Environmental Microbiology &
Chemistry
CEB 20303
Muhammad Ahmad
© 2013
Inorganic
O2
Carbon
source
Fe(III)
NO3
Commensalism
SO4
Parasitism
Predator
Electron acceptors
Symbiosis
Metabolism
Organic
Competitor
Energy
source
Inorganic
Mutualism
Plate count
MPN
Radiation
PCR
Growth
conditions
Radiation
T
Abiotic
Factors
Biochemical
Epifluorescence
ATP
HA H+ + A-
Biotic
Factors
H2O
Hydro~
Biosphere
Mole
Man-made
Molar
Mass
Chemical
equations
Stoichiometry
ENVIRONMENT
CHEMISTRY
Zero
Litho~
Natural
Phospholipids
Acid Base
Gas
Atmo~
Phylochip
Light ~
p
pH
Genetic
Microscope
Pollution
Processes
Kinetics
Chemical
Equilibrium
Order
First
Acid-base
Chemical
reactions
Precipitation
Spectrophoto
metry
Second
Biogeochemical
Physicochemical
Biochemical
AAS
MS
Spectroscopy
XRay
C-H-O-N-P-S-metal cycles
Biodegradation
Weathering
Analytical
Chromatography
GC
Redox
Voltammetry
HPLC
2.0 Environmental
Chemistry
Objectives
Study nature of Redox reactions
Familiarize with concept of oxidation states
Learn rules of assigning oxidation states
Know how to balance redox equations
2.1 Chemical reactions
2.1.1 Redox Reactions
Redox = Reduction Oxidation
Many important chemical reactions involve oxidation and
reduction:
• Oxidation of sugars, fats and proteins in unicellular and
multicellular organisms provides the energy necessary for life
2.1.1 Redox Reactions
• Combustion reactions provide most of the energy to power our
civilization
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + energy
2.1.1 Redox Reactions
• Energy storage (batteries)
Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq) → 2PbSO4(s) + 2H2O(l)
2.1.1 Redox Reactions
• Corrosion of metals (iron, copper)
4Fe2+(aq) + O2(g) + (4+2n)H2O (l) → 2Fe2O3 · nH2O(s) + 8H+(aq)
2.1.1 Redox Reactions
• Manufacturing of industrial materials such as aluminum,
chlorine and sodium hydroxide
2Al2O3 + 3C → 4Al + 3CO2
2.1.1 Redox Reactions
In ALL reactions shown previously, electrons are transferred
from a donor to an acceptor.
Example:
2Na(s) + Cl2(g) → 2NaCl(s)
In this reaction, an electron is transferred from a sodium
atom (Na) to a chlorine atom (Cl), yielding a Na+ and a Clion.
Reactions like this one, in which one or more electrons are
transferred, are called oxidation – reduction reactions, or
redox reactions.
How to work out that electrons
actually flowed in a reaction??
Answer: Concept of oxidation states!
Provides a way to keep track of electrons in redox
reactions.
Oxidation states are defined by a set of rules, most
of which describe how to divide up the shared
electrons in compounds containing covalent bonds.
Covalent bonds are formed by electron sharing
between atoms, for example O2, H2, H2O
Covalent bonds can be polar or non-polar.
Distribution of electrons in a bond
From our chemistry classes we know that oxygen has a
greater attraction for electrons than does hydrogen, causing
the O-H bonds in the water molecule to be polar.
Because oxygen has a greater attraction for electrons, the
shared electrons tend to spend more time close to the
oxygen than to either of the hydrogens thus giving the
oxygen atom a slight excess of negative charge, and the
hydrogen atoms become slightly positive.
Non-metals with the highest attraction for shared electrons
are in the upper right-hand corner of the periodic table.
They are fluorine, oxygen, nitrogen and chlorine.
The relative ability of these atoms to attract shared electrons is
F > O > N ≈ Cl
Rules for assigning oxidation states
•
•
•
•
•
•
•
The oxidation state in elemental form is always zero e.g Cl,Cl2,O,O2,O3,F2,Na,P4,S8.
The Alkali metals Na,K,Li ,Rb atc have always +1 in their compounds like Na2SO4,
KCl etc
The Alkaline earth metals Group IIA of periodic table have +2 oxidation state in their
compounds CaCl2, MgSO4 etc.
H has zero oxidation state in H and H2, in all other compounds it Has +1 or -1
oxidation state, generally with nonmetals +1 (HCl) and with metals -1 (MgH2, NaH),
depends on electronegativity of other element.
