The document discusses oxidation-reduction (redox) reactions, where there is a transfer of electrons between reactants. It defines oxidation as the loss of electrons and reduction as the gain of electrons. An example redox reaction and its net ionic form are provided. The document explains how to determine the oxidation states of elements and identifies common oxidation states of nonmetals. It describes how to write half-reactions by separating a redox reaction into its oxidation and reduction components.

1. OXIDATION-
REDUCTION
REACTIONS
Settle in, this is going to take a while…

2. What is redox?
 Reaction where there is a transfer of electrons
between reactants
 Oxidation involves the loss of electrons (OIL)
 Oxidation number/state of the element increases
 Oxidized element is the reducing agent
 Reduction involves the gain of electrons (RIG)
 Oxidation number/state of the element decreases
 Reduced element is the oxidizing agent

3. Example
Complete Reaction:
Mg + Zn(NO3)2  Mg(NO3)2 + Zn
Net-ionic Reaction:
Mg + Zn2+  Mg2+ + Zn
The magnesium metal was oxidized by the zinc
and the zinc was reduced by the magnesium.

4. Do what?!?!
 The oxidation state of the magnesium
changed from 0 to +2
 Oxidation state increased = oxidation
 Because magnesium gave its electrons away, it is
the reducing agent
 The oxidation state of zinc changed from +2 to
0
 Oxidation state decreased = reduction
 Because zinc took the electrons, it is the oxidizing
agent

5. How do you know oxidation
states?
 The oxidation number for any pure element is
zero.
 Group 1 metals form +1 ions, group 2 metals
form +2 ions, group 13 metals form +3 ions.
 Transition metals can be all kinds of oxidation
numbers (ranging from +1 to +7)
 Transition metal oxidation states can be
determined based on the nonmetal(s) it’s
bonded to…

6. Nonmetal oxidation states
 Fluoride is ALWAYS -1, the other halides are
usually -1.
 Oxide is usually -2, except when it’s in the
peroxide ion (-1) or bonded to fluorine (+2)
 Hydrogen is +1, unless it is the hydride ion (-1)

7. Putting it all together
 The total charge on a compound is zero, so all
oxidation numbers must cancel out.
 The total charge of elements in a polyatomic
ion must add to the charge on the ion

8. Practice
What is the oxidation number of each element in
the following compounds?
1. Zn(NO3)2
2. H2SO4
3. KMnO4
4. N2O4
5. PCl3

9. What’s the point?
 When an element gains electrons, another
element must accept those electrons
(Newton’s 3rd law).
 If you separate the reaction into half-reactions,
you can exploit this electron transfer to
generate electricity.
 The study of this is electrochemistry, but more
on that later…

10. Half-Reactions?
 You can separate a redox reaction into the
reduction reaction and the oxidation reaction.
 First you have to identify which element is
oxidized and which is reduced.
 So let’s practice identification first:

11. Practice
Determine the oxidation states of all elements in
the following reactions and then identify which
element is oxidized and which is reduced.
N2 + 3H2  2NH3
2MnO2 + Zn + 2H2O  2MnO(OH) + Zn(OH)2
AgNO3 + Cu  Cu(NO3)2 + Ag

12. N2 + 3H2  2NH3

13. 2MnO2 + Zn + 2H2O  2MnO(OH) +
Zn(OH)2

14. AgNO3 + Cu  Cu(NO3)2 +
Ag

15. Separating reactions
 Once the oxidized and reduced elements have
been identified, separate the reactions.
 Use net ionic reactions instead of complete
reactions
2AgNO3 + Cu  Cu(NO3)2 + 2Ag
2Ag+1 + Cu  Cu2+ + 2Ag

16. 2Ag+1 + Cu  Cu2+ + 2Ag
 The silver is reduced, so that is the reduction
reaction:
2Ag+1  2Ag
 The masses are balanced, but the charges are
not, so add the electrons being transferred:
2Ag+1 + 2e-  2Ag
 Notice that the reduction half reaction has
electrons as reactants

17. 2Ag+1 + Cu  Cu2+ + 2Ag
 The copper is oxidized, so that is the oxidation
reaction:
Cu  Cu2+
 The masses are balanced, but the charges are
not, so add the electrons being transferred:
Cu  Cu2+ + 2e-
 Notice that the oxidation half reaction has
electrons as products