Inorganic Chemistry I
Reduction-Oxidation (REDOX)
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
• 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 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
• 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.
• 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?
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
• 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
• 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
) factor common
• Many reactions occur in either acidic or basic solutions
The Ion-Electron Method in Acidic Solution:
1) Divide the equation into two half-reactions. 2) Balance atoms other than H and O.
3) Balance O by adding water.
4) Balance H by adding hydrogen ion.
5) Balance net charge by adding electrons.
6) Make electron gain and loss equal: add half-reactions.
• 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
can. you
that O
H any Cancel
10)
O. H
form to
OH and
H Combine 9)
. H are there
as ions OH
of
number same
he equation t the
of sides both
to Add 8)
2
2
-
Example: Balance the following in basic solution: solution t
• Metals more active than hydrogen (H2) dissolve in oxidizing acids
O Reduction
• Some examples:
• More active metals will displace a less active metal from its compound
• This often occurs in solution and is called a
single replacement reaction
• An activity series arranges metals according to their ease of oxidation
• They can be used to predict reactions
Zinc is a more active metal than copper. Copper ions
Activity Series for Some Metals and Hydrogen
• A given element will be displaced from its compounds by any element below it in the table Oxidation
Element
• Oxygen reacts with many substances • The products depends, in part, on how
much oxygen is available
• Combustion of hydrocarbons
• Organic compounds containing O also produce carbon dioxide and water
• Organic compounds containing S produce sulfur dioxide
• Many metals corrode or tarnish when exposed to oxygen
• Most nonmetals react with oxygen directly
• Redox reactions are more complicated than most metathesis reactions
2
Corrosion
• 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
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
O 4H 2MnO
3Sn 8H
2MnO
• Use the balanced equation to define equivalence relations and determine the mass of Sn in the original sample
• Convert to percentage
Sn g 719 0.0
sol MnO L 808 0.00
Sn mol 00 1.0 Sn g .7 118
Sn mol 1 Sn mol 1 MnO mol 2
Sn mol 3
sol MnO L 0 1.0
MnO mol 500 0.0 -4
2
-4
2
-4
-4
ore in
Sn %
0 . 24 %
100 Sn
