Chapter 4
Aqueous solutions
Types of reactions
Parts of Solutions
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Solution – Homogeneous mixture.
Solute – What gets dissolved.
Solvent – What does the dissolving.
Soluble – Can be dissolved.
Insoluble – Can’t be dissolved
Miscible – Liquids dissolve in each other.
Immiscible – Liquids that don’t dissolve
in each other
Aqueous solutions
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Dissolved in water.
Water is a good solvent because the
molecules are polar.
The oxygen atoms have a partial
negative charge.
The hydrogen atoms have a partial
positive charge.
Hydration
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Hydration is the process of breaking the
ions of salts apart.
Ions have charges and attract the
opposite charges on the water molecules.
An ion is surrounded by other ions of the
opposite charge.
Hydration
Also check out the animation on the website.
H
H
H
H
H
Solubility
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How much of a substance will dissolve in a
given amount of water.
Usually g of solute / 100 grams of H2O
Varies greatly, but if they do dissolve the
ions are separated, and they move around.
Water can also dissolve non-ionic
compounds if they have polar bonds, such
as soap.
Electrolytes
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Electricity is moving charges.
The ions that are dissolved can move.
Solutions of ionic compounds can conduct
electricity.
Solutions are classified three ways in their
ability to conduct electricity.
Types of solutions

Strong electrolytes - completely dissociate
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Many ions - Conduct well.
Weak electrolytes - Partially split into ions.
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Examples: NaCl
Examples: Vinegar, ammonia
Few ions - Conduct slightly.
Non-electrolytes - Don’t split apart.
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Examples: Sand, plastic
No ions - Don’t conduct.
Types of solutions - Acidic
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Acids – form H+1 ions when dissolved.
Strong acids ionize completely.
The following are strong acids
H2SO4 HNO3 HCl HClO4 HBr HI HClO3
Weak acids – partial ionization
Example: HC2H3O2
Types of Solutions - Basic
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Bases - form OH-1 ions when dissolved.
Strong bases – ionize completely.
Any Group 1 hydroxide is considered to
be a strong base.
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Examples: NaOH and KOH
Transition metal hydroxides will be
weaker bases.

Fe(OH)2 and CuOH
Point of Emphasis
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Strong v. Weak & Concentrated v. Dilute
These terms often get misapplied or
confused.
Strong or Weak is a description of how much
a species will dissociate into ions and is an
intensive property.
Concentrated or Dilute is a description of
how much solute is used by the
experimenter.
Some examples
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Strong and Concentrated means that
the substance will dissociate fully and
the experimenter used a lot of it – it is
in a high concentration. Ex. 8 M HCl
Strong and Dilute means that the
substance will dissociate fully and the
experimenter used a small
concentration. Ex. .001 M HCl
Some more examples
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Weak and concentrated means that the
substance doesn’t dissociate fully and the
experimenter used a lot of it.
Ex. 8 M HC2H3O2 (acetic acid)
Weak and dilute means that the substance
doesn’t dissociate fully and the
experimenter used a small concentration.
Ex. .001 M HC2H3O2
Measuring Solutions
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Concentration - how much is dissolved.
Molarity = Moles of solute
n
Liters of solution L
abbreviated M (Capital M)
1 M = 1 mole solute / 1 liter solution
Calculate the molarity of a solution with
34.6 g of NaCl dissolved in 125 mL of
solution.
Solved Answer
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Calculate the molarity of a solution with
34.6 g of NaCl dissolved in 125 mL of
solution.
First – find moles of NaCl  .592 moles
Second – set up formula M = n / L
M = .592 / .125 L
 M = 4.736 M
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Molarity
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How many grams of HCl would be
required to make 50.0 mL of a 2.7 M
solution ?
What would the concentration be if you
used 27g of CaCl2 to make 500. mL of
solution ?
What is the concentration of each ion?
Solved Answer

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How many grams of HCl would be required
to make 50.0 mL of a 2.7 M solution ?
Use M = n / L
2.7 = n / .050 L
 n = .135 moles
 Use n = g / M  4.921 g of HCl

