1

Chapter 5 Chemical Reactions

5.3Oxidation-Reduction Reactions

2

Oxidation and Reduction

An oxidation-reduction reaction

• provides us with energy from food.• provides electrical energy in

batteries.• occurs when iron rusts.

4Fe(s) + 3O2(g) 2Fe2O3(s)

3

An oxidation-reduction reaction

• transfers electrons from one reactant to another.

• loses electrons in oxidation. (LEO) or (OIL)

Zn(s) Zn2+(aq) + 2e- (loss of electrons)

• gains electrons in reduction. (GER) or (RIG)Cu2+(aq) + 2e- Cu(s) (gain of electrons)

Electron Loss and Gain

4

Oxidation and Reduction

5

Zn and Cu2+

Zn(s) Zn2+(aq) + 2e- oxidationSilvery metal

Cu2+(aq) + 2e- Cu(s) reduction Blue orange

6

Electron Transfer from Zn to Cu2+

Oxidation: electron loss

Reduction: electron gain

7

Identify each of the following as 1) oxidation or 2) reduction.

__A. Sn(s) Sn4+(aq) + 4e−

__B. Fe3+(aq) + 1e− Fe2+(aq)

__C. Cl2(g) + 2e− 2Cl-(aq)

Learning Check

8

Identify each of the following as 1) oxidation or 2) reduction.

1 A. Sn(s) Sn4+(aq) + 4e−

2 B Fe3+(aq) + 1e− Fe2+(aq)

2 C. Cl2(g) + 2e− 2Cl-(aq)

Solution

9

Write the separate oxidation and reduction reactions for the following equation.

2Cs(s) + F2(g) 2CsF(s)

A cesium atom loses an electron to form cesium ion.

Cs(s) Cs+(s) + 1e− oxidation

Fluorine atoms gain electrons to form fluoride ions.

F2(s) + 2e- 2F−(s) reduction

Writing Oxidation and Reduction Reactions

10

In light-sensitive sunglasses, UV light initiatesan oxidation-reduction reaction.

uv lightAg+ + Cl− Ag + Cl

A. Which reactant is oxidized?

B. Which reactant is reduced?

Learning Check

11

Solution

In light-sensitive sunglasses, UV light initiatesan oxidation-reduction reaction.

uv lightAg+ + Cl− Ag + Cl

A. Which reactant is oxidized? Cl− Cl + 1e−

B. Which reactant is reduced? Ag+ + 1e− Ag

12

Learning Check

Identify the substances that are oxidized and reduced ineach of the following reactions.

A. Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)

B. 2Al(s) + 3Br2(g) 2AlBr3(s)

13

Solution

A. Mg is oxidized Mg(s) Mg2+(aq) + 2e−

H+ is reduced2H+ + 2e− H2

B. Al is oxidized Al Al3+ + 3e−

Br is reducedBr + e− Br −

14

Common uses of the terms oxidization and reduction

Term Meaning

Oxidation To combine with oxygenTo lose hydrogenTo lose electronsTo increase in oxidation number

Reduction To lose oxygenTo combine with hydrogenTo gain electronsTo decrease in oxidation number

15

Oxidation Number (O.N.)

Oxidation number or oxidation states – Positive or negative number assigned to the elements in chemical formulas according to a specific set of rules.

There are 7 rules that you need to know in order to determine the O.N.

Reducing Agent – The substance that contains an element that is oxidized during a chemical reaction.

Oxidizing Agent – The substance that contains an element that is reduced during a chemical reaction. (note: When one element in a molecule or ion is the oxidizing or reducing agent, the convention is to refer to the entire molecule or ion by the appropriate term.)

16

Oxidation Number Rules

Rule 1 – The oxidation number (O.N.) of any uncombined element is 0.

examples: Al(0), O2(0), Br2(0), and Na(0) Rule 2 – The O.N. of a simple ion is equal to the

charge of the ion. examples: Na+ (+1), Mg2+ (+2), S2- (-2) and Br- (-1) Rule3 – The O.N.s of group IA(1) and IIA(2) elements

are +1 and +2, respectively. examples: Na2CO3 (Na = +1), Sr(NO3)2 (Sr = +2), and CaCl2 (Ca = +2) Rule 4 – The O.N. of hydrogen is +1. examples: HCl (H = +1) and H3PO4 (H=+1)

17

Oxidation Number Rules Cont’.

Rule 5 – The O.N. of oxygen is -2 except in peroxides, where it is -1. examples: CaO (O = -2), H2SO4 (O = -2), H2O (O = -2), and H2O2 (O = -1) Rule 6 – The algebraic sum of the O.N.s of all atoms in a

complete compound formula equals zero. ex: K2CO3: 2(O.N. of K) + (O.N. of C) + 3(O.N. of O) = 0 2(+1) +4 3(-2) = 0 +2 +4 -6 = 0 Rule 7 – The algebraic sum of the O.N.s of all atoms in a

polyatomic ion is equal to the charge of the ion. ex. MnO4

- : (O.N. of Mn) + 4(O.N. of O) = -1 +7 4(-2) = -1 +7 -8 = -1

18

Assign oxidization numbers to the each element in the following:

A. CO2

B. Mg(NO3)2

Learning Check

19

Solution

A. O.N.s O is -2 (Rule 5) and C is +4 (Rule 6)

(O.N. of C) + 2(O.N. of O) = 0 (O.N. of C) + 2(-2) = 0 (O.N. of C) + (-4) = 0 (O.N. of C) = +4

B. O.N.s Mg is +2 (Rule 3), O is -2 (Rule 5), and N +5 using (Rule 6)

(O.N. of Mg) + 2(O.N. of N) + 6(O.N. of O) = 0 (+2) + 2(O.N. of N) + 6(-2) = 0 (+2) + 2(O.N. of N) + (-12) = 0 2(O.N. of N) – 10 = 0 2(O.N. of N) = +10 (O.N. of N) = +10/2 = +5

20

Determine oxidation number for each atom represented in the following equations and identify the oxidizing and reducing agents:

A. 4Al(s) + 3O2(g) 2Al2O3(s)

B. CO(g) + 3H2(g) H2O(g) + CH4(g)

Learning Check

21

Solution

A. 4Al(s) + 3O2(g) 2Al2 O3(s) 0 0 +3 -2

The O.N. of Al has changed from 0 to +3. Therefore, Al has been oxidized and is the reducing agent. The O.N. of O has decreased from 0 to -2. The oxygen has been reduced and is the oxidizing

agent.

B. C O(g) + 3H2(g) H2 O(g) + C H4(g) +2 -2 0 +1 -2 -4 +1

The O.N. of H2 increased from 0 to +1. H2 has been oxidized and is the reducing agent. The O.N. of C has decreased from +2 to -4. CO has been reduced and is the oxidizing agent.

22

Find the element with the highest oxidation number in each of the following formulas:

A. Na2Cr2O7

B. P2O5

C. HClO2

Learning Check

23

Solution

A. Cr (+6)

B. P (+5)

C. Cl (+3)