OXIDATION-REDUCTION CHAPTER 19. OXIDATION-REDUCTION REACTIONS OXIDATION is the process by which a substance loses one or more electrons. Iron loses three

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OXIDATION-REDUCTION REACTIONS

OXIDATION is the process by which a substance loses one or more electrons. Iron loses three electrons from reactant to product.

0 +3

4 Fe + 3 O2 2 Fe2 O3

REDUCTION is the process by which a substance gains one or more electrons. Iron gains three electrons from reactant to product.

+3 0

2 Fe2 O3 + 3 C 4 Fe + 3 CO2


The oxidation number of an atom in a substance is equal to the charge that the atom would have if the electrons in each bond belonged to the most electronegative atom.

The more electronegative atom gains electrons from the other atom because it is treated as if it were reduced.

The less electronegative atom is oxidized.


You can use one of two mnemonic devices to remember oxidation-reduction: a. OIL RIG (Oxidation Is Loss of electrons and Reduction Is Gain of electrons.)b. LEO GER (Lose Electrons - Oxidation; Gain Electrons - Reduction.)

OXIDIZING AND REDUCING AGENTS

An OXIDIZING AGENT causes the oxidation of another substance by accepting electrons from that substance. The oxidizing agent contains the atom that shows a decrease in oxidation number. The oxidizing agent is itself reduced. +3 0 0 +4

2 Fe2 O3 + 3 C 4 Fe + 3 CO2

C: 0 +4; C is oxidized, C is the reducing agent

OXIDIZING AND REDUCING AGENTS

A REDUCING AGENT causes the reduction by providing electrons to another substance. The reducing agent contains the atom that shows an increase in oxidation number. The reducing agent is oxidized.

+3 0 0 +4

2 Fe2 O3 + 3 C 4 Fe + 3 CO2

Fe: +3 to 0;

Fe is reduced, Fe2O3 is the oxidizing agent

ANOTHER EXAMPLE

0 0 +3 -2

4 Fe + 3 O2 2 Fe2 O3

Fe: 0 to +3; Fe is oxidized, Fe is the reducing agent

O: 0 to -2; O is reduced, O2 is the oxidizing agent

The substance that is oxidized or reduced is always a single element.

The substance that is the oxidizing agent or the reducing agent, is always whatever the reactant is – element or compound.

A CLOSER LOOK

Okay, we will come back to oxidation and reduction. First we have to learn how to determine oxidation numbers. Get out your periodic table.

Watch the video - Copy and paste this into the URL window in Mozilla Firefox:

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A CLOSER LOOK

In order to identify a redox reaction, you must be able to write the charges for each element. This is known as the oxidation number. I have done these three for you. You write the oxidation number above the compound and do the bookkeeping below.

K2Cr2O7 SO4-2 Al(NO3)3

+1 +6 -2 +6 -2

+3 +5 -2

K2Cr2O7 SO4-2 Al(NO3)3

+2 +12 -14 = 0 +6 -8 = -2

+3 +15 -18 = 0

OXIDATION NUMBER RULES

1. The oxidation number for any uncombined (free) element is zero (ex. O2, Na, Mg)

2. The oxidation number for a monatomic ion equals its ionic charge. (ex. Ca+2, O-2)

3. The oxidation number of the more electronegative atom is equal to the charge it would have if it were an ion. Remember that fluorine is the most electronegative element, followed by oxygen and nitrogen.

OXIDATION NUMBER

4. Some elements’ oxidation number corresponds to their position on the periodic table:

Elements in group 1A = +1 Elements in group 2A = +2 Aluminum is always +3 Fluorine is always –1 (most electronegative) Hydrogen has an oxidation number of +1 when

combined with nonmetals. If bonded to a metal, it is a –1 because it is more electronegative.

Oxygen has an oxidation number of -2 in most compounds and ions. Peroxides (O2

-1) are the exception and if oxygen is bonded to fluorine, the oxygen’s charge is a +2.

