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Chapter 4
The Major Classes of Chemical Reactions
4-1
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Figure 4.1
4-2
Electron distribution in molecules of H2 and H2O.
A. In H2, the identical nuclei
attract e-s equally. B. In H2O,
the O nucleus attracts the shared
e-s more strongly than the H
nuclei.
C. Model shows the polar covalent
bond from plus to neg. D. The
vector addition for the overall
molecular dipole moment.
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AQUEOUS SOLUTIONS
WATER: a polar molecule which can dissolve
many compounds (what kind of compounds?)
Process is called hydration
Solubility: amount of solute dissolved in
specified amount of solvent
NATURE OF AQUEOUS SOLUTIONS:
Hydrated particles in water
Hydrated means surrounded by water molecules
4-3
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DEFINITIONS:
Electrolytic solutions:
- can conduct electricity (current)
- have charged ions that can move
Strong acid/base: stronger electrolytes
Weak acid/base: weaker electrolytes
Salts: strong electrolytes (if they dissolve at
all!)
Covalent solutes do not have charges and
therefore are nonelectrolytes
4-4
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• See informational handout and lecture
outline for table called “Types of Aqueous
Solutions”
4-5
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DEFINITIONS:
STRONG ACIDS: dissociate 100%
Memorize formulas and names of these: HCl,
HBr, HI, HNO3, H2SO4, HCl04
WEAK ACIDS: all others, including organic
acids (-COOH). Memorize acetic (HC2H3O2 or
CH3COOH), phosphoric (H3PO4), hydrofluoric
(HF), and carbonic (H2CO3), among others.
STRONG BASES: dissociate 100%
Group IA and IIA hydroxides and oxides
WEAK BASES:
Ammonia (NH3) and all organic bases (amines)
are weak. Know methanamine CH3NH2.
4-6
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The dissolution of an ionic compound.
Figure 4.2
4-7
Particles of the ionic cmpd are attracted to water, separate from the
crystal and become surrounded by H2O molecules. Notice orientation of
H2O molecules to the cation vs the anion.
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CH3COOH
dissociates < 10%.
HCl dissociates
completely.
4-8
BaCl2 dissociates
completely.
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Figure 4.3
4-9
The electrical conductivity of ionic solutions.
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ACIDS & BASES
Acids produce H+ (H3O+) in aqueous solutions
Strong: HX(aq) + H2O(l)  H3O+(aq) + X-(aq)
Weak: HX(aq) + H2O(l)  H3O+(aq) + X-(aq)
Bases produce OH- in aqueous solutions
Strong: MOH(s)  M+(aq) + OH-(aq)
Weak: NR3(aq) + H2O(l)  NR3H+(aq)+OH-(aq)
H+ is also the hydrated proton H3O+
R can be a H or a hydrocarbon group
4-10
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Figure 4.4
from 4th
ed.
4-11
The hydrated proton.
In water, the H+ ion attaches to a water molecule immediately, making
the H3O+ ion, called the hydronium ion.
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Finding [H+] in Strong Acid
Solutions
Strong acids dissociate 100%, so determine
[H+] using solution inventory
What is [H+] in 1.1 M HNO3?
It is 1.1 M
See sample problem 4.4 and follow-up
problem 4.4 in 2nd ed.
How many moles of H+(aq) are in 451 mL of
3.20 M hydrobromic acid?
4-12
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Writing Equations for Aqueous Ionic Reactions
The regular equation (not really all molecules)
shows all of the reactants and products as intact, undissociated
compounds.
The total ionic equation
shows all of the soluble ionic substances (salts, SA, SB) dissociated
into ions.
The net ionic equation
eliminates the spectator ions and shows the actual
chemical change(s) taking place.
4-13
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How do you know a chemical reaction
has occurred?
Five indications:
gas formed (see bubbles or smell odor)
precipitate formed
temperature changed
color changed
chemical indicator color changed (pH)
Video demo of the first four indications:
4-14
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Gas-forming reactions can be found in
more than one type of reaction
You may see gas-forming reactions in
decompostion, single replacement and double
replacement reactions (we don’t include
combustion since gases always form in those)
MEMORIZE: H2SO3, H2CO3, NH4OH going to gas
and water in equilibrium
H2SO3(aq) ↔ H2O(l) + SO2(g)
H2CO3(aq) ↔ H2O(l) + CO2(g)
NH4OH(aq) ↔ H2O(l) + NH3(g)
4-15
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TYPES OF REACTIONS:
double replacement has three types
Double replacement reactions can be precipitation,
neutralization (acid-base), or gas-forming
PRECIPITATION REACTIONS AND SOLUBILITY:
STRONG interparticle forces affect whether a
compound is soluble in water
Salts (ionic solids) that are INSOLUBLE "stay
together" because they have a stronger
attraction to each other than to the surrounding
water molecules
4-16
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Figure 4.4
Regular
4-17
A precipitation reaction and its equation.
