Chemistry and Chemical Reactivity
6th Edition
1
John C. Kotz
Paul M. Treichel
Gabriela C. Weaver
CHAPTER 5
Reactions in Aqueous Solution
Lectures written by John Kotz
©2006
2006
Brooks/Cole
Thomson
©
Brooks/Cole
- Thomson
Reactions
in Aqueous Solution
Many reactions involve ionic compounds,
especially reactions in water — aqueous
solutions.
KMnO4 in water
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K+(aq) + MnO4-(aq)
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3
CCR, page 177
An Ionic Compound,
CuCl2, in Water
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Aqueous Solutions
How do we know ions are
present in aqueous
solutions?
The solutions conduct
electricity!
They are called
ELECTROLYTES
HCl, CuCl2, and NaCl are
strong electrolytes.
They dissociate
completely (or nearly so)
into ions.
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Aqueous Solutions
HCl, CuCl2, and NaCl are strong
electrolytes. They dissociate completely
(or nearly so) into ions.
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Aqueous Solutions
Acetic acid ionizes only to a small extent, so
it is a weak electrolyte.
CH3CO2H(aq)
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--->
CH3CO2-(aq) + H+(aq)
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Aqueous
Solutions
Acetic acid ionizes only
to a small extent, so it
is a weak
electrolyte.
CH3CO2H(aq) --->
CH3CO2-(aq) + H+(aq)
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Aqueous
Solutions
Some compounds
dissolve in water but
do not conduct
electricity. They are
called nonelectrolytes.
Examples include:
sugar
ethanol
ethylene glycol
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Water Solubility of Ionic Compounds
If one ion from the “Soluble
Compd.” list is present in a
compound, the compound is
water soluble.
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Water Solubility of Ionic Compounds
Common minerals are often formed with
anions that lead to insolubility:
sulfide
fluoride
carbonate
oxide
Iron pyrite, a sulfide
Azurite, a copper
carbonate
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Orpiment,
arsenic sulfide
ACIDS
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An acid -------> H+ in water
Some strong acids are
HCl
H2SO4
HClO4
HNO3
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hydrochloric
sulfuric
perchloric
nitric
HNO3
ACIDS
An acid -------> H+ in water
HCl(aq) ---> H+(aq) + Cl-(aq)
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The
Nature of
Acids
HCl
Cl-
H 2O
hydronium
ion
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H 3O+
Weak Acids
WEAK ACIDS = weak
electrolytes
CH3CO2H
acetic acid
H2CO3
carbonic acid
H3PO4
phosphoric acid
HF
hydrofluoric acid
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Acetic acid
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ACIDS
Nonmetal oxides can be acids
CO2(aq) + H2O(liq)
---> H2CO3(aq)
SO3(aq) + H2O(liq)
---> H2SO4(aq)
and can come from burning
coal and oil.
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BASES
see Screen 5.9 and Table 5.2
Base ---> OH- in water
NaOH(aq)
NaOH is
a strong
base
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---> Na+(aq) + OH-(aq)
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Ammonia, NH3
An Important Base
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BASES
Metal oxides are
bases
CaO(s) + H2O(liq)
--> Ca(OH)2(aq)
CaO in water. Indicator
shows solution is basic.
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Know the strong
acids & bases!
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Net Ionic
Equations
Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq)
We really should write
Mg(s) + 2 H+(aq) + 2 Cl-(aq) --->
H2(g) + Mg2+(aq) + 2 Cl-(aq)
The two Cl- ions are SPECTATOR IONS —
they do not participate. Could have used NO3-.
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Net Ionic Equations
Mg(s) + 2 HCl(aq)
--> H2(g) + MgCl2(aq)
Mg(s) + 2 H+(aq) + 2 Cl-(aq)
---> H2(g) + Mg2+(aq) + 2 Cl-(aq)
We leave the spectator ions out —
Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq)
to give the NET
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IONIC EQUATION
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Chemical Reactions in Water
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Sections 5.2 & 5.4-5.6—CD-ROM Ch. 5
We will look at
EXCHANGE
REACTIONS
AX + BY
AY + BX
The anions
exchange places
between cations.
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Pb(NO3) 2(aq) + 2 KI(aq)
----> PbI2(s) + 2 KNO3 (aq)
Precipitation Reactions
The “driving force” is the formation of an
insoluble compound — a precipitate.
