Chem 40S Uunit 1 Notes

Outcome 1-01
Explain observed examples of solubility and precipitation at the molecular and symbolic
levels.

Solubility
&
Precipitation Compare and Contrast

Solubility
vs
Precipitation

Outcome 1-01
levels.

• When a solute dissolves in a solvent,
The the individual particles of the solute
Solution separate from the other particles of
Process the solute and move between the
spaces of the solvent particles. The
solvent particles collide with the
A Review from solute particles and forces of
Chemistry 30S attraction between solute and solvent
particles "hold" the solute particles in
the spaces.
• For a solute to be dissolved in a
solvent, the attractive forces between
the solute and solvent molecules
must be greater than the forces of
attraction between the solute
molecules.

Outcome 1-01
levels.

1. The solvent particles must move apart to
The make room for solute particles. This
process requires energy to overcome
Solution forces of attraction between solvent
Process particles. The first step in the dissolving
process is endothermic.
A 3 Step 2. The solute particles must separate form
the other solute particles. This process
Process also requires energy to overcome the
forces of attraction between the solute
particles. The second step in the dissolving
process is endothermic.
3. When the solute particles move between
the solvent particles the forces of attraction
between solute and solvent take hold and
the particles "snap" back and move closer.
This process releases energy. The final
step in the dissolving process is
exothermic.

Outcome 1-01
levels.

• The total heat change in the
Energy dissolving process is the sum of the
Changes three heat changes. If the sum of
During the heat absorbed in the first two
Dissolving steps of the dissolving process is
greater than the heat released in the
last step, the dissolving of that
substance will be endothermic. If
the dissolving process for a
substance is endothermic, the
container will feel cooler as the
substance dissolves.
• In some cases, the exothermic
process is larger than the sum of
the two endothermic processes. In
these cases, as the solid dissolves
large amounts of heat are evolved.

Outcome 1-01
levels.

• When we talk about the mixing of two
Solubility or more substances together in solution
we must consider solubility. Simply
defined, it is a measure of how much
solute will dissolve into the solvent. Not
all substances will dissolve in all
solvents.
• Example 1: NaCl(s) dissolved in water
Molecular level:

NaCl (solid) + H2O (liquid) NaCl(aq)

Outcome 1-01
levels.

Student TASK
Solubility
• Create a series of drawings that outline
what happens in the solution process
involving NaCl dissolving in water.

Outcome 1-01
levels.

• Example 1: NaCl(s) dissolved in water
Solubility
Symbolic level:

NaCl(s) → NaCl(aq)

Outcome 1-01
levels.

• A precipitation reaction is a common
Precipitation type of chemical reaction in solution
chemistry. Basically, two or more
solutions are combined resulting in a
reaction that produces an insoluble
product, a precipitate. Typically, these
types of reactions involve ionic
compounds in aqueous solution. The
precipitation reaction is said to occur
because of the strong attractive forces
certain ions have for each other. These
ions combine and fall out of solution in
the formation of a solid.
• The ability to predict precipitate
formation is based upon solubility.
When the product of a reaction is
insoluble, it precipitates.

Outcome 1-01
levels.

• Example 1: NaCl(s) and AgNO3 combined together
Precipitation Molecular level:

•

Outcome 1-01
levels.

Student TASK
Precipitation
• Create a series of 3 diagrams that
outline what happens when NaCl and
AgNO3 are combined in a precipitation
reaction.

Diagram 1 – draw Na+ and Cl- ions
circulating amidst the water molecules
Diagram 2 – AgNO3 is drawn with the
Ag+ and NO3- ions floating around the
water molecules
Diagram 3 – Shows the mixing of the
two solutions. The Ag+ and Cl-
combine to form a precipitate.

Outcome 1-01
levels.

Precipitation

• Symbolic level:

Molecular equation:
NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)

Ionic equation:
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) → AgCl (s) + Na+ (aq) + NO3- (aq)

Net ionic equation:
Ag+ (aq) + Cl- (aq) → AgCl (s)

Outcome 1-01
levels.

Precipitation • When certain solutions of electrolytes
are mixed, the cation of one solute may
combine with the anion of the other
How to write solute to form an insoluble compound.
chemical This leads to the formation of a
equations for precipitate.
precipitation • A precipitate is an insoluble solid that
reactions settles from a solution. For instance,
when a solution of Pb(NO3)2 is mixed
with a solution of HCl, a precipitate of
insoluble PbCl2(s) forms. This occurs
because combining the solutions
results in a new solution containing
more PbCl2(aq) than is soluble. A
condition referred to as
supersaturation. In this case, solid
PbCl2 must settle out of the solution.

