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Electrochemistry
Reduction-Oxidation
Oxidation
• Historically means “to combine with oxygen”
• Reactions of substances with oxygen, ie Combustion,
Rusting (corrosion)
Oxidation
• Substances other than oxygen could cause reactions
with characteristics of oxidation, so we say:
• Oxidation is the loss of electrons
• A reactant that oxidizes another reactant is called an
oxidizing agent.
• Example:
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
Reduction
• Reduction is the gain of electrons.
• A reactant that reduces another reactant is called a
reducing agent.
• Example:
Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
Reduction-Oxidation Reaction
• Reduction-Oxidation reactions, or redox reactions,
are reactions in which both oxidation and reduction
occur.
3Fe(s) + Al2(SO4)3(aq)  2Al(s) + 3FeSO4 (aq)
Write the total and net ionic equation for the reaction.
Try to determine which species undergo
oxidation/reduction and which are oxidizing/reducing
agents.
Mnemonic Device
OIL RIG
• Oxidation is loss
• Reduction is gain
• http://educationportal.com/academy/lesson/balancing-redoxreactions-and-identifying-oxidizing-and-reducingagents.html#lesson
Write a balanced net ionic equation for
calcium with aqueous aluminum sulfate.
Identify the reactant oxidized and reduced as
well as the reducing and oxidizing agent.
Practice
P715 #1-4
14-5-23
Half-Reactions
• To monitor the transfer of electrons in a redox
reaction, you can represent the oxidation and
reduction separately, these reactions are called halfreactions.
• Example: Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
Oxidation half-reaction: Zn(s)  Zn2+(aq) + 2eReduction half-reaction: Cu2+(aq) + 2e-  Cu(s)
Disproportionation
• A disproportionation reaction is a reaction in which a
single element undergoes both oxidation and
reduction in the same reaction.
• Example: 2Cu+(aq)  Cu(s) + Cu2+(aq)
Practice
p 716 #5-8
Oxidation Numbers
• Oxidation numbers are actual or hypothetical
charges, assigned using a set of rules, used to
describe redox reactions with covalent compounds.
• They are also used to identify redox reactions, and to
identify oxidizing and reducing agents.
Oxidation Numbers of an
Element
• Some elements (transition metals) have more than one possible
charge.
• Vanadium can form V2+, V3+,V4+, V5+. Oxidation numbers help
to determine the oxidation state.
• Ie. Vanadium is oxidized by removing 2 e-s, it has an oxidation
number of +2.
V  V2+ + 2e-
• If this ion is oxidized again, it has an oxidation number of +3.
V2+  V3+ + e-
Note: every time you oxidize vanadium by removing another electron,
its oxidation number increases by 1.
Oxidation Numbers of an
Element
• If you add electrons to an element, the oxidation
number will drop.
• Example: If you add 2 electrons to sulfur, the
oxidation number is -2.
S + 2e- S2Rules have been established for determining oxidation
numbers of elements within compounds.
Rules for Assigning Oxidation Numbers
1. A pure element has an oxidation number of 0.
2. The oxidation number of a monoatomic ion is equal to its charge.
3. The oxidation number of hydrogen in its compounds is +1, except in metal
hydrides, where it is -1.
4. The oxidation number of oxygen in its compounds is usually -2. Exceptions
include peroxides and OF2.
5. In covalent compounds that do not contain oxygen or hydrogen, the
oxidation number of the less electronegative element is positive and the
more electronegative element is negative.
6. The algebraic sum of the oxidation numbers in the formula of a compound
must equal zero.
7. The algebraic sum of the oxidation numbers in the formula of a polyatomic
ions must equal the charge of the ion.
Determine the oxidation number of each
atom, in each of the following formulas.
a) Cl2 b) CH4 c) NaCl d) OF2
Using Oxidation Numbers
• During a redox reaction, if the oxidation number
increases, there was a loss of electrons.
• If the oxidation number decreases, there was a gain
of electrons.
• Determining the oxidation number of each atom in
an equation can help to identify the species being
oxidized/reduced and the oxidizing/reducing agent.
Use oxidation numbers to determine whether
or not the reaction is a redox reaction and to
identify all species.
CH4(g) + Cl2(g)  CH3Cl(g) + HCl (g)
Use oxidation numbers to determine whether
or not the reaction is a redox reaction and to
identify all species.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Practice
P 726 #9-12
P 728 #13-16
Try the end of section review on page 728-729.
Half-Reaction Method for Acidic
Solutions
1. Step 1 Write an unbalanced half-reaction that shows the
formulas of the given reactant(s) and product(s).
2. Step 2 Balance any atoms other than oxygen and hydrogen
first.
3. Step 3 Balance any oxygen atoms by adding water
molecules.
4. Step 4 Balance any hydrogen atoms by adding hydrogen
ions.
5. Step 5 Balance the charges by adding electrons.
Write a balanced half-reaction that shows the reduction of
permanganate
ions, MnO4-, to manganese (II) ions in an acidic solution.
Step 1
MnO4− → Mn2+
Step 2 Balance the atoms. Here, the Mn atoms are already balanced.
Step 3 Add water molecules to balance the oxygen atoms.
MnO4− → Mn2+ + 4H2O
Step 4 The reaction occurs in acidic solution, add hydrogen ions to
balance the hydrogen atoms.
MnO4− + 8H+ → Mn2+ + 4H2O
Step 5 The net charge on the left side is 7+, the net charge on the right
side is 2+. Add five electrons to the left side to balance the charges.
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
Half-Reaction Method for Basic
Solutions
1. Use the half-reaction method for acidic solutions to
get the final balanced equation as if H+ ions were
present.
2. To both sides of the equation, add the # of OHions that is equal to the number of H+ ions present.
3. Combine H+ and OH- ions to form H2O on the side
containing both and eliminate the H2O molecules
that appear on both sides of the equation.
4. Check that elements and charges are balanced.
Write a balanced half-reaction that shows the oxidation of
thiosulfate ions, S2O32−, to sulfite ions, SO32−, in a basic solution.
Step 1
S2O32− → SO32−
Step 2 