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This is added with enough water to make 50 mL of soln.
More Solved Answers
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What would the concentration be if you
used 27 g. of CaCl2 to make 500. mL of
solution ? (FW of CaCl2 = 110.98 g)
Find moles  27 / 110.98 = .243 moles
 Use M = n / L
 M = .243 / .5 L
 M = .487 M (can be read as .487 Molar)
Molarity
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Calculate the concentration of a solution
made by dissolving 45.6 g of Na2SO4 to
475 mL.
(FW of Na2SO4 = 142.06)
What is the concentration of each ion ?
A balanced equation is a good place to start
Na2SO4  2 Na+1 + SO4-2
This shows 2 Na+1 ions for every sulfate.
A Little Stoichiometry
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Calculate the concentration of a solution made
by dissolving 45.6 g of Na2SO4 to 475 mL.
Solve as before:
45.6 / 142.06  .321 moles
Use M = n / L .321 / .475 = .676 M
Since the Sodium Ions are in a 2 to 1
ratio there would be 1.352 M Na+1 and
there would be .676 M SO4-2
Look at the Stoichiometry

Let’s look at the dissociation
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Na2SO4  2 Na+1 + SO4-2
We see the 2 : 1 ratio of the two ions, so
that is how we get the values from the
slide above.
Making solutions # 1
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Describe how to make 100.0 mL aliquot of a
1.0 M K2Cr2O4 solution.
The formula wt. of the solute is 246.2 g
Using M = n / L  1.0 = n / .1 L
So we need .1 moles of solute.
Using n = g / m  .1 = g / 246.2
We need 24.62 grams of solute
Making Solutions # 2
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Describe how to make 250. mL of an 2.0 M
copper (II) sulfate solution.
The formula wt. of the solute is 159.6 g.
Using M = n / L  2 = n / .25
So we need .5 moles of solute
Using n = g / M  .5 = g / 159.6
We need 79.8 grams of solute
Dilution
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Adding more solvent to a known solution.
The moles of solute stay the same.
moles = M x L
M1 V1 = M2 V2 (this is a very handy formula)
moles = moles
Stock solution is a solution of known
concentration used to make more dilute
solutions
Dilution # 1
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What volume of a 1.7 M solutions is needed
to make 250 mL of a 0.50 M solution ?
1. Given info. M1 = 1.7 M2 = .5 V2 = .25
2. Formula
M1 V1 = M2 V2
3. Substitute
1.7 V1 = .5 (.25)
4. Solve
.0735 L (or 73.5 mL)
Dilution # 2
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18.5 mL of 2.3 M HCl is added to 250 mL of
water. What is the concentration of the
solution ?
1. Given info. V1 = 18.5 M1 = 2.3 V2 = .25
2. Formula
M1 V1 = M2 V2
3. Substitute
18.5 (2.3) = x (.25)
4. Solve
.1702 M
Dilution
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You have a 4.0 M stock solution. Describe
how to make 1.0 L of a .75 M solution.
Using M1V1 = M2V2
1.0 (.75) = 4.0 V2
V2 = .188 L (or 18.75 mL)
Take 18.75 mL of the 4.0 M Stock soln. and
add enough water to make 1.0 L
Types of Reactions
Precipitation reactions
 When aqueous solutions of ionic
compounds are poured together a solid
forms.
 A solid that forms from mixed solutions is a
precipitate
 If you’re not a part of the solution, your
part of the precipitate (old chem joke)
Precipitations Reactions
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Only happens if one of the products is
insoluble.
Otherwise all the ions stay in solutionnothing has happened.
Need to memorize the rules for
solubility.
Some Solubility Rules
We need to know these.
 All nitrates are soluble
 Alkali metals ions and NH4+1 ions are
soluble
 Halides are soluble except Ag+1, Pb+2, and
Hg2+2
 Most sulfates are soluble, except Pb+2,
Ba+2, Hg+2,and Ca+2
Solubility Rules
 Most hydroxides are slightly soluble
(insoluble) except NaOH and KOH
 Sulfides, carbonates, chromates, and
phosphates are generally insoluble
Precipitation reactions

NaOH(aq) + FeCl3(aq) NaCl(aq) + Fe(OH)3(s)
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Na+1(aq)+OH-1(aq) + Fe+3(aq) + Cl-1(aq) 
Na+1(aq) + Cl-1 (aq) + Fe(OH)3 (s)
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is really
So all that really happens is
3 OH-1(aq) + Fe+3(aq)  Fe(OH)3(s)
Essentially a more complex double replacement
reaction.
Precipitation reaction
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We can predict the products by knowing
types of reactions and the rules for
solubility
AgNO3 (aq) + KCl (aq) 
Zn(NO3)2 (aq) + BaCr2O7 (aq) 
CdCl2 (aq) + Na2S (aq) 
Answers

AgNO3 (aq) + KCl (aq)  KNO3 (aq) + AgCl (s)
We know the KNO3 is soluble because all Group 1
metals are soluble.