Outcome 1-01
levels.

• In the molecular equation the
Precipitation formulas of the compounds are written
as their usual chemical formulas.

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

• The precipitation reaction is more
accurately represented by the ionic
equation. The ionic equation shows
the compounds as being dissociated in
solution.

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

Outcome 1-01
levels.

Precipitation Pb2+ (aq) + 2NO3- (aq) + 2H+ (aq) + 2Cl– (aq)
PbCl2 (s) + 2H+ (aq) + 2NO3- (aq)
• If you examine the ionic equation, you
will notice that the H+ ions and NO3-ions
are not involved in the formation of the
precipitate. We refer to the ions which
are not involved in the reaction as
spectator ions. Identical species on
both sides of the equation can be
omitted from the equation. The net
ionic equation shows only the species
that actually undergo a chemical
change.
Pb2+ (aq) + 2Cl– (aq) PbCl2 (s)
• In order to predict whether a precipitate
will form when two solutions are mixed,
you need to know the solubility rules for
ionic compounds.

Outcome 1-01
levels.

Precipitation

Outcome 1-02
Perform a lab to develop a set of solubility rules.
Outcome 1-03
Use a table of solubility rules to predict the formation of a precipitate.

Solubility
Rules

Outcome 1-02
Perform a lab to develop a set of solubility rules.
Outcome 1-03
Use a table of solubility rules to predict the formation of a precipitate.

Process
Notes for
Writing Net
Ionic
Equations
Rules

Outcome 1-04
Write balanced neutralization reactions involving strong acids and bases.
Outcome 1-05
Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong base and a
strong acid.

Acid/Base Definitions of acids and bases
Review • Arrhenius
acid: generates [H+] in solution
base: generates [OH-] in solution
• normal Arrhenius equation:
acid + base salt + water

• example: HCl + NaOH NaCl + H2O

Outcome 1-04
Outcome 1-05
strong acid.

Acid/Base • A strong acid is defined as an
Review acid that completely dissociates
into ions. This means that if there
are 100 molecules of HCl
dissolved in water, 100 ions of H+
and 100 ions of Cl- are produced.
• It should be stressed that there
are only 6 strong acids. These
are:
– hydrochloric acid (HCl),
– hydrobromic acid (HBr),
– hydroiodic acid (HI),
– sulfuric acid (H2SO4),
– nitric acid (HNO3),
– perchloric acid (HClO4).

Outcome 1-04
Outcome 1-05
Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a
strong acid.

Acid/Base • A strong base is defined as a
Review base that completely dissociates
into ions. This means that if
there are 100 formula units of
NaOH dissolved in water, 100
ions of Na+ and 100 ions of OH-
are produced.
• Strong bases include any ionic
compound that contains the
hydroxide (OH-) ion.
• When combined with the
hydroxide ion, elements found in
groups 1 and 2 form strong
bases.

Outcome 1-04
Outcome 1-05
strong acid.

Naming Acids • To name a binary acid follow the
steps given:
1. the prefix "hydro" is used
2. the root of the anion is used
3. the suffix "ic" is used
4. the word "acid" is used as the
second word in the name
• Example of naming a binary acid:
HCl
Step 1: hydro-
Step 2: -chloride
Step 3: -chloric
Step 4: hydrochloric acid

Outcome 1-04
Outcome 1-05
strong acid.

Naming Acids • Naming polyatomic acids follow a different
set of rules. Many of the oxygen-rich
polyatomic negative ions form acids that are
named by replacing the suffix -ate with -ic
and the suffix -ite with -ous.
• To name oxyacids (acids containing the
element oxygen) you recognize them by the
general formula HaXbOc where X represents
an element other than hydrogen or oxygen. If
enough H+ ions are added to a (root)ate
polyatomic ion to completely neutralize its
charge, the (root)ic acid is formed
• Examples of polyatomic acids:
– If one H+ ion is added to nitrate, NO3-
– HNO3 is formed. This is named nitric
acid.
– If two H+ ions are added to sulfate,
SO42-
– H2SO4 is formed. This is named
sulfuric acid.