F in its compounds have only -1 oxidation state in its compounds while other
halogens can have-1 , +1 to +7 oxidation states e.g HClO , HClO2, HClO3, HClO4
etc.
Oxygen has -2 oxidation state in most of the compounds except
peroxides(H2O2,Na2O2) superoxide (KO2) and in binary compounds with flourine e.g
OF2.
Certain elements have variable oxidation states like S in H2S, H2SO3 and H2SO4 , Cr
in Cr2O3, K2Cr2O7 , P in PCl3 and PCl5 etc
Rules for assigning oxidation states
§1
The oxidation state of an atom in an element is 0. For example, the oxidation state of
each atom in the substances Na(s), O2(g), O3(g) and Hg(l) is 0.
§2
The oxidation state of a mono-atomic ion is the same as its charge. For
example, the oxidation state of the Na+ ion is +1.
§3
In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of
+1. For example, in the compounds HCl, NH3, H2O, and CH4, hydrogen is assigned
an oxidation state of +1.
§4
In binary compounds the element with the greater attraction for the electrons in
the bond is assigned a negative oxidation state equal to its charge in its ionic
compound. For example, fluorine is always assigned an oxidation state of -1.
That is, for purposes of counting electrons, fluorine is assumed to be F-.
Nitrogen is usually assigned -3. For example, in NH3, nitrogen is assigned an
oxidation state of -3; in H2S, sulfur is assigned an oxidation state of -2; in HI,
iodine is assigned an oxidation state of -1, and so on.
§5
The sum of the oxidation states must be zero for an electrically neutral
compound and must be equal to the overall charge for an ionic species. For
example, the sum of the oxidation states for the carbon and oxygen atoms in
CO32- is -2; and the sum of oxidation states for the nitrogen and hydrogen
atoms in NH4+ is +1.
Assign oxidation states to all atoms in the following:
a) CO2
c) NO3-
b) SF6
Solution:
a) The rule that takes precedence here is that oxygen is
assigned an oxidation state of -2. The oxidation state for
carbon can be determined by recognizing that since CO2
has no charge, the sum of the oxidation states for oxygen
and carbon must be 0. Since each oxygen is -2 and there
are two oxygen atoms, the carbon atom must be assigned
an oxidation state of +4:
CO2
+4
-2
Solution:
b) Since fluorine has the greater attraction for the shared
electrons, we assign its oxidation state first. Since its
charge in ionic compounds is 1-, we assign -1 as the
oxidation state of each fluorine atom. The sulfur must then
be assigned an oxidation state of +6 to balance the total of 6 from the fluorine atoms:
SF6
+6
-1
Solution:
c) Since oxygen has a greater attraction than does nitrogen for
the shared electrons, we assign its oxidation state of -2 first.
Because the sum of the oxidation states of the three
oxygens is -6, and the net charge on the NO3- ion is 1-, the
nitrogen must have an oxidation state of +5:
NO3+5
-2
Assign oxidation states to underlined atoms
P2O5
SO3
H3PO4
FeCl3
Cr2O7-2
MnO
4
-1
Next, let us consider the oxidation
states of the atoms in Fe3O4…
Fe3O4 is the main component in
magnetite, an iron ore that
accounts for the reddish color of
many types of rocks and soils.
We assign each oxygen atom its usual oxidation state of -2.
The three iron atoms must yield a total of +8 to balance the
total of -8 from the four oxygens.
Thus, each iron atom has an oxidation state of 8/3.
A noninteger value for oxidation state may seem strange since
its charge is expressed in whole numbers.
…Fe3O4 cont’d
The rules assume that all of the iron atoms are equal, when in
fact this compound can best be viewed as containing four (4)
O2- ions, two (2) Fe3+ and one (1) Fe2+ ion per formula unit.
Characteristics of OxidationReduction Reactions
Consider the combustion of methane (main component of
biogas):
-4 +1
0
+4 -2
+1 -2
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Note that carbon undergoes a change in oxidation state from 4 to +4 in CO2. Such a change can be accounted for by a loss
of eight electrons:
CH4(g) → CO2(g) + 8eNo change occurs in the oxidation state of hydrogen, so it is not
formally involved in the electron transfer process.
Characteristics of OxidationReduction Reactions (cont’d)
With this background, we can now define some important
terms. Oxidation is an increase in oxidation state (a loss of
electrons). Reduction is a decrease in oxidation state (gain of
electrons).
Concerning our methane combustion example we can say:
• Carbon is oxidised because there is an increase in its
oxidation state (carbon has formally lost electrons).
