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OME General Chemistry
Lecture 6: Chemical Reactions
Dr. Hartwig Pohl
Office: Beyer-Bau 122e
Email: [email protected]
Phone: +49 351 463 42576
1
Empirical & Molecular Formulas
For any new chemical, chemists determine its elemental composition,
the empirical formula, and its structure to understand its properties
Empirical formula, the simplest positive integer ratio of atoms present in
a compound, can be determined from the mass percentage
Molecular formula, gives the actual number of atoms of each kind
present per molecule
Law of definite proportions, a chemical compound always contains the
same proportion of elements by mass
Mass percentage, divide mass of each atom in the molecule by the total
molecular mass
Combustion analysis, common way to determine the elemental
composition of an unknown hydrocarbon
2
The Mole & Molar Masses
Molecular Mass (amu), sum of the average masses of the atoms in
one molecule of the substance, calculated by summing the atomic
masses of the elements in the substance
Formula Mass (amu), used for ionic compounds because they do
not have an identifiable molecular unit, sum of the atomic masses
of all the elements in the empirical formula
The Mole (mol), Avogadro’s number, the number of constituent
particles, atoms or molecules, of a given substance per mol
(6.022e23 mol–1)
Molar Mass (g mol–1), the mass of 6.022e23 atoms, molecules, or
formula units of that substance
3
Chemical Equations
A chemical equation is an expression that gives the identities and
quantities of the substances in a chemical reaction
Chemical formulas and other symbols are used to indicate the starting
material(s) or reactant(s), which are written on the left side of the
equation, and the final compound(s) or product(s), which are written on
the right side. An arrow, read as yields or reacts to form, points from the
reactants to the products.
A balanced chemical equation is when both the numbers of each type
of atom and the total charge are the same on both sides. A chemical
reaction represents a change in the distribution of atoms but not in the
number of atoms
Stoichiometry is a collective term for the quantitative relationships
between the masses, numbers of moles, and numbers of particles
(atoms, molecules, and ions) of the reactants and products in a balanced
reaction.
4
Mass Relationships in Chemical Eqns
A balanced chemical equation gives the identity of the reactants and
products and the accurate number of molecules or moles of each that
are consumed or produced.
A stoichiometric quantity is the amount of product or reactant
specified by the coefficients in a balanced chemical equation.
Converting amounts of substances to moles, and vice versa, is the key
to all stoichiometry problems.
5
Limiting Reactants & Percent Yield
A limiting reactant is the reactant that restricts the amount of product
obtained. The reactant that remains after a reaction has gone to
completion is present in excess
• Determines the maximum amount of product that can be formed
from the reactants
Theoretical yield, the amount you would obtain if the reaction
occurred perfectly and your method of purifying the product was 100%
efficient
Actual yield, the measured mass of products obtained from a reaction,
is less than the theoretical yield
Percent yield, the ratio of the actual yield to the theoretical yield
6
General Reaction Types
Synthesis (Combination), two or more simple substances combine to form a
more complex substance.
A + B → AB
where A and B are elements or compounds, and AB is a compound consisting
of A and B. Examples:
2Na + Cl2 → 2NaCl
formation of table salt
S + O2 → SO2
formation of sulfur dioxide
4Fe + 3O2 → 2Fe2O3
iron rusting
CO2 + H2O → H2CO3
carbon dioxide reacting with water
to form carbonic acid
7
General Reaction Types
Decomposition, opposite of a synthesis reaction, the separation of a
chemical compound into elements or simpler compounds.
AB → A + B
where A and B are elements or compounds, and AB is a compound consisting
of A and B. Examples:
2H2O → 2H2 + O2
electrolysis of water
H2CO3 → H2O + CO2
carbonic acid forming CO2 in sodas
MCO3 → MO + CO2
metal carbonates + heat gives metal
oxides and CO2
2MClO3 → 2MCl + 3O2
metal chlorates + heat gives metal
chlorides and oxygen
8
General Reaction Types
Single Displacement, one element is replaced by another in a compound.
A + BC → AC + B
where A, B, and C are elements or compounds; A and B must be different
metals, hydrogen, or halogens; C is either an anion or cation and is called a
spectator ion. Examples:
Mg + 2HCl → MgCl2 + H2
Cl2 + 2NaBr → NaCl + Br2
Reaction proceeds if A is more reactive than B, if A is more likely to donate its
electrons than B, see reactivity series chart. For common metals: slide 11. For
halogens: F > Cl > Br > I.