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Figure 4.6
The reaction of Pb(NO3)2 and NaI.
Unbalanced equation:
NaI(aq) + Pb(NO3)2 (aq)
PbI2(s) + NaNO3(aq)
Regular 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:
2 I-(aq) + Pb2+ (aq)
PbI2(s)
Double replacement reaction: precipitation
When aqueous solutions of these ionic compounds are mixed,
the yellow solid PbI2 forms.
4-18
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Predicting Whether a Precipitate Will Form
1. Note the ions present in the reactants.
2. Consider the possible cation-anion combinations.
3. Decide whether any of the ion combinations is insoluble.
See Table 4.1 (next slide) for solubility rules.
4-19
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Table 4.1 Solubility Rules for Ionic Compounds in Water (plus
see back of large periodic table)
Soluble Ionic Compounds
1. All common compounds of Group 1A(1) ions (Li+, Na+, K+, etc.) and
ammonium ion (NH4+) are soluble.
2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2-) and most
perchlorates (ClO4-) are soluble.
3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble,
except those of Ag+, Pb2+, Cu+, and Hg22+.
4. All common sulfates (SO42-) are soluble except those of Ca2+, Sr2+,
Ba2+, Ag+, and Pb2+.
Poorly soluble Ionic Compounds
1. All common metal hydroxides are insoluble, except those of Group 1A(1)
and the larger members of Group 2A(2)(beginning with Ca2+).
2. All common carbonates (CO32-) and phosphates (PO43-) are insoluble,
except those of Group 1A(1) and NH4+.
4-20
3. All common sulfides are insoluble except those of Group 1A(1), Group
2A(2) and NH4+.
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PRACTICE: show physical states as
well as balancing the following
1. __NaNO3(aq) + __BaBr2(aq)  ____________
2. __K2SO4(aq) + __Pb(NO3)2(aq)___________
3. __NaOH(aq) + __FeCl3(aq) ________________
Then go back and write the net ionic equations for
each (if there was really a reaction)
4-21
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Stoichiometry with Double
Replacement Ppt Reactions
If you have the balanced chemical equation
and some knowledge about the reactants,
you can do some stoichiometry!
4-22
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Precipitation Reaction and
Stoichiometry Practice
EXAMPLE: What is the concentration of a silver nitrate solution if it
takes 33.31 mL of 0.2212 M NaCl to titrate 25.00mL of silver nitrate
in solution?
Step 1: NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
33.31mL 25.00mL
0.2212M
??M
STEP 2: 33.31 mL*1L/103 mL*.2212 mol/L
= 7.368x10-3 mol NaCl
STEP 3: 7.368x10-3 mol NaCl * 1 mol AgNO3/1 mol NaCl
= SAME
STEP 4: 7.368x10-3 mol AgNO3/0.02500 L = .2947 mol/L
(Could also find mass of AgCl present: moles*169.873 g/mol=0.7948 g
4-23
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TYPES OF REACTIONS: double replacement
reactions include neutralization (acid/base)
Acid-base titration (and precipitation)
reactions are used for:
Volumetric analysis and/or
gravimetric analysis
4-24
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ACID-BASE REACTIONS
MEMORIZE: ACID IS PROTON DONOR AND
BASE IS PROTON ACCEPTOR (H+ is proton)
TITRATIONS: volumetric quantitative analysis
using a standardized solution to find out
something about another solution
Uses burettes containing titrant to analyze the
analyte in a flask or beaker; titrant is added to
either the endpoint (indicator) or the
equivalence point (exact stoichiometric point)
4-25
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ACID-BASE REACTIONS
Must know:
- balanced chemical equation,
- endpoint color or indication,
- volume of titrant,
- “something” about analyte
4-26
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Figure 4.7
Start of titration
Excess of acid
4-27
An acid-base titration.
