1 Oxidation-Reduction Chapter 17 Hein and Arena Eugene Passer Chemistry Department Bronx Community...

1

Oxidation-Reduction Chapter 17

Oxidation-Reduction Chapter 17

Hein and Arena

Eugene PasserChemistry DepartmentBronx Community College

© John Wiley and Sons, Inc.

Version 1.1

2

Chapter Outline17.1 Oxidation Number

17.2 Oxidation-Reduction

17.3 Balancing Oxidation- Reduction Equations 17.6 Electrolytic and Voltaic

Cells

17.5 Activity Series of Metals

17.4 Balancing Ionic Redox Redox Equations

3

Oxidation NumberOxidation Number

4

The oxidation number (oxidation state) of an atom represents the number of electrons lost,

gained, or unequally shared by an atom.

5

Oxidation numbers can be zero, positive or negative.

6

An oxidation number of zero means the atom has the same number of electrons assigned to it as there are in the free neutral atom.

7

A positive oxidation number means the atom has fewer electrons assigned to it than in the neutral atom.

8

A negative oxidation number means the atom has more electrons assigned to it than in the neutral atom.

9

The oxidation number of an atom that has gained or lost electrons to form an ion is the same as the positive or negative charge of the ion.

NaCl

Sodium has lost an electron.

Chlorine has gained an electron.

The charge on sodium is +1.

The oxidation number of

sodium is +1.

The charge on chlorine is –1.

The oxidation number of

chlorine is -1.

10

In covalently bonded substances, oxidation numbers are assigned by an arbitrary system based on relative electronegativities.

11

For symmetrical covalent molecules each atom is assigned an oxidation number of 0 because the bonding pair of electrons is shared equally between two like atoms of equal electronegativity.

Electronegativity 2.1


Oxidation Number 0

Oxidation Number0

12

For symmetrical covalent molecules each atom is assigned an oxidation number of 0 because the bonding pair of electrons is shared equally between two like atoms of equal electronegativity.



Oxidation Number0

Oxidation Number0

13

When the covalent bond is between two unlike atoms, the bonding electrons are shared unequally because the more electronegative element has a greater attraction for them.



shared pair of electrons

unequal electron sharing

both shared electrons are assigned to chlorine

there is a partial transfer of an

electron to chlorine

after assignment chlorine has one more electron than neutral chlorine

Oxidation Number -1

after assignment hydrogen has one less electron than neutral chlorine

Oxidation Number+1

14

N2 N2O NO N2O3 NO2 N2O5

N oxidation number

0 +1 +2 +3 +4 +5 +5

Many elements have multiple oxidation numbers

-3NO

17

Step 1 Write the oxidation number of each known atom below the atom in the formula.

Step 2 Multiply each oxidation number by the number of atoms of that element in the compound.

Step 3 Write an expression indicating the sum of all the oxidation numbers in the compound. Remember: The sum of the oxidation numbers in a compound must equal zero.

Rules for Determining the Oxidation Number of an Element Within a Compound

18

Determine the oxidation number for sulfur in sulfuric acid.

H2SO4Step 1 -2+1

Write the oxidation number of each known atom below the atom in the formula.

Multiply each oxidation number by the number of atoms of that element in the compound.

Step 2 4(-2) = -82(+1) = +2

Step 3 +2 + S + (-8) = 0

Write an expression indicating the sum of all the oxidation numbers in the compound.

Step 4 S = +6 (oxidation number for sulfur)

19

Determine the oxidation number for sulfur in sulfuric acid.

Step 1 -2

Write the oxidation number of each known atom below the atom in the formula.

Multiply each oxidation number by the number of atoms of that element in the compound.

Step 2 4(-2) = -8

Step 3 2C + (-8) = -2 (the charge on the ion)

Write an expression indicating the sum of all the oxidation numbers in the compound.

2-2 4C O

Step 4 2C = +6 C = +3 (oxidation number for sulfur)

20

Oxidation-ReductionOxidation-Reduction

21

Oxidation-reduction (redox) is a chemical process in which the oxidation number of an

element is changed.

22

Redox may involve the complete transfer of electrons to form ionic bonds or a partial

transfer of electrons to form covalent bonds.