Outcome 1-04
Outcome 1-05
strong acid.

Neutralization • Most common acid-base
Reactions reactions take place in water
solutions (commonly referred to
as aqueous solutions).
• In the reaction of an acid with a
base in aqueous solution, the
hydrogen ions of the acid react
with the hydroxide ions of the
base to give water. The second
product is a salt, which is
composed of the positive metal
ion from the base and the
negative ion from the acid.
• In general, the reaction for a
neutralization reaction is given by
acid + base → salt + water

Outcome 1-04
Outcome 1-05
strong acid.

Neutralization • For example,
Reactions • HCl(aq) + KOH(aq) → H2O(l) + KCl(aq)
• Since HCl(aq) and KOH(aq) are fully
ionized in solution, the preceding
equation can be written as
• H+ (aq) + Cl− (aq) K+(aq) + OH−(aq) → H2O (l)
K+(aq) + Cl−(aq)
• Ions common to both sides (spectator
ions) can be canceled to yield
• H+(aq) + OH−(aq) → H2O(l)
• This is referred to as the net ionic
equation for the neutralization
reaction. If H3O+ is substituted for H+
(aq) the neutralization equation
becomes
• H3O+(aq) + OH−(aq) → 2 H2O (l)

Outcome 1-04
Outcome 1-05
strong acid.

Process
Notes for
Writing Net
Ionic
Equations
Rules

Chemistry 40S
Outcome 1-04 Unit 3 – Acid/Base
Equilibria neutralization reactions involving strong acids and bases.
Write balanced
Outcome 1-05
strong acid.

Titrations • A titration is a carefully
controlled neutralization
reaction.
• To perform a titration, a
standard solution is required. A
standard solution is a solution
of a strong acid or base of
known concentration. To
determine the concentration of
an acid, a basic standard
solution is required, and vice
versa for an unknown base. The
standard solution is usually
added to a sample of unknown
concentration, until
neutralization has occurred.

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titrations • A buret is used to control the
additions of acids and bases in a
titration. The buret determines the
volumes of acids and bases added
in a titration.
• If you recall, when an acid or base
is just neutralized, the moles of
hydronium ions from the acid and
moles of hydroxide ions from the
base are equal. The point at which
the amount of standard acid or
base solution added just neutralizes
the unknown sample is called the
equivalence point.

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titrations • The equivalence point can be
determined by adding a pH indicator to
the unknown sample or by using a pH
meter to measure the pH change as
standard solution is added to the
sample. If an indicator is used, the point
in the titration at which the desired
colour forms is called the endpoint of
the titration. The indicator is chosen so
that the endpoint and the equivalence
point are very close.
• The endpoint and the equivalence point
are NOT the same thing. The
equivalence point is a single point
defined by the reaction stoichiometry.
The endpoint is determined by the
choice of indicators.

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titrations Titration Procedure
• When performing an acid-base titration on an
unknown solution, both the standard and
unknown solutions are placed into separate
burets. For this example, we will determine
the concentration of a hydrochloric acid
solution using a 0.100 mol/L solution of
NaOH.
• Each buret is filled and the initial volume
readings are taken from the bottom of the
meniscus (the smile-shaped surface of the
solution in the container).
• A sample or aliquot of the acid solution is
released into an Erlenmeyer flask. The acid
volume in the buret is read and called the
"final acid volume". The volume of the aliquot
is equal to:
final volume reading - initial volume reading

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titrations Titration Procedure - continued
• We will use phenolphthalein as our
indicator. The endpoint will be reached
when the sample turns a light pink. Even
though phenolphthalein turns pink at a pH
of about 8.2, this is not a significant
problem, as we will see later.
• We slowly add the standard base solution
to the flask, with stirring or swirling, until
the solution in the flask turns (and stays) a
light pink. We take the reading on the
base buret and call this the "final base
volume".
• This procedure is repeated until there are
several pieces of consistent data.

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration
Procedure

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration
Video
Demo

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Determining Example 1
Concentratio
n • A titration was performed using a standard
solution of 0.100 mol/L NaOH into and
unknown HCl solution. The following data was
collected:

Base Acid
Final Volume Reading 14.45 mL 12.57 mL
Initial Volume Reading 0.62 mL 1.13 mL

Determine the concentration of the acid.

