balanced equation
showing intact compounds (not dissociated).
Molecular Equation
Na2SO4(aq) + Sr(NO3)2 (aq)
Molecular Equation
NH4ClO4(aq) + NaBr (aq)
NH4Br (aq) + NaClO4(aq)
2NaNO3(aq) + SrSO4(s)
2. Dissociate the formula equation to an ionic equation
showing cations and anions with charge and phase.
Ionic Equation
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.
2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3-(aq)
2Na+(aq) +2NO3-(aq)+ SrSO4(s)
Ionic Equation
3. Cancel the spectator ions to obtain the net ionic equation
Net Ionic Equation
SO42-(aq)+ Sr2+(aq)
NH4+ + ClO4+(aq) + Na+ + Br- (aq)
NH4+ + Br -(aq) +
Na+ + ClO4 -(aq)
SrSO4(s)
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.
Write the net ionic equation for the reaction of
silver nitrate with sodium chromate.
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
2. Acid-Base Reactions
There are 3-classes of chemical reactions that
occur in aqueous solution.
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
The effects of acid rain on a statue of George Washington
taken in 1935 (left) and 2001 (right) marble.
substances react with react with substances that accept
electrons.
Precipitation
Involves Substances
Cations
Anions
Acid-Base
Neutralization
Oxidation
Reduction
Involves Substances
Involves Substances
H+ ions
OH- ions
Combine to Form
Combine to Form
Insoluble
Precipitate
Salt and H2O
Oxidation Reduction
Which is the
Loss of e-
Predicted by
Which is called
Solubility
Rules
Chemistry uses 3-definitions of acids and bases:
1) Arrehenius
2) Bronsted-Lowry 3) Lewis
Gain of e-
Reducing
Agent
Oxidizing
Agent
Acid: substance that has a covalent H atom in its formula,
and releases a proton H+ when dissolved in water.
H 2O
Chem 7
HA(g) ==> H+(aq) + A-(aq)
Chem 11
Name
Acid Definition
Base Definition
Arrhenius
Substance that
increases H+
Substances that
increase OH-
BrønstedLowry
Substances that
donate H+
Substances that
accept H+
Lewis
Electron-pair
acceptor
Electron-pair
donar
HCl(g), HBr(g), HI (g), HNO3(l), H2SO4(l), HClO4(l)
HCl(aq), HBr(aq), HI (aq), HNO3(aq), H2SO4(aq), HClO4(aq)
Chemists symbolize the reactions of acids with
water two confusing ways. Don’t let it bug you.
H+ representation
H 2O
H3O+ Representation
HCl(g) ===> H+ + Cl-
HCl(g) + H2O => H3O+ + Cl-
H 2O
HNO3(l) ===> H+ + NO3-
HNO3(g) + H2O => H3O+ + NO3-
Base: a substance that contains OH in its formula, and
releases hydroxide ions (OH-) when dissolved in water.
H 2O
MOH(s) ==> OH- + M+
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)
Acids and bases are classified as either strong or
weak. We use arrows to symbolize the difference.
100% ionized = strong electrolyte = ==> arrow
Strong Acid
H2SO4(aq) ==> 2H+(aq) + SO42-(aq)
Strong Base
KOH(aq) ==> OH-(aq) + K+ (aq)
<20% ionized = weak electrolyte = <==> arrow
H 2O
HA(g) ===> H+ + AGeneralized Acid HA
HA(g) + H2O => H3O+ + AGeneralized Acid HA
Weak Acid
HNO2(aq) <==> H+(aq) + NO2-(aq)
Weak Base
NH4OH(aq) <==> OH-(aq) + NH4+ (aq)
Strong acids and strong bases dissociate
completely, conduct electricity and are strong
electrolytes.
