I II III IV V

Ch. 15 – Oxidation Reduction Equations

I. Oxidation and Reduction Defined

Redox Reactions

Historically meant reacting with oxygen

Oxidation : the loss of electrons:

Ca Ca2+ + 2 e –

The opposite reaction Reduction: the gain of electrons

Br2 + 2 e – 2 Br 1-

Oxidation & Reduction are 2 opposite processes which usually occur at the same time.

Redox Reactions When calcium and bromine get together redox ocurs: Ca + Br2 Ca2+

+ 2 Br1– One atom is oxidized the other is reduced Easy to remember which reaction gains and loses electrons my

remembering:

O I L R I G X S O E S A I S D I D S U N A C T T I I O O N N

Oxidation State One can determine how much oxidized by memorizing a simple set of

rules to determine the ‘oxidation state’.Rules for Determining Oxidation State

1. Any neutral element is zero2. Any alkali metal is + 13. Any alkaline earth in a compound is + 24. Oxidation state of oxygen in a compound is always -2, except in peroxides

where it is -15. Hydrogen in a compound is +1, except in hydrides where it is -1.

Rules for Determining Oxidation State

6. Variable Oxidation of atoms Sum of oxidation states must equal zero for neutral molecules OR must equal the value of the charge for ions

Examples: Sodium phosphate: Na2PO4

Oxidation state: +1 ? 2- Totals: +2 ? 8- =0 Charge on Phosphorous must be: +6

Sulfite: SO32-

Oxidation state: ? 2- Totals: ? 6- = 2- Charge on Sulfur must be: +4

Rules for Determining Oxidation State

7. More than one unknown oxdation state Determine which unknown atom is closest to the upper right hand corner of the periodic table. Assign that atom a negative oxidation state equal to the number of electrons it would need to have the electron configuration of a noble gas. Example:

NaSClO2

Oxidation state: +1 ? ? 2- Most upper right: +1 ? 1- 2- Totals: +1 ? 1- 4- = 0 Charge on Sulfur must be: +4

Examples:1) sodium nitrate NaNO3

2) ammonia NH3

3) zinc oxide ZnO

4) chromate ion CrO42-

5) calcium hydride CaH2

6) carbon dioxide CO2

7) nitrogen N2

8) sodium sulfate Na2SO4

9) aluminum hydroxide Al(OH)3

10) nitric acid HNO3

1) +1, +5, 2-

2) 3-, +1

3) 2+, 2-

4) +6, 2-

5) +2, 1-

6) 4+, 2-

7) 0

8) +1, +6, 2-

9) +3, -2, +1

10) +1, +5, 2-

Balancing Half Reactions Half reactions are equations that represent either

a loss or a gain of electrons from one compound. Steps:

1) Balance atoms other than hydrogen or oxygen.

2) Balance the oxygen by adding water molecules to one side.

3) Balance the hydrogen by adding hydrogen ions to one side.

4) Balance the charge by adding electrons to one side.

Example 1 Balance the half reaction: N2O2 NH4

+

1st atoms other than O & H: N2O2 2 NH4

+

2nd balance oxygen: N2O2 2 NH4

+ + 2 H2O

3rd balance hydrogen:

12H+ + N2O2 2 NH4+

+ 2 H2O 4th balance the charge (1) add up the charge on both sides

(2) add enough electrons to the most positive side to make the charge on that side equal the charge on the other side

10 e- + 12H+ + N2O2 2 NH4+

+ 2 H2O

+12 +2

Balance the half reactions ! 1. SO4

2- S2O32-

8 e- + 10H+ + 2 SO42- S2O3

2- + 5 H2O

2. NH4+1 NO2

1-

2 H2O + NH4+1 NO2

1- + 8 H+ + 6 e-

3. HSO31- S8

32 e- + 40H+ + 8 HSO31- S8

+ 24 H2O

4. Cl2 ClO31-

6 H2O + Cl2 2 ClO31- + 12 H+ + 10 e-

5. H3PO4 PH3

8 e- + 8H+ + H3PO4 PH3 + 4 H2O

Balancing Redox Equations by Half Reactions

An oxidation-reduction equation may be balanced by breaking it up into two half reactions, balancing each half reaction, and putting the two half reactions back together again.

Sample Problem: SO2 + MnO4

1- + H2O SO42- + MnO2 + H+

Solution: First pick the two half reactions. Each half reaction will

generally contain an element other than hydrogen and oxygen.

»SO2 SO42-

»MnO4 MnO2

Balancing Redox Equations Second, balance each half reaction as we did in the last section:

2 H2O + SO2 SO42- + 4 H+ + 2 e-

3 e- + 4H+ + MnO41- MnO2

+ 2 H2O

Next, multiply each half reaction by the lowest numbers that will make the electrons the same & add the 2 half reactions together. (2 H2O + SO2 SO4

2- + 4 H+ + 2 e-) x 3

(3 e- + 4H+ + MnO41- MnO2

+ 2 H2O) x 2

6 H2O + 3SO2 + 6e- + 8H+ + 2MnO41-

3SO42- + 12 H+ + 6e- + 2 MnO2

+ 4 H2O

+

Balancing Redox Equations

Finally simplify by canceling like terms (notice H2O, e-, & H+

on both sides)

6 H2O + 3SO2 + 6e- + 8H+ + 2MnO41-

3SO42- + 12 H+ + 6e- + 2 MnO2

+ 4 H2O

Solution: 2 H2O + 3SO2

+ 2MnO41-

»3SO42- + 4 H+ + 2 MnO2

Redox Problems Balance each of these redox reactions by the half

reaction method demonstrated in the previous sample

1) Cu + H+ + NO3- Cu2+ + NO2 + H2O

(Cu Cu2+ + 2 e-) x 1 (1 e- + 2H+ + NO3

1- NO2 + H2O) x 2

Cu +4H+ +2e-+ 2NO3- Cu2+ + 2e-+ 2NO2 + 2H2O

Solution: 1

Cu +4H+ + 2NO3- Cu2+ + 2NO2 + 2H2O

oxidized

reduced

reduced

oxidized

Redox Problems 2) Mn2+ + H+ + PbO2 MnO4

1- + Pb2+ + H2O

(4H2O + Mn2+ MnO41- + 8H+ + 5 e-) x 2

(2e- + 4H+ + PbO2 Pb2+ + 2H2O) x 5

8H2O + 2Mn2+ + 10e- + 20H+ + 5PbO2 2MnO41- +

16H+ + 10e- + 5Pb2+ + 10H2O

Solution: 2Mn2+ + 4H+ + 5PbO2 2MnO4

1- + 5Pb2+ + 2H2O

Redox Problems 3) I2 + HClO + H2O IO3

1- + Cl1- + H+

(6H2O + I2 2 IO31- + 12H+ + 10 e-) x 1

(2e- + H+ + HClO Cl1- + H2O) x 5

6H2O + I2 + 10e- + 5H+ + 5HClO

2IO31- + 12H+ + 10e- + 5Cl1- + 5H2O

Solution: H2O + I2 + 5HClO 2IO3

1- + 7H+ + 5Cl1-

oxidized

reduced