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Determini Example 2
ng the
Mass of • A student receives a solid
an sample of sulfamic acid (molar
unknown mass is 97.09 g/mol), a
Sample monoprotic acid, and dissolves
the sample in enough water to
make 100.0 mL of solution. The
students takes a 12.00 mL
aliquot and titrates with 0.0985
mol/L sodium hydroxide. If
13.38 mL of the base is needed
to reach the endpoint, what is
the mass of the sample of acid?

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration • If pH readings are taken during a
Curves titration and those pH values are
plotted, a titration curve is
generated. The general shape of the
curves generated may be grouped
into families, according to the solution
titrated and the titrating solutions. We
will discuss four classes of titrations:
–1. Strong acid titrated with strong
base.
–2. Strong base titrated with
strong acid.
–3. Weak acid titrated with strong
base.
–4. Weak base titrated with strong
acid.

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration 1. Strong Acid Titrated with a Strong Base
Curves

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration 2. Strong Base Titrated with a Strong Acid
Curves

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration 3. Titration of a Weak Acid with a Strong Base
Curves

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration 4. Titration of a Weak Base with a Strong Acid
Curves

Chemistry 40S
Write balanced
Outcome 1-05
strong acid.

Titration
Curves

Chemistry 40S Unit 3 – Acid/Base
Equilibria1-08
Outcome
Define oxidation and reduction.
Include: gain and loss of electrons, oxidizing agent, reducing agent.

Oxidation What is oxidation and reduction?
Reduction
• The term “oxidation” was first applied
to the combining of oxygen with other
elements (ex. rusting iron or burning
carbon or methane). Burning is
another name for rapid oxidation.
• The term “reduction” originally meant
the removal of oxygen from a
compound. The term reduction comes
from the fact that the free metal has a
lower mass than its oxide compound.
There is a decrease or reduction in
the mass of the material as the
oxygen is removed.

Equilibria1-08
Outcome

Reduction
• Chemists recognized that other
nonmetallic elements unite with
substances in a manner similar to that
of oxygen (ex. hydrogen, antimony,
and sodium will burn in chlorine; iron
will burn in fluorine).
• Therefore, the term oxidation was
redefined as the process by which
electrons are removed from an
atom or ion.
• Reduction was then defined as the
process by which any atom or ion
gains electrons.

Equilibria1-08
Outcome

Oxidation
Reduction

Equilibria1-08
Outcome

Reduction
• Oxidation occurs when an atom loses
one or more electrons and reduction
is when an atom gains one or more
electrons. There are two mnemonics
to help us remember these terms:

• "OIL RIG": Oxidation Is Losing
electrons, Reduction Is Gaining
electrons
Or

• "LEO says GER": Losing Electrons is
Oxidation, Gaining Electrons is
Reduction

Equilibria1-08
Outcome

Oxidation- What are REDOX Reactions?
Reduction • Oxidation-Reduction reactions, or
Reactions Redox reactions, are defined as
chemical changes that occur when
electrons are transferred from one
reactant to another.
• For example,
2 Mg(s) + O2(g) → 2 MgO(s)
• If this reaction is written in ionic form it
becomes:
2 Mg0 + O20 → Mg2+ O2-
• Magnesium and oxygen gases are
both elements and have no charge
• Non-scientists usually refer to this
reaction as burning or combustion but
as scientists, we refer to this reaction
as oxidation.

Equilibria1-08
Outcome

Reduction • For example,
Reactions 2 Mg(s) + O2(g) → 2 MgO(s)
• We say that the magnesium has been
oxidized to MgO by the reaction with
oxygen gas
• Considering the charges, the metal
has gone from a 0 charge to 2+ and
the non-metal from 0 charge to 2-
• In this reaction magnesium begins as
a neutral atom and loses two
electrons to become a Mg2+ ion in
MgO. Magnesium is oxidized. Oxygen
begins as a neutral atom and gains
the two electrons from magnesium to
become an O2-¯ion in MgO. Oxygen is
reduced.

Equilibria1-08
Outcome

Reduction • For example,
Reactions
2 Mg(s) + O2(g) → 2 MgO(s)

• If we look at the change in ion charge
as a function of electrons, the
following relationships can be written:
• 2 Mg → 2Mg 2+ + 4 e- in this equation,
charge is conserved, 0 charge on both
sides
• O2 + 4 e- → 2O 2- in this equation,
charge is conserved, 4- charge on
both sides

Equilibria1-08
Outcome

Reduction
Reactions • Similarly, the reaction of magnesium
combining with chlorine to produce
magnesium chloride is also a redox
reaction.