23

• Oxidation occurs when the oxidation number of an element increases as a result of losing electrons.

• Reduction occurs when the oxidation number of an element decreases as a result of gaining electrons.

• In a redox reaction oxidation and reduction occur simultaneously, one cannot occur in the absence of the other.

24

• Oxidizing agent The substance that causes an increase in the oxidation state of another substance.

• Reducing agent The substance that causes a decrease in the oxidation state of another substance.– The reducing agent is oxidized in a redox

reaction.

– The oxidizing agent is reduced in a redox reaction.

25

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

o + 2- 2+ 2- o4 4 2Zn + 2H + SO Zn + SO + H

The electron transfer is more clearly expressed as

The reaction of zinc with sulfuric acid is a redox reaction.

Zinc is oxidized.Zinc is the reducing agent.Zinc transfers electrons to hydrogen.Hydrogen is reduced.Hydrogen is the oxidizing agent.Hydrogen accepts electrons from zinc.

26

17.1

27

Balancing Oxidation-Balancing Oxidation-Reduction EquationsReduction Equations

28

Change-In-OxidationNumber Method

Change-In-OxidationNumber Method

29

oxidation number of tin increases

oxidation number of nitrogen decreases

Balance the equation Sn + HNO3 → SnO2+ NO2+H2O

Step 1 Assign oxidation numbers to each element to identify the elements being oxidized and those being reduced. Write the oxidation numbers below each element to avoid confusing them with ionic charge.

Sn + HNO3 → SnO2 + NO2 + H2O

0 +1 +5 -2 +4 -2 +4 -2 +1 -2

30


Step 2 Write two new equations, using only the elements that change in oxidation number. Then add electrons to bring the equations into electrical balance.

Sno → Sn4+ + 4e-

4N5+ + 1e- → N4+

+ 4e-

+ 1e-

31


Step 3 Multiply the two equations by the smallest whole numbers that will make the electrons lost by oxidation equal to the number of electrons gained by reduction.

Sno → Sn4+ + 4e-

4N5+ + 4e- → 4N4+4 4 4

32


Step 4 Transfer the coefficient in front of each substance in the balanced oxidation-reduction equations to the corresponding substances in the original equation.

Sn + 4HNO3 → SnO2 + 4NO2 + H2O4 4

Sno → Sn4+ + 4e-

4N5+ + 4e- → 4N4+4 4 4

33


Step 5 Balance the remaining elements that are not oxidized or reduced to give the final balanced equation.

Sn + 4HNO3 → SnO2 + 4NO2 + 2H2O2

34

Balancing Ionic Balancing Ionic Redox EquationsRedox Equations

35

In ionic redox equations both the numbers of atoms and the charges on both sides of the equation must be the same.

36

The Ion-Electron MethodThe Ion-Electron Method

37

Step 2 Balance the elements other than hydrogen and oxygen.

Write the two half-reactions that contain the elements being oxidized and reduced.

Step 1

38

Step 3 Balance hydrogen and oxygen.Acidic solution: • For reactions in acidic solution, use H+ and

H2O to balance oxygen and hydrogen.

• For each oxygen needed use one H2O.

• Then add H+ as needed to balance the hydrogen atoms.

39

Step 3 Balance hydrogen and oxygen.Basic solution: • For reactions in alkaline solutions, first

balance as thought the reactions were in an acid solution, using Steps 1-3.

• Then add as many OH- ions to each side of the equation as there H+ ions in the equation.

• Combine OH- ions into water.

• Rewrite the equation, canceling equal numbers of water molecules that appear on opposite side of the equation.

Example: 4H+ + 4OH- → 4H2O

40

Step 5 Since the loss and gain of electrons must be equal, multiply each half-reaction by the appropriate number to make the number of electrons the same in each half-reaction.

Add electrons (e-) to each half-reaction to bring them into electrical balance.

Step 4

41

Step 6 Add the two half-reactions together, canceling electrons and any other identical substances that appear on opposite sides of the equation.

42

Balance the equation

Step 1 Write two half-reactions, one containing the element being oxidized and the other the element being reduced (use the entire molecule or ion).