Strong acids and strong bases completely
dissociate or completely ionize in water:
HA
H+ + A-
HA(g) + H2O(l)
Before
dissociation
Uni-directional arrow used
to heavily product-favored
H3O+(aq) + A-(aq)
After
dissociation
AH3O+
HCl(aq)
H+(aq) + Cl-(aq)
HNO3(aq)
H+ (aq) + NO3-(aq)
H2SO4(aq)
Ca(OH)2(aq)
Na+(aq) + OH-(aq)
Ca2+(aq) + 2OH-(aq)
HA H3O+ A-
We must memorize common strong acids and
strong bases. All are strong electrolytes that
dissociate completely in solution.
Strong Bases
Strong Acids
hydrochloric acid HCl
hydrobromic acid HBr
lithium hydroxide:
LiOH
sodium hydroxide:
NaOH
hydroiodic acid
HI
nitric acid
HNO3
sulfuric acid
H2SO4
perchloric acid HClO4
potassium hydroxide: KOH
calcium hydroxide
Ca(OH)2
strontium hydroxide Sr(OH)2
barium hydroxide
H-X acids
Oxide containing acids
Ba(OH)2
Group I and II Hydroxides
Weak acids and weak bases dissociate only to a
slight extent in water Ka << 1.
HA
HA(aq) + H2O(l)
H+ + AH3O+(aq) + A-(aq)
Remember anything not strong is weak!
Common Weak Acids and Their Anions
Acid
HF
CH3COOH
HCN
HNO2
H2CO3
H2SO3
H3PO4
(COOH)2
Anion
FCH3COOCNNO2CO32SO32PO43(COO)22-
All not strong is weak!
After
dissociation
HNO2 (aq) <=> NO2- (aq) + H+(aq)
CH3CO2H(aq) <=> CH3CO2- (aq) + H+
HA
HA
HA
HA
HA
HA
HA
HA H3O+ A-
Anion Name
fluoride ion
acetate ion
cyanide ion
nitrite ion
carbonate ion
sulfite ion
phosphate ion
oxalate ion
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).
reactant-favored
Before
dissociation
Strong
H+ (aq) + HSO4-(aq)
NaOH(aq)
HA
Strong
H2SO3 (aq) <=> HSO3- (aq) + H+(aq)
Acids can have one, two or three acidic protons
depending on their structure.
Monoprotic acids--only one H+ available
HCl
H+ + ClHNO3
H+ + NO3CH3COOH
H+ + CH3COO-
Acids and bases have distinct properties.
Acids:
•
•
Strong electrolyte, strong acid
•
Strong electrolyte, strong acid
•
Weak electrolyte, weak acid
Diprotic acids--two acidic H+ available for reaction
H2SO4
H+ + HSO4-
Strong electrolyte, strong acid
HSO4-
H+ + SO42-
Weak electrolyte, weak acid
Acid
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 200 Million MT H2SO4
Base
Bases:
Triprotic acids--three acidic H+
H3PO4
H2PO4HPO42-
H+
+ H2PO4
H+ + HPO42H+ + PO43-
•
50 Million MT NaOH/yr
Bitter taste
3 million containers
Slippery feel
Changes plant dye litmus from red to blue
•
React and neutralizes the effects of acids
•
-
•
Weak electrolyte, weak acid
Acids and bases are everywhere.
Common Acids
, soft drinks
Most anti-perspirants, water
treatment plants, paper
Determining the Molarity of H+ Ions in Aqueous
Solutions of Acids
Nitric 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?
Common Bases
Of H+ .466 M H3PO4?
Of 2.5 M NaOH ?
Determining the Molarity of H+ Ions in Aqueous
Solutions of Strong Acids or Bases
Nitric 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?
One mole of H+(aq) is released per mole of nitric acid (HNO3)
HNO3(l)
H 2O
H+(aq) + NO3-(aq)
Acids react with bases in a chemical reaction called
a “neutralization reaction” forming a salt and water.
NaOH (aq) + HCl (aq) ==> H2O + Na+ + Clbase
+ acid
==> Water + salt
1.4M HNO3(aq) is 1.4M H+(aq).
