Redox Reactions

Oxidation Reduction ( Redox Reactions )

Objective A Given the formula for a neutral or charged molecule, determine the oxidation number for each atom in the molecule.

Oxidation Number: a charge assigned to an atom according to a set of rules. Its purpose is to help you keep track of electrons as they move from one atom or molecule to the next. They don't necessarily represent a true charge.

Rules:

1. The oxidation number of an atom of any free element is ZERO.

Element	Oxidation Number of each atom
sodium metal,Na	0
helium gas, He	0
Si	0

The oxidation number of any atom in a molecule of a single element is also ZERO.

Element	Oxidation Number of each atom
chlorine, Cl2	0
sulfur, S8	0
Phosphorous, P5	0

2. The oxidation number of hydrogen in H-containing compound is +1.

Compound	Oxidation number of each H atom	Total contribution by H
H2O	+1	+2
HNO3	+1	+1
H₂O₂	+1	+2

Exception: in metallic hydrides the oxidation number of hydrogen is -1.

a metal hydride has hydrogen bonded to a metal.

Compound	Oxidation Number of each H atom	Total contribution by H
CaH2	-1	-2
NaH	-1	-1
AlH3	-1	-3

A nice little mixture of what we have seen so far:

Compound	Oxidation Number of each H atom	Total contribution by H
H2S	+1	+2
MgH2	-1	-2
H2	0	0

3. The oxidation number of oxygen is -2.

Compound	Oxidation Number of each O atom	Total contribution by O
H2O	-2	-2
NO2	-2	-4
NO₃^-1	-2	-6

Exception: In peroxides which contain one extra oxygen, each oxygen is assigned an oxidation number of -1.

normal oxide: H₂O

peroxide: H2O2

4. The oxidation of a monoatomic ion is equal to its charge. For members of metallic families, the periodic table can be consulted. Also, halides ( a metal + a halogen) will be -1.

Compound	Oxidation Number of each metal atom	Total contribution by metal ion
Na2O	+1 = Na	+2
Al₂S₃	+3 = Al	+6
CaCl₂	+2 = Ca	+2

5. The sum of the oxidation numbers of all the atoms in a compound is ZERO.

The sum of all the oxidation numbers of all the atoms in a polyatomic ion is equal to the charge of the polyatomic ion.

Example: H₂O

Each hydrogen = +1; oxygen is -2.

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

Example: SO₄^-2

Each O = -2; S is unknown.

X + 4(-2) = -2.

X = 6. So sulfur's oxidation number is 6.

Additional Examples:

Use the five rules to assign oxidation numbers to each element in the following:

NO₂^-1

EachO = -2

X +2(-2) = -1

X = 3. So N's oxidation number is +3.

b. KMnO4

K = +1

O = -2

Mn = x

1 + x + 4(-2) =0

x = 7 = Mn's oxidation number

c. MnO₂

x + 2(-2) = 0

x = 4. Note how Mn's oxidation number ca vary from one compound to another.

C3H8

3x + 8(⁺1) = 0

x = -8/3 ( Yes. Oxidation numbers can be fractional!)

Exercises:

1. Use the five rules to assign oxidation numbers to each element in the following:

a. MnO4-1

b. C2H6

c. Xe

d. Fe3O4

e. OF2 (fluoride's rule takes precedence)

f. H2O2

g. CH4

h. Cr2O7-2

i. C2H6O

j. C2H2

k. XeOF4

l. Na2C2O4

m. Ca(NO3)2

n. UO2+2

o. NaBiO3

p. NH3

q. H3AsO4

r. Al(OH)-1

s. N2