Zn + 2AgNO3 → Zn(NO3)2 + 2Ag
Ag + Cu(NO3)2 → no reaction
9
General Reaction Types
Double Displacement, a chemical process involving the exchange of bonds
between two reacting chemical species, which results in the creation of
products with similar or identical bonding affiliations.
AB + CD → AD + CB
Precipitation, usually when AB and CD are aq. ionic compounds (or acids),
the dissolved cations and anions switch partners resulting in the formation of
two new ionic compounds AD and CB, one of which is in the solid state
Pb(NO3)2 (aq) + 2KCl (aq) → 2KNO3 (aq) + PbCl2 (s)
Neutralization, when AB is an acid and BC is a base, cations and anions switch
partners, resulting in the formation of water and a new ionic compound (salt)
H2SO4 (aq) + 2LiOH (aq) → Li2SO4 (aq) + 2H2O
10
Solubility Rules & Reactivity Series
Solubility Rules
Reactivity Series
1. Alkali metal compounds, acetates, nitrates,
and ammonium compounds are all soluble.
K
Na
Ca
Mg
Al
C
Zn
Fe
Sn
Pb
H
Cu
Ag
Au
Pt
2. Hydroxides of alkali metals and NH4+1, Ca+2,
Sr+2, and Ba+2 are soluble.
3. All halides are soluble except for those
containing Ag+1, Pb+2, and Hg2+2.
4. Most sulfates are soluble, except for BaSO4,
SrSO4, Ag2SO4, PbSO4, and CaSO4.
5. Most phosphates, carbonates, chromates
and sulfides are insoluble.
6. All acids are soluble!
Potassium
Sodium
Calcium
Magnesium
Aluminum
Carbon
Zinc
Iron
Tin
Lead
Hydrogen
Copper
Silver
Gold
Platinum
11
Special Cases
Acid-Base, can have different definitions depending on the concept
employed, generally involves a transfer of protons (H+) from one species
(the acid) to another (the base).
Oxidation-Reduction, reactions in which a transfer of electrons occurs from
one species to another resulting in a change in oxidation state.
Substitution Reaction, a single replacement reaction in which one
functional group in an organic compound is replaced by another (will be
addressed later!).
Complexation, a reaction in which several ligands react with a metal atom
to form a coordination complex.
Catalysis, a reaction that does not proceed directly but through a third
substance that is not consumed during the reaction.
12
Definition of Acids & Bases
Arrhenius definition: An acid is a substance that dissociates in water to
produce H+ ions (protons), and a base is a substance that dissociates in
water to produce OH– ions (hydroxide); an acid-base reaction involves
the reaction of a proton with the hydroxide ion to form water
Brønsted–Lowry definition: An acid is any substance that can donate a
proton, and a base is any substance that can accept a proton; acid-base
reactions involve two conjugate acid-base pairs and the transfer of a
proton from one substance (the acid) to another (the base)
Lewis definition: A Lewis acid is an electron-pair acceptor, and a Lewis
base is an electron-pair donor
13
Water: Acid & Base
Water is amphiprotic: it can act as an acid by donating a proton to a base to
form the hydroxide ion (OH–), or as a base by accepting a proton from an
acid to form the hydronium ion (H3O+), one water molecule can react with
another:
2H2O(l) ⇋ H3O+ (aq) + OH– (aq)
equilibrium constant
K = [H3O+] [OH–]
[H2O]2
By treating [H2O] as a constant, a new equilibrium constant, the
ion-product constant of liquid water (Kw), can be defined:
Kw = K[H2O]2 = [H3O+] [OH–] = 1.00e–14 (at 25 ºC)
The acidic, basic, or neutral condition of an aqueous solution depends on
the relative concentrations of H3O+ and OH– ions. Aqueous solutions are
neutral when [H+] = [OH–] , acidic when [H+] > [OH–] , and basic when [H+] <
[OH–].
14
The pH Scale
The pH scale provides a convenient way of expressing the hydrogen ion (H+)
concentration of a solution and describes acidity or basicity quantitatively.
pH = – log [H+]
(pOH = – log [OH–], pH = 14 – pOH)
Pure water is neutral at pH 7 on a scale from 0 to 14.
pH decreases with increasing [H+]  adding an acid to pure water increases the
[H+] and decreases the [OH–], pH < 7 = acidic
pH increases with decreasing [H+]  adding a base to pure water increases the
[OH–] and decreases the [H+], pH > 7 = basic
Can be measured with pH paper (paper dyed with colored molecules that
change w/certain pH) or, more accurately, using a pH meter (glass electrode
whose voltage depends on the H+ conc.)