What is the molarity of H+ in a 0.70 M H2SO4?
What is the molarity of H+ in a 0.466 M H3PO4?
A salt’s cation comes from a parent base and anion
comes from a parent acid!
Chemists use general symbols to represent the
“neutralization reaction” to form a salt and water.
NaOH (aq) + HCl (aq) ==> H2O + Na+ + Clbase
+ acid
==> Water + salt
Learning check: write the molecular, ionic and net
equation for the neutralization between calcium
hydroxide and sulfuric acid.
1. The molecular equation
shows all reactants and products as undissociated compounds.
Ca(OH)2 (aq) + H2SO4 (aq) ===> 2H2O + CaSO4
2. The total ionic equation
MOH (aq) + HX (aq) ==> H2O +
Alkali
Metal Cation
M+ +
X-
Halide
Anion
Shows all soluble ionic substances dissociated into ions.
Ca2+ + 2OH- + 2H+ + SO42- ===> 2H2O + Ca2+ + SO42Ca2+ and SO42- are spectator ions---they just watch!
3. The net ionic equation
Eliminate spectator ions and show actual chemical change
H+(aq) + OH- (aq) ===> H2O(l)
Learning check: Write molecular, ionic and net
equations for the following acid base neutralization
reactions.
Writing Ionic Equations for Acid-Base Reactions
strontium hydroxide(aq) + perchloric acid(aq)
(a) Sr(OH)2(aq)+2HClO4(aq)
strontium hydroxide(aq) + perchloric acid(aq)
2H2O(l)+Sr(ClO4)2(aq)
(b) Sr2+(aq)+2OH-(aq)+ 2H+(aq)+2ClO4-(aq)
2H2O(l)+Sr2+(aq)+2ClO4-(aq)
barium hydroxide(aq) + sulfuric acid(aq)
Nitric acid(aq) + barium hydroxide(aq)
(c)
2OH-(aq)+
2H+(aq)
barium hydroxide(aq) + sulfuric acid(aq)
(b) Ba(OH)2(aq) + H2SO4(aq)
Acetic acid(aq) + potassium
2H2O(l)
2H2O(l) + BaSO4(aq)
Ba2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq)
2H2O(l)+Ba2+(aq)+SO42-(aq)
(c) 2OH-(aq)+ 2H+(aq)
Weak acids dissociate to a very small extent and
this fact is reflected in their equations as well
using a double-arrow (<==>).
2H2O(l)
Learning Check: Acids and Bases
1. Name and characterize each as a base, acid or salt?
HF(g), HI(aq), LiOH(aq), Mg(OH)2, Na2SO4 CH3COONH4
Molecular equation
2. Name and classify the following as strong, weak acid or base?
NaOH(aq) + CH3COOH(aq)
CH3COONa(aq) + H2O
Weak acid is not dissociated!
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)
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:
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
3. What is the correct formula of the salt formed in the
neutralization reaction of hydrochloric acid with calcium
hydroxide?
Name the type of reaction and write M/I/NI
ionic equations for the following:
1.
2.
3.
4.
5.
6.
7.
hydrochloric acid and potassium hydroxide
sodium carbonate and hydrochloric acid
aluminum nitrate and sodium phosphate
potassium chloride + iron(II) nitrate
Iron sulfide and hydrochloric acid
acetic acid and magnesium hydroxide
sodium sulfite and hydrochloric acid
4. What is the chemical formula of the salt produced by the
neutralization of sodium hydroxide with sulfurous acid?
Reactions between acids and carbonate containing
compounds yield gaseous CO2 and H2O + salt.