Here are some common oxidation numbers. Notice that nonmetals tend to have multiple oxidation numbers.

OXIDATION NUMBER

4. The sum of the oxidation numbers of all the atoms in a particle must equal the charge of that ion. (ex. SO4

-2)

5. The sum of all oxidation numbers for all atoms in a neutral compound is zero.

PRACTICE

Write the oxidation numbers for the following compounds. Write the numbers above each element.

1.HNO3 8. NO

2.H3PO4 9. CrI3

3.AgNO3 10. SO2

4.Cu(NO3)2 11. K2SO4

5. 2 Fe 12. H2S2O7

6.H2SO4 13. P4O6

7.KH2PO4 14. AsO4-3

ANSWERS

+1 +5 -2 +2 +5 -2 +1 +1 +5 -2

1.HNO3 4. Cu(NO3)2 7. KH2PO4

+1+5 -6=0 +2 +10 -12=0

+1+2 +5 -8=0

+1 +5 -2 0 +2 -

2

2. H3PO4 5. 2 Fe 8. NO

3. +3 +5 -8 =0 +2 -2 =0

+1 +5 -2 +1 +6 -2 +3 -1

3. AgNO3 6. H2SO4 9. CrI3

+1+5 -6=0 +2 +6 -8 =0 +3 -3 =0

ANSWERS

+4 -2

10. SO2

+4 -4 =0

+1 +6 -2 +3 -2

11. K2SO4 13. P4O6

+2+6 -8=0 +12 -12=0

+1 +6 -2 +5 -2

12. H2S2O7 14. AsO4-3

+2 +12 -14=0 +5 -8 = -3

REVIEW

Now, put it all back together – see how to determine what is oxidized and what is reduced.

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NOW, PRACTICE THIS

A. Determine the oxidation number for each atom.

B. State what is being oxidized and what is being reduced.

C. State the oxidizing agent and the reducing agent.

1. Cu + 2 AgNO3 Cu(NO3) 2 + 2 Ag

2. 3 H2S + 2 HNO3 3 S + 2 NO + 4 H2O

**REMEMBER: you must have something oxidized and something reduced in all redox equations

PRACTICE

Cu + 2 AgNO3 Cu(NO3) 2 + 2 Ag

PRACTICE

3 H2S + 2 HNO3 3 S + 2 NO + 4 H2O

ANSWERS

0 +1 +5 -2 +2 +5 -2 0

A. Cu + 2 AgNO3 Cu(NO3) 2 + 2 Ag

B. Cu 0 +2 lost electrons, oxidized

Ag +1 0 gained electrons, reducedC. Cu is oxidized; Cu is the reducing

agent Ag is reduced; AgNO3 is the oxidizing agent

ANSWERS

+1 -2 +1 +5 -2 0 +2 -2 +1 -2

A. H2S + 2 HNO3 3 S + 2 NO + 4 H2O

B. S -2 0 lost electrons, oxidized N +5 +2 gained electrons, reducedC. S is oxidized; H2S is the reducing

agent N is reduced; HNO3 is the oxidizing agent

DO NOW

Write the oxidation numbers for the following compounds. Write the charges above the elements:

1. H2SO4 2. Ca2(OH)PO4 3. ClO4-1

4. NaHCO3

State the charge changes, which compound lost or gained electrons, which is oxidized, which is reduced, and name the oxidizing and reducing agents:

