So far:

• Unit I. Energy Transfer: energy (heat) was moved from one substance ( the one that was hot) to another substance (the one that was cold).

• Unit II. Proton Transfer: Proton(s) were moved from one substance (Acid) to another substance (Base).

Now:• Unit III. Electron Transfer: Electron(s) will be moved

from one substance (Reducing Agent) to another substance (Oxidizing Agent).

An Example of Oxidation - Reduction

I. Cu(s) + Ag+(aq) Cu2+

(aq) + Ag(s)

This equation is balanced for atoms but not for charge.

To balance the charge, the equation becomes:

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

(aq) + 2Ag(s)

In this reaction, two things have taken place:

A. Solid copper has changed to copper ions, and

B. Silver ions have changed to metallic silver

Review ion formation

Or (in symbol form)A. Cu(s) → Cu 2+

(aq)

B. Ag+(aq) → Ag(s)

These reactions are not balanced. Recall ion formation. In order to balance these reactions we must add one or more electrons to each equation.

+ 2e-

+ e-

These half reactions may be added together, just like we have done in the past. When adding half reactions, the number of electrons given up must be equal to the number of electrons received.

Called HALF REACTIONS

Cu(s) → Cu 2+(aq) + 2e-

Ag+(aq) + e- → Ag (s)2 [

]2 2 2

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

(aq) + 2 Ag(s)

OXIDATION HALF REACTION – the one in which electron(s) are given up (lost).

OXIDATION

REDUCTION HALF REACTION – the one in which electron(s) are received (gained).

REDUCTION

NET IONIC EQUATION

REDOX or NET IONIC EQUATION – abbreviation for complete oxidation / reduction reaction. These must occur together.

How can I remember which is oxidation and which is reduction?

O I L R I G

My name is

LEO the lion says GER

Now, rewrite the same equation in reverse order.

II. Cu 2+(aq) + Ag(s) → Cu(s) + Ag +

(aq)

This reaction is made up of the following two half reactions.

A. Cu2+(aq) + 2 e- → Cu(s)

B. 2 [ Ag(s) → Ag+(aq) + e- ] 2 Ag(s) → 2 Ag+ (aq) + 2 e-

Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag +

(aq)

Only one of the above two reactions (I & II) actually occurs. We will compare the two reactions using values from the data book.

This is the first reaction we looked at.

I. Cu(s) + Ag+(aq)

→ Cu2+(aq) + Ag(s)

Ag+(aq) + e- → Ag(s)

Cu (s) → Cu2+(aq) + 2e-

Cu(s) + 2Ag+(aq)

→ Cu2+(aq) + 2Ag(s)

Positive potential, spontaneous

+ 0.80 V- 0.34 V

2 x[ ]

+ 0.46 V

2 Ag+(aq) + 2e- → 2Ag(s)

This reaction has a positive net potential. This means that the reaction will occur as written. These are called

Spontaneous Reactions.

This is the second (reverse) reaction.

II. Cu2+(aq) + Ag(s) → Cu(s) + Ag+

(aq)

Cu2+

(aq) + 2e- → Cu(s)

+ 0.34 V

Ag(s) → Ag+(aq)

+ e- -0.80 V

Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+

(aq) -0.46 V

Negative potential, non-spontaneous

+

This reaction has a negative net potential. It will not occur aswritten. This type of reaction is known as a non-spontaneousreaction. It can be forced with an outside energy source.

2[ ]

NOTE: Half reaction potentials are NEVER doubled or tripled as per balancing.

OXIDIZING AGENTS & REDUCING AGENTS

OXIDIZING AGENTS – Accept (gain) electron(s).

- These substances cause an oxidation.

- They are reduced in the reaction.

REDUCING AGENTS – Give up (lose) electron(s).

- These substances cause a reduction.

- They are oxidized in the reaction.

Oxidizing agents and reducing agents are ALWAYS labelled on the reactant side of the equation.

Data Booklet

Oxidizing agents Reducing Agents

Strongest O.A.

Weakest O.A. Strongest R.A.

Weakest R.A.

An example

Cu(s) + Ag+(aq) Cu2+

(aq) + Ag(s)

O.A. accepts electrons

(gets reduced)

R.A. gives up electrons

(gets oxidized)

The reaction that takes place will always be between the strongest oxidizing agent present and the strongest reducing agent present.