Oxidation
Reduction
Involves simultaneous
Molecular equation
Oxidation
NaHCO3(aq) + CH3COOH(aq)
Reduction
Which is the
CH3COONa(aq) + CO2(g) + H2O(l)
Loss of eTotal ionic equation
Gain of e-
Which is called
Na+(aq)+ HCO3-(aq) + CH3COOH(aq)
Reducing
Agent
CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)
Oxidizing
Agent
Which
Net ionic equation
Increases
oxidation
number
HCO3-(aq) + CH3COOH(aq)
CH3COO-(aq) + CO2(g) + H2O(l)
Redox reactions trends can be understood by
applying the properties of elements in the periodic
table.
Non-metals tend to gain
electrons (be reduced)
forming anions which
are oxidizing agents.
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(s) + O2 (g)
2MgO (s)
Oxidized
Reduced
Oxidizing Perspective
Mg loses electron(s)
Metals tend to loose electrons
(oxidized) forming cations
Metals are reducing agents
Decreases
oxidation
number
Mg(s) is oxidized
Mg is the reducing
agent
Mg increases its
oxidation number
We learn rules
to figure out
what is
oxidized and
what is
reduced!
Reduction Perspective
O gains electron(s)
O is reduced
O is the oxidizing agent
O decreases its
oxidation number
Oxidation and reduction occur simultaneously.
2Mg(s) + O2 (g)
2MgO (s)
A set of rules called “assigning oxidation numbers”
is used to identify what species is being oxidized and
reduced in a chemical reaction.
1. The oxidation state of a free elements in atomic or
molecular for are assigned an oxidation number of 0.
Oxidized
Reduced
Na, Be, K, Pb, H2, O2, P4 = 0
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2Mg
O2 +
4e-
2O2-
Reduction half-reaction (gain
e-)
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
2Mg + O2 + 4e2Mg + O2
2Mg2+ + 2O2- + 4e2MgO
Ionic Equation
Net Equation
4. Fluorine has an oxidation state of –1 in its compounds
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
Quick Table Assigning Oxidation Numbers
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.
1. The oxidation state of any free element = 0
6. The oxidation number of oxygen is usually –2. In
H2O2 and O22- it is –1 (exceptions).
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.
7. When there is a conflict the lower numbered rules
take priority over the higher numbered rules.
Oxidation numbers of all
the elements in HCO3- ?
HCO3O = -2 H = +1
3x(-2) + 1 + ? = -1
C = +4
Assign oxidation numbers to the following ions
and the elements within a compound.
IF7
HCO3O = -2
H = +1
3x(-2) + 1 + ? = -1
C = +4
7x(-1) + ? = 0
I = +7
K2Cr2O7
O = -2
F = -1
K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
NaIO3
Na = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
2. In monatomic ions, the oxidation number is equal to
the charge on the ion (Use Group Number)
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 O22- it is –1 (exceptions).
7. When there is a conflict the lower numbered rules
take priority over the higher numbered rules.
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
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.
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)
0
-4 +1
CH4 + 2O2
+2 -2
2HgO
-4 -2
+1 -2
CO2 + 2H2O
0
0
2Hg + O2
+2
0
+2
0
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Identify the oxidizing agent and reducing
agent in each of the following:
Zn (s) + CuSO4 (aq)
Identify the oxidizing agent and reducing
agent in each of the following:
ZnSO4 (aq) + Cu (s)
+2 +6 -2
0
Zn (s) + CuSO4 (aq)
+2 +6 -2
0
ZnSO4 (aq) + Cu (s)
Oxidized
1. Assign oxidation numbers
2. 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.
Reduced
Zn
Zn2+ + 2e- Zn is oxidized
Cu2+ + 2e-
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?
Zn is the reducing agent
Cu Cu2+ is reduced Cu2+ is the oxidizing agent
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)
0
+1
Cu (s) + 2AgNO3 (aq)
Oxidized
Cu is oxidized
Ag+ is reduced
+2
0
Cu(NO3)2 (aq) + 2Ag (s)
Reduced
Cu is the reducing agent
Ag+ is the oxidizing agent
Identify the oxidizing agent and reducing
agent in each of the following:
Identify the oxidizing agent and reducing
agent in each of the following:
+2 -2
(a) 2Al(s) + 3H2SO4(aq)
Al2(SO4)3(aq) + 3H2(g)
(b) PbO(s) + CO(g)
(c) 2H2(g) + O2(g)
Pb(s) + CO2(g)
2H2O(g)
+2 -2
(b) PbO(s) + CO(g)
0
+4 -2
Pb(s) + CO2(g)
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.
