Redox Reactions - Reactions that involve the transfer of electrons
Oxidation – Occurs when a substance loses electrons
Reduction - Occurs when a substance gains electrons
LEO-GER – A way to remember this by; Loss of Electrons is Oxidation, Gain of Electrons is Reduction
Reduction and oxidation occur together, hence the term Redox (reduction and oxidation) reactions
The number of electrons lost during the oxidation reaction must equal the number of electrons gained during the reduction reaction
Oxidizing Agent or oxidant – The substance that is gaining electrons and being reduced, oxidizes the other substance in the reaction
Substances with very electronegative elements like oxygen are usually good oxidizing agents, hence the name
They can accept extra electrons easily and remain stable
Reducing Agent or reductant– The substance that is losing electrons and being oxidized, reduces the other substance in the reaction
Substances containing excess electrons or ones that can form stable cations are usually good reducing agents
Activity Series – a list of metals in order of decreasing ease of oxidation
The metals at the top of the table are most easily oxidized, known as active metals
The metals at the bottom are not easily oxidized, they are known as noble metals
Can be used to predict whether a substance will be oxidized or reduced
Any metal on the list can be oxidized by the ions of elements below it
Oxidation numbers – a method of keeping track of the electrons during a redox reaction
When an element is oxidized, its oxidation number increases.
When an element is reduced, its oxidation number decreases.
Rules for Assigning Oxidation numbers:
For an atom in its elemental form the oxidation number is zero.
For a monatomic ion the oxidation number equals the charge on the ion.
The oxidation number of oxygen is usually -2 except in peroxides.
The oxidation number of hydrogen is +1 except when bonded to metals.
The oxidation number of fluorine is -1.
Group 1 metals always have an oxidation number of +1.
Group 2 metals always have an oxidation number of +2.
The sum of the oxidation numbers of all atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion equals the charge of the ion.
Balancing Redox Reactions – To balance redox reactions, use the following steps:
Assign oxidation numbers.
Divide the redox reaction into two half-reactions, the oxidation and reduction reactions.
Balance the atoms of the element being oxidized or reduced.
Balance oxidation numbers by adding electrons. For a reduction half-reaction, electrons are added to the reactants. For an oxidation half-reaction, add electrons to the product side.
Balance the charge with H+ if in acidic solution or
Balance the H+ or
Check to make sure oxygen is balanced.
Combine the two half-reactions. Balance the number of electrons lost and the number of electrons gained.
Some Types of Redox Reactions
A reaction is a redox reaction if there is a change in oxidation number of an atom on the reactant side to the same atom on the product side
One species must be losing electrons and another must be gaining electrons
Combustion reactions – elemental oxygen is converted to compounds of oxygen
Displacement reactions – an element reacts with a compound and displaces an element from it
Example: Zn + CuCl2 à ZnCl2 + Cu
Zinc is oxidized from an oxidation state of 0 to +2
Copper is reduced from an oxidation state of +2 to 0
Corrosion – a metal is oxidized by substances in its environment
Table of Reduction Potentials
Every half-reaction has a potential, or voltage, associated with it
Potentials are given as reduction half-reactions
Read them in reverse and change the sign on the voltage to get oxidation potentials
The more positive the Eºred value for a half-reaction, the more likely it is for the reactant of the half-reaction to be reduced and to oxidize another species
The more positive the Eºox value for a half-reaction, the more likely it is for the reactant of the half-reaction to be oxidized and to reduce another species
Substances that are most easily reduced and therefore strong oxidizing agents are the reactants at the top left of the table
Their products do not oxidize easily and are weak reducing agents
Substances that do not reduce easily and are weak oxidizing agents are the reactants in the bottom left of the table
Their products oxidize easily and are strong reducing agents
Spontaneity of Redox Reactions
Add reduction and oxidation half-reaction potentials to determine if a redox reaction will be spontaneous
Eº = Eºred + Eºox
If Eº is positive the reaction is spontaneous
If Eº is negative the reaction is nonspontaneous
Example of Balancing a Redox Reaction:
Balance the following redox reaction:
Fe2+(aq) + MnO4-(aq) à Fe3+(aq) + Mn2+(aq) in acidic solution
Step 1: Assign oxidation numbers.
Fe2+(aq) Oxidation Number = +2
MnO4-(aq) Oxidation Number of Mn = +7 Oxidation Number of O = -2
Fe3+(aq) Oxidation Number = +3
Mn2+(aq) Oxidation Number = +2
Since Fe has an increase in oxidation number, it is the species being oxidized. Mn has a decrease in oxidation number, so it is reduced.