15
Acids & Bases in Aqueous Solutions
Strong acids/bases react essentially completely with water to give H+
and the corresponding anion (acids) or OH– and the corresponding
cation (bases)
 acids donate a proton to a water molecule that acts as a
proton acceptor or base.
 bases accept a proton from water, so small amounts of OH–
are produced (water acts like an acid)
Neutralization Reaction, a reaction in which a strong acid and a strong
base react in stoichiometric amounts to produce water and a salt
Weak acids/bases are only slightly dissociated in water to give ions
that are at equilibrium with their molecular form, the dominant
species in the case of weak acids/bases is the molecular form
16
Strength of Acids & Bases
Strong acids undergo complete dissociation
Weak acids undergo partial dissociation
• Depends on strength of acid/base
• Described by equilibrium constant
17
Acid-Base Equilibrium Constants
The magnitude of the equilibrium constant for an ionization reaction can
be used to determine the relative strengths of acids and bases
Ionization of a weak acid in water:
HA(aq) + H2O(l) ⇋ H3O+(aq) + A–(aq)
Equilibrium Constant:
K = [H3O+] [A–]
[H2O] [HA]
The concentration of water is constant for all reactions in aqueous
solution, so [H2O] can be incorporated into a new quantity, the acid
ionization constant (Ka):
Ka = K[H2O] = [H3O+] [A–]
[HA]
 larger Ka = stronger acid and higher [H+] at equilibrium
18
Acid-Base Equilibrium Constants
A very similar situation can be considered for bases…
Ionization of a weak base in water:
B(aq) + H2O(l) ⇋ BH+(aq) + OH–(aq)
Equilibrium Constant:
K = [BH+] [OH–]
[H2O] [B]
Base Ionization Constant:
Kb = K[H2O] = [BH+] [OH–]
[B]
 larger Kb = stronger base and higher [OH–] at equilibrium
19
Acid-Base Equilibrium Constants
Equilibrium constants often expressed as…
pKa = – log Ka and pKb = – log Kb
Important Trends:
1. Smaller pKa  larger acid ionization constants  stronger acids
• moderate/weak acids pKa ≈ 0 to 14, strong acids pKa < 0
2. Smaller pKb  larger base ionization constants  stronger bases
• moderate/weak bases pKb ≈ 0 to 14, strong bases pKb < 0
3. Related by: pKa + pKb = 14
20
Conjugate Acid-Base Pairs
Two species that differ by only a proton constitute
a conjugate acid-base pair
Example: In the reaction of HCl with water, HCl, the parent acid, donates
a proton to a water molecule, the parent base, forming Cl–; HCl and Cl–
constitute a conjugate acid-base pair
In the reverse reaction, the Cl– ion in solution acts as a base to accept a
proton from H3O+, forming H2O and HCl; H3O+ and H2O constitute a
second conjugate acid-base pair
HCl
parent acid
CH3COOH
parent acid
+
+
H2O ⇋
H3O+
+
Cl–
parent base conjugate acid conjugate base
H2O
⇋
H3O+
+
parent base
conjugate acid
CH3COO–
conjugate base
Any acid-base reaction must contain two conjugate acid-base pairs!
21
Strength of Conjugate Pairs
There is an inverse relationship between the strength of the parent acid and
the strength of the conjugate base  the conjugate base of a strong acid is a
weak base, and the conjugate base of a weak acid is a strong base
• However, a weak acid/base will not necessarily have a strong
conjugate base/acid
One can use the relative strengths of acids and bases to predict the direction
of an acid-base reaction by following a simple rule:
An acid-base equilibrium always favors the side with the weaker acid/base
Important relationship or conjugate pairs:
Kw = KaKb
22
Structure & Acid/Base Strength
The acid-base strength of a molecule depends strongly on its structure!
The stronger the A–H or B–H+ bond, the less likely the bond is to break
to form H+ ions, and thus the less acidic the substance.
