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:
Each O = -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.
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