REDOX AND ELECTROCHEMISTRY
Oxidation Number
A. Convenient way for keeping track of
the number of electrons transferred in a
chemical reaction
B. Oxidation numbers are either: + , -, or
0
Rules for assigning oxidation numbers
1. Each atom of a free element has an
oxidation number of zero
Ex: Na  Nao
Na in NaCl is not a free element and does
not have a charge of zero
H2 each H-atom has a charge of zero
2. For any element in which only one
oxidation number is present use that
number.
3. Oxygen has an oxidation number of –2
EXCEPT in peroxides (H2O2) where it is
–1 and in compounds with fluorine where
it may be +1 or +2
Example in H2SO4 oxygen has an oxidation
state of –2
A peroxide may occur when hydrogen or a
group one or two reacts with oxygen, in
this case there is more oxygen than in a
normal metal oxide and the oxidation
state is -1
4. Hydrogen has a +1 oxidation state in all
compounds EXCEPT metal hydrides (LiH,
CaH2) where its oxidation state is -1
Example in H2SO4 H has a charge of +1
5.
If the compound is ionic use the first
negative oxidation state for the
nonmetal.
example: Fex Cl3-1
x+3(-1)=0
x=+3 oxidation state of Fe is
+3
HOW TO ASSIGN OXIDATION NUMBERS
IN COMPOUNDS
1. Use the “rules” to identify the oxidation
numbers of all atoms for which a rule
exists
2. Multiply the oxidation number of each
atom by its subscript
3. The sum of all oxidation number times
the subscripts must be zero
4. Example: Na2S
Na is group 1 and has an oxidation number
of +1 and there is no rule for S
Na2 +1 Sx
2(+1) + x =0
x=-2 so the oxidation state of S is -2
5. For polyatomic ions follow the same rules
and procedure used for a compound
except the sum of the (oxidation
numbers X subscript) must equal the
charge on the ion.
6. Example (SO4) -2
Oxygen has an oxidation number of –2 so
SxO4 -2
x-8=-2
x=+6 so the oxidation state of S
in (SO4) -2 is +6
If a compound consists of 2 polyatomic ions
break it into the two ions and solve
separately.
Example
(NH4)2SO4
(NH4)+1
(SO4)-2
NxH4+1
SxO4-2
X+4=+1
x-8=-2
X=-3
x=+6
So the oxidation state of N is -3 and of
S is +6
REDOX
 Oxidation and Reduction result from the
competition for electrons between atoms
A. Oxidation
1. Represents a loss of electrons
2. Refers to any chemical change in
which there is an increase in
oxidation number
3. The particle that increases in oxidation
number is said to be oxidized.
4. Reducing Agent -the substance that is
oxidized
5. Examples
Ca + Cl2 CaCl2
Assign oxidation numbers
Ca0 +Cl2 0  Ca+2Cl2-1
 Calcium, Ca0, loses electrons to the
chlorine atom, its oxidation state increases
from 0 to +2, therefore the calcium, Ca0
is oxidized or it is said to be the
reducing agent
B. Reduction
1. Represents a gain of electrons
2. Refers to any chemical change in
which there is a decrease in the
oxidation number
3. The particle that decreases in oxidation
number is said to be reduced.
4. Oxidizing Agent –The substance that
is reduced
5. Examples
Ca + Cl2  CaCl2
Assign oxidation numbers
Ca0 +Cl2 0  Ca+2Cl2-1
The Cl0 gains electrons from Ca0, its
oxidation state decreases from 0 to –1,
therefore Cl0 is reduced or it is said to be
the oxidizing agent
Redox Reactions- Reactions that involve
both oxidation and reduction
All types of reactions EXCEPT double
replacement reactions can be a redox
reaction.
Double replacement reactions are NEVER
redox reactions.
How to analyze a redox reaction.
1. Assign oxidation numbers to all atoms
Example
2H2 + O2 2H2O
0
0 +1 –2
2. Show the changes in oxidation number
for the atoms involved. Use a line to
connect the atoms undergoing oxidation
and those undergoing reduction
ox
RED
2H2 + O2 2H2O
0
0
+1 –2
3. Indicate which element is oxidized and
which is reduced
 H2- increases its oxidation number
therefore it is oxidized so it is the reducing
agent
 O2- decreases its oxidation number
therefore it is reduced so it is the oxidizing
agent


ONLY REACTANTS CAN BE THE
SUBSTANCE OXIDIZED OR REDUCED.
The best reducing agents are group I and
2 metals because they want to lose
electrons the most. Using Table J the best
reducing agents are found at the top of
the table.

The best oxidizing agents are the most
electronegative elements such as
fluorine and oxygen because they want to
gain electrons the most. Using table J the
best oxidizing agents are found at the
bottom of table J
ELECTROCHEMISTRY
A. Half –Reactions
1. Every redox reaction consists of 2
parts each called a half reaction
a. Oxidation half reaction shows an
atom or ion losing one or more
electrons
b. Reduction half reaction shows an
atom or ion gaining one or more
electrons
2. Separate electron equations can be
written for each
3. Like other chemical reactions, half
reactions must follow the law of
conservation of mass. Atoms must
be balanced.
4. In addition to conservation of mass,
there must also be a conservation of
charge.
a. The net charge must be the same on
both sides of the equation but it does not
necessarily equal zero
b. Charge is balanced by adding
electrons to one side of the half reaction
5. Example:
1. Assign oxidation states and identify the
substance oxidized and reduces.
Mg + Cl2  MgCl2
0
0
+2 -1
2. Write oxidation half reaction
a. balance for mass by adding
coefficients
b. balance for charge by adding
electrons
Oxidation half reaction: Mgo  Mg+2 + 2e3. Write the reduction half reaction
a. balance for mass by adding
coefficients
b. balance for charge by adding
electrons
Reduction half reaction: Cl2o + 2e-  2Cl-
B. Balancing simple redox reactions
1. Write the oxidation and reduction half
reactions (see previous notes)
Example
__Pb +__ Cr+3 __ Pb+2 +__ Cr
Oxidation half reaction:
Pbo  Pb+2 + 2eReduction half-reaction:
Cr+3 + 3e-  Cro
2. Balance the half reactions by adding
coefficients such that the number of
electrons in both equations are equal
3(Pbo Pb+2+2e-) = 3Pbo3Pb+2 + 6e2(Cr+3 + 3e- Cro) = 2Cr+3 + 6e- 2Cro
3. Add the two half reactions together
3Pbo3Pb+2 + 6e+2Cr+3 + 6e-  2Cro
3Pbo + 2Cr+3  3Pb+2 + 2Cro
D. Calculating Net Potential (voltage) of a
Redox Reaction
1. Write both the oxidation and reduction
half reactions
2. Go to the table of Reduction potential
3. For the reduction half reaction used
the E0 value as written
4. For the oxidation half reaction negate
the E0 value
5. Add the E0 values
6. Meaning of E0
a. A positive E0 means the reaction is
spontaneous in the direction written
b. A negative E0 means the reaction is not
spontaneous in the direction it is written
c. An E0 equal to zero indicates the reaction is
in equilibrium


Examples Calculate E0 for the following
Zn + CuSO4  ZnSO4 + Cu