Zn(NO3)2 (aq) + BaCr2O7 (aq)  Ba(NO3)2 (aq) + ZnCr2O7 (s)
The nitrates are soluble.
CdCl2 (aq) + Na2S (aq)  NaCl (aq) + CdS (s)
The Group 1 Sodium ion is soluble.

Precipitation Stoichiometry
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25 mL of 0.67 M H2SO4 is added to 35 mL
of 0.40 M CaCl2 . What mass CaSO4 Is
formed ?
First we need a balanced equation and
moles of each reactant.
H2SO4 + CaCl2  CaSO4 + 2 HCl
.0168
.014
Precipitation Stoichiometry
H2SO4 + CaCl2  CaSO4 + 2 HCl
.0168
.014
The stoichiometry is 1:1 so CaCl2 is the limiting
reagent.
The ratio would be:
1 n CaCl2 / 1 CaSO4 = .014 / x
x = .014 n of CaSO4
Convert to grams
n g / M .014 = g / 136.14 g = 1.9 grams
Three Types of Equations
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Molecular Equation - written as whole
formulas, not the ions.
K2CrO4 (aq) + Ba(NO3)2 (aq) 
Complete Ionic Equation show dissolved
electrolytes as the ions.
2 K+1 + CrO4-2 + Ba+2 + 2 NO3-1 
BaCrO4 (s) + 2 K+1 + 2 NO3-1
Spectator ions are those that don’t react.
Three Type of Equations
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Net Ionic Equations show only those ions
that react, not the spectator ions
Ba+2 + CrO4-2  BaCrO4 (s)
Write the three types of equations for the
reactions when these solutions are mixed.
a. Iron (III) sulfate and potassium sulfide
b. Lead (II) nitrate and sulfuric acid.
Three Types of Equations

a. Iron (III) sulfate and potassium sulfide
Fe2(SO4)3 (aq) + K2S (aq)  K2SO4 (aq) + Fe2S3 (s)
Fe+3 + SO4-2 + K+1 + S-2  K+1 + SO4-2 + Fe2S3 (s)
Fe+3

(aq)
+ S-2
(aq)
 Fe2S3
(s)
b. Lead (II) nitrate and sulfuric acid.
Pb(NO3)2 (aq) + H2SO4 (aq)  HNO3 (aq) + PbSO4 (s)
Pb+2 + NO3-1 + H+1 + SO4-2  NO3-1 + H+1 + PbSO4
Pb+2 + SO4-2  PbSO4
(s)
(s)
Stoichiometry of Precipitation
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Exactly the same as regular stoichiometry,
except you are finding moles in a different
way.
Concepts of Limiting Reagent and materials
in excess are also the same as before.
Stoichiometry of Precipitation
– Problem # 1
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What mass of solid is formed when 100.00 mL of 0.100 M
Barium chloride is mixed with 100.00 mL of 0.100 M
sodium hydroxide ?
Balanced Equation:
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BaCl2 + 2 NaOH  Ba(OH)2 + 2 NaCl
Calculate Moles of each reactant
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BaCl2  M = n / L .1 = n / .1 L n = .01 n
NaOH  same .01 n
Stoichiometry of Precipitation –
Problem # 1
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Look at ratios to determine limiting
reagent.
BaCl2 + 2 NaOH  Ba(OH)2 (s) + 2 NaCl (aq)

