REDOX REACTIONS
(reduction/oxidation)
REDOX REACTIONS
 Involve
the exchange of
electrons in a chemical reaction
 Electrons are lost by one
substance and gained by
another substance
 One substance goes through
oxidation, while the other
substance goes through
reduction
REDUCTION
 gain
of electrons
 the gain of hydrogen
 the loss of oxygen
 a decrease in oxidation number for
the substance being reduced
OXIDATION
 loss
of electrons
 the gain of oxygen
 the loss of hydrogen
 an increase in oxidation number for
the substance being oxidized
REDUCING AGENT
 experiences
oxidation
 is an electron donor
OXIDIZING AGENT
 experiences
reduction
 is an electron acceptor
OXIDATION STATES
Also known as oxidation numbers
 Allows understanding of what is
oxidized and what is reduced
 Imaginary charges that atoms would
have if shared electrons were
divided equally in a covalent bond
 Or real charges that monatomic ions
have in an ionic bond

ASSIGNING OXIDATION STATES
 Written
directly above a symbol with
the sign and then the
number…unlike charges
ASSIGNING OXIDATION STATES
 The
oxidation state of an atom of an
element in its natural state is zero.
0
Na(s)
0
Br2(l)
0
Cl2(g)
0
C(s)
ASSIGNING OXIDATION STATES
 The
oxidation state of a monatomic
ion is equal to its charge.
+1
+1
+2
+3
Al3+(aq)
Na1+(aq)
Fe2+(aq)
-1
NaCl(g)
ASSIGNING OXIDATION STATES
 Oxygen
is assigned an oxidation
number of -2 in compounds; an
exception is found in the
peroxide ion, O22-, where each
oxygen is assigned an oxidation
number of -1.
+1
-2
Na2O
+3
-2
Fe2O3
ASSIGNING OXIDATION STATES
 Hydrogen
is assigned an oxidation
number of +1 in its covalent
compounds with nonmetals. In
compounds with metals, the
oxidation number of hydrogen is
-1.
+1 -2
H2O
+1 -1
HCl
ASSIGNING OXIDATION STATES
 The
sum of the oxidation states in a
compound must be zero, as a
compound’s charge is zero.
+1 -2
+3 -2
Na2O
Fe2O3
+1 -2
+1 -1
H2O
HCl
ASSIGNING OXIDATION STATES
 The
sum of the oxidation states in a
polyatomic ion must be equal to
that ion’s charge.
-2 +1
OH1+2-3
CN1-
+4 -2
SO32+7-2
ClO41-
ASSIGNING OXIDATION STATES
 Non-integer
oxidation states do
exist and indicate the average
division of electrons among the
elements.
+8/3 -2
Fe3O4
ASSIGNING OXIDATION STATES
+2 -2
+4 -2
+2 -2
+4 -2
CO
NO
+3 -2
NO21-
CO2
NO2
0
N2
OXIDATION STATES METHOD OF
BALANCING
 Assign
oxidation numbers to
every atom in the reaction
 Connect the atoms involved in
oxidation with a line above the
reaction.
 Connect the atoms involved in
reduction with a line beneath
the reaction.
OXIDATION STATES METHOD OF
BALANCING
 Write
the change in oxidation
number on each line with
parentheses around it.
 Determine the least common
multiple of the changes.
 Use multipliers to achieve the
LCM.
 The multipliers guide you in
determining your coefficients.
OXIDATION STATES METHOD OF
BALANCING
1 (+2)
+2 -2
+2-2
0
+4-2
PbO(s) + CO(g)  Pb(s) + CO2(g)
1 (-2)
OXIDATION STATES METHOD OF
BALANCING
1 (+2)
+4 -1
+2 -1
+3 -1
+4 -1
2 CeCl4 + SnCl2  2 CeCl3 + SnCl4
2 (-1)
OXIDATION STATES METHOD OF
BALANCING
3 (+1)
+1-1
+2 -1
+1 +5-2
+3 -1
+2 -2
+1 -2
+1 -1
HCl +3 FeCl2 + KNO3  3 FeCl3 + NO + H2O + KCl
1 (-3)
OXIDATION STATES METHOD OF
BALANCING
4 HCl+3 FeCl2+KNO3 3FeCl3+NO+2 H2O+KCl
HALF-REACTION METHOD OF BALANCING
 Assign
oxidation numbers to
every atom in the reaction
 Identify what is oxidized and
what is reduced. Anything
whose oxidation number does
not change is a spectator.
HALF-REACTION METHOD OF BALANCING
Write a ½ reaction for the reduction
reaction without the spectators.
 If an atom being reduced is part of a
solid, a polyatomic ion or part of a
molecular compound, you will keep
that ion or cpd together when you
bring it down for the ½ rxn.

