Inorganic Chemistry I
Reduction-Oxidation (REDOX)
Kun Sri Budiasih
Oxidation-Reduction Reactions
• Reactions that involve the transfer of
electrons are called oxidation-reduction or
redox reactions
• Oxidation is the loss of electrons by a
reactant
• Reduction is the gain of electrons by a
reactant
• Oxidation and reduction always occur
together
• The total number of electrons lost by one
substance is the same as the total number of
electrons gained by the other
• For a redox reaction to occur, something
must accept the electrons that are lost by
another substance
• The substance that accepts the electrons is
called the oxidizing agent
• The substance that lost the electrons is
called the reduction agent
• Note that the oxidizing agent is reduced
and the reducing agent is oxidized
• For example:
2 Na + Cl2  2 NaCl
– Na is the reducing agent because it lost
electrons and was oxidized
– Cl2 is the oxidizing agent because it gained
electrons and was reduced
•
Oxidation numbers provide a way to
keep track of electron transfers :
1) The oxidation number of any free element is zero.
2) The oxidation number of any simple, monoatomic
ion is equal to the charge on the ion.
3) The sum of all oxidation numbers of the atoms in
a molecule or polyatomic ion must equal the
charge on the particle.
4) In its compounds, fluorine has an oxidation
number of –1.
5) In its compounds, hydrogen has an oxidation
number of +1.
6) In its compounds, oxygen has an oxidation
number of –2.
• If there is a conflict between two rules
apply the rule with the lower number and
ignore the conflicting rule
• In binary ionic compounds with metals, the
nonmetals have oxidation numbers equal to
the charges on their anions
Example: What is the oxidation number of Fe
in Fe2O3?
ANALYSIS: This binary compound is ionic.
Apply rule 3 and 6
Fe: 2x
O: 3(-2) = -6
0 = 2x + (-6) or x = +3 = ox. number of Fe
• Note that fractional values of oxidation
numbers are allowed
• In terms of oxidation numbers:
– Oxidation is an increase in oxidation number
– Reduction is a decrease in oxidation number
• This provides a simple way to follow redox
reactions
• Many redox reactions take place in aqueous
solution
• A procedure called the ion-electron
method provides a way to balance these
equations
• The oxidation and reduction are divided
into equations called half-reactions
• The half-reactions are balanced separately,
then combined into the fully balanced net
ionic equation
• Both mass and charge must be balanced
• Charge is balanced by adding electrons to
the side of the equation that is more positive
or less negative
Example: Balance the following skeleton
equation
Al( s )  Cu 2 (aq)  Al 3 (aq)  Cu( s )
ANALYSIS : This is a redox reaction.
SOLUTION :
Oxidation : Al( s )  Al 3  3e Reduction : Cu 2 (aq)  2e -  Cu( s )
The least common factor is 6, combining
2 Al( s)  3Cu 2 (aq)  2Al 3 (aq)  3Cu( s )
•
Many reactions occur in either acidic or
basic solutions
The Ion-Electron Method in Acidic Solution:
1)
2)
3)
4)
5)
6)
Divide the equation into two half-reactions.
Balance atoms other than H and O.
Balance O by adding water.
Balance H by adding hydrogen ion.
Balance net charge by adding electrons.
Make electron gain and loss equal: add halfreactions.
7) Cancel anything that’s the same on both sides of
the equation.
• The simplest way to balance reactions in
basic solution is to first balance them as if
they were in acidic solution, then “convert”
to basic solution:
Additional Steps for Basic Solutions
8) Add to both sides of the equation t he same number
of OH- ions as there are H  .
9) Combine H  and OH- to form H 2O.
10) Cancel any H 2 O that you can.
Example: Balance the following in basic solution:
MnO -4  C 2 O 24-  MnO 2  CO32ANALYSIS : Balance as if in acidic solution t hen " convert".
SOLUTION :
C 2 O 24-  2H 2 O  2CO32-  4H   2e MnO -4  4H   3e -  MnO 2  2H 2 O
Net ionic : 3C 2 O 24-  2H 2 O  2MnO -4  6CO32  4H   2MnO 2
Add OH3C 2 O 24-  2H 2 O  4OH-  2MnO -4  6CO32  4(H   OH  )  2MnO 2
Form H 2 O
3C 2 O 24-  2H 2 O  4OH-  2MnO -4  6CO32  4H 2 O  2MnO 2
Simplify
3C 2 O 24-  4OH-  2MnO -4  6CO32  2H 2 O  2MnO 2
• Metals more active than hydrogen (H2)
dissolve in oxidizing acids
Nonoxidizi ng acids : HCl( aq), cold dilute H 2SO 4 (aq),
H 3PO 4 (aq), and most organic acids.