• The larger the atom to which H is bonded, the weaker the bond 
acid strengths of binary hydrides increase as we go down a column of
the periodic table
The conjugate base (A– or B) contains one more lone pair of electrons
than the parent acid (AH or BH+)…
• Any factor that stabilizes the lone pair on the conjugate base favors
dissociation of H+ and makes the parent acid a stronger acid  acid
strengths of binary hydrides increase as we go from left to right
across a row of the periodic table
23
Polyprotic Acids & Bases
Polyprotic acids contain more than one ionizable proton, and the
protons are lost in a stepwise manner.
The fully protonated species is always the strongest acid because it is
easier to remove a proton from a neutral molecule than from a
negatively charged ion; the fully deprotonated species is the strongest
base
The strengths of the conjugate acids and bases are related by pKa + pKb
= pKw, and equilibrium favors formation of the weaker acid-base pair.
Example: Sulfuric Acid, H2SO4
H2SO4(aq) + H2O(l) ⇋ H3O+(aq) + HSO4–(aq)
HSO4–(aq) + H2O(l) ⇋ H3O+(aq) + SO42–(aq)
pKa(H2SO4) < pKa(HSO4–)
pKb(SO42–) < pKb(HSO4–)
24
Example: Acid/Base Strength
What is the pH of the resulting solution if you add one drop (0.05 mL) of 2 M
HCl (pKa = –7) to 100 mL of water?
1. Strong or weak acid?  pKa = –7 strong acid!
• Strong acid  all HCl is dissociated into H3O+
2. Determine amount amount of HCl added = [H3O+]
3. Calculate pH.
0.05 mL * 1 L * 2 mol * 1 * 1000 mL = 10–3 M = [H3O+]
1000 mL L
100 mL
1L
pH = –log[H3O+] = 3
25
Example: Acid/Base Strength
What is the pH of a 0.125 M HClO solution? Ka = 3.5e–8
Strong or weak acid?  pKa = 7.46 weak acid!
1) Write the balanced equation and the expression for Ka
2) Solve for the hydronium ion concentration
HClO(aq) + H2O(aq) ⇋ ClO–(aq) + H3O+(aq)
[HClO]
[ClO–] [H3O+
]
i
0.125
0
0
Δ
–x
+x
+x
Ka = [ClO–][H3O+]
[HClO]
Ka =
x2 _
0.125
f 0.125–x +x
+x
= Ka* 0.125 M = 0.438e–8 M
weak acid
x = [H3O+] = 6.6e–5 M
x2
pH = –log[H3O+] = 4.18
26
Example: Conjugate Acid-Base Pairs
The Ka of formic acid (HCOOH) is 1.8e–4. What is the pH of 0.35 M solution of
sodium formate (NaHCOO)?
1. Is NaHCOO a strong or weak base?
NaHCOO = conjugate base of HCOOH, Kb(NaHCOO) = Kw/Ka(HCOOH) =
5.6e–11
pKb (NaHCOO) = 10.3  weak base
2. Write the balanced equation and the expression for Kb
NaHCOO(aq) + H2O(aq) ⇋ HCOOH(aq) + NaOH(aq)
[NaHCOO] [HCOOH] [NaOH]
Kb = [HCOOH][NaOH] = Kw = 1.0e–14
i
0.35
0
0
[NaHCOO]
Ka 1.8e–4
Δ
–x
+x
+x
x2 / 0.35 = 5.6e–11
f
0.35–x
+x
+x
–
x = 4.4e–6 M = [OH ]
weak base
pOH = –log[OH–] = 5.36  pH = 14 – pOH = 8.64
27
Acid-Base Titration
An acid–base titration is the determination of the concentration of an
acid or base by neutralizing the acid or base with an acid or base of
known concentration. This allows for quantitative analysis of the
concentration of an unknown acid or base solution. It makes use of the
neutralization reaction that occurs between acids and bases.
Problems are a mix of stoichiometry and equilibrium…
Stoichiometry: Add a strong base (acid) to a weak acid (base), and
determine the extent of neutralization using stoichiometric concepts.
Equilibrium: The concentration of weak base/acid conjugate species is
determined, and the equilibrium expression used to determine [H3O+],
or pH.
28
Titrations of Acids & Bases
In acid-base titrations, a buret is used to
deliver measured volumes of an acid or
base solution of known concentration
(the titrant) to a flask that contains a
solution of a base or an acid of
unknown concentration.
Plotting the pH changes that occur
during an acid-base titration against the
amount of acid or base added produces
a titration curve; the shape of the curve
provides important information about
what is occurring in solution during the
titration.
