Oxidation
 When a substance loses electrons, it
undergoes oxidation.
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g)
 Ca lost 2 electrons to the H+ ions.
 The metal Ca became an ion: Ca2+
 Ca has been oxidized.
 H+ was the oxidizing agent. (It made
oxidation happen.)
Mullis
1
Oxidation examples
 Originally, oxidation referred to the
combination of a substance with oxygen.
4Fe + 3O2  2Fe2O3
Oxidation
state of Fe
0
 +3
2CO + O2  2CO2
Oxidation
state of C
+2
 +4
Mullis
2
Reduction
 When a substance loses electrons, it
undergoes oxidation.
2Ca(s) + O2(g) 2CaO (s)
 O gained 2 electrons to become O2- in CaO.
 The neutral O2 became an ion: O2 O2 has been reduced.
 In all reduction-oxidation reactions, one
species is reduced at the same time as
another is oxidized.
Mullis
3
Reduction examples
Originally, reduction referred to the removal of
oxygen from a compound. Oxide ores are
reduced to metals—a real reduction in
mass.
WO3 + 3H2  W + 3H2O
Oxidation
state of W
+6
 0
Mullis
4
Recognizing a Redox
Reaction
 Analyze oxidation numbers.
 If no elements change in oxidation
numbers from reactant side to product
side, the reaction is NOT redox.
 If changes occur, the reaction is redox.
 Another clue that the reaction is redox
are the words “in acidic
conditions/solution” or “in basic
Mullis
conditions/solution.”
5
LEO
says
GER
(Identifying ½ Reactions)
 Lose Electrons Oxidation / Gain Electrons Reduction
 Assign oxidation numbers 1st
 Ex. H2 + Cl2  2H Cl
0
0
+1 -1
 Pick the element that goes down in Oxidation Number:
Here, this is Cl (0-1). Its half reaction is:
 Cl2 + 2e-  2ClCl is reduced
 Pick the element that goes up in Oxidation Number:
Here, this is H (0+1). Its half reaction is:
 H2  2H+ + 2eH is oxidized
Mullis
6
Water
 Oxidation of water:
2H2O  O2 +4H+ + 4e Reduction of water:
2H2O + 2e- H2 +2OH-
Mullis
7
Redox Equation Example
SbCl5 +
+5 -1
KI
+1 -1
 KCl + I2 +
+1 -1
0
LEO:
2I-  I2 + 2eGER: Sb5+ + 2e-  Sb3+
SbCl3
+3 -1
(ox # increased)
(ox # decreased)
_________________________________________________
Sb5+ + 2I-
 Sb3+ + I2
Mullis
8
Balancing Redox Equations
1.
2.
3.
4.
5.
6.
7.
Write the two half-reactions.
Balance all elements except O and H.
Balance O with waters.
Balance H with H+.
Balance the charge with electrons (e-1).
Multiply by a factor to make the electrons equal for both equations.
Add the two equations together and combine like terms.
If a basic solution, do number 8.
8. Add hydroxides (OH-) to both sides equal to the H +. On the side with
the H+, addition of OH- will produce water molecules.
(H+ + OH- H2O). Combine waters if necessary.

Hint: Balance all redox equations as if they are in acidic solution. If
they are in a basic solution, convert it over at the end using step 8.
Mullis
9
Redox Equation Example
(acidic solution)
H3PO4 +
+1 +5 -2
HNO2
+1 +3 -2
 N 2O 4+
+4 -2
H3PO3
+1 +3 -2
LEO: HNO2  N2O4
(ox # increased)
2HNO2  N2O4 + 2H+ + 2e- (balance with H and e )
GER: H3PO4  H3PO3
(ox # decreased, acid was reduced)
H3PO4  H3PO3 + H2O
(balance O with H O )
H3PO4 + 2H++ 2e-  H3PO3 + H2O (balance with H and e )
+
-
2
+
-
_________________________________________________
H3PO4 + 2HNO2  N2O4+ H3PO3 + H2O (add ½ reactions)
Mullis
10
Common Oxidizing Agents
 Free halogen (Cl2)









Metal-ic(high) Fe 3+
MnO 4- (acid)
MnO4- (base)
MnO2 (acid)
Cr2O7 2- (acid)
HNO3 (conc.)
HNO3 (dilute)
H2SO4 (hot)
H 2O 2
Halide ion (Cl-)
Metal-ous(low) Fe2+
Mn 2+
MnO2
Mn 2+
Cr 3+
NO2
NO
SO2
H 2O
Mullis
11
Common Reducing Agents






Halide ion (Cl-)
Metal-ous(low) Fe2+
Free metal (Cu)
Sulfite ion(SO3 2-)
Nitrite ion(NO2 -)
C2O4- (oxalate ion)
Free halogen (Cl2)
Metal-ic(high) Fe 3+
Metal ions (Cu 2+ )
Sulfate ion(SO4 2-)
Nitrate ion(NO3 -)
CO2
Mullis
12