.01
.01
2 / 1 = .01 NaOH / x x = .005 moles
So NaOH is the limiting reagent
Proportion
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2 n NaOH / 1 n Ba(OH)2 = .01 / x x = .005 moles
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Stoichiometry of Precipitation –
Problem # 2
What volume of 0.204 M HCl is needed to
precipitate all the silver from 50.ml of 0.0600
M silver nitrate solution ?
 HCl (aq) + AgNO3 (aq)  AgCl (s) + HNO3 (aq)
.003 n
It is a 1:1 ratio so the moles are the same.
Use M = n / L .204 = .003 / L
L = .0147 L
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Types of Reactions
Acid-Base
+1 donor.
 For our purposes an acid is a H
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A base is a OH-1 donor.
A neutralization is when there is equal
numbers of moles of acid and moles of
base.
A titration is the lab procedure used to
achieve this.
Acid – Base Reactions
Always remember the following when
dealing with Acid-Base Reactions
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Acid + Base  salt + water
Net Ionic Form: H+1 + OH-1  H2O
Dissociation of Water: H2O  H+1 + OH-1
General Formula:
Acid - Base Reactions
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Often called a neutralization reaction
Because the acid neutralizes the base.
Often titrate to determine concentrations.
Solution of known concentration (titrant),
Is added to the unknown (analyte),
Until the equivalence point is reached
where enough titrant has been added to
neutralize it.
Titration
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Where the indicator changes color is the
endpoint.
Not always at the equivalence point.
A 50.00 mL sample of aqueous Ca(OH)2
requires 34.66 mL of 0.0980 M Nitric acid
for neutralization. What is [Ca(OH)2 ] ?
# of H+1 = # of OH-1
Titration Solved Problem
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50.00 mL sample of aqueous Ca(OH)2 requires
34.66 mL of 0.0980 M Nitric acid for
neutralization. What is [Ca(OH)2 ] ?
Ca(OH)2 + 2 HNO3  Ca(NO3)2 + H2O
.098 = n / .03466 n = .0034 moles H+1
Ca(OH)2  Ca+2 + 2 OH-1
.0017
.0034
Acid – Base Reaction
This is an important problem
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75 mL of 0.25 M H2SO4 is mixed with 225 mL
of 0.055 M NaOH . What is the
concentration of the excess H+1 or OH-1 ?
Think about what you are being asked to
solve and about the process that is going on.
In other words, don’t just dive in before you
plan the problem.
Acid – Base Reaction

First lets get a balanced equation.
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
Notice this follows the general form that we
talked about a few slides back.
Acid + Base  Salt + Water
Acid – Base Reaction

We need to determine moles of each
reactant.
H2SO4: M = n / L .25 = n / .075 n = .0188 n

NaOH: same format

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.0124 moles
We need to set up a proportion to determine
which species is the Limiting Reagent and
which is in excess.
Acid – Base Reaction
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1 n H2SO4 / 2 n NaOH : x / .0124 n NaOH
Solving for x gives .0062 n of H2SO4
When we compare that to the given we see
that we have a lot more of the acid than
what is needed so we can conclude that
H2SO4 is in excess and the NaOH is the
limiting reagent.
So what is the next step ?
Acid – Base Reaction
Since we are given .0188 n of H2SO4 and
only need .0062 moles we are left with
.0188 - .0062 = .0126 n of H2SO4 in excess

However, each molecule of H2SO4 liberates 2
ions of H+1 so we need to factor that in.
H2SO4  2 H+1 + SO4-2
.0126
.0252 .0126
Acid – Base Reaction
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This is a limiting reagent problem as well as
an acid base problem.
After the reaction takes place and accounting
for the dissociation of the acid we have the
following: .0252 n of H+1
But we were asked for concentration.
Clearly we need to use M = n / L , but there
is a catch. Solve for M in your binder.
Acid – Base Reaction
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We need moles and we just solved for that
and got .0252 moles of H+1 .
What about the volume ?
Both reagents are in the solution together, so
the volumes need to be combined. The total
volume is 75 + 225 mL.
M = .0252 / .3  M = .084 M H+1
Forgetting to add the volumes together is a
common mistake in these types of problems.
Types of Reaction - Redox
Oxidation-Reduction called Redox
 Ionic compounds are formed through the
transfer of electrons.
 An Oxidation – Reduction reaction involves
the transfer of electrons.
 We need a way of keeping track.
Oxidation States
A way of keeping track of the electrons.
 You need to memorize the rules for
assigning oxidation states.
 The oxidation state of elements in their
standard states is zero.
 Oxidation state for monoatomic ions are the
same as their charge.