0
0
+1 -2
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)
Using oxidation numbers
(c) 2H2(g) + O2(g)
2H2O(g)
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.
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.
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.
Ion-Electron Method To Balance Redox Rxtns
1. Divide the equation into two half-reactions-oxidation and
reduction
2. Balance all atoms other than H and O
3. Balance O by adding H2O
4. Balance H by adding H+
5. Balance net charge by adding e6. Balance e- gain to e- loss and add the half-reactions
7. Add the half reactions and cancel common factors.
8. Add to both sides of the equation the same number of OHas there are H+
9. Combine H+ and OH- to form H2O and cancel common H2O
10. Cancel common H2O
• 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”
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
1. Determine the oxidation states of the various species.
Example 5-6
4. Balance O by adding H2O to the side that needs O:
4+
6+
7+
2+
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
2. Identify the oxidizing and reducing agents. Write the halfreactions
4+
6+
SO32-(aq) → SO42-(aq)
7+
2+
MnO4-(aq) → Mn2+(aq)
H2O(l) + SO32-(aq) → SO42-(aq)
MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
5. Balance hydrogen by adding H+:
H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq)
8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
3. Balance atoms other than H and O:
Already balanced for elements S and Mn
Balancing Redox Equations
6. Add
sides.
e-
to balance charge. Charge should be equal on both
H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) + 2 e-(aq)
5 e-(aq) + 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
Balancing Redox Reactions
8. Add both equations and simplify:
5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) →
9. Check the result
5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l)
10. If a basic solution is desired then add OH- to neutralize the
H+ on both sides of the equation. Cancel common water.
7. Multiply the half-reactions to balance all e- in both
equations:
5 SO32-(aq) + 2 MnO4-(aq) + 6H+ + 6OH- (aq) →
5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l) + 6OH-
5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e-(aq) + 10 H+(aq)
16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)
5 SO32-(aq) + 2 MnO4-(aq) + 3H2O (aq) →
5 SO42-(aq) + 2 Mn2+(aq) + 6OH-
Balancing Redox Reactions
x2
4H+ + MnO4- +3e-
x3
C2O42- + 2H2O
• Method used to balance redox reactions
3+
2+
2+
2CO32- + 4H+ + 2e-
4+
Fe3+(aq) + Sn2+ → Fe2+ + Sn4+
Oxidation:
Sn2+ → Sn4+(aq) + 2 e-
Reduction:
Fe3+(aq) + 1 e- → Fe2+
Atoms and mass are balanced but charge is not. Let’s multiply by
2 to get charge to cancel
Sn2+ → Sn4+(aq) + 2 e2 Fe3+(aq) + 2 e- → 2Fe2+
Overall:
MnO2+ 2H2O
Fe3+(aq) + Sn2+ → Fe2+ + Sn4+
8H+ + 2MnO4- + 6e3C2O42- + 6H2O
2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l)
+ 4OH-
2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq)
2MnO2+ 4H2O
6CO32- + 12H+ + 6e-
2MnO2(s) + 6CO32-(aq) + 4H+(aq)
+ 4OH-
2MnO2(s) + 6CO32-(aq) + 2H2O(l)
The redox reaction between dichromate and iodide anions
Cr2O72-
Cr3+ + I2
I-
Balancing redox reactions by the half-reaction
method
PROBLEM:
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) + C2O42-(aq)
PLAN:
MnO2(s) + CO32-(aq)
Proceed in acidic solution and then neutralize with base.