• Oxygen is reduced as shown by the decrease in its oxidation
state (oxygen has formally gained electrons).
• CH4 is the reducing agent, while O2 is the oxidizing agent.
Balancing Oxidation-Reduction
Equations
Redox reactions are often complicated, which means that it
can be difficult to balance their equations by simple
inspection. Two methods for balancing redox reactions are
commonly used
a) Oxidation States Method
b) Half-reaction Method (ion electron method).
In the following couple of slides we are going to familiarize
ourselves with the half-reaction method.
Balancing Oxidation-Reduction
Equations (cont’d)
Consider the unbalanced equation for the oxidation-reduction
reaction between cerium(IV) ion and tin(II) ion:
Ce4+(aq) + Sn2+(aq) → Ce3+(aq) + Sn4+(aq)
This reaction can be separated into a half-reaction involving
the substance being reduced,
Ce4+(aq) → Ce3+(aq)
and one involving the substance being oxidized,
Sn2+(aq) → Sn4+(aq)
Balancing Oxidation-Reduction
Equations (cont’d)
• The general procedure is to balance the equations
for the half-reactions separately and then to add
them to obtain the overall balanced equation.
• The half-reaction method for balancing redox
equations differs slightly depending on whether the
reaction takes place in acidic or basic solution.
Balancing redox equations in acidic
solution
§ 1 Write the equations for the oxidation and reduction
half-reactions.
§ 2 For each half-reaction:
a)
b)
c)
d)
Balance all of the elements except hydrogen and
oxygen.
Balance oxygen using H2O.
Balance hydrogen using H+
Balance the charge using electrons.
§ 3 If necessary, multiply one or both balanced half-
reactions by integers to equalize the number of
electrons transferred in the two half-reactions.
§ 4 Add the half-reactions, and cancel identical species.
§ 5 Check to be sure that the elements and charges
balance.
Consider following reaction between permanganate and
iron(II) ions in acidic solution:
MnO4-(aq) + Fe2+(aq) → Fe3+(aq) + Mn2+(aq)
(This reaction is used to analyze iron ore for its iron content.)
Solution:
§1 Identify and write equations for the half-reactions.
The oxidation states for the half-reactions involving the
permanganate ion show that manganese is reduced:
+7 -2
+2
MnO4- → Mn2+
This is the reduction half-reaction.
Solution:
§1 Identify and write equations for the half-reactions.
The other half-reaction involves the oxidation of iron(II) to
iron(III) ion and is the oxidation half-reaction:
Fe2+ → Fe3+
§2 Balance each half-reaction
For the reduction reaction, we have
MnO4-(aq) → Mn2+(aq)
• The manganese is balanced.
Solution:
§2 Balance each half-reaction
b) We balance oxygen by adding 4H2O to the right side of the
equation
MnO4-(aq) → Mn2+(aq) + 4H2O(l)
c) Next, we balance hydrogen by adding 8H+ to the left side:
8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)
Solution:
§2 Balance each half-reaction
d) All of the elements have been balanced, but we need to
balance the charge by using electrons. At this point we have
following charges for reactants and products in the
reduction half-reaction:
8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)
7+
2+
We can equalize the charges by adding five (5) electrons to
the left side:
5e- + 8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)
2+
2+
Solution:
§2 Balance each half-reaction
For the oxidation reaction,
Fe2+(aq) → Fe3+(aq)
The elements are balanced, and we must simply balance the
charge:
Fe2+(aq) → Fe3+(aq)
3+
2+
One electron is needed on the right side to give a net 2+
charge on both sides:
Fe2+(aq) → Fe3+(aq) + e2+
2+
Solution:
§3 Equalize the electron transfer in the two half-reactions.
Since the reduction half-reaction involves a transfer of five
electrons and the oxidation half-reaction involves a transfer
of only one electron, the oxidation half-reaction must be
multiplied by 5:
5Fe2+(aq) → 5Fe3+(aq) + 5e§4 Add the half-reactions.
The half-reactions are added to give:
5e- + 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) + 5eNote that the electrons cancel (as they MUST) to give the final balanced
equation:
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Solution:
§5 Check that the elements and charges balance.
• Elements balance: 5 Fe, 1 Mn, 4 O, 8 H → 5 Fe, 1 Mn, 4 O, 8 H
• Charges balance: 5(2+) + (1-) + 8(1+) = 17+
→ 5(3+) + (2+) + 0 = 17+
The equation is balanced!