Pb(NO3)2(aq) + 2 KI(aq) ----->
2 KNO3(aq) + PbI2(s)
Net ionic equation
Pb2+(aq) + 2 I-(aq) ---> PbI2(s)
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Acid-Base Reactions
• The “driving force” is the formation of water.
NaOH(aq) + HCl(aq) --->
NaCl(aq) + H2O(liq)
• Net ionic equation
OH-(aq) + H+(aq)
---> H2O(liq)
• This applies to ALL reactions
of STRONG acids and bases.
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Acid-Base Reactions
CCR, page 191
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Acid-Base Reactions
• A-B reactions are sometimes called
NEUTRALIZATIONS because the solution is
neither acidic nor basic at the end.
• The other product of the A-B reaction is a SALT, MX.
HX + MOH ---> MX + H2O
Mn+ comes from base & Xn- comes from acid
This is one way to make compounds!
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Gas-Forming Reactions
This is primarily the chemistry of metal
carbonates.
CO2 and water ---> H2CO3
H2CO3(aq) + Ca2+ --->
2 H+(aq) + CaCO3(s) (limestone)
Adding acid reverses this reaction.
MCO3 + acid ---> CO2 + salt
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Gas-Forming
Reactions
CaCO3(s) + H2SO4(aq) --->
2 CaSO4(s) + H2CO3(aq)
Carbonic acid is unstable and forms CO2 & H2O
H2CO3(aq)
---> CO2 + water
(Antacid tablet has citric acid + NaHCO3)
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Quantitative Aspects
of
Reactions in Solution
Sections 5.8-5.10
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Terminology
In solution we need to define the
• SOLVENT
the component whose
physical state is
preserved when
solution forms
• SOLUTE
the other solution component
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Concentration of Solute
The amount of solute in a solution
is given by its concentration.
moles solute
Molarity(M) =
liters of solution
Concentration (M) = [ …]
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1.0 L of water
was used to
make 1.0 L of
solution. Notice
the water left
over.
CCR, page 206
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Preparing a Solution
Active Figure 5.18
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PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O
in enough water to make 250 mL of solution.
Calculate molarity.
Step 1: Calculate moles
of NiCl2•6H2O
1 mol
5.00 g •
= 0.0210 mol
237.7 g
Step 2: Calculate molarity
0.0210 mol
= 0.0841 M
0.250 L
[NiCl2•6 H2O ] = 0.0841 M
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The Nature of a CuCl2 Solution:
Ion Concentrations
CuCl2(aq) -->
Cu2+(aq) + 2 Cl-(aq)
If [CuCl2] = 0.30 M,
then
[Cu2+] = 0.30 M
[Cl-] = 2 x 0.30 M
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USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a
0.0500 M solution?
Because
Conc (M) = moles/volume = mol/V
this means that
moles = M•V
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USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a 0.0500 M
solution?
moles = M•V
Step 1: Calculate moles of acid required.
(0.0500 mol/L)(0.250 L) = 0.0125 mol
Step 2: Calculate mass of acid required.
(0.0125 mol )(90.00 g/mol) =
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1.13 g
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Preparing Solutions
• Weigh out a solid solute
and dissolve in a given
quantity of solvent.
• Dilute a concentrated
solution to give one
that is less
concentrated.
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Preparing a Solution by
Dilution
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PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH.
What do you do?
Add water to the 3.0 M solution to lower
its concentration to 0.50 M
Dilute the solution!
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PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?
But how much water
do we add?
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PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?
How much water is added?
The important point is that --->
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution
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PROBLEM: You have 50.0 mL of 3.0 M NaOH and
you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution =
M•V
=
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must also =
0.15 mol NaOH
Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or
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300 mL
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PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?
Conclusion:
add 250 mL
of water to
50.0 mL of
3.0 M NaOH
to make 300
mL of 0.50 M
NaOH.
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Preparing Solutions by
Dilution
A shortcut
Cinitial • Vinitial = Cfinal • Vfinal
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Concentration of Solute
The amount of solute in a solution
is given by its concentration.
moles solute
Molarity(M) =
liters of solution
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pH, a Concentration Scale
pH: a way to express acidity -- the concentration of
H+ in solution.
Low pH: high [H+]
Acidic solution
Neutral
Basic solution
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High pH: low [H+]
pH < 7
pH = 7
pH > 7
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The pH Scale
pH = log (1/ [H+])
= - log [H+]
In a neutral solution,
[H+] = [OH-] = 1.00 x 10-7 M at 25 oC
pH = - log [H+] = -log (1.00 x 10-7)
- [0 + (-7)]
= 7
See CD Screen 5.17 for a tutorial
See book Appendix A.3 for more on logs
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=
[H+] and pH
If the [H+] of soda is 1.6 x 10-3 M,
the pH is ____?