SOLUTION:
+7
MnO44H+ + MnO4+ 3e-
MnO4-
R
C2O42-
MnO2
O
+4
MnO2
+3
C2O42-
MnO2 + 2H2O
C2O42- + 2H2O
CO32-
+4
2 CO 23
+
2CO32- + 4H
+ 2e-
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
14H+(aq) + Cr2O72-(aq) + 6I-(aq)
3. Balance O by adding H2O (count)
2Cr3+(aq) + 3I2(aq) + 7H2O(l)
4. Balance H by adding H+
5. Balance net charge by adding e6. 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.
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)
2-
14H2O + Cr2O7 + 6
I-
2Cr3+
+ 3I2 + 7H2O +
14OH-
Show a balanced equation for the oxidation of Fe2+ to
Fe3+ by Cr2O72- (to Cr3+) in acid solution?
1. Write the unbalanced equation for the reaction ion ionic form.
Fe2+ + Cr2O72-
2. Identify what is being oxidized and reduced using oxidation
numbers. Separate the equation into two half-reactions.
+2
Oxidation:
❐
Reconcile the number of water molecules.
7H2O + Cr2O72- + 6 I-
2Cr3+ + 3I2 + 14OH-
Do a final check on atoms and charges.
Fe3+ + Cr3+
Reduction:
+3
Fe2+
Fe3+
Cr2O72-
Cr3+
+6
+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.
Cr2O722Cr3+ + 7H2O
+
214H + Cr2O7
2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the
charges on the half-reaction.
Fe2+
Fe3+
+
2Cr3+ + 7H2O
6e- + 14H+ + Cr2O72-
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).
1. Translate words to chemical formula HCl + Fe(s).
2. What will be oxidized and reduced by looking at the activity series
3. Write the half-reactions maintaining mass and charge balance
4. Sum the two half-reactions
2H+ (s) + 2e- (g)
Fe(s) +
2H+(aq)
Fe2+(aq) + 2 e+ H2(g)
Binary Ionic Compound
2NaCl
2. Non-Metal + Non-metal
0
S + O2
3. Compound + Element
+3
0
PCl3+ Cl2
4. Compound + Compound
+2
-2
+1 -2
CaO(s)+ H2O
Reduction: 6e- + 14H+ + Cr2O7214H+ + Cr2O72- + 6Fe2+
2Cr3+ + 7H2O
6Fe3+ + 2Cr3+ + 7H2O
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.
There are 5-Main Types of Redox Reactions
1. Combination
A + B => AB
-produce fewer products than reactants.
i. Element + Element
=> Compound
ii. Element + Compound
=> New Compound
iii. Compound + Compound => New Compound
2. Decomposition
AB => A + B
3. Displacement
AB + C => AC + B
-produce more products than reactants.
-methathesis (switching places)
element in compound is both oxidized and reduced
Fe2+(aq)
Combination reactions produce fewer products than reactants.
0
6Fe3+ + 6e-
4. Disproportionation--common element or
H2
1. Combination Reactions: A + B => C
1. Metal + Non-metal
2Na(s)+ Cl2
6Fe2+
8. Verify that the number of atoms and the charges are balanced.
2Cr3+ + 7H2O
Fe(s)
Oxidation:
1e-
6. If necessary, equalize the number of electrons in the two halfreactions by multiplying the half-reactions by appropriate
coefficients.
6Fe2+
6Fe3+ + 6e6e- + 14H+ + Cr2O72-
7. Add the two half-reactions together and balance the final
equation by inspection. The number of electrons on both
sides must cancel.
Binary Molec Cmpd
5. Combustion
CH4 + O2 => CO2 + H2O
2. Decomposition Reactions
--produces more products than reactants through the
action of heat (thermal decomposition) or electricity
(electrolytic decomposition).