Procedure for writing net Redox Reactions

• List all species present. Include water. List all ionic salts and acids in ionic form. Include all states of matter.

• Using your data book, look up the SOA present (left side of the table starting at the top). Copy this reaction out exactly as it appears in the table (forward) including the potential. This is your reduction half reaction.

• Using the right side of the table (starting at the bottom), locate the SRA present. Copy this reaction out in reverse order and change the sign of the potential. This is your oxidation half reaction.

• Make the number of electrons given up equal to the number of electrons gained.

• Now, add the two reactions together. Add the potentials as well. This is the net ionic equation or the redox reaction.

ExamplesFor each of the following; write the half reactions & the net ionic equation, label the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent, calculate the net potential, decide if the reaction is spontaneous and indicate the favored side.

1. A piece of zinc is placed in a solution of nickel(II)nitrate.

Species list

Zn(s)

H2O(l)

Ni2+(aq)

NO3-(aq)

SOA

SRA

2. A tin rod is placed into a solution of acidified KMnO4(aq) .

3. A solution of PbSO4 (aq) is stored in a magnesium bottle.

4. SnCl4(aq) is mixed with HCl(aq) in a copper cup.

5. Copper wire placed in HNO3(aq) .

Worksheets: 15-1,15-30,15-23

Sheet 15-23 Species Lists

1. Cu2+(aq)

H2O(l)

Cu(s)

Cl2(g)

Br2(l)

SO42-

(aq)

b. Al(s)

Fe(s)

H2O(l)

Cr3+(aq)

c. Sn(s)

H2O(l)

H+(aq)

NO3-(aq)

d. Cu(s)

H2O(l)

H+(aq)

Cl-(aq)

O2(g)

e. Hg(l)

H+(aq)

SO42-

(aq)

K+(aq)

MnO4–

(aq)

H2O(l)

f. Na(s)

H2O(l)

Which side is favored? Reactants or Products

1. All spontaneous reactions favor products.

(The reaction will run until all reactants are used up.)

2. All non spontaneous reactions favor reactants.

(No reaction takes place so all materials remain as reactants.)

Short cut procedure to determine if a reaction is spontaneous or not

1. Locate the strongest oxidizing agent present (left side of table) and the strongest reducing agent present (right side of table).

2. If the oxidizing agent is higher up on the page than the reducing agent reaction is spontaneous.

3. If the oxidizing agent is lower down on the page than the reducing agent reaction is non spontaneous.

Carry out the lab or view the demonstration for building an

activity series

Assigning Oxidation Numbers

• Oxidation number → +ve or –ve number assigned to an atom according to set of arbitrary rules.

• Written charge first – number second. Eg. +2, -3, etc.• Help us to identify oxidizing agents and reducing agents.• Help us to identify redox reactions.• Help us to balance complex redox equations.

The Rules

These rules are so important…I think that I might have a special show just

about them!

1. The oxidation number for all Standard State Elements is equal to 0. Examples:

K(s) H2(g), Na(s), Fe(s) , S8(s) , Br2(l) , Mg(s) etc.0 0 0 0 0 0 0

2. The oxidation number for all Simple Ions is equal to the charge of the ion. Examples:

Mg2+(aq), Fe3+

(aq), O2-(aq), K+

(aq), Cl-(aq) etc.

+2 +3 -2 +1 -1

3. The oxidation number for oxygen in a compound is = -2

(except in peroxides where it is -1) Examples:

H2O(l) , CO2(g) , C6H12O6(s) , AgNO3(aq) , H2O2(l) etc.-2 -2 -2 -2 -1

4. The oxidation number for Hydrogen in a compound is +1.

(except in metallic hydrides where it is -1) Examples:

A peroxide

H2O(l), C6H12O6(s), HCl(aq), Ca(OH)2(aq), NaH(s) etc.

+1 +1 +1 +1 -1Metallic hydride

5. The sum of the oxidation numbers in a compound is always equal to 0.

6. The sum of the oxidation numbers in a complex ion is always equal to the charge of the ion.

Example1. CO2(g)

-2 Rule 3

Now: solve for the oxidation number of Carbon.