Oxidation states
 Oxygen is assigned an oxidation state of -2 in its
covalent compounds except as a peroxide.
 In compounds with nonmetals hydrogen is
assigned the oxidation state +1.
 In its compounds fluorine is always –1.
6 The sum of the oxidation states must be zero in
compounds or equal the charge of the ion.
Oxidation States
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
Assign the oxidation states to each
element in the following.
CO2
NO3-1
H2SO4
Fe2O3
Oxidation States

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Assign the oxidation states to each
element in the following.
CO2
C=+4
O = -2
NO3-1
N=+5
O = -2
H2SO4 H = + 1 S = + 6 O = -2
Fe2O3
Fe = + 3 O = -2
Oxidation-Reduction
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
Transfer electrons, so the oxidation states
change.
2 Na + Cl2  2 NaCl
CH4 + 2 O2  CO2 + 2 H2O
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
OIL RIG or LEO GER or OLE
Oxidation-Reduction

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Oxidation means an increase in oxidation
state - lose electrons.
Reduction means a decrease in oxidation
state - gain electrons.
The substance that is oxidized is called the
reducing agent.
The substance that is reduced is called the
oxidizing agent.
Redox Reactions
Redox Vocabulary

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
Oxidizing agent gets reduced.
Gains electrons.
More negative oxidation state.
Reducing agent gets oxidized.
Loses electrons.
More positive oxidation state.
Both have to happen in a reaction for it to
occur.
Identify the following


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Oxidizing agent & reducing agent
Substance oxidized & Substance reduced
for the following reactions
Fe (s) + O2(g)  Fe2O3(s)
Fe2O3(s)+ 3 CO(g)  2 Fe(s ) + 3 CO2(g)
SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2
Answers

Fe (s) + O2(g)  Fe2O3(s)
Iron is oxidized, Oxygen is reduced
Iron is the reducing agent, Oxygen is the oxidizing agent

Fe2O3(s)+ 3 CO(g)  2 Fe(s ) + 3 CO2(g)
Fe2O3 is reduced, CO is oxidized
Fe2O3 is the oxidizing agent, CO is the reducing agent

SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2
S is oxidized, Mn is reduced
S is the reducing agent, Mn is the oxidizing agent
Balancing Half-Reactions is being
de-emphasized for the AP Test.
(But it is good to know the theory behind it)


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
All redox reactions can be thought of as
happening in two halves.
One produces electrons - Oxidation half.
The other requires electrons - Reduction
half.
Write the half reactions for the following.
Na + Cl2  Na+1 + Cl-1
SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2
Balancing Half-reactions

Na + Cl2  Na+1 + Cl-1
Na  Na+1 + e-1 (oxidation ½ reaction)
Cl2 + 2 e-1  2 Cl-1 (reduction ½ reaction)



SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2
SO3-2 + H2O  SO4-2 + 2 H+1 + 2 e-1
8 H+1 + MnO4-1 + 5 e-1  Mn+2 + 4 H2O
Redox Theory

You need to remember that while these
are studied separately, the oxidation halfreaction and the reduction half-reaction
must always happen together. You can’t
have a species gaining electrons unless
you have a species there to lose those
same electrons.
Balancing Redox Equations
In aqueous solutions the key is the number
of electrons produced must be the same as
those required.
 For reactions in acidic solution an 8 step
procedure.
 Write separate half reactions
 For each half reaction balance all reactants
except H and O
 Balance O using H2O

Acidic Solution
 Balance H using H+1
 Balance charge using e-1
 Multiply equations to make electrons
balance.
 Add equations and cancel identical species.
 Check that charges and elements are
balanced.
Summary of Rules
in Acidic Medium

Balance the atoms in each half-reaction.




