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OXIDATION AND REDUCTION REACTION

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Types of Reactions

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Redox Reaction

Is the reaction that involve the transfer of electrons from a reducing agent to an oxidizing agent

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TYPE OF REDOX REACTION

OUT OF BODY IN THE BODY

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Electrochemistry

Is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy

Electrochemical processes are redox reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electrical energy is used to cause a nonspontaneous reaction to occur

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Definition of Oxidation and Reduction

Oxidation:

Is defined as the

removal of electrons

Reduction:

is defined as the gain

of electrons

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Common uses of the terms oxidation and reduction

Term Meaning

Oxidation

Reduction

To Combine with oxygenTo lose hydrogenTo lose electronTo increase in oxidation numberTo lose oxygenTo combine with hydrogenTo gain electronTo decrease in oxidation number

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Oxidation of Ferrous to ferric ion

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OXIDATION NUMBER

Oxidation numbers are the charges atoms in a compound would have if the electrons of each bond belonged to the more electronegative atoms.

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Properties of oxidizing and reducing agents

oxidizing agents reducing agents

- Gains electrons- Oxidation number decrease- Becomes reduced

- Loses electron- Oxidation number increase- Becomes oxidized

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THE RULES FOR ASSIGNING

OXIDATION NUMBER

1. Atoms of any element not combined with atoms from another element have oxidation numbers of zero.

Examples: The oxidation number of the atom in N2, Br2, Cl2, P4 and S2 are zero

continued

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2. The oxidation numbers of monatomicions equal their ionic charge.

Examples:

The oxidation numbers of Na+ and

K+ are +1, of Ca2+, Cu2+, and Mg2+

are +2, and Cl¯ and Br¯ are –1.

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3. In their compounds, the oxidation number of any atoms of the

- Group I A elements is +1 (e.g., Na+, K+) - Group II A elements is +2

(e.g., Cu2+, Mg2+) - Group III A is +3

continued

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4. The oxidation number of any nonmetal in its binary compounds with metals equals the charge of the monatomic anion

e.g., The oxidation number of Br in

CrBr3 is –1 because the monatomic ion of Br is the bromide ion, which

has a charge of 1-.

continued

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5. In compounds, the oxidation number of - O is almost always –2. (Exceptions occur only when the rules for H or F would be violated) - H is almost always +1. (the exceptions are binary compounds with metals, like NaH, in which the H has the oxidation number –1.) - F is always –1. (No exceptions. Fluorine is the most electronegative of all elements).

continued

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6. The sum of the oxidation numbers

of all the atoms in the formula of

the atom, ion, or molecule must

equal the overall charge given for

the formula – the sum rule

continued

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Problem No. 1Calomel, long use in medicine, has theformula Hg2Cl2. What are the oxidation numbers on the atoms in this compound?

The rule for assigning oxidation number: No. 4 & 6

4. The oxidation number of any nonmetal in its binary compounds with metals equals the charge of the monatomic anion

6. The sum of the oxidation numbers of all the atoms in the formula of the atom, ion, or

molecule must equal the overall charge given for the formula – the sum rule

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The oxidation number of Hg (x)

Hg2Cl2

Rule 4 : Cl 2 atoms X (-1) = -2 Hg 2 atoms X (x) = 2xRule 6 : sum = 0The value of x comes from the sum, 2x + (-2) = 0 2x = +2 x = +1The oxidation number of Hg in the Hg2Cl2 is +1

_

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BALANCING REDOX REACTIONS

O x id a tio n rea ction R e du ction rea ction

R e do x re a ction

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BALANCING REDOX REACTIONS BY THE ION-ELECTRON METHOD

Simply list the steps to balance a redox

reaction in an acidic (or neutral) medium.

Balancing redox reactions by the

ion-electron method has 8 steps.

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Those are:1. Write a skeletal equation that shows only the ions or molecules involved in the reaction.2. Divide the skeletal equation into two half- reactions.3. Balance all atoms that are not H or O.4. Balance O by adding H2O.5. Balance H by adding H+ (not H or H2 or H¯, but H+).6. Balance the net charge by adding e¯. (remember its minus sign.)

continued

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7. Multiply Entire Half-reactions by

simple whole numbers, as needed,

to get the gain of e¯ in one half-

reaction to match the loss of e¯ in

the other. Then add the half-reactions.

8. Cancel whatever is the same on

both sides of the arrow.

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The oxidation of methyl alcohol, CH3OH, to formic acid, HCHO2, using the dichromate ion,

Cr2O72¯ in an acidic medium.

As this reaction proceeds, the chromium inCr2O7

2¯ change to Cr3+.Step 1. Write a skeletal equation showing reactants and products as given. CH3OH + Cr2O7

2¯ HCHO2 + Cr3+

Step 2. Divide the skeletal equation into two

half-reactions. Except for H and O, the same elements must appear on both sides of each half-reaction.