29
Titrations: Strong Acid w/ Strong Base
Before addition of any strong acid, the initial [H3O+] equals the concentration
of the strong acid; addition of strong base decreases the [H3O+] because
added base neutralizes some of the H3O+ present.
Equivalence point, the point at which the number of moles of base (or acid)
added equals the number of moles of acid (or base) originally present
Addition of a strong base after the
equivalence point causes an excess of OH–
and produces a rapid increase in pH.
For the titration of a monoprotic strong
acid with a monobasic strong base, the
volume of base needed to reach the
equivalence point is:
moles of base = moles of acid
(volume)b(molarity)b = (volume)a(molarity)a
VbMb = VaMa
30
Titrations: Weak Acids w/ Strong Bases
The shape of the titration curve for a weak acid depends dramatically on the
identity of the acid or base and the corresponding value of Ka or Kb. Generally:
Before base is added, the pH of the weak acid is higher than the pH of the strong
acid, pH changes more rapidly during the first part of the titration, due to the
higher starting pH
The pH changes much more gradually
around the equivalence point, the pH of the
weak acid at the equivalence point is
greater than 7.00
Above the equivalence point, the two
curves are identical; once acid has been
neutralized, the pH of the solution is
controlled only by the amount of excess of
OH– present, regardless of whether the acid
is weak or strong
31
Example: Titration
Calculate the pH during the titration of 20 mL of 0.25 M nitrous acid (HNO2;
Ka = 4.5e–4) after adding the following volumes of 0.25 M NaOH: (a) 0 mL,
(b) 10 mL, (c) 20 mL.
(a) Initial pH of the solution before addition of base.
HNO2 + H2O ⇋ H3O+ + NO2–
Ka = [NO2–][H3O+]
[HNO2]
Ka =
x2 _
0.25
[HNO2] [NO2–] [H3O+
]
i
0.25
0
0
Δ
–x
+x
+x
f
0.25–x
+x
+x
x2 = Ka* 0.25 M = 1.13e–4 M
x = [H3O+] = 0.01 M
pH = –log[H3O+] = 1.97
32
Example: Titration
(b) Stoichiometry of neutralization reaction… OH– from NaOH consumes
initially added HNO2 converting it to H2O and NO2–.
HNO2 + OH– → H2O +NO2–
10 mL NaOH * 0.25 mol NaOH = 0.0025 mol NaOH
103 mL NaOH
20 mL HNO2 * 0.25 mol HNO2 = 0.005 mol HNO2
103 mL HNO2
0.005 mol HNO2 – 0.0025 mol NaOH = 0.0025 mol HNO2 left unreacted
0.0025 mol NO2– produced
New concentrations = 0.0025 mol / 30 mL = 0.083 M = [HNO2] = [NO2–]
33
Example: Titration
(b) Determine new pH.
HNO2 + H2O ⇋ H3O+ + NO2–
Ka = [NO2–][H3O+]
[HNO2]
Ka = (0.083)x
0.083
[HNO2]
[NO2–]
[H3O+]
i
0.083
0.083
0
Δ
–x
+x
+x
f
0.083–x 0.083+x
+x
x = [H3O+] = Ka = 4.5e–4 M
pH = –log[H3O+] = 3.35
34
Example: Titration
(c) Stoichiometry of neutralization reaction… OH– from NaOH consumes
initially added HNO2 converting it to H2O and NO2–.
HNO2 + OH– → H2O +NO2–
20 mL NaOH * 0.25 mol NaOH = 0.005 mol NaOH
103 mL NaOH
20 mL HNO2 * 0.25 mol HNO2 = 0.005 mol HNO2
103 mL HNO2
0.005 mol HNO2 – 0.005 mol NaOH =
0 mol HNO2 left unreacted
0.005 mol NO2– produced
New concentration = 0.005 mol / 0.04 L = 0.123 M = [NO2–]
pH is determined by the NO2– equilibrium now!
35
Example: Titration
(b) Determine new pH.