This refers to atoms other than H and O
Balance the Oxygen by adding H2O where
needed.
Balance the Hydrogen by adding H+1 where
needed.
Balance by charge by adding e-1
Balancing in Acidic Medium
Problem # 1

Fe (s) + O2(g)  Fe2O3(s)
Determine the charge on each species

0
0
+3 -2
Fe (s) + O2(g)  Fe2O3(s)
So: Fe  Fe+3 (Fe is oxidized)
&
O2  O-2 (O2 is reduced)

Balancing in Acidic Medium
Problem # 1
Fe  Fe+3 (Fe is oxidized)
Fe  Fe+3 + 3 e-1
O2  O-2 (O2 is reduced)
O2  2 O-2
4 e-1 + O2  2 O-2
Balancing in Acidic Medium
Problem # 1


The final step is to combine the two ½
reactions.
But first we must get rid of the electrons
4 ( Fe  Fe+3 + 3 e-1 )
3 ( 4 e-1 + O2  2 O-2 )
We get: 4 Fe + 3 O2  4 Fe+3 + 6 O-2
Balancing in Acidic Medium
Problem # 2

Fe2O3(s)+ 3 CO(g)  2 Fe(s ) + 3 CO2(g)
+3





-2
+2 -2
0
+4 -2
Fe2O3(s)+ 3 CO(g)  2 Fe(s ) + 3 CO2(g)
Fe+3 + 3 e-1  Fe (Fe is reduced)
a. CO  CO2
b. H2O + CO  CO2
c. H2O + CO  CO2 + 2 H+1
d. H2O + CO  CO2 + 2 H+1 + 2 e-1
Balancing in Acidic Medium
Problem # 2






Fe+3 + 3 e-1  Fe (Fe is reduced)
The CO is oxidized. Here are the steps.
Atoms: CO  CO2
Oxygen (with water) H2O + CO  CO2
Hydrogen (with H+1) H2O + CO  CO2 + 2 H+1
Charge (with e-1)
H2O + CO  CO2 + 2 H+1 + 2 e-1
Balancing in Acidic Medium
Problem # 2




Once again, the final step is to factor out the
electrons and combine the two ½ reactions.
2 ( Fe+3 + 3 e-1  Fe )
3 (H2O + CO  CO2 + 2 H+1 + 2 e-1 )
2 Fe+3 + 3 H2O + 3 CO  2 Fe +3 CO2 + 6 H+1
Balancing in Acidic Medium
Problem # 3

SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2

+4 -2

SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2

+1
+7 -2
+6 -2
+1 -2
+2
We can see that the S is being oxidized and
the Mn is being reduced.
Balancing in Acidic Medium
Problem # 3
Oxidation ½ reaction







+4 -2
+1
+7 -2
+6 -2
+1 -2
+2
SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2
a. SO3-2  SO4-2
b. H2O + SO3-2  SO4-2
c. H2O + SO3-2  SO4-2 + 2 H+1
d. H2O + SO3-2  SO4-2 + 2 H+1 + 2 e-1
Sulfur is oxidized. How can you tell ?
Balancing in Acidic Medium
Problem # 3
Reduction ½ reaction

+4 -2

SO3-2 + H+1 + MnO4-1  SO4-2 + H2O + Mn+2





+1
+7 -2
+6 -2
+1 -2
+2
a. MnO4-1  Mn+2
b. MnO4-1  Mn+2 + 4 H2O
c. 8 H+1 + MnO4-1  Mn+2 + 4 H2O
d. 5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O
Factor out the electrons and then combine the
two half – reactions.
Balancing in Acidic Medium
Problem # 3



H2O + SO3-2  SO4-2 + 2 H+1 + 2 e-1
5 e-1 + 8 H+1 + MnO4-1  Mn+2 + 4 H2O
Multiply the first ½ rxn. By 5 and the second
½ rxn. by 2 to factor out the e-1.

5 H2O + 5 SO3-2  5 SO4-2 + 10 H+1 + 10 e-1
10 e-1 + 16 H+1 + 2 MnO4-1  2 Mn+2 + 8 H2O

We can factor out species other than e-1

Balancing in Acidic Medium
Problem # 3








5 H2O + 5 SO3-2  5 SO4-2 + 10 H+1 + 10 e-1
10 e-1 + 16 H+1 + 2 MnO4-1  2 Mn+2 + 8 H2O
We can start with the water and electrons.
5 H2O + 5 SO3-2  5 SO4-2 + 10 H+1 + 10 e-1
10 e-1 + 16 H+1 + 2 MnO4-1  2 Mn+2 + 8 3 H2O
Next we can factor out the H+1 ions.
5 SO3-2  5 SO4-2 + 10 H+1
16 6 H+1 + 2 MnO4-1  2 Mn+2 + 3 H2O
Balancing in Acidic Medium
Problem # 3