Point of
neutralization
Slight excess of
base
Phenolphthalein indicator changes around pH of 8; want
very pale pink color.
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Figure 4.8
An aqueous strong acid-strong base reaction on the atomic scale.
The net ionic equation for all SA/SB reactions is:
4-28
H3O+(aq) + OH-(aq)  2 H2O(l)
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STANDARDIZATION OF A
SOLUTION
Two methods – titrate it using a pure dry solid primary standard or
using a purchased aqueous primary standard
Method 1 Example: A student used 0.250 grams of dried solid
sodium carbonate to determine exact concentration of a solution of
hydrochloric acid. (NOTE: this is also a gas-producing reaction.) It
took 25.76 mL of the acid to reach "endpoint" - where the chemical
indicator changed color indicating a pH change.
Na2CO3(s) + 2 HCl(aq)  2 NaCl(aq) + H2O(l) + CO2(g)
Moles base = 0.250 g/(106.0 g/mol) = 0.00236 mol
Moles acid = mol base (2 HCl/1 Na2CO3) = 0.00472 mol
Molarity of acid = mol/vol = 0.00472 mol/0.02576 L
= 0.183 M
4-29
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STANDARDIZATION OF A
SOLUTION
Method 2 Example: A student used a 0.100 M HCl
primary standard solution to standardize a NaOH
solution. Exactly 25.00 mL of the base took 32.56 mL of
acid to reach the endpoint.
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Moles acid = 0.100 mol/L x 0.03256 L
= 0.003256 mol HCl
Moles base = mol acid (1/1) = 0.003256 mol NaOH
Molarity of base = mol/vol = 0.003256 mol/0.02500 L
= 0.130 M NaOH
4-30
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ACID-BASE TITRATIONS
In acid-base reactions, the Endpoint is
FOUND with chemical indicator.
EQUIVALENCE point is when # of H+ = # of
OH-. Usually found with a graph of pH vs.
volume
4-31
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Figure 4.10
from 4th ed.
An acid-base reaction that forms a gaseous product.
Regular equation
NaHCO3(aq) + CH3COOH(aq)
CH3COONa(aq) + CO2(g) + H2O(l)
Total ionic equation
Na+(aq)+ HCO3-(aq) + CH3COOH(aq)
CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)
Net ionic equation
HCO3-(aq) + CH3COOH(aq)
CH3COO-(aq) + CO2(g) + H2O(l)
4-32
Note that the acetic acid remains unionized! It is a
weak acid and dissociates < 10%.
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Acids, bases and pH
NOTE – NOT IN CHP 4 BUT YOU HAVE TO
KNOW IT!
pH = - log [H3O+]
pOH = - log [OH-]
[H3O+] = 10-pH
[OH-] = 10-pOH
pH scale from 0 to 14
0 – 7 acidic, 7 = neutral, 7-14 basic
pH + pOH = 14
4-33
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Acids, bases and pH
We can calculate pH based on concentration for
strong acids and strong bases (WA/WB require
equilibrium calcs)
A 0.015 M HCl solution will have what pH?
Do solution inventory to find [H+], which is
0.015 M.
Take the log and change the sign:
pH = - log (0.015) =1.82
Decimal places in pH = # of sig figs in
concentration
4-34
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Acids, bases and pH
Practice: Find the pH of a 0.00050 M NaOH
solution.
First note that this is a base, therefore the pH
should be above 7.
(Yes, pH and pOH will be on the quiz and the test)
4-35
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OXIDATION-REDUCTION
OXIDATION STATE OR NUMBER:
designated number of electrons an atom
has lost, gained or used to make a
compound or polyatomic ion.
NOT THE SAME AS CHARGE!!
4-36
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Table 4.3 Rules for Assigning an Oxidation Number (O.N.)
General rules
1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 0
2. For a monoatomic ion: O.N. = ion charge
3. The sum of O.N. values for the atoms in a compound equals zero. The
sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge.
Rules for specific atoms or periodic table groups
1. For Group 1A(1):
O.N. = +1 in all compounds
2. For Group 2A(2):
O.N. = +2 in all compounds
3. For hydrogen:
O.N. = +1 with nonmetals (-1 metals)
4. For fluorine:
O.N. = -1 always in all compounds
5. For oxygen:
O.N. = -1 in peroxides (very rare)
O.N. = -2 in all other compounds(except with F)
O.N. = -1 with metals, nonmetals (except O), and
other halogens lower in the group
6. For Group 7A(17):
4-37
In a polyatomic ion, the more electronegative element will have a
negative oxidation number.