2. Cu + AgNO3 Cu(NO3)2 + Ag

2. H2S + HNO3 S + NO + H2O

DO NOW ANSWERS

Oxidation Numbers +1 +6 -2 +7 -2

1. H2SO4 3. ClO4-1

+2 +6 -8 = 0 +7 -8 = -1

+2 -2 +1 +5 -2 +1 +1+4-2

2.Ca2(OH)PO4 4. NaHCO3

+4 -2 +1 +5 -8 = 0 +1 +1+4-6 = 0

DO NOW ANSWERS

0 +1+5-2 +2 +5 -2 0

1.Cu + AgNO3 Cu(NO3)2 + Ag

Cu 0 +2 lose e-, oxidized Cu = reducing agent

Ag +1 0 gain e-, reduced AgNO3 = oxidizing agent

DO NOW ANSWERS

+1 -2 +1+5-2 0 +2-2 +1 -2

2.H2S + HNO3 S + NO + H2O

S -2 0 lose e-, oxidized H2S = reducing

agent

N +5 +2 gain e-, reduced HNO3 = oxidizing

agent

APPLICATIONS

1. Corrosion This is the oxidation of a

metal caused by a reaction between the metal and some substance in the environment.

The best example is rusting.

Corrosion is a problem for it results in the loss of structural strength of the metal.

Can be prevented by coating the metal with paint, plastic or another metal.

APPLICATIONS

2. BleachingA chemical substance that is

used to eliminate unwanted color from fabrics and other materials.

Bleaches are oxidizing reagents because they remove electrons from the pigments that cause color.

Common bleaches are chlorine (Cl2), hydrogen peroxide (H2O2), and hypochlorite ion (ClO-1).

APPLICATIONS

3. Fuels and ExplosivesFuels release energy as they are

oxidized.

Common fuels (gasoline, natural gas) are composed largely of carbon and hydrogen.

Once ignited, they are oxidized by oxygen, forming water and carbon dioxide.

Nitroglycerin is both an oxidizer (hydrogen and oxygen) and a reducer (nitrogen).

APPLICATIONS

4. PhotographyPhotography involves the

capturing of a light image on a light-sensitive medium and the processing of the image to make a permanent record.

The process is based on the redox of silver halides, such as AgBr.

APPLICATIONS

5. Electrochemical cells (page 664)These involve a transfer of electrons from an

oxidized substance to a reduced substance.

If a zinc strip is in contact with copper(II) sulfate solution, the zinc strip (the anode) loses electrons to the copper(II) ions (the cathode) in solution.

The copper(II) ions accept the electrons and drop out of solution as copper atoms (now neutral).

APPLICATIONS

As the electrons are transferred, energy is released in the form of heat and the temperature rises.

If the two metals cannot touch and are separated, a transfer of electrical energy instead of heat accompanies the electron transfer.

To complete the circuit, electrons flow in one direction from metal to metal.

Then a salt bridge, or some other set-up, allows the ions to pass from one side to another, completing the electrical circuit.

APPLICATIONS

An electrochemical cell converts chemical energy to electrical energy by a spontaneous redox reaction. It was invented by Allesandro Volta and is also called a voltaic cell.

BALANCING REDOX EQUATIONS

Many redox reactions are easy to balance by the usual methods.

KClO3 KCl + O2

Some are harder.

Cu + HNO3 Cu(NO3)2 + NO2 + H2O


The OXIDATION NUMBER METHOD allows you to apply your knowledge of oxidation numbers in order to balance equations.

The fundamental principle in balancing redox equations is that the number of electrons lost in an oxidation process (increase in oxidation number) must equal the number of electrons gained in the reduction process (decrease in oxidation number).


Two methods:

1. Oxidation Number/Bracket Method

This will not always work if the reaction is complicated.

2. Half Reaction Method

This method always works but takes longer.

OXIDATION NUMBER METHOD

STEP ONE:

Assign oxidation numbers to all atoms in the equations. Write them above the element.

S + HNO3 SO2 + NO + H2O


STEP TWO:

Identify the element oxidized and the element reduced. Determine the change in oxidation number of each oxidized and reduced element.

STEP THREE:

Connect the atoms that change oxidation number using a bracket. Write the change in oxidation number at the midpoint.


STEP FOUR:

Choose coefficients that make the total increase in oxidation number equal the total decrease.

STEP FIVE:

Balance the remaining elements by inspection using the conventional method and check the final equation.