CH3OH HCHO2

Cr2O72¯ Cr3+ continued

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Step 3. Balance all atoms that are not H or O CH3OH HCHO2 (No change, yet.) Cr2O7

2¯ 2Cr3+ (Balances Cr atoms.)

Step 4. Balance O by adding H2O CH3OH + H2O HCHO2 (Cs and Os balance.) Cr2O7

2¯ 2Cr3++ 7H2O (Crs and Os balance)

Step 5. Balance H by adding H+

CH3OH + H2O HCHO2 + 4H+ (All atoms now balance.) Cr2O7

2¯ + 14H+ 2Cr3+ + 7H2O (All atoms now balance.)

continued

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LOOK AT THE STEP 5, AND THEN MAKE THE NET CHARGE ON THE LEFT SIDE OF THE ARROW EQUAL TO THE NET CHARGE ON THE RIGHT SIDE.

Step 5. Balance H by

adding H+

(a) CH3OH + H2O

HCHO2 + 4H+

(b) Cr2O72¯ + 14H+

2Cr3+ + 7H2O

Step 6. Balance the net charge by adding e¯.

(a) CH3OH + H2O

HCHO2 + 4H++ 4e¯

(b) Cr2O72¯ + 14H++ 6e¯

2Cr3+ + 7H2O continued

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Step 7. Multiply half-reaction by whole

numbers so that the electrons will

cancel when the half-reactions are

added.

3 X [CH3OH + H2O HCHO2 + 4H++ 4e¯]

2 X [Cr2O72¯ + 14H++ 6e¯ 2Cr3+ + 7H2O]

Sum:

3CH3OH + 2Cr2O72 ¯+ 3H2O + 28H++ 12e¯

3HCHO2 + 4Cr3++ 12H+ + 14H2O +12e¯

+

continued

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Step 8. Cancel everything that can be canceled

a). The 12 electrons on each side obviously

cancel.

b). Water molecule: there are 3 on the left and

14 on the right, so we can strike those on

the left and change those on the right to 11.

…. + 3H2O +…. …. + 14H2O +….

becomes: …. .… + 11H2O + ….

c). And then also cancel some H+

…. + 28 H+ + …. …. + 12 H+ + ….

becomes: …. + 16 H+ + …. ….. continued

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3CH3OH(aq) + 2Cr2O72 ¯(aq) + 16H+(aq)

3HCHO2(aq) + 4Cr3+(aq) +11H2O

Check to see that both material and electrical balance exist.

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Home work

Problem No. 1What is the oxidation number of carbon in

ethane, C2H6?Problem No. 2What are the oxidation numbers of the

atoms in the nitrate ion, NO3¯?Problem No.3Balance the following equation, which occurs in an acidic medium.

Cu(s) + NO3¯(aq) Cu2+(aq) + NO2(g)

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MnO4¯(aq) + SO3

2¯(aq) MnO2(s) + SO42¯(aq)

Step 1. Through 8 for acidic solutions2MnO4

¯ + 3SO32¯ + 2H+ 2MnO2 + 3SO4

2¯ + H2O

Step 9. Add as many OH as there are H+ to both sides of equation. There are two H+ on the left.So we add 2OH¯ to both sides.2OH¯ + 2MnO4

¯ + 3SO32¯ + 2H+

2MnO2 + 3SO42¯ + H2O + 2OH¯

To Balance a Redox Equation When the Medium is Basic.

1. First Balance it for an Acid Medium and

2. Then Neutralize the Acid

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Step 10. When they occur on the same

side of the arrow, combine H+

and OH¯ into H2O.

2OH¯ + 2MnO4¯ + 3SO3

2¯ + 2H+

2MnO2 + 3SO42¯ + H2O + 2OH¯

The left side has 2OH¯ and 2H+, so we

combine them into 2H2O.

The equation:

2H2O + 2MnO4¯ + 3SO3

2¯

2MnO2 + 3SO42¯ + H2O + 2OH¯

continued

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Step 11. Cancel H2O molecules as possible

Step 10.

2H2O + 2MnO4¯ +

3SO32¯

2MnO2 + 3SO42¯

+ H2O + 2OH¯

Step 11.

Cancel H2O molecule

as possible.

The final equation:

H2O + 2MnO4¯(aq) +

3SO32¯(aq)

2MnO2(s)+ 3SO42¯(aq)

……… + 2OH¯(aq)

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Reduction Potentials or Redox Potentials

Half Reaction

E° (volts)

Fe3+ (aq) + e¯ ⇌ Fe2+(aq) Fe2+(aq) + 2e¯ ⇌ Fe(s)½ O2 + 2H+ + 2 e¯ ⇌ H2ONAD+ + H+ + 2 e¯ ⇌ NADH

+ 0.77 - 0,44+ 0.815 - 0.315

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Rule for Combining Reduction Half-Reactions

When two reduction half-reactions are combined into a full redox reaction the one with the more positive E° always runs as written, as a reduction, and it forces the other, with the less positive E°, to run in reverse, as an oxidation.