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REVIEW WITH SOME EXAMPLES:
S8
Al2O3
MnO4-
H2O2
CH4
C2H6
C6H6
H3PO4
MgO
BaF2
Sr(ClO4)2
N2O
PCl3
SO42-
SO32-
SO3(g)
Hg(NO3)2
NaHCO3
Look at Dry Lab III in your lab manual. We will be doing most
of those during a lab session.
4-38
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Figure 4.10
Highest and lowest
oxidation numbers of
reactive main-group
elements. (except F
always -1; heavy metals
in 3A can be +5)
4-39
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OXIDATION-REDUCTION
Single Replacement reactions: ALL are redox
A + BC  B + AC
Practice: write the copper metal and silver nitrate
reaction, and determine oxidation states. Also
write the net ionic equation.
What is happening to copper?
Write its half-reaction.
What is happening to silver?
Write its half-reaction.
4-40
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OXIDATION-REDUCTION
Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
0
+1,+5,-2
The net ionic equation is:
0
+2,+5,-2
Cu(s) + 2 Ag+(aq)  2 Ag(s) + Cu2+(aq)
What is happening to copper? It is losing e-s
Write its half-reaction: Cu  Cu2+ + 2eWhat is happening to silver? It is gaining an eWrite its half-reaction: Ag+ + e-  Ag
4-41
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OXIDATION-REDUCTION
REDOX REACTIONS INVOLVE THE TRANSFER OF
ELECTRONS
How do you know if electrons are lost or gained, if
an atom has been oxidized vs. reduced?
Must see a change in oxidation numbers from
reactants to products
LEO says GER (Loss of Electrons is Oxidation; Gain
of Electrons is Reduction)
4-42
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Figure 4.9
The redox process in compound formation.
A, In forming the ionic compound MgO, each Mg atom transfers two electrons to each O atom. (Note that atoms
become smaller when they lose electrons and larger when they gain electrons.) The resulting Mg 2+ and O2- ions
aggregate with many others to form an ionic solid. B, In the reactants H2 and Cl2, the electron pairs are shared
equally (indicated by even electron density shading). In the covalent product HCl, Cl attracts the shared electrons
more strongly than H does. In effect, the H electron shifts toward Cl, as shown by higher electron density (red) near
4-43
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Figure 4.11
A summary of terminology for oxidation-reduction
(redox) reactions.
e-
X
4-44
transfer
Y
or shift of
electrons
X loses electron(s)
Y gains electron(s)
X is oxidized
Y is reduced
X is the reducing agent
Y is the oxidizing agent
X increases its
oxidation number
Y decreases its
oxidation number
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OXIDATION-REDUCTION
If one species is oxidized, another must be reduced
The oxidizing agent causes oxidation, therefore it is
reduced.
Vice versa, the reducing agent causes reduction
therefore it is oxidized.
Must know if oxidation numbers change to determine if it
is a redox reaction.
Most double replacement reactions are not redox
Therefore always do oxid numbers for all species in the
equation before proceeding to do anything else with a
problem.
4-45
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Sample Problem 4.8
Recognizing Redox Reactions
PROBLEM: Identify if each of the following is a redox reaction or not:
(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)
Assign an O.N. for each atom and see which atom gained and which atom
lost electrons in going from reactants to products.
An increase in O.N. means the species was oxidized (and is the reducing
agent) and a decrease in O.N. means the species was reduced (is the
oxidizing agent).
SOLUTION:
0
+1 +6 -2
(a) 2Al(s) + 3H2SO4(aq)
+3 +6 -2
0
Al2(SO4)3(aq) + 3H2(g)
The O.N. of Al increases; it is oxidized; it is the reducing agent.
The O.N. of H decreases; it is reduced; H2SO4 is the oxidizing agent.
4-46
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Sample Problem 4.7
continued
+2 -2
Recognizing Redox Reactions
+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
(c) 2H2(g) + O2(g)
+1 -2
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.
4-47
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Figure 4.14
from 4th ed.
4-48
A redox titration.
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Sample Problem 4.9
from 4th ed.