NO2– + H2O ⇋ OH– + HNO2
Kb = [NO2–][H3O+]
[HNO2]
Kb =
x2
0.125
[NO2–]
[HNO2]
[OH–]
i
0.125
0
0
Δ
–x
+x
+x
+x
+x
f 0.125–x
Kb = Kw/Ka = 1.0e–14/4.5e–4 = 2.2e–11
x = [OH–] = 1.65e–6 M
pOH = –log[OH–] = 5.78
pH = 14 – pOH = 8.22
36
Oxidation-Reduction Reactions
The term oxidation was first used to describe reactions in which metals
react with oxygen in air to produce metal oxides:
• Metal acquires a positive charge by transferring electrons to the neutral
oxygen atoms of an oxygen molecule  oxidation is the loss of
electrons, oxygen atoms acquire a negative charge and form oxide ions
(O2–)  reduction is the gain of electrons
Oxidation states of each atom in a compound is the charge that atom
would have if all of its bonding electrons were transferred to the atom
with the greater attraction for electrons (larger EN). Atoms in their
elemental form are assigned an oxidation state of zero. An atom is oxidized
when its oxidation number increases and an atom is reduced when its
oxidation number decreases
Oxidation-reduction reactions are called redox reactions, in which there is
a net transfer of electrons from one reactant to another. The total number
of electrons lost must equal the total number of electrons gained.
37
Example of a Redox Reaction
The reaction between hydrogen and fluorine is an example of an
oxidation-reduction reaction:
H2 + F2 → 2HF
The overall reaction may be written as two half-reactions:
H2 → 2H+ + 2e− (the oxidation reaction)
F2 + 2e− → 2F− (the reduction reaction)
There is no net change in charge in a redox reaction so the excess
electrons in the oxidation reaction must equal the number of electrons
consumed by the reduction reaction. The ions combine to form hydrogen
fluoride:
H2 + F2 → 2H+ + 2F− → 2HF
38
Rules for Assigning Oxidation States
1. The OS of an individual atom is 0
2. The total OS of all atoms in: a neutral species is 0 and in an ion is equal
to the ion charge
3. Group 1 metals have an OS of +1 and group 2 an OS of +2
4. The OS of fluorine is –1, when in compounds
5. Hydrogen generally has an OS of +1 in compounds
6. Oxygen generally has an OS of –2 in compounds
7. In binary metal compounds, group 17 elements (halogens) have an OS
of –1, group 16 of –2, and group 15 of –3.
39
Examples: Oxidation States
Determine the oxidation states of the elements in the following reactions:
a) Fe(s) + O2(g) → Fe2O3(g)
b) Fe2+
c) Ag(s) + H2S → Ag2S(s) + H2(g)
ANSWERS:
a) Fe and O2 are neutral elements, therefore they have an OS of "0"
according to Rule #1. The product has a total OS equal to "0" and
following Rule #6, O has an OS of –2, which means Fe has an OS of +3.
b) The OS of Fe corresponds to its charge, therefore the OS is +2.
a) Ag has an OS of 0, H has an OS of +1 according to Rule #5, S has an OS
of –2 according to Rule #7 and hence Ag in Ag2S has an OS of +1.
40
Examples: Oxidation States
Determine the OS of the bold element in each of the following:
a) Na3PO3
b) H2PO4–
ANSWERS:
a) The oxidation numbers of Na and O are +1 and –2. Since sodium
phosphite is neutral, the sum of the oxidation numbers must be
zero. Letting x be the oxidation number of phosphorus then, 0= 3(+1) +
x + 3(–2), x = oxidation number of P = +3
b) Hydrogen and oxygen have oxidation numbers of +1 and –2. The ion
has a charge of –1, so the sum of the oxidation numbers must be –1.
Letting y be the oxidation number of phosphorus, –1= y + 2(+1) +4(–2),
y = oxidation number of P = +5
41
Oxidants & Reductants
Oxidants (oxidizing reagents), compounds that are capable of accepting
electrons, they can oxidize other compounds
• An oxidant is reduced in the process of accepting electrons
Reductants (reducing agents), compounds that are capable of donating
electrons, they can cause the reduction of another compound
• A reductant is oxidized in the process of donating electrons
oxidant + reductant  oxidation–reduction
gains e–
loses e–
redox reaction
is reduced
is oxidized
Remember: An atom is oxidized when its oxidation number increases and
an atom is reduced when its oxidation number decreases.