Simple Oxidation-Reduction Titrations
• Oxidation-reduction reactions are
commonly used as the basis for
volumetric analytical procedures.
• For example, a reducing substance
can be titrated with a solution of a
strong oxidizing agent.
• Three of the most frequently used
oxidizing agents are aqueous solutions
of potassium permanganate (KMnO4),
potassium dichromate (K2Cr2O7) and
cerium hydrogen sulfate [Ce(HSO4)4].
Simple Oxidation-Reduction Titrations
• Permanganate has the advantage of being its own
indicator – the MnO4- ion is intensely purple, and the
Mn2+ ion is almost colorless.
• As long as some reducing agent remains in the
solution being titrated, the solution remains
colourless, since purple MnO4- is converted to
colourless Mn2+.
Final reading
Initial reading
Placing the flask
with sample
Adding titrant...
... to the sample
One final squirt
Slightly pink - the end point
Simple Oxidation-Reduction Titrations
Iron ores often involve a mixture of oxides and contain both
Fe2+ and Fe3+ ions. Such an ore can be analyzed for its iron
content by dissolving it in acidic solution, reducing all the
iron to Fe2+ ions, and then titrating with a standard solution
of KMnO4. In the resulting solution, MnO4- is reduced to
Mn2+, and Fe2+ is oxidized to Fe3+. A sample of iron ore
weighing 0.3500g was dissolved in acidic solution, and all of
the iron was reduced to Fe2+. Then the solution was titrated
with a 1.621 x 10-2 M KMnO4 solution. The titration required
41.56 mL of the permanganate solution reach the light
purple (pink) end point. Determine the mass % of iron in the
iron ore.
Solution:
First, we write the unbalanced equation for the reaction:
H+(aq) + MnO4-(aq) + Fe2+(aq) → Fe3+(aq) + Mn2+(aq) + H2O(l)
Using the half-reaction method, we balance the equation:
8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
The number of moles of MnO4- ion required in the titration is
found from the volume and concentration of permanganate
solution used:
nMnO  cMnO  VMnO
4
nMnO
4
4
4
1.62110  2 mol MnO4
1L

 41.56mL 
 6.737 10  4 mol MnO4
L
1000mL
Solution:
The balanced equation shows that five times as much Fe2+ as
MnO4- is required:
nFe  nMnO  X molar fraction 
4
5 mol Fe2
3
2
nFe  6.737 10 mol MnO 

3
.
368

10
mol
Fe
1 mol MnO4
4

4
Thus, the 0.3500-g sample of iron ore contained 3.368x10-3
mol of iron. The mass of iron present is:
m
 m  n  M .W .
M .W .
mFe  nFe  M .W .Fe
n
55.85 g Fe
mFe  3.368  10 mol Fe 
 0.1881 g Fe
1 mol Fe
3
The mass percent of iron in the iron ore is:
0.1881 g
 100%  53.7%
0.3500 g
1.Tell which of the following are oxidation-reduction reactions.
For those that are, identify the oxidizing agent, the reducing
agent, the substance being oxidized, and the substance
being reduced.
a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
b) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
c) Cr2O72-(aq) + 2OH-(aq) → 2CrO42-(aq) + H2O(l)
d) O3(g) + NO(g) → O2(g) + NO2(g)
e) HCl(g) + NH3(g) → NH4Cl(s)
2. Balance the following oxidation-reduction reactions, which
occur in acidic solution, using the half-reaction method.
a) Cu(s) + HNO3(aq) → Cu2+(aq) + NO(g)
b) Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s)
c) Mn2+(aq) + NaBiO3(s) → Bi3+(aq) + MnO4-(aq)
d) As2O3(s) + NO3-(aq) → H3AsO4(aq) + NO(g)
e) Fe(s) + HCl(aq) → HFeCl4(aq) + H2(g)
3. A 50.00 mL sample of solution containing Fe2+ ions is
titrated with a 0.0216 M KMnO4 solution. It required 20.62
mL of the KMnO4 solution to oxidize all of the Fe2+ ions to
Fe3+ ions by the reaction
MnO4
-(aq)
+
Fe2+(aq)
→ Mn2+(aq) + Fe3+(aq) (unbalanced)
acidic
a) What was the concentration of Fe2+ ions in the sample
solution?
b) What volume of 0.0150 M K2Cr2O7 solution would it take to
do the same titration? The reaction is
Cr2O7
2-(aq)
+
Fe2+(aq)
→ Cr3+(aq) + Fe3+(aq) (unbalanced)
acidic