PROBLEM:
Finding an Unknown Concentration by a Redox
Titration
To measure the Ca2+ concentration, 1.00mL of human blood was
treated with Na2C2O4 solution. The resulting CaC2O4 precipitate
was filtered and dissolved in dilute H2SO4. This solution required
2.05mL of 4.88x10-4M KMnO4 to reach the end point. The
unbalanced equation is:
KMnO4(aq) + CaC2O4(s) + H2SO4(aq)
MnSO4(aq) + K2SO4(aq) + CaSO4(s) + CO2(g) + H2O(l)
(a) Balance the equation.* Calculate the amount (mol) of Ca2+.
(b) Calculate the amount (mol) of Ca2+ ion concentration expressed
in units of mg Ca2+/100mL blood.
PLAN:
(a)
volume of KMnO4 soln
multiply by M
mol of KMnO4
mol of CaC2O4
molar ratio
4-49
mol of Ca2+
ratio of elements in formula
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Sample Problem 4.9
Finding an Unknown Concentration by a Redox
Titration
continued
2.05mL soln
SOLUTION:
L
4.88x10-4mol KMnO4
103 mL
1.00x10-6mol KMnO4
= 1.00x10-6mol KMnO4
L
5mol CaC2O4
= 2.50x10-6 mol CaC2O4
2mol KMnO4
2.50x10-6 mol CaC2O4
1mol Ca2+
= 2.50x10-6 mol Ca2+
1mol CaC2O4
PLAN:
(b)
mol Ca2+/1mL blood
multiply by 100
mol Ca2+/100mL blood
multiply by M
g Ca2+/100mL blood
SOLUTION:
2.50x10-6 mol Ca2+ x100 =2.50x10-4 mol Ca2+
1mL blood
100mL blood
2.50x10-4 mol Ca2+ 40.08g Ca2+ mg
100mL blood
mol Ca2+
10-3g
10-3g = 1mg
=10.0mg Ca2+/100mL blood
mg Ca2+/100mL blood
4-50
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Figure 4.15 Combining elements to form an ionic compound.
from 4th ed.
4-51
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Figure 4.16
from 4th ed.
4-52
Decomposing a compound to its elements.
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Figure 4.12
An active metal displacing hydrogen from water.
Lithium displaces hydrogen from water in a vigorous reaction that yields an aqueous solution of lithium
hydroxide and hydrogen gas, as shown on the macroscopic scale (top), at the atomic scale (middle),
and as a balanced equation (bottom). (For clarity, the atomic-scale view of water has been
greatly simplified, and only water molecules involved in the reaction are
colored red and blue.)
4-53
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Figure 4.18
from 4th ed.
The displacement of H from acid by nickel.
O.N. increasing
O.N. decreasing
oxidation occurring
reduction occurring
reducing agent
oxidizing agent
0
+1
Ni(s) + 2H+(aq)
4-54
+2
0
Ni2+(aq) + H2(g)
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Figure 4.13
4-55
Displacing one metal with another.
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Figure 4.14
The activity series
This list of metals (and H2)
is arranged with the most
active metal (strongest
reducing agent) at the top
and the least active metal
(weakest reducing agent) at
the bottom. The four metals
below H2 cannot displace it
from any source.
4-56
strength as reducing agents
of the metals.
Li
K
Ba
Ca
Na
Mg
Al
Mn
An
Cr
Fe
Cd
Co
Ni
Sn
Pb
H2
Cu
Hg
Ag
Au
can displace H
from water
can displace H
from steam
can displace H
from acid
cannot displace H
from any source
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Practice Using the Activity Series
Decide if these reactions will occur, then
complete and balance them:
___Mg(s) + ___H2O(l) 
___Fe(s) + ___HCl(aq) 
___Al(s) + ___MnSO4(aq) 
___Cu(s) + ___MgCl2(aq)
4-57
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Sample Problem 4.8
PROBLEM:
Identifying the Type of Redox Reaction
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)
(b) hydrogen peroxide(l)
magnesium nitride (aq)
water(l) + oxygen gas
(c) aluminum(s) + lead(II) nitrate(aq)
aluminum nitrate(aq) + lead(s)
PLAN: Combination reactions produce fewer products than reactants.
Decomposition reactions produce more products than reactants.
Displacement reactions have the same number of products and reactants.
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continued
0
(a) Combination
3Mg(s) +
0
+2 -3
N2(g)
Mg3N2 (aq)
Mg is the reducing agent; N2 is the oxidizing agent.