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Na(s) and Cl2(g).Will sodium react with chlorine?

Na+(aq) + e¯ ⇌ Na(s) E°= -2,71 V Cl2(g) + 2 e¯ ⇌ 2Cl¯ E°= +1.36V

When two reduction half-reactions are combined into a full redox reaction the one with the more positive E° always runs as written, as areduction, and it forces the other, with the lesspositive E°, to run in reverse, as an oxidation.

Na(s) → Na+(aq) + e¯ (oxidation) Cl2(g) + 2 e¯ → 2Cl¯ (aq) (reduction) Continued next slide

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to get the net reaction, multiply the coefficients

of the first half-reaction by 2.

2X [Na(s) → Na+(aq) + e¯ (oxidation)]

2 Na(s) → 2Na+(aq) + 2e¯ (oxidation)

Cl2(g) + 2e¯ → 2Cl¯ (aq) (reduction)

Sum: 2Na(s) + Cl2(g) 2Na+(aq) + 2Cl¯ (aq)

Thus sodium and chlorine spontaneously react.

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TYPE OF REDOX REACTION

OUT OF BODY IN THE BODY

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Biological Oxidation

The processes of oxidation are essential for maintaining life because oxidation and the simultaneously occuring reduction supply the free energy for the vital work

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Organic Reaction Mechanisms

Organic reactions mechanisms has classified biochemical reactions into four categories:

1. Group-transfer reactions 2. Oxidations and reductions 3. Eliminations, isomerizations, and

rearrangements 4. Reactions that make or break carbon-

carbon bonds

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Covalent bonds

A covalent bond consists of an electron pair shared between two atoms.

In breaking such a bond, the electron pair can either remain with one of the atoms (heterolytic bond cleavage) or separate such that one electron accompanies each of the atoms (homolytic bond cleavage).

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Homolytic Bond Cleavage

Homolytic bond cleavage, which usually produces unstable radicals, occurs mostly in oxidation-reduction reactions

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Heterolytic Bond Cleavage

Heterolytic C-H bond cleavage involves either carbanion and proton (H+) formation or carbocation (carboniun ion) and hydride ion (H-) formation.

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Biologically Importan Nucleophillic Groups

Nucleophiles are the conjugate bases of weak acids

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1. Group-transfer reactions

Types of metabolic group-transfer: Acyl group transfer involves addition of a nucleophile (Y) to the electrophilic carbon atom of an acyl compound to form a tetrahedral intermediate. The original acyl carrier (X) is then expelled to form a new acyl compound.

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2. Oxidations and reductions

NAD+ IS AN ELECTRON ACEPTOR (OXIDATOR); TWO ELECTRON FROM GENERALBASE ARE TRANSFERRRED TO AN ELECTRON ACCEPTOR SUCH AS NAD+.

General Base Alcoho

lNAD+

General acid Keton

eNAD

H

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In Living Systems

The electron-transfer process connecting these half-reactions occursthrough a multistep pathway thatharnesses the liberated free energy toform ATP.

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The principal use of oxygen is in respiration. Which may be defined as the process by with cells derive energy in the form of ATP from the controlled reaction of hydrogen with oxygen to form water.

For Example: Biologic Oxidation

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For Example: oxidation of primary alcohols in the Body

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Toxicity test

ACCUT TOXICITY

CHRONIC TOXICITY

SUBCHRONIC TOXICITY

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FREE ENERGY CHANGES CAN BE EXPRESSED IN TERMS OF REDOX POTENTIAL

In reaction involving oxidation and reduction, the free energy change is proportionate to the tendency of reactants to donate or accept electrons. Thus, in addition to expressing free energy change in terms of ΔG0.

It is possible, in an analogous manner, to express it numerically as an oxidation-reduction or redox potential (E0).

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Oxidation of a Metabolite catalyzed by an oxidase forming H2O

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Oxidation of a metabolite by Hydrogenases and Finally by an Oxidase in a Respiratory Chain.

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The Citric Acid Cycle

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The complete oxidation of glucose

by molecular oxygen

C6H12O6 + 6O2 6CO2 + 6H2O Break this equation down into two half- reactions

1). C6H12O6 + 6H2O 6CO2 + 24H+ + 24e¯

The glucose carbon atoms are oxidized

2). 6O2 + 24H+ + 24e¯ 12H2O

Molecular oxygen is reduced

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The sites of electron transfer that form NADH and FADH2 in glycolysis and the citric acid cycle.

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Cytochrome P450 hydroxylase cycle in microsomes

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A sugar that can be oxidized

by Cu2+ solutions