2 (-2) + 1(x) = 0

x = +4

+4

Example 2. C6H12O6(s)

Rule 4 Rule 3

+1 -2

Solve for the oxidation number of Carbon.

12(+1) + 6(-2) + 6(x) = 0

x = 0

0

Example 3. SO4 2-(aq)

-2 (rule 3)

Now, solve for the oxidation number for sulphur.

4(-2) + 1(x) = -2

x = +6

+6

Provide oxidation numbers for each of the following

• P2O5

• NH4+

• Na2Cr2O7

• Ca(OH)2

• AgNO3

• Cu(s)• NaClO2

• Cu(NO3)2

• NH3

• N2H4

• SO22-

• H2SO4

• C3H8

• CO32-

• NaHSO4

• SnCl4• MnO2

• NH2OH

If you have 2 unknowns?First ionize the substance in water. Then work out the two resulting ions separately.

Example: CuSO4(aq)

CuSO4(s) → Cu2+(aq) + SO4

2-(aq)

Simple ion

+2

Solve as a complex ion

4(-2) + 1(x) = -2

x = +6

+6 -2

Provide oxidation numbers for each of the following

• P2O5

• NH4+

• Na2Cr2O7

• Ca(OH)2

• AgNO3

• Cu• NaClO2

• Cu(NO3)2

• NH3

• N2H4

• SO22-

• H2SO4

• C3H8

• CO32-

• NaHSO4

• SnCl4• MnO2

• NH2OH

ANSWERS

• P2O5

• NH4+

• Na2Cr2O7

• Ca(OH)2

• AgNO3

• Cu

• NaClO2

• Cu(NO3)2

• NH3

• N2H4

• SO22-

• H2SO4

• C3H8

• CO32-

• NaHSO4

• SnCl4

• MnO2

• NH2OH

+5 -2

-3 +1

+1 +6 -2

+1 +5 -2

+2 -2 +1

0

+1 +3 -2

+2 +5 -2

-3 +1

-2 +1

+2 -2

+1 +6 -2

+4 -2

+1 +1+6 -2

+4 -1

+4 -2

-1 +1 -2 +1

-8/3 +1

Oxidation number changes in Chemical Reactions

1. In any chemical equation, an increase in the oxidation number of an atom indicates that it is oxidized in the reaction.

2. Similarly, a decrease in oxidation number indicates that the atom is reduced in the reaction.

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

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

1. The oxidation number for copper increases. (it is oxidized)

2. The oxidation number for silver decreases. (it is reduced)

Use oxidation number changes to identify which element is oxidized and which is reduced.

C(s) + O2(g) → CO2(g)

0 0 +4 -2

Carbon is oxidized (RA) / Oxygen is reduced (OA)

Cl2(g) + HBr(aq) → HCl(aq) + Br2(l)

0 +1-1 +1-1 0

Bromine is oxidized (RA)/ Chlorine is reduced (OA)

Zn(s) + MnO2(s) + NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + NH3(g) + H2O(l)

0 +4 -2 -3 +1 -1 +2 -1 +3 -2 -3 +1 +1 -2

Zinc is oxidized (RA) / Manganeese is reduced (OA)

And some more!!

• N2O5

• MnO4-

• Al2(SO4)3

• Au(NO3)3

• Cr2O72-

• NaIO3

• CaCO3

• KH• H2O2

More Answers• N2O5

• MnO4-

• Al2(SO4)3

• Au(NO3)3

• Cr2O72-

• NaIO3

• CaCO3

• KH (metallic hydride)

• H2O2 (peroxide)

+5 -2

+7 -2

+3 +6 -2

+3 +5 -2

+6 -2

+1 +5 -2

+2 +4 -2

+1 -1

+1 -1

Identifying Redox Reactions….Which of these are Redox reactions?

1) N2O4(g) → 2NO2(g)

2) Cl2(g) + 2NaBr(aq) → 2 NaCl(aq) + Br2(l)

3) PbCl2(aq)+ K2SO4(aq) → 2KCl(aq) + PbSO4(aq)

4) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

5) 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

+4 -2 +4 -2

0 +1 -1 +1 -1 0

NO

YES

+2 -1 +1 +6 -2 +1 -1 +2 +6 -2

NO

+1 -2 +1 +1 +6 -2 +1 +6 -2 +1 -2

NO

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

YES