1. Compound
+4 -2
Element A + Element B
electricity
SO2
2H2O
New Compound
+5 -1
PCl5
New Compound
+2
-2 +1
Ca(OH)2
2HgO
heat
2H2 + O2
2Hg + O2
2. Compound
+1 +5 -2
2KClO3
Compound A + Element B
heat
+1 -1
0
2KCl + 3O2
Activity Series For Metals
3. Displacement Reactions
Pure metals higher in the series will be
oxidized by a dissolved elemental ion
below it in the series which is reduced.
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.
double
A + BC
AC + B Methathesis = displacement
+1 -2
0
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
+2 -2 +1
0
Ca(s) + 2H2O(aq) → Ca(OH)2 + H2
H2 Displacement
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
+4 -1
0
0
+2 -1
0
TiCl4(aq) + 2Mg(s)
Ti + 2MgCl2
+2
+2
0
Zn(s) + CuSO4 (aq) => ZnSO4 (aq) + Cu (s)
METAL DISPLACEMENT
More 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)
Metal
Displacement
Metal Displacement (Activity Series)
Zn Rod
Predict whether the following
reactions will occur as written.
1. Fe(s) + 3Cu2+ => 2Fe3+ +
Cu(s)
2. Sn(s) + Ca2+ => Sn2+ + Ca(s)
CuSO4
3. Au(s) + 3AgNO3(aq) => Au(NO3)3
CuSO4
Solid Zn rod
dunked
A
solid Zn rod
into
a solution
dunked
into a of
solution4.of copper
CuSO
sulfate (CuSO4)
+ 3 Ag(s)
Use activity series:
Zn metal is higher
than Cu in the series.
It will displace Cu2+
from solution.
The Zn is oxidized
and the Cu2+ is
reduced to solid Cu(s)
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Strength as oxidizing agent
A more active halogen will displace a less active
(heavier) halide from their binary salts.
0
F2
+1 -1
0
+1 -1
Cl2 + 2NaI ===> I2 + 2NaCl
Br2 + 2NaI ===> I2 + 2NaBr
I2(s) + 2F- ===> No reaction
Cl2
Br2
I2
Cl2 and Br2 both are higher and will displace Ifrom 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. 2 Al(s) + Cr2O3(s) ! Al2O3(s) + 2 Cr(s)
5. Pt(s) + 4 HCl(aq) ! PtCl4(aq) + 2 H2(g)
More active metal + Salt of less active metal =>
Reduced Less active metal + Salt of more active
metal.
4. Disproportionation Reaction
A single element in a compound is both oxidized and
reduced (both oxidizing and reducing agent in one)
+1 -1
-2 +1
+1 -2
H2O2 + 2OH-
0
-2 +1
Cl2 + 2OH-
0
2H2O + O2
+1 -2
-1
+1 -2
ClO- + Cl- + H2O
Identifying the Type of Redox Reaction
5. Combustion Reaction
In 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).
Classify 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)
+4 -1
0
+4 -2
CH4 + 2O2
(b) hydrogen peroxide(l)
+1 -2
CO2 + 2H2O
2C4H10 + 13O2
(c) aluminum(s) + lead(II) nitrate(aq)
8CO2 + 10H2O
Identifying the Type of Redox Reaction
0
0
+2 -3
3Mg(s) + N2(g)
Mg3N2 (aq)
(a) Combination
Mg is the reducing agent; N2 is the oxidizing agent.
+1 -1
+1 -2
H2O2(l)
2 H2O2(l)
0
H2O(l) + 1/2 O2(g)
or
(b)Decomposition
2 H2O(l) + O2(g)
H2O2 is the oxidizing and reducing agent.
0
+2 +5 -2
Al(s) + Pb(NO3)2(aq)
2Al(s) + 3Pb(NO3)2(aq)
+3 +5 -2
0
(c)Displacement
Al(NO3)3(aq) + Pb(s)
2Al(NO3)3(aq) + 3Pb(s)
Pb(NO3)2 is the oxidizing and Al is the reducing agent.
magnesium nitride (aq)
water(l) + oxygen gas
aluminum nitrate(aq) + lead(s)