PRACTICE

1. Cu + HNO3 Cu(NO3)2 + NO2 + H2O

PRACTICE

2. Cl2 + KBr KCl + Br2

BALANCING REDOX REACTIONS

HALF REACTIONSThese reactions are used to balance redox equations. The reaction is broken into two parts – the oxidation part and the reduction part. The half –reaction method allows you to apply your knowledge of oxidation numbers in order to balance equations.

HALF REACTION METHOD

STEPS TO FOLLOW:

1. Write out the equation. Then change it to a net ionic equation if it is not one already, omit the spectator ions, and assign oxidation numbers to all atoms in the equation. Write them above the element.

HS-1 + IO3-1 I-1 + S + H2O


2. Write the separate half-reactions.

HS-1 S IO3-1 I-1

3. Balance all the elements except O and H (already balanced in this one).

HS-1 S IO3-1 I-1


4. If the oxygen is unbalanced, add enough water (H2O) to the side deficient in oxygen.

HS-1 S IO3-1 I-1 + 3H2O

5. Add sufficient hydrogen ions (H+) to the side deficient in hydrogen to balance the hydrogen.

HS-1 S + H+1 6H+1 + IO3-1 I-1 +

3H2O


6. Write the electrons in each half reaction.

-2 0

HS-1 S + H+1 + 2e-

+5 -1

6e- + 6H+1 + IO3-1 I-1 + 3H2O


7. Determine the least common multiple and multiply each to get it so that the number of electrons gained equals the number of electrons lost.

(x3) 3HS-1 3S + 3H+1 + 6e-

(x1) 6e- + 6H+1 + IO3-1 I-1 + 3H2O


8. Add the two half reactions together and return the spectator atoms. Delete anything that exactly occurs on both sides. (Notice how the water showed back up!)

3HS-1 + 6e- + 6H+1 + IO3-1 3S +

3H+1 + 6e-

+ I-1 + 3H2O

becomes 3H+1

3HS-1 + 3H+1 + IO3-1 3S + I-1 + 3H2O


PRACTICE (ACIDIC):

MnO4-1 + H2SO3 Mn+2 + HSO4

-1 + H2O

BALANCING REDOX REACTIONS

BASIC: Basic solutions add OH-1 to get rid of the H+1. Balance them the same way that you balance acidic solutions (H2O and then H+1) only add one more step.

Anywhere that you added an H+1, you need to add an OH-1 to both sides. Combine any OH-1 with any H+1 on the same side to made water.


PRACTICE (Basic):

I-1 + OCl-1 I2 + Cl-1 + H2O

PRACTICE (BASIC)

-1 -2 +1 0 -1 +1 -2

I-1 + OCl-1 I2 + Cl-1 + H2O

+1 0 -1 0

2e- + 2H+ + OCl-1 Cl-1 + H2O 2I-1 I2 + 2e-

2e- + 2H+ + OCl-1 + 2I-1 Cl-1 + H2O + I2 + 2e-

2H+ + OCl-1 + 2I-1 Cl-1 + H2O + I2


-1 -2 +1 0 -1 +1 -2

I-1 + OCl-1 I2 + Cl-1 + H2O +1 0 -1 0

2e- + 2H+ + OCl-1 Cl-1 + H2O 2I-1 I2 + 2e-

2e- + 2H+ + OCl-1 + 2I-1 Cl-1 + H2O + I2 + 2e-

2H+ + OCl-1 + 2I-1 Cl-1 + H2O + I2BASIC 2OH- + 2H+ + OCl-1 + 2I-1 Cl-1 + H2O +

I2 + 2OH-

2H2O

H2O + OCl-1 + 2I-1 Cl-1 + I2 + 2OH-

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OXIDATION-REDUCTION CHAPTER 19. OXIDATION-REDUCTION REACTIONS OXIDATION is the process by which a substance loses one or more electrons. Iron loses three