Figure 4.11 A summary of terminology for oxidation-reduction (redox) reactions.

X Y

e-

transfer or shift of electrons

X loses electron(s) Y gains electron(s)

X is oxidized Y is reduced

X is the reducing agent Y is the oxidizing agent

X increases its oxidation number

Y decreases its oxidation number

2

Oxidation-Reduction Reactions(electron transfer reactions)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-

2Mg + O2 2MgO

3

Table 4.3 Rules for Assigning an Oxidation Number (O.N.)

1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 02. For a monoatomic ion: O.N. = ion charge3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge.

General rules

Rules for specific atoms or periodic table groups

1. For Group 1A(1): O.N. = +1 in all compounds

2. For Group 2A(2): O.N. = +2 in all compounds

3. For hydrogen: O.N. = +1 in combination with nonmetals

4. For fluorine: O.N. = -1 in combination with metals and boron

6. For Group 7A(17): O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group

5. For oxygen: O.N. = -1 in peroxidesO.N. = -2 in all other compounds(except with F)

Sample Problem 4.6 Determining the Oxidation Number of an Element

PROBLEM: Determine the oxidation number (O.N.) of each element in these compounds:

(a) zinc chloride (b) sulfur trioxide (c) nitric acid

PLAN:

SOLUTION:

The O.N.s of the ions in a polyatomic ion add up to the charge of the ion and the O.N.s of the ions in the compound add up to zero.

(a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1.

(b) SO3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6.

(c) HNO3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.

Figure 4.10

Highest and lowest oxidation numbers of reactive main-group elements.

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

Zn is oxidizedZn Zn2+ + 2e-

Cu2+ is reducedCu2+ + 2e- Cu

Zn is the reducing agent

Cu2+ is the oxidizing agent

Copper wire reacts with silver nitrate to form silver metal.What is the oxidizing agent in the reaction?

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

Cu Cu2+ + 2e-

Ag+ + 1e- Ag Ag+ is reduced Ag+ is the oxidizing agent

8

NaIO3

Na = +1 O = -2

3x(-2) + 1 + ? = 0

I = +5

IF7

F = -1

7x(-1) + ? = 0

I = +7

K2Cr2O7

O = -2 K = +1

7x(-2) + 2x(+1) + 2x(?) = 0

Cr = +6

What are the oxidation numbers of all the elements in each of these compounds? NaIO3 IF7 K2Cr2O7

9

Types of Oxidation-Reduction Reactions

Combination Reaction

A + B C

2Al + 3Br2 2AlBr3

Decomposition Reaction

2KClO3 2KCl + 3O2

C A + B

0 0 +3 -1

+1 +5 -2 +1 -1 0

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Types of Oxidation-Reduction Reactions

Combustion Reaction

A + O2 B

S + O2 SO2

0 0 +4 -2

2Mg + O2 2MgO0 0 +2 -2

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Displacement Reaction

A + BC AC + B

Sr + 2H2O Sr(OH)2 + H2

TiCl4 + 2Mg Ti + 2MgCl2

Cl2 + 2KBr 2KCl + Br2

Hydrogen Displacement

Metal Displacement

Halogen Displacement

Types of Oxidation-Reduction Reactions

0 +1 +2 0

0+4 0 +2

0 -1 -1 0

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The Activity Series for Halogens

Halogen Displacement Reaction

Cl2 + 2KBr 2KCl + Br2

0 -1 -1 0

F2 > Cl2 > Br2 > I2

I2 + 2KBr 2KI + Br2

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The same element is simultaneously oxidized and reduced.

Example:

Disproportionation Reaction

Cl2 + 2OH- ClO- + Cl- + H2O

Types of Oxidation-Reduction Reactions

0 +1 -1

oxidized

reduced

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Ca2+ + CO32- CaCO3

NH3 + H+ NH4+

Zn + 2HCl ZnCl2 + H2

Ca + F2 CaF2

Precipitation

Acid-Base

Redox (H2 Displacement)

Redox (Combination)

Classify each of the following reactions.

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Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =moles of solute

liters of solution

What mass of KI is required to make 500. mL of a 2.80 M KI solution?

volume of KI solution moles KI grams KIM KI M KI

500. mL = 232 g KI166 g KI

1 mol KIx

2.80 mol KI

1 L solnx

1 L

1000 mLx

16

Preparing a Solution of Known Concentration

17

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

Dilution

Add Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=

18

How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

Mi = 4.00 M Mf = 0.200 M Vf = 0.0600 L Vi = ? L

Vi =MfVf

Mi

= 0.200 M x 0.0600 L4.00 M = 0.00300 L = 3.00 mL

Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.