Step 2: Divide into two half-reactions.
Oxidation: Fe2+(aq) à Fe3+(aq)
Reduction: MnO4-(aq) à Mn2+(aq)
Step 3: Balance the atoms of the element being oxidized or reduced.
Fe2+(aq) à Fe3+(aq) Already balanced.
MnO4-(aq) à Mn2+(aq) Already balanced (one atom of Mn on both sides)
Step 4: Balance oxidation numbers by adding electrons.
Fe2+(aq) à Fe3+(aq) + e- (Iron is oxidized from an oxidation number of +2 to +3, so one electron is added to the right side).
MnO4-(aq) + 5e-à Mn2+(aq) (Manganese is reduced from an oxidation number of +7 to +2, so five electrons are added to the left side).
Step 5: Balance the charge with H+ if in acid or
Fe2+(aq) à Fe3+(aq) + e- There is a charge of +2 on both sides, so it is balanced.
MnO4-(aq) + 5e- à Mn2+(aq) There is a charge of -6 on the reactant side and +2 on the product side. To balance, add eight H+ ions
MnO4-(aq) + 5e- + 8H+à Mn2+(aq) on the left to have a charge of +2 on both sides.
Step 6: Balance the H+ or
Fe2+(aq) à Fe3+(aq) + e- Already balanced.
MnO4-(aq) + 5e- + 8H+à Mn2+(aq) To balance the eight H+ ions on the left, add
MnO4-(aq) + 5e- + 8H+à Mn2+(aq) + 4H2O four H2O molecules to the right.
Step 7: Check to make sure oxygen is balanced.
Fe2+(aq) à Fe3+(aq) + e- Already balanced.
MnO4-(aq) + 5e- + 8H+à Mn2+(aq) + 4H2O There are four oxygen atoms on each side, so it is balanced.
Step 8: Combine the two balanced half-reactions. Balance the number of electrons lost and the number of electrons gained.
Fe2+(aq) à Fe3+(aq) + e- 1 electron lost.
MnO4-(aq) + 5e- + 8H+à Mn2+(aq) + 4H2O 5 electrons gained.
Multiply the oxidation half-reaction by five to make sure the number of electrons lost equals the number of electrons gained. Then add it to the reduction half-reaction. The result is the balanced redox reaction.
5 [Fe2+(aq) à Fe3+(aq) + e-]
5 Fe2+(aq) à 5 Fe3+(aq) + 5e-
+ MnO4-(aq) + 5e- + 8H+à Mn2+(aq) + 4H2O
5 Fe2+(aq) + MnO4-(aq) + 8H+(aq)à 5 Fe3+(aq) + Mn2+(aq) + 4H2O
Questions:
Why are elements like gold, platinum and silver commonly used to make coins and jewelry?
Balance the following redox reaction if in acidic solution.
SO2(aq) + Cr2O72-(aq) à SO42-(aq) + Cr3+(aq)
Balance the following redox reaction if in basic solution.
MnO4-(aq) + Br -(aq) à MnO2(s) + BrO3-(aq)
For the reaction Cl2(g) + 2NaBr(aq) à 2NaCl(aq) + Br2(l) write:
The unbalanced oxidation and reduction half-reactions
The oxidizing agent and reducing agent
The changes in oxidation numbers of the atoms of the oxidizing and reducing agents
If the reaction is spontaneous as written. Justify your answer.
Answers:
These elements are at the bottom of the activity series of metals. This means they are very unreactive and are not oxidized easily. Therefore, they will not be affected as much in the presence of oxygen or water. They do not experience corrosion or tarnish easily, which are useful properties to have and make them common choices to use for coins and jewelry.
Cr2O72-(aq) + 3 SO2(aq) + 2 H+(aq) à 2 Cr3+(aq) + 3 SO42-(aq) + H2O(l)
2MnO4-(aq) + Br -(aq) + H2O(l) à 2OH-(aq) + 2MnO2(s) + BrO3-(aq)
4. a) Oxidation: 2Br -(aq) à Br2(l) Reduction: Cl2(g) à 2Cl-(aq)
b) Oxidizing agent: Cl2(g) Reducing agent: Br -(aq)
c) Oxidizing agent: Cl2(g) Cl changes from 0 to -1
Reducing agent: Br -(aq) Br changes from -1 to 0
The reaction is spontaneous. Using the formula Eº = Eºred + Eºox, and the table of reduction potentials, the equation becomes Eº = 1.36V + -1.07V = .29V. Since Eº is positive, the reaction is spontaneous.
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