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 :
The oxidation number of any free element is zero.
The oxidation number of any simple, monoatomic ion is equal to the charge on the ion.
The sum of all oxidation numbers of the atoms in a molecule or polyatomic ion must equal the charge on the particle.
In its compounds, fluorine has an oxidation number of –1.
In its compounds, hydrogen has an oxidation number of +1.
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
Many reactions occur in either acidic or basic solutions
The Ion-Electron Method in Acidic Solution:
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 half-reactions.
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
Example: Balance the following in basic solution:
Metals more active than hydrogen (H2) dissolve in oxidizing acids
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
Activity Series for Some Metals and Hydrogen
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
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
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
Use the balanced equation to define equivalence relations and determine the mass of Sn in the original sample
Convert to percentage