Mg(s) + Cl2(g) → MgCl2(s)

• Magnesium loses two electrons to
become a Mg2+ ion and is oxidized.
Each chlorine atom gains one electron
from magnesium to become Cl- ions.
Each chlorine atom is reduced.

Equilibria1-08
Outcome

Reduction
Reactions

Equilibria1-08
Outcome

Reducing What are Reducing Agents?
Agents

Equilibria1-08
Outcome

Oxidizing What are Oxidizing Agents?
Agents

Equilibria1-08
Outcome

Reducing What are Reducing & Oxidizing
& Agents?
Oxidizing
Agents • Any substance which causes the
reduction of another substance is
called the reducing agent.

• Any substance which causes the
oxidation of another substance is
called an oxidizing agent.

2 Mg(s) + O2(g) → 2 MgO(s)

Equilibria1-09
Outcome
Determine the oxidation numbers for atoms in compounds and ions.

Oxidation Assigning Oxidation Numbers
Numbers
• The oxidation number represents the
charge the atom would have if every
bond were ionic. Not every bond is
ionic, but chemists assume they are
for this system. The oxidation number
is not always the actual charge, but it
is very helpful to follow electrons in
redox reactions.

Equilibria1-09
Outcome

Numbers – Oxidation numbers are always
assigned PER ATOM.
– The oxidation numbers of all
uncombined elements is zero. (ex. O2,
K(s), H2,
– The oxidation number of monatomic
ions equals the charge of that ion.
– In compounds, the oxidation number
for alkali metals (ex. Li, Na, K, etc.) is
always +1.
for the alkaline earth metals (ex. Be,
Mg, Ca, etc.) is always +2.
of aluminum is always +3.

Equilibria1-09
Outcome

Numbers – In compounds, the oxidation number of
fluorine is -1.
– In compounds, the oxidation number of
hydrogen is +1. An exception is in metal
hydrides, such as NaH or MgH2, when
hydrogen is -1.
– In compounds, the oxidation number of oxygen
is -2. An exception is in peroxides, such as
H2O2 or Na2O2, when its oxidation number is
-1.
– For any neutral compound, the sum of the
oxidation numbers for each atom must be
zero.
– For a poly atomic ion, the sum of the oxidation
numbers for each atom must be the charge of
that ion.

Note the difference between ion charge and
oxidation numbers:
for example, the magnesium ion, Mg 2+
ion charge = 2+
oxidation number = +2

Equilibria1-10
Outcome
Identify and describe reactions as redox or non-redox.
Include: oxidizing agent, reducing agent, oxidzed substance, and reduced substance

Redox Recognizing Redox Reactions
Reactions
• In order to recognize a redox
reaction, we must follow the
electrons. We can follow the
electrons by determining if
the oxidation numbers
change. If no oxidation
numbers change, the reaction
is NOT a redox reaction.

Equilibria1-10
Outcome

Reactions
Example
– Is the reaction below a redox reaction?
SO2 + H2O → H2SO3

Solution
– If we assign oxidation numbers to each atom,
we find

– the oxidation number of S remains at +4 in the
reactants and in the products,
– O remains unchanged in the products at -2
and H remains unchanged at +1.
– The oxidation numbers do not change, so
electrons were not transferred. Therefore, this
is not a redox reaction.

Equilibria1-10
Outcome

Reactions
Example
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)

Solution
– Assign oxidation numbers to each atom,

– Cu is zero in the reactants and +2 in the
products.
– Ag is +1 in the reactants and zero in the
products.
– N = +5 in both products and reactants.
– O = -2 in both products and reactants.

Equilibria1-10
Outcome

Reactions
Example
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)

Solution

– This is redox reaction, since the oxidation
numbers of Cu and Ag change.
– Cu is oxidized because its oxidation number
becomes more positive, indicating it has lost
electrons. .
– The oxidation number of Ag becomes more
negative, indicating a gain of electrons. Ag is
reduced.

Equilibria1-11
Outcome
Balance oxidation-reduction reactions using redox methods.
Include acidic and basic solutions

Balancing Balancing Redox Reactions
Redox
Reactions • Redox reactions transfer electrons
from one substance to another.
Balancing a redox reaction balances
the number of electrons lost and
gained, then ensures the reaction
obeys the Law of Conservation of
Mass.