Ox. Acid
HNO 3
Reduction Reaction
(conc.) NO3-  2H   e -  NO2 ( g )  H 2 O
(dilute) NO3-  4H   3e -  NO(g )  4H 2 O
(very dilute, with strong reducing agent)
NO3-  10H   8e -  NH 4  3H 2 O
H 2SO 4
(hot, conc.) SO 24-  4H   3e -  SO 2 ( g )  2H 2 O
(hot conc., with strong reducing agent)
SO -4  10H   8e -  H 2S( g )  4H 2 O
• Some examples:
Concentrat ed HNO 3 :
Cu( s )  4HNO3 (aq)  Cu(NO 3 ) 2 (aq)  2NO2 ( g )  2H 2 O
Dilute HNO 3 :
3Cu( s)  8HNO3 (aq)  3Cu(NO 3 ) 2 (aq)  2NO(g )  4H 2 O
Hot concentrat ed H 2SO 4 :
Cu( s )  2H 2SO 4 (aq)  CuSO 4 (aq)  SO 2 ( g )  2H 2 O
• More active metals will displace a less
active metal from its compound
• This often occurs in solution and is called a
single replacement reaction
Zinc is a more active metal
than copper. Copper ions
(blue) collide with zinc metal
(gray) picking up electrons.
Copper ions become copper
atoms (red-brown) and stick
to the zinc surface. Zinc ions
(yellow) replace the copper
ions in solution.
• An activity series arranges metals according
to their ease of oxidation
• They can be used to predict reactions
Activity Series for Some Metals and Hydrogen
Element Oxidation Product
Least active Gold
Au 3
Silver
Ag 
Copper
Cu 2
Hydrogen H 
Most active
Sodium
Na 
Cesium
Cs 
(See Table
6.2 for a
more
extensive
list.)
• A given element will be displaced from its
compounds by any element below it in the
table
• Oxygen reacts with many substances
• The products depends, in part, on how much
oxygen is available
• Combustion of hydrocarbons
O 2 plentiful :
CH 4  2O 2  CO2  2H 2 O
O 2 limited :
2CH 4  3O 2  2CO  4H 2 O
O 2 very limited : CH 4  O 2  C  2H 2 O
• Organic compounds containing O also produce
carbon dioxide and water
C2 H5OH  3O2  2CO2  3H 2O
• Organic compounds containing S produce sulfur
dioxide
2C2 H5SH  9O2  4CO2  6H 2O  2SO2
• Many metals corrode or tarnish when exposed to
oxygen
Corrosion of iron : 4Fe  3O 2  2Fe2O3
Tarnishing of silver : 4Ag  O 2  2Ag 2O
• Most nonmetals react with oxygen directly
O 2 plentiful : C  O 2  CO2
O 2 limited : 2C  O 2  2CO
• Redox reactions are more complicated than
most metathesis reactions
• In general, it is not possible to balance a
redox reaction by inspection
• This is especially true when acid or bases
are involved in the reaction
• Once balanced, they can be used for
stoichiometric calculations
• Redox titrations are common because they
often involve dramatic color changes
• Mole-to-mole ratios are usually involved
Example: A 0.3000 g sample of tin ore was dissolved
in acid solution converting all the tin to tin(II). In a
titration, 8.08 mL of 0.0500 M KMnO4 was required
to oxidize the tin(II) to tin(IV). What was the
percentage tin in the original sample?
ANALYSIS: This is a redox titration in acidic
solution.
SOLUTION:
• Form skeleton equation and use the ion-electron
method to produce a balanced equation
3Sn 2  2MnO -4  8H   3Sn 4  2MnO 2  4H 2O
• Use the balanced equation to define equivalence
relations and determine the mass of Sn in the
original sample
Sn 2
1.000 mol Sn  0.0719 g Sn
 11mol

mol Sn
118.7 g Sn
0.00808 L MnO sol  0.0500 mol MnO 4 
4
-
1.00 L MnO -4 sol
-
2 mol MnO -4
3 mol Sn 2
• Convert to percentage
g
% Sn  0.0719
0.3000 g 100%  24.0% Sn in ore