Chapter Outlinepostonp/ch222/pdf/Ch08-w15-pt3.pdf1 Chapter Outline 8.1 Solutions and Their...

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1 Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange

Transcript of Chapter Outlinepostonp/ch222/pdf/Ch08-w15-pt3.pdf1 Chapter Outline 8.1 Solutions and Their...

Page 1: Chapter Outlinepostonp/ch222/pdf/Ch08-w15-pt3.pdf1 Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization

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Chapter Outline

8.1 Solutions and Their Concentrations

8.2 Dilutions

8.3 Electrolytes and Nonelectrolytes

8.4 Acids, Bases, and Neutralization Reactions

8.5 Precipitation Reactions

8.6 Oxidation-Reduction Reactions

8.7 Titrations

8.8 Ion Exchange

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Section 8.6 - Oxidation-Reduction Reactions (electron transfer reactions)

Oxidation-Reduction Reactions (Redox):

Characterized by gain or loss of electrons by atoms involved in the reaction.

Oxidation:

Historical definition = gain of oxygen (sometimes with a simultaneous loss of hydrogen)

e.g. C2H6 + O2 CO2 + H2O

Modern definition = loss of electrons

Reduction:

Historical definition = loss of oxygen (sometimes with a simultaneous gain in hydrogen)

e.g. Fe2O3 + 3 CO 2 Fe + 3 CO2

Modern definition = gain of electrons

“oil rig”

Table 8.4 - Oxidation Number Rules

The charge the atom would have in a molecule (or an ionic

compound) if electrons were completely transferred, i.e.

applies to both ionic and covalent compounds.

1. The O.N. before a reaction for pure elements = 0

(including diatomics).

2. O.N = the charge on monovalent ions

3. O.N. of fluorine = -1 for all of its compounds

H H Mg

Mg

Cl

O O

Mg2+ + 2 e + e

Cl

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4. O.N. of oxygen = -2 in nearly all of its compounds

5. O.N. of hydrogen = +1 in nearly all of its compounds

6. O.N. values of the atoms in a neutral molecule sum up

to zero

7. O.N. values of the atoms in a polyatomic ion sum up to

the charge on the ion

O O + 2 e

O

2- Exceptions are peroxides

(O.N. = -1) and superoxides

(O.N. = -1/2), e.g. H2O2, KO2

H H 2 H+ + 2 e Exception = hydrides, e.g. CaH2

CaCl2

PO4

3-

+2 + 2(Cl) = 0 so Cl = -1

P + 4(-2) = -3 so P = +5

What are the oxidation states for sulfur in (a) S8, (b) SO2, (c)

Na2S, and (d) CaSO4 ?

Sample Exercise 8.9

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Electron Transfer in Redox Reactions

oxidation lose n electrons O.N. = +n

reduction gain n electrons O.N. = -n

What follows is a method that keeps track of the flow of electrons

within the atoms and molecules taking part in a chemical reaction.

Pb0 Pb2+ + 2 e

O.N. = final O.N. - initial O.N.

Cu2+ + 2 e Cu0

O.N. = (2 – 0) = +2

O.N. = (0 – 2) = -2

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What is the net ionic equation for the reaction of zinc metal plus cupric sulfate?

Oxidizing Agent =

Reducing Agent =

oxidizes something else while being reduced (Cu2+ Cu0)

reduces something else while being oxidized (Zn Zn2+)

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Sample Exercise 8.10: Identifying Oxidizing and Reducing Agents and Determining

Number of Electrons Transferred

Energy released by the reaction of hydrazine and

dinitrogen tetroxide is used to orient and maneuver

spacecraft and to propel rockets into space. Identify

the elements that are oxidized and reduced, the

oxidizing agent, and the reducing agent, and

determine the number of electrons transferred in

the balanced chemical equation.

2N + 4(-2) = 0

So N = +4

2N + 4(+1) = 0

So N = -2

(-2) (+1) (+4) (-2)

Element in its

natural state

(0)

O.N. rules

H = +1

O = -2

(+1) (-2)

(-2) (0) (+4)

O.N. = 0 – (-2) = +2

lost electrons oxidation

O.N. = (0 – 4) = -4

gained electrons reduction

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Net number of electrons transferred during the reaction:

= (#molecules) x (#atoms in molecule) x O.N.

2 x 2 x +2 = +8 (oxidation)

O.N. from

previous slide

1 x 2 x -4 = -8 (reduction)

O.N. from

previous slide

So the net number of electrons flowing is 8

Balancing Redox Reactions in Acid and Base (We’re going to use a simpler method than the book

called the Method of Half Reactions)

Examples of Half Reactions:

copper wire immersed in a silver nitrate solution:

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

(+1) (+2) (+5) (+5)

(0) (0)

(-2) (-2)

O.N. Cu = +2 – (0) = +2 oxidation

O.N. Ag = 0 – (1) = -1 reduction

The half-reactions are:

Cu(s)

AgNO3(aq)

(clear soln)

Ag(s)

Cu(NO3)2(aq)

(blue soln)

ox: Cu(s) Cu2+(aq) + 2 e

red: Ag+(aq) + e Ag(s)

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Examples of Half Reactions (cont’d):

copper wire immersed in a silver nitrate solution:

ox: Cu(s) Cu2+(aq) + 2 e

red: Ag+(aq) + e Ag(s)

The total reaction is obtained by adding the half reactions together, after

balancing electrons -

x 2

ox: Cu(s) Cu2+(aq) + 2 e

red: 2 Ag+(aq) + 2 e 2 Ag(s)

Net: Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

The next topic is how to balance a redox reaction that occurs in acid or base.

1. Determine the oxidation numbers for the reactants and compare to the products.

2. Write the oxidation and reduction half-reactions without electrons (yet)

3. Balance everything but oxygen and hydrogen

4. Balance oxygen by adding water

5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions,

continue as if in acidic solution, but at the end each H+ ion will be neutralized by

adding OH- ions

6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons

will be on the right, for the reduction half-reaction, the electrons will appear on the

left

7. Make sure the number of electrons in each half-reaction are the same. Then add

the half reactions together

8. Make sure that the equation is balanced for mass and for charge

NOTE: sometimes you have to cancel H2O’s that are on each side, as well

as H+ or OH-

Procedure for balancing a redox reaction in acid or base.

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Sample Exercise 8.11

Balancing Redox Equations I: Acidic Solutions

Microorganisms called denitrifying bacteria that grow in waterlogged soil convert

NO3- ions into N2O gas as they feed on dead plant tissue (empirical formula =

CH2O), converting it into CO2 and H2O. Write a balanced net ionic equation

describing this conversion of dissolved nitrates to N2O gas. Assume the reaction

occurs in slightly acidic water.

NO3-(aq) + CH2O(s) N2O(g) + CO2(g) + H2O(l) Given:

1. Determine the oxidation numbers for the reactants and compare to the products.

(+5) (0) (+1) (-2) (-2) (+1) (+4) (+1) (-2) (-2) (-2)

O.N. C = +4 – (0) = +4 oxidation

O.N. N = 0 – (5) = -4 reduction

2. Write the oxidation and reduction half-reactions without electrons (yet)

ox: CH2O(s) CO2(g)

red: NO3-(aq) N2O(g)

3. Balance everything but oxygen and hydrogen

4. Balance oxygen by adding water

ox: H2O + CH2O(s) CO2(g)

red: 2 NO3-(aq) N2O(g) + 5 H2O

ox: CH2O(s) CO2(g)

red: 2 NO3-(aq) N2O(g)

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5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions,

continue as if in acidic solution, but at the end each H+ ion will be neutralized by

adding OH- ions

ox: H2O + CH2O(s) CO2(g) + 4 H+(aq)

red: 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O

6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons

will be on the right, for the reduction half-reaction, the electrons will appear on the

left

ox: H2O + CH2O(s) CO2(g) + 4 H+(aq) + 4 e

red: 8 e + 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O

x 2

7. Make sure the number of electrons in each half-reaction are the same. Then add

the half reactions together

ox: H2O + CH2O(s) CO2(g) + 4 H+(aq) + 4 e

red: 8 e + 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O

Net: 2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 5H2O

NOTE: sometimes you have to cancel H2O’s that are on each side, as well

as H+ or OH-

2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 3H2O

2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 5H2O

2CH2O(s) + 2H+(aq) + 2NO3-(aq) 2CO2(g) + N2O(g) + 3H2O

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8. Make sure that the equation is balanced for mass and for charge

2CH2O(s) + 2H+(aq) + 2NO3-(aq) 2CO2(g) + N2O(g) + 3H2O

Mass: C = 2 C = 2

H = 6 H = 6

O = 8 O = 8

N = 2 N = 2

Charge: 2(+1) + 2(-1) = 0 = 0

Homework Problem 98(c)

Balancing Redox Equations II: Basic Solutions

Permanganate ion is used in water purification to remove oxidizable substances.

Complete and balance the reaction for the removal of sulfite ion.

MnO4-(aq) + SO3

2-(aq) MnO2(s) + SO42-(aq)

1. Determine the oxidation numbers for the reactants and compare to the products.

(+7) (+4) (+6) (+4)

2. Write the oxidation and reduction half-reactions without electrons (yet)

ox: SO32-(aq) SO4

2-(aq)

red: MnO4-(aq) MnO2(s)

O.N. S = +6 – (+4) = +2 oxidation

O.N. Mn = +4 – (+7) = -3 reduction

(-2) (-2) (-2) (-2)

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3. Balance everything but oxygen and hydrogen

4. Balance oxygen by adding water

5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions,

continue as if in acidic solution, but at the end each H+ ion will be neutralized

by adding OH- ions

ox: SO32-(aq) SO4

2-(aq)

red: MnO4-(aq) MnO2(s)

S and Mn are

already balanced

ox: H2O + SO32-(aq) SO4

2-(aq)

red: MnO4-(aq) MnO2(s) + 2H2O

ox: H2O + SO32-(aq) SO4

2-(aq) + 2H+(aq)

red: 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O

6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons

will be on the right, for the reduction half-reaction, the electrons will appear on the

left

7. Make sure the number of electrons in each half-reaction are the same. Then add

the half reactions together

ox: H2O + SO32-(aq) SO4

2-(aq) + 2H+(aq)

red: 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O

+ 2 e

3 e +

ox: H2O + SO32-(aq) SO4

2-(aq) + 2H+(aq) + 2 e

red: 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O 3 e +

x 3

x 2

Net: 3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4

-(aq)

3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O

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NOTE: sometimes you have to cancel H2O’s that are on each side, as well

as H+ or OH-

3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4

-(aq) 3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O

3SO32-(aq) + 8H+(aq) + 2MnO4

-(aq) 3SO42-(aq) + 6H+(aq) + 2MnO2(s) + H2O

3SO32-(aq) + 2H+(aq) + 2MnO4

-(aq) 3SO42-(aq) + 2MnO2(s) + H2O

2H2O

3SO32-(aq) + H2O(l) + 2MnO4

-(aq) 3SO42-(aq) + 2MnO2(s) + 2OH-(aq)

+ 2OH- + 2OH-

8. Make sure that the equation is balanced for mass and for charge

Mass: S = 3 S = 3

O = 18 O = 18

H = 2 H = 2

Mn = 2 Mn = 2

Charge: 3(-2) + 2(-1) = -8 3(-2) +2(-1) = -8

3SO32-(aq) + H2O(l) + 2MnO4

-(aq) 3SO42-(aq) + 2MnO2(s) + 2OH-(aq)