HALF-REACTION METHOD OF BALANCING
 Balance
the ½ rxn by first
balancing the non-H, non-O
atoms.
 Then balance the H’s and O’s
using the following guide:
 Use
H1+ and H2O if the rxn occurs
in an acidic medium.
 Use OH1- and H2O if the rxn
occurs in a basic medium.
HALF-REACTION METHOD OF BALANCING
 Balance
the ½ rxn electrically
(charge-wise) by adding
electrons (e-) to the left side
since a reduction rxn goes GER
HALF-REACTION METHOD OF BALANCING
Write a ½ reaction for the oxidation
reaction without the spectators.
 If an atom being oxidized is part of
a solid, a polyatomic ion, or a
molecular compound, you will keep
that solid, ion, or cpd together when
you bring it down for the ½ rxn.

HALF-REACTION METHOD OF BALANCING
 Balance
the ½ rxn by first
balancing the non-H, non-O
atoms.
 Then balance the H’s and O’s
using the following guide:
 Use
H1+ and H2O if the rxn occurs
in an acidic medium.
 Use OH1- and H2O if the rxn
occurs in a basic medium.
HALF-REACTION METHOD OF BALANCING
 Balance
the ½ rxn electrically
(charge-wise) by adding
electrons (e-) to the right side
since an oxidation rxn is LEO
HALF-REACTION METHOD OF BALANCING
 Normalize
the electrons in each
½ rxn by finding the LCM of the
two numbers of electrons and
distributing the multipliers
through each entire ½ rxn.
HALF-REACTION METHOD OF BALANCING
 Add
the two ½ rxns and cancel
any like atoms, ions, or cpds
that appear on both sides.
 Add the spectators back into
the rxn and balance the rest by
inspection.
½ RXN METHOD OF BALANCING
+2 -2
+2-2
0
+4-2
PbO(s) + CO(g)  Pb(s) + CO2(g)
½ RXN METHOD OF BALANCING
PbO(s) + 2H1+ + 2e-  Pb(s) + H2O
½ RXN METHOD OF BALANCING
CO(g) + H2O  CO2(g) + 2H1+ + 2e-
½ RXN METHOD OF BALANCING
1 PbO(s) + 2H1+ + 2e-  Pb(s) + H2O
1 CO(g) + H2O  CO2(g)+ 2H1+ + 2e-
½ RXN METHOD OF BALANCING
1 PbO(s) + 2H1+ + 2e-  Pb(s) + H2O
1 CO(g) + H2O  CO2(g)+ 2H1+ + 2ePbO(s) + 2H++ 2e- + CO(g) + H2O  Pb(s) + H2O + CO2 + 2H++ 2e-
½ RXN METHOD OF BALANCING
PbO(s) + CO(g)  Pb(s) + CO2(g)
Ta-da!
½ RXN METHOD OF BALANCING
+4 -1
+2 -1
+3 -1
+4 -1
CeCl4 + SnCl2  CeCl3 + SnCl4
½ RXN METHOD OF BALANCING
Ce4+ + 1e-
 Ce3+
½ RXN METHOD OF BALANCING
Sn2+
 Sn4++ 2e-
½ RXN METHOD OF BALANCING
2 Ce4+ + 1e-
1 Sn2+
 Ce3+
 Sn4+ + 2e-
½ RXN METHOD OF BALANCING
2 Ce4+ + 1e-
1 Sn2+
 Ce3+
 Sn4+ + 2e-
2Ce4+ + Sn2++ 2e-  2Ce3+ + Sn4+ + 2e-
½ RXN METHOD OF BALANCING
2CeCl4 + SnCl2  2CeCl3 + SnCl4
Ta-da!
½ RXN METHOD OF BALANCING
HCl + FeCl2 + KNO3  FeCl3 + NO + H2O + KCl