• There are two methods we will use to
balance redox reactions:

– Oxidation Number Method
– Half-Reactions Method

Equilibria1-11
Outcome

Balancing Oxidation Number Method
Redox
Reactions • We will learn this method by balancing
the following reaction:
K2Cr2O7 + H2O + S SO2 + KOH + Cr2O3

• Step 1: Assign oxidation numbers.

• Eventually, you will be able to
recognize those just substances that
are oxidized and reduced. You can
just label those substances.

Equilibria1-11
Outcome

Redox K2Cr2O7 + H2O + S SO2 + KOH + Cr2O3
Reactions
• Step 2: Identify substances oxidized
and reduced, then balance electrons
lost and gained.

• S is oxidized and Cr is reduced.
• Notice that there are two Cr atoms
and each gains three electrons, for a
total of six.

Equilibria1-11
Outcome

Reactions
• Step 2 (continued): We then
determine the lowest common multiple
of 6 and 4, which is 12. We multiply
the S atoms by 3 and the Cr atoms by
2, so the total number of electrons lost
and gained is 12.

Equilibria1-11
Outcome

Reactions
• Step 3: Balance all other atoms by
inspection.
• We need 4 K atoms on the left, and 4 H
atoms on the right.

• The final balanced equation is

2K2Cr2O7 + 2H2O + 3S 3SO2 + 4KOH + 2Cr2O3

Equilibria1-11
Outcome

Balancing Half-Reactions Method
Redox • Half-reactions, as their name suggests, show the
Reactions oxidation or reduction reaction.
• For example:
Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)
• The oxidation half reaction shows the oxidation of
copper to copper (II) ions and the loss of electrons:
Cu(s) Cu2+(aq) + 2e¯
• The loss of electrons is indicated in the equation by
putting electrons on the product side of the
equation.
• The reduction half reaction shows the reduction of
silver ions to metallic silver and the gaining of
electrons:
Ag+ (aq) + 1e¯ Ag (s)
• Gaining electrons is indicated in the equation by
placing electrons on the reactant side of the
equation.

Equilibria1-11
Outcome

Redox • Redox reactions often occur in the
Reactions presence of acids or bases. The half-
reaction method is the preferred
method for balancing these types of
reactions.
• In general, the method requires the
entire equation to be separated into
the oxidation and reduction half-
reactions. The electrons are balanced
and the overall balanced equation is
produced by adding the two balanced
half-reactions.

Equilibria1-11
Outcome

Redox • Steps for Balancing Redox Reactions in
Reactions Acidic Solutions
• This method will be demonstrated by
balancing the following reaction in an acidic
solution.
Cr2O72¯(aq) + SO3 2¯(aq) Cr3+(aq) + SO42-(aq)
• Step 1: Assign oxidation numbers. Identify
then write the oxidation and reduction half-
reactions.
• Step 2: Balance all elements except hydrogen
and oxygen. Add electrons lost and gained.
• Step 3: Balance oxygen atoms by using H2O.
• Step 4: Balance hydrogen atoms using H+
ions.
• Step 5: Balance the number of electrons lost
and gained.
• Step 6: Add the two half-reactions.

Equilibria1-11
Outcome

Half-Reactions Method
Balancing
Redox • The first part of balancing redox
Reactions reactions in basic solutions follows the
same steps as that for acidic
solutions. Before adding the two half-
reactions, you add the same number
of hydroxide ions as hydrogen ions to
BOTH sides of the equation. We
eliminate the hydrogen ions by
forming water
(H+ + OH- H2O)

Equilibria1-11
Outcome

Half-Reactions Method
Balancing
Redox • For example, balance the following
Reactions reaction in a basic solution.
MnO4¯ + C2O42¯ CO2 + MnO2

• Step 1: Assign oxidation numbers.
Identify then write the oxidation and
reduction half-reactions.
• Step 2: Add water and hydrogen ions to
balance oxygen and hydrogen.
• Step 3: Add hydroxide ions to each side
for each hydrogen ion.
• Step 4: Change the hydroxides and
hydrogens to waters .
• Step 5: Balance electrons lost and
gained .
• Step 6: Add the two half-reactions.

Chem 40S Uunit 1 Notes

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