Because pH = - log [H+]
then
pH= - log (1.6 x 10-3)
pH = -{log (1.6) + log (10-3)}
pH = -{0.20 - 3.00)
pH = 2.80
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pH and [H+]
If the pH of Coke is 3.12, it is ____________.
Because pH = - log [H+] then
log [H+] = - pH
Take antilog and get
[H+] = 10-pH
[H+] = 10-3.12 =
7.6 x 10-4 M
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SOLUTION STOICHIOMETRY
Section 5.10
• Zinc reacts with
acids to produce H2
gas.
• Have 10.0 g of Zn
• What volume of
2.50 M HCl is
needed to convert
the Zn completely?
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GENERAL PLAN FOR STOICHIOMETRY
CALCULATIONS
Mass
HCl
Mass
zinc
Moles
zinc
Stoichiometric
factor
Moles
HCl
Volume
HCl
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Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
Step 1: Write the balanced equation
Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)
Step 2: Calculate amount of Zn
1.00 mol Zn
10.0 g Zn •
= 0.153 mol Zn
65.39 g Zn
Step 3: Use the stoichiometric factor
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Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
Step 3: Use the stoichiometric factor
2 mol HCl
0.153 mol Zn •
= 0.306 mol HCl
1 mol Zn
Step 4: Calculate volume of HCl req’d
1.00 L
0.306 mol HCl •
= 0.122 L HCl
2.50 mol
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ACID-BASE REACTIONS
Titrations
H2C2O4(aq) + 2 NaOH(aq) --->
acid
base
Na2C2O4(aq) + 2 H2O(liq)
Carry out this reaction using a TITRATION.
Oxalic acid,
H2C2O4
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Setup for titrating an acid with a base
Active Figure 5.23
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Titration
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1. Add solution from the
buret.
2. Reagent (base) reacts
with compound (acid) in
solution in the flask.
3. Indicator shows when
exact stoichiometric
reaction has occurred.
4. Net ionic equation
H+ + OH- --> H2O
5. At equivalence point
moles H+ = moles OH-
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LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately
determine its concentration.
1.065 g of H2C2O4 (oxalic acid) requires
35.62 mL of NaOH for titration to an
equivalence point. What is the
concentra-tion of the NaOH?
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1.065 g of H2C2O4 (oxalic acid) requires 35.62
mL of NaOH for titration to an equivalence
point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
1 mol
1.065 g •
= 0.0118 mol
90.04 g
Step 2: Calculate amount of NaOH req’d
2 mol NaOH
0.0118 mol acid •
= 0.0236 mol NaOH
1 mol acid
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1.065 g of H2C2O4 (oxalic acid) requires 35.62
mL of NaOH for titration to an equivalence
point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
= 0.0118 mol acid
Step 2: Calculate amount of NaOH req’d
= 0.0236 mol NaOH
Step 3: Calculate concentration of NaOH
0.0236 mol NaOH
 0.663 M
0.03562 L
[NaOH] = 0.663 M
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LAB PROBLEM #2:
Use standardized NaOH to determine
the amount of an acid in an unknown.
Apples contain malic acid, C4H6O5.
C4H6O5(aq) + 2 NaOH(aq) --->
Na2C4H4O5(aq) + 2 H2O(liq)
76.80 g of apple requires 34.56 mL of 0.663 M
NaOH for titration. What is weight % of malic
acid?
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76.80 g of apple requires 34.56 mL of
0.663 M NaOH for titration. What is
weight % of malic acid?
Step 1:
C•V =
=
Step 2:
Calculate amount of NaOH used.
(0.663 M)(0.03456 L)
0.0229 mol NaOH
Calculate amount of acid titrated.
1 mol acid
0.0229 mol NaOH •
2 mol NaOH
= 0.0115 mol acid
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76.80 g of apple requires 34.56 mL of
0.663 M NaOH for titration. What is
weight % of malic acid?
Step 1:
=
Step 2:
=
Calculate amount of NaOH used.
0.0229 mol NaOH
Calculate amount of acid titrated
0.0115 mol acid
Step 3: Calculate mass of acid titrated.