5 SO3-2  5 SO4-2
6 H+1 + 2 MnO4-1  2 Mn+2 + 3 H2O

Combining the two gives us:


6 H+1 + 2 MnO4-1 + 5 SO3-2 
2 Mn+2 + 3 H2O + 5 SO4-2
Basic Solution




Do everything you would with acid, but add
one more step.
Add enough OH-1 to both sides to neutralize
the H+1
Determine the species undergoing redox in
the problems below.
CrI3 + Cl2  CrO4-1 + IO4-1 + Cl-1
Redox – Basic Solutions




CrI3 + Cl2  CrO4-1 + IO4-1 + Cl-1
Cr+3  Cr+7 (CrO4-1) Cr is oxidized
I-1  I+7
I is oxidized
Cl2  Cl-1
Cl is reduced
Practice





Cr(OH)3 + OCl-1 + OH-1  CrO4-2 + Cl-1 + H2O
MnO -1 + Fe+2 Mn+2 + Fe+3
4
Cu + NO3-1  Cu+2 + NO(g)
Pb + PbO2 + SO4-2  PbSO4
Mn+2 + NaBiO3  Bi+3 + MnO4-1
Balancing in Basic Medium
Problem # 1





Mn+2 + NaBiO3  Bi+3 + MnO4-1
You must know the procedure to do this or
any problem.
Get the oxidation states.
Mn+2 + NaBiO3  Bi+3 + MnO4-1
+2
+1 +5 -2
+3
+7
-2
Balancing in Basic Medium
Problem # 1

Mn+2 + NaBiO3  Bi+3 + MnO4-1

+2





+1 +5 -2
+3
+7
-2
Do the ½ reactions
Mn+2  MnO4-1
4 H2O + Mn+2  MnO4-1 + 8 H+1
8 OH-1 + 4 H2O + Mn+2  MnO4-1 + 8 H2O
Notice how the OH-1 has been added above
8 OH-1 + 4 H2O + Mn+2  MnO4-1 + 8 H2O + 5 e-1
Balancing in Basic Medium
Problem # 1

BiO3-1  Bi+3
6 H+1 + BiO3-1  Bi+3 + 3 H2O
6 H2O + BiO3-1  Bi+3 + 3 H2O + 6 OH-1

2 e-1 + 6 H2O + BiO3-1  Bi+3 + 3 H2O + 6 OH-1


Balancing in Basic Medium
Problem # 1






We now combine the two ½ reactions.
8 OH-1 + 4 H2O + Mn+2  MnO4-1 + 8 H2O + 5 e-1
2 e-1 + 6 H2O + BiO3-1  Bi+3 + 3 H2O + 6 OH-1
We need to take care of the electrons.
16 OH-1 + 8 H2O + 2 Mn+2  2 MnO4-1 + 16 H2O
30 H2O + 5 BiO3-1  5 Bi+3 + 15 H2O + 30 OH-1
Balancing in Basic Medium
Problem # 1


16 OH-1 + 8 H2O + 2 Mn+2  2 MnO4-1 + 16 H2O
30 H2O + 5 BiO3-1  5 Bi+3 + 15 H2O + 30 OH-1
Combine
 16 OH-1 + 38 H2O + 2 Mn+2 + 5 BiO3-1 
2 MnO4-1 + 31 H2O + 5 Bi+3 30 OH-1
Reduce

7 H2O + 2 Mn+2 + 5 BiO3-1  2 MnO4-1 + 5 Bi+3 + 14 OH-1
Redox Titrations



Same as any other titration.
The permanganate ion is used often
because it is its own indicator. MnO4-1 is
purple, Mn+2 is colorless. When reaction
solution remains clear, MnO4-1 is gone.
Chromate ion is also useful, but color
change, orange yellow to green, is harder
to detect.
Example

The iron content of iron ore can be
determined by titration with standardized
KMnO4. The iron ore is dissolved in HCl, and
the iron reduced to Fe+2 ions. This solution
is then titrated with KMnO4 soln., producing
Fe+3 and Mn+2 ions in acidic solution. If it
requires 41.95 mL of 0.205 M KMnO4 to
titrate a solution made with 0.6128 g of iron
ore, what percent of the ore was iron ?