Environmental Microbiology & Environmental Microbiology & ChemistryChemistry

CEB 20303

Muhammad Ahmad

© 2013

CHEMISTRYCHEMISTRYENVIRONMENTENVIRONMENT

Processes

Biotic Factors

Growth Growth conditionsconditions Microscope

Epi-fluorescence

Light ~

Chemical Equilibrium

Kinetics

Order

Second

Zero

First

Acid Base

HA H+ + A-

pAbiotic Factors

Biosphere

Pollution

Atmo~

Hydro~

Litho~

Natural

Man-madeBiochemical

Physico-chemical

Biodegradation WeatheringC-H-O-N-P-S-metal cycles

Biogeochemical

Metabolism

Electron acceptors

Energy source

Radiation

pH

T

GasH2O

Carbon source

Radiation

Organic

In-organic

In-organic

O2 Fe(III) NO3 SO4

Predator

Competitor

Symbiosis

Parasitism

Plate count

MPN

GeneticPCR

PhylochipBiochemical

ATPPhospho-

lipids

Commensalism

Mutualism

Stoichiometry

Chemical reactions

Analytical

Precipitation

Redox

Acid-base

Spectroscopy

Chromatography

SpectrophotometryAAS

MS

X-Ray

Voltammetry

GC HPLC

Mole Molar Mass

Chemical equations

2.0 Environmental 2.0 Environmental ChemistryChemistry

ObjectivesObjectives

Study nature of Redox reactionsStudy nature of Redox reactionsFamiliarize with concept of oxidation statesFamiliarize with concept of oxidation statesLearn rules of assigning oxidation statesLearn rules of assigning oxidation statesKnow how to balance redox equationsKnow how to balance redox equations

2.1 Chemical reactions2.1 Chemical reactions

2.1.1 Redox Reactions2.1.1 Redox ReactionsRedox = 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 Reactions2.1.1 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 Reactions2.1.1 Redox Reactions• Energy storage (batteries)

Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq) → 2PbSO4(s) + 2H2O(l)

2.1.1 Redox Reactions2.1.1 Redox Reactions• Corrosion of metals (iron, copper)

4Fe2+(aq) + O2(g) + (4+2n)H2O (l) → 2Fe2O3 · nH2O(s) + 8H+(aq)

2.1.1 Redox Reactions2.1.1 Redox Reactions• Manufacturing of industrial materials such as aluminum, chlorine and sodium hydroxide

2Al2O3 + 3C → 4Al + 3CO2

2.1.1 Redox Reactions2.1.1 Redox ReactionsIn 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 Cl- ion.

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 How to work out that electrons actually flowed in a reaction??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 CO3

2- 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 b) SF6 c) NO3-

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

MnO4

-1

Next, let us consider the oxidation Next, let us consider the oxidation states of the atoms in Festates of the atoms in Fe33OO44……

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 is the main component in

magnetitemagnetite, an iron oreiron ore that accounts for the reddish color of many types of rocks and soils.

……FeFe33OO44 cont’d 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 Oxidation-Characteristics of Oxidation-Reduction ReactionsReduction Reactions

Consider the combustion of methane (main component of biogas):

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)-4 +1 +4 -20 +1 -2

No change occurs in the oxidation state of hydrogen, so it is not formally involved in the electron transfer process.

CH4(g) → CO2(g) + 8e-

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:

Characteristics of Oxidation-Characteristics of Oxidation-Reduction Reactions (cont’d)Reduction 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 Balancing Oxidation-Reduction EquationsEquations

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 Balancing Oxidation-Reduction Equations (cont’d)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 Balancing Oxidation-Reduction Equations (cont’d)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) Balance all of the elements except hydrogen and oxygen.

b) Balance oxygen using H2O.

c) Balance hydrogen using H+

d) 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:

MnO4- → Mn2+

This is the reduction half-reaction.

+7 -2 +2

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)

We can equalize the charges by adding five (5) electrons to the left side:

5e- + 8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)

7+ 2+

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)

One electron is needed on the right side to give a net 2+ charge on both sides:

Fe2+(aq) → Fe3+(aq) + e-

2+ 3+

2+ 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) + 5e-

Note 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 TitrationsSimple 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 TitrationsSimple 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+.

Initial reading

Placing the flask with sample

Adding titrant... ... to the sample One final squirt Slightly pink - the end point

Final reading

Simple Oxidation-Reduction TitrationsSimple Oxidation-Reduction Titrations

Carrying out a titration.flv - Shortcut.lnk

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:

444

2

10737.61000

156.41

10621.14

444

MnOmolmL

LmL

L

MnOmoln

Vcn

MnO

MnOMnOMnO

Solution:

The balanced equation shows that five times as much Fe2+ as MnO4

- is required:

Thus, the 0.3500-g sample of iron ore contained 3.368x10-3 mol of iron. The mass of iron present is:

The mass percent of iron in the iron ore is:

23

4

2

44 10368.3

1

510737.6

4

FemolMnOmol

FemolMnOmoln

fractionmolarXnn

Fe

MnOFe

FegFemol

FegFemolm

WMnm

WMnmWM

mn

Fe

FeFeFe

1881.01

85.5510368.3

..

....

3

%7.53%1003500.0

1881.0

g

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) → 2CrO4

2-(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)

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

Cr2O72-(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq) (unbalanced)

acidic

acidic