134 g
0.0115 mol acid •
= 1.54 g
mol
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76.80 g of apple requires 34.56 mL of
0.663 M NaOH for titration. What is
weight % of malic acid?
Step 1:
=
Step 2:
=
Step 3:
=
Calculate amount of NaOH used.
0.0229 mol NaOH
Calculate amount of acid titrated
0.0115 mol acid
Calculate mass of acid titrated.
1.54 g acid
Step 4: Calculate % malic acid.
1.54 g
• 100% = 2.01%
76.80 g
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Oxidation-Reduction Reactions
Section 5.7
Thermite reaction
Fe2O3(s) + 2 Al(s)
---->
2 Fe(s) + Al2O3(s)
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EXCHANGE: Precipitation Reactions
EXCHANGE
Gas-Forming
Reactions
REACTIONS
REDOX
REACTIONS
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EXCHANGE
Acid-Base
Reactions
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REDOX REACTIONS
REDOX = reduction & oxidation
2 H2(g) + O2(g) ---> 2 H2O(liq)
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REDOX REACTIONS
REDOX = reduction & oxidation
Corrosion of aluminum
2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s)
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REDOX REACTIONS
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
In all reactions if
something has been
oxidized then
something has also
been reduced
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REDOX REACTIONS
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
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Why Study Redox Reactions
Batteries
Corrosion
Manufacturing metals
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Fuels
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REDOX REACTIONS
Redox reactions are characterized by
ELECTRON TRANSFER between an
electron donor and electron acceptor.
Transfer leads to—
1. increase in oxidation number
of some element = OXIDATION
2. decrease in oxidation number
of some element = REDUCTION
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OXIDATION NUMBERS
The electric charge an element
APPEARS to have when electrons are
counted by some arbitrary rules:
1. Each atom in free element has ox. no. = 0.
Zn
O2
I2
S8
2. In simple ions, ox. no. = charge on ion.
-1 for Cl+2 for Mg2+
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OXIDATION NUMBERS
3. O has ox. no. = -2
(except in peroxides: in H2O2, O = -1)
4. Ox. no. of H = +1
(except when H is associated with a metal as in
NaH where it is -1)
5. Algebraic sum of oxidation numbers
= 0 for a compound
= overall charge for an ion
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OXIDATION NUMBERS
NH3
N =
ClO-
Cl =
H3PO4
P =
MnO4-
Mn =
Cr2O72C3H8
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Cr =
C =
Oxidation
number of F
in HF?
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Recognizing a Redox Reaction
Corrosion of aluminum
2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s)
Al(s) --> Al3+(aq) + 3 e• Ox. no. of Al increases as e- are donated by the metal.
• Therefore, Al is OXIDIZED
• Al is the REDUCING AGENT in this balanced halfreaction.
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Recognizing a Redox Reaction
78
Corrosion of aluminum
2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s)
Cu2+(aq) + 2 e- --> Cu(s)
• Ox. no. of Cu decreases as e- are accepted by the ion.
• Therefore, Cu is REDUCED
• Cu is the OXIDIZING AGENT in this balanced halfreaction.
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Recognizing a Redox Reaction
Notice that the 2 half-reactions add up to
give the overall reaction
—if we use 2 mol of Al and 3 mol of Cu2+.
2 Al(s) --> 2 Al3+(aq) + 6 e3 Cu2+(aq) + 6 e- --> 3 Cu(s)
----------------------------------------------------------2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s)
Final eqn. is balanced for mass and charge.
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Examples of Redox Reactions
Metal + halogen
2 Al + 3 Br2 ---> Al2Br6
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Examples of Redox Reactions
Nonmetal (P) + Oxygen
---> P4O10
Metal (Mg) + Oxygen
---> MgO
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Recognizing a Redox
Reaction
See Table 5.4
Reaction Type
Oxidation
Reduction
In terms of oxygen
gain
loss
In terms of halogen
gain
loss
In terms of electrons
loss
gain
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Common Oxidizing and
Reducing Agents
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See Table 5.4
Metals
(Cu) are
reducing
agents
Metals
(Na, K,
Mg, Fe)
are
reducing
agents
HNO3 is an
oxidizing
agent
Cu + HNO3 -->
2 K + 2 H2O -->
Cu2+ + NO2
2 KOH + H2
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Examples of Redox Reactions
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Metal + acid
Mg + HCl
Mg = reducing agent
H+ = oxidizing agent
Metal + acid
Cu + HNO3
Cu = reducing agent
HNO3 = oxidizing agent
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