Lesson #1: Assigning Oxidation Numbers.
Oxidation Numbers: the real or apparent charge of an atom or ion when all bonds are
assumed to be __________.
Q: Why assume bonds are ionic if they are not?
A: Oxidation numbers allow us to determine if a substance has __________ or
__________ _____________. Oxidation numbers also allow us to determine the
___________ of ____________ that have been transferred.
Rules for Assigning Oxidation Number
1. The oxidation number of an atom in the elemental state is zero.
e.g.,
______________________________________________________________________.
2. The oxidation numbers in binary ionic compounds are simply the charges.
e.g.,
______________________________________________________________________
______________________________________________________________________.
3. The sum of the oxidation numbers in a compound add to zero.
e.g.,
______________________________________________________________________.
4. The sum of the oxidation numbers in an ion add to the total charge on the ion.
e.g.,
______________________________________________________________________.
5. The oxidation number of hydrogen is +1, except when hydrogen forms
compounds called hydrides. In the case of hydrides the oxidation number of
hydrogen is –1.
e.g.,
______________________________________________________________________
______________________________________________________________________.
6. The oxidation number of oxygen is –2 except in peroxides when it is –1.
e.g.,
______________________________________________________________________.
Calculating Oxidation Numbers
•
Be sure to assign the oxidation numbers of Oxygen and Hydrogen first before
determining the apparent charges of the remaining atoms.
Determine the oxidation number for each atom:
1. SO3-2
2. H2SO4
Atoms
Total oxidation value
1S
1(X) = X
3O
SO3
3(-2) = -6
-2
Atoms
Total oxidation value
___________________________________
3. H4B3O7-1
Atoms
Total oxidation value
___________________________________
4. KMnO4
Atoms
Total oxidation value
___________________________________
Oxidation Reduction and Half Cells
Oxidation: When a chemical species loses electrons. When oxidation occurs, the
oxidation number of the species increases.
Reduction: When a chemical species gains electrons. When reduction occurs, the
oxidation number of the species decreases.
Consider the following single replacement reaction
Zn(s) + CuSO4(aq) -------------------
ZnSO4(aq) + Cu(s)
Spectator ions do not participate in either the oxidation or reduction process, so let’s
eliminate them for simplicity.
Each net ionic equation can be written as two _______ ____________. By writing halfequations, we can observe the _____________ that are lost (_______________) or
gained (_______________) in the REDOX reaction.
Since Zinc starts with an oxidation number of _______ and ends with _______, the zinc
must have been ______________.
Since Copper starts with an oxidation number of _______ and ends in _______, the
copper must have been _____________.
Memory Aid:
GER says LEO the lion.
Rules for Half Cells
1. A Reduction half-reaction must always be accompanied by an Oxidation
half-reaction.
Q: Why?
A:
2. The number of electrons gained in a REDOX reaction must always be equal to the
number lost.
Q: Why?
A:
Exercises: Write the half-reactions for the following net ionic equations.
1.
Pb(s) + Cu+2(aq) ----------------
2.
Zn(s) + 2Ag+(aq) ------------------
Pb+2(aq) + Cu(s)
Zn+2(aq) + 2Ag(s)