+1 -1
(b) Decomposition
+1 -2
0
H2O2(l)
H2O(l) + 1/2 O2(g)
2 H2O2(l)
2 H2O(l) +
or
O2(g)
H2O2 is the oxidizing and reducing agent.
(c) Displacement
0
+2 +5 -2
Al(s) + Pb(NO3)2(aq)
2Al(s) + 3Pb(NO3)2(aq)
+3 +5 -2
Al(NO3)3(aq) + Pb(s)
2Al(NO3)3(aq) + 3Pb(s)
Pb(NO3)2 is the oxidizing and Al is the reducing agent.
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0
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PROCESS FOR BALANCING REDOX REACTIONS*
BY THE HALF REACTION METHOD IN ACID
METHOD (different from your textbook, but this
always works)
1. Assign oxidation numbers to all atoms in the
reaction and determine which species has been
oxidized and which has been reduced. Separate
these species into two starting half-reactions.
Determine the number of electrons transferred.
2. For each half-reaction:
a. Balance all atoms other than O and H.
b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding electrons.
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*From chp 21, but required in Chem 1.
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PROCESS FOR BALANCING REDOX REACTIONS BY
THE HALF REACTION METHOD IN ACID METHOD
(different from your textbook, but this always
works)
3. Make electrons lost equal to electrons
gained by using least common
denominator on the two half reactions.
4. Add the two half-reactions, canceling
electrons and other common species that
appear on both sides.
5. Check the charge balance.
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Practice redox with acidic solution
___Cr2O72- + ___CH3CH2OH  ___ Cr3+ + ___CH3COOH
Step 1 Reactants’ oxidation numbers: Cr is +6, O -2,
H+1, O-2, C -2. Products’ on’s: Cr is +3, H+1, O -2, C
is 0. Cr is reduced from +6 to +3 with a gain of 3 e-/Cr,
C is oxidized from -2 to 0 with a loss of 2 e-/C. The
initial half-rxns are:
Red: Cr2O72-  Cr3+ and Oxid: CH3CH2OH  CH3COOH
Step 2a, the red ½ rxn needs a 2 in front of Cr3+.
Step 2b, add H2O to balance O.
Cr2O72-  2 Cr3+ + 7 H2O; CH3CH2OH + H2O  CH3COOH
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Practice redox with acidic
solution
Step 2c, add H+ to balance H.
14 H+ + Cr2O72-  2 Cr3+ + 7 H2O;
CH3CH2OH + H2O  CH3COOH + 4 H+
Step 2 d, add e-s you determined in step 1, to
balance electrical charge.
6 e- + 14 H+ + Cr2O72-  2 Cr3+ + 7 H2O;
CH3CH2OH + H2O  CH3COOH + 4 H+ + 4 e-
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Practice redox with acidic
solution
Step 3 make e-s equal
2x (6 e- + 14 H+ Cr2O72-  2 Cr3+ + 7 H2O)
= 12 e- + 28 H+ 2 Cr2O72-  4 Cr3+ + 14 H2O
3x (CH3CH2OH + H2O  CH3COOH + 4 H+ + 4 e-)
= 3 CH3CH2OH + 3H2O  3 CH3COOH + 12 H+ + 12 e-
Step 4: cancel the 12 e-s, 12 of the H+ and 3 of the H2O,
leaving:
16H+ + 2Cr2O72- + 3CH3CH2OH  4Cr3+ + 3CH3COOH + 11 H2O
Charge balance: +16 -4 = +12; 4(+3) = + 12. Charges equal on both
sides.
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Half-reaction Method in Base
Go thru same procedure as acid, but add OHequal to the number of H+ to both sides.
Where both H+ and OH- are present on the same
side, put them together to make water.
Cancel water that appears on both sides.
Rewrite the reaction neatly.
DISPROPORTIONATION: one species is both
oxidized and reduced (see practice sheet in Dry
Lab III)
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Practice redox with basic solution
BASE:
MnO4-(aq) + C2O42-(aq)  MnO2(s) + CO32-(aq)
Answer:
2 MnO4- + 3 C2O42- + 4 OH-  2 MnO2 + 6 CO32+ 2 H2O
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Now do follow-up to sample problem 21.1
__Mn04-(aq) + __I-(aq)  __MnO42-(aq) + __ IO3-(aq) ( in base)
And don’t despair! We will spend one lab session practicing
this stuff in Dry Lab III.
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