42
Examples: Oxidation / Reduction
Determine which element is oxidized and which element is reduced in
the following reactions (be sure to include the OS of each):
a) Zn + 2H+ → Zn2+ + H2
b) 2Al + 3Cu2+→2Al3+ +3Cu
c) CO32– + 2H+→ CO2 + H2O
ANSWERS:
a) Zn is oxidized (oxidation number: 0 → +2); H+ is reduced (+1 → 0)
b) Al is oxidized (oxidation number: 0 → +3); Cu2+ is reduced (+2 → 0)
a) This is not a redox type because each element has the same oxidation
number in both reactants and products: O = –2, H = +1, C = +4
43
Types of Redox Reactions
Combination reactions are some of the simplest redox reactions and as
the name suggests involves the "combining" of elements to form a
chemical compound. As usual, oxidation and reduction occur together.
General Equation:
Example:
Oxidation States:
A + B → AB
H2 + O2 → H2O
0 + 0 → (2)(+1)+(–2) = 0
Decomposition reactions are the reverse of combination reactions,
meaning they are the breakdown of a chemical compound into the
individual elements.
General Equation:
Example:
Oxidation States:
AB → A + B
H2O → H2 + O2
(2)(+1)+(–2) → 0 + 0
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Types of Redox Reactions (cont.)
Combustion reactions always involve oxygen, in the form of O2 and are
almost always exothermic, meaning they produce heat.
General Equation:
Example:
Oxidation States:
CxHy + O2 → CO2 + H2O
CH4 + 2O2 → CO2 + 2H2O
((–4)+(4)(+1)) + 0
→ ((+4)+(2)(–2)) + ((2)(+1)+(–2))
Disproportionation Reactions: In some redox reactions substances can be
both oxidized and reduced.
General Equation:
Example:
Oxidation States:
2A → A' + A''
2H2O2(aq) → 2H2O(l) + O2(g)
((2)(+1)+(2)(–1)) → ((2)(+1)+(–2)) + 0
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Types of Redox Reactions (cont.)
A single replacement reaction involves the "replacing" of an element in
the reactants with another element in the products.
General Equation:
Example:
Oxidation States:
A + BC → AB + C
Cl2 + 2NaBr → 2NaCl + Br2
0 + ((+1)+(–1)) → ((+1)+(–1)) + 0
A double replacement reaction is similar to a single replacement
reaction, but involves "replacing" two elements in the reactants, with two
in the products.
General Equation:
Example:
Oxidation States:
AB + CD → AD + CB
Fe2O3 + 6HCl → 2FeCl3 + 3H2O
((2)(+3)+(3)(–2)) + ((+1)+(–1))
→ ((+3)+(3)(–1)) + ((2)(+1)+(–2))
Not actually a redox reaction... No change in oxidation state!
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Complexation
A coordination complex or metal complex consists of a central atom or
ion, which is usually metallic and is called the coordination center, and a
surrounding array of bound molecules or ions, that are in turn known as
ligands or complexing agents. Many metal-containing compounds,
especially those of transition metals, are coordination complexes.
Coordination refers to the "coordinate covalent bonds" (dipolar bonds,
the ligands are donating electrons from a lone electron pair into an empty
metal orbital on the coordination center) between the ligands and the
central atom.
The final charge of the complex is the sum of all the charges of the
individual components. Thus a complex in solution can have a positive,
negative or zero charge.
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Ligand Complexation Example
A ligand that is commonly used in chemical
analyses is ethylenediamine tetra acetic acid
(EDTA). It possesses lone pairs on 4 oxygen
atoms and 2 nitrogen atoms that it can share
with metal atoms to for octahedral complexes.
Ligands can have several different descriptors,
however the commonality for all complexing
agents is that they can contribute an electron
pair to a vacant orbital of a central atom.
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Properties of Ligands
1. They are Lewis bases (donate an electron pair).
2. They contain at least one atom that is not a hydrogen or carbon.
3. They can have multiple atoms and charge centers like EDTA or they
can be single ions like chloride.
4. They can be negatively charged or neutral; for example, water is a
ligand.
5. The charge density of the resulting complex will be small due to the
increase of volume of the complex over the individual ions.
6. The concentration of the complex can be calculated using a
thermodynamic constant called the formation constant.
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Catalysts
A catalysts is a substance that participates in a reaction and causes it
to occur more rapidly but that can be recovered unchanged at the
end of the reaction and reused
Catalysts are not involved in the stoichiometry of the reaction and are
usually shown above the arrow in a net chemical equation
A biological catalyst is called an enzyme
Note: chemical processes in industry rely heavily on catalysts, which
are added to a reaction mixture in trace amounts
Can be classified as either homogeneous (uniformly dispersed
throughout the reaction mixture) or heterogeneous (in a different
physical state from the reactants)
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