- 2- 2+ o4MnO + S Mn + S (acidic solution)

2- oS S

- 2+4MnO Mn

Balance elements other than oxygen and hydrogen. (Step 2 is unnecessary, since these elements are already balanced).

Step 2

43

Step 3 Balance O and H. The solution is acidic. The oxidation requires neither O nor H, but the reduction equation needs 4H2O on the right and 8H+ on the left.

2- oS S

4+

2- 2+8 + MnO Mn + H 4H O+ 8H + 2+ 4H O

4 water molecules are necessary to balance the 4 oxygens in MnO4.

-4MnO .

8H+ balance the 8 hydrogens of 4 water molecules. MnO4.




44


Step 4 Balance each half-reaction electrically with electrons:

net charge = +2 on each side


net charge = -2 on each side

2- o -S S e 2 -+ 2e

- - 2

2+4

+5 + + MnO Mn + 8H 4 He O- 5e +

balanced oxidation half-reaction

balanced reduction half-reaction

45

2- o -S S e 2


Step 5 Equalize loss and gain of electrons. In this case, multiply the oxidation equation by 5 and the reduction equation by 2.

o -2-5S 5S e 10

- 2+4

+ 2

- + + 2MnO 10e 2Mn + 16H 8H O


- - 2

2+4

+5 + + MnO Mn + 8H 4 He O

46


Step 6 Add the two half-reactions together, canceling the 10e- from each side, to obtain the balanced equation.

o -2-5S 5S e 10

- 2+4

2

- + + 2MnO 10e + 16 2Mn +H 8H O- 2- 2+4

+ o21 + 2MnO + 5S 2Mn + 5S +H 6 8H O


The charge on both sides of the balanced equation is +4

47


Step 1 Write two half-reactions, one containing the element being oxidized and the other the element being reduced (use the entire molecule or ion).

2-4 2 3 3CrO + Fe(OH) Cr(OH) + Fe(OH) (basic solution)

2 3Fe(OH) Fe(OH)

2-4 3CrO Cr(OH)

Step 2 Balance elements other oxygen and hydrogen (Step 2 is unnecessary, since these elements are already balanced).

48

Balance the equation Step 3 Balance O and H as though the solution were

acidic. Use H2O and H+.


Since the solution is basic, add 1 OH- to each side.

Combine H+ and OH- as H2O and rewrite, canceling H2O on each side.

2 23-

2H O OHFe(OH) + Fe(OH) O+ H+ -

2 3Fe(OH) + OH Fe(OH)

2 3+

2H O Fe(OH) + Fe(OH) + H ++ H2+ H O

2+

3- -

2H O OH Fe(OH) + Fe(OH+ H ) + + OH-+ OH -+ OH

oxidation half-reaction

For each unbalanced O add one H2O to the other side of the equation.

For each unbalanced H add one H+ to the other side of the equation.

49



2 3-2-

4CrO + Cr(OH) +4H O 5OH Rewrite, canceling 1 H2O from each side:

Step 3 Balance O and H as though the solution were acidic. Use H2O and H+.

2 +4 3 2-CrO + 5H Cr(OH) + H O++ 5H 2+ H O

Since the solution is basic, add 5 OH- to each side.

2-4 3

+ - -2CrO + + Cr(OH) 5H 5OH H O + + 5OH-+ 5OH -+ 5OH

2- -4 2 23CrO + Cr(5H O H O OH) + + 5OH

Combine 5H+ + 5OH- → 5H2O

For each unbalanced H add one H+ to the other side of the equation.

For each unbalanced O add one H2O to the other side of the equation.

oxidation half-reaction

50


Step 4 Balance each half-reaction electrically with electrons:




-2

2-4 3

-3eCrO + + Cr(OH) +4H O 5OH

2 3--Fe(OH) + OH Fe(OH) + e -+ e

-+ 3e

balanced oxidation half-reaction

balanced reduction half-reaction

51


Step 5 Equalize loss and gain of electrons.


2 3--Fe(OH) + OH Fe(OH) + e

-2

2-4 3


2 3--3Fe(OH) + 3OH 3Fe(OH) + 3e

Multiply the oxidation reaction by 3.

It is not necessary to multiply the reduction equation.

52

-2

2-4 3



Step 6 Add the two half-reactions together, canceling 3e- and 3OH- from each side of the equation.

-2 2 3

2-4 3+ 3Fe(OH) + 4H O Cr(OH) + 3FCrO e(OH) + 2OH

The charge on both sides of the balanced equation is -2


2 3--3Fe(OH) + 3OH 3Fe(OH) + 3e

53

Activity SeriesActivity Seriesof Metalsof Metals

54

activity series A listing of metallic elements in descending order of reactivity.

55

Sodium (Na) will displace any element below it from one of its compounds.

56

incr

easi

ng

acti

vity

Mg(s) + PbS(s) MgS(s) + Pb(s)

PbH2

KBaCaNaMgAlZnCrFeNiSn

Cu

Magnesium is above lead in the activity series.

Magnesium will displace lead from its compounds.

57

incr

easi

ng

acti

vity

Silver is below copper in the activity series.

Silver will not displace copper from one of its compounds.

Ag(s) + CuCl2(s) no reactionBa

PbH2

NaMgAlZnCrFeNiSn

CuAgHg

58

Electrolytic andElectrolytic andVoltaic CellsVoltaic Cells

59

electrolysis The process whereby electrical energy is used to bring about a chemical change.

electrolytic cell An electrolysis apparatus in which electrical energy from an outside source is used to produce a chemical change.

60

cathode The negative electrode.anode The positive electrode.

61

Electrolysis ofHydrochloric Acid

Electrolysis ofHydrochloric Acid

62

In an electrolytic cell electrical energy from the voltage source is used to bring about nonspontaneous redox reactions.

63

17.3

H3O+ + 1e- → Ho + H2O

Ho + Ho → H2

Cathode Reaction

Hydronium ions migrate to the cathode and are reduced.

64

17.3

Cl-→ Clo + e-

Clo + Clo→ Cl2

Anode Reaction

Chloride ions migrate to the anode and are oxidized.

65

The hydrogen and chlorine produced when HCl is electrolyzed have more potential energy than was present in the hydrochloric acid before electrolysis.

electrolysis2HCl(aq) H2(g) + Cl2(g) electrolysis

66

The Zinc-Copper Voltaic Cell

The Zinc-Copper Voltaic Cell

67

voltaic cell A cell that produces electrical energy from a spontaneous chemical reaction. (Also known as a galvanic cell).

68

When a piece of zinc is put in a copper(II) sulfate solution, the zinc quickly becomes coated with metallic copper. This occurs because zinc is above copper in the activity series.

69

Zn(s) + CuSO4(s) ZnSO4(s) + Cu(s)in

crea

sin

g ac

tivi

ty

Zinc is above copper in the activity series.

Zinc will displace copper from one of its compounds.

PbH2

KBaCaNaMgAlZnCrFeNiSn

Cu

70

If this reaction is carried out in a voltaic cell, an electric current is produced.

72

Zno(s) → Zn2+(aq) + 2e-anode oxidation

Cu2+(aq) + 2e- → Cuo(s) reductioncathode

loss of electrons

gain of electrons

Zno(s) + Cu2+(aq) → Zn2+(aq) + Cuo(s)

Net ionic reaction

Zno(s) + CuSO4(aq) → ZnSO4(aq) + Cuo(s)

Overall equation

73

The reactions of the zinc-copper cell are spontaneous. Spontaneous reactions occur in all voltaic cells. Reactions that occur in electrolytic cells are nonspontaneous. Spontaneity is the crucial difference between all voltaic and electrolytic cells.

74

Key Concepts

17.1 Oxidation Number

17.2 Oxidation-Reduction

17.3 Balancing Oxidation- Reduction Equations 17.6 Electrolytic and Voltaic

Cells

17.5 Activity Series of Metals

17.4 Balancing Ionic Redox Redox Equations

1 Oxidation-Reduction Chapter 17 Hein and Arena Eugene Passer Chemistry Department Bronx Community...

Documents

Transcript of 1 Oxidation-Reduction Chapter 17 Hein and Arena Eugene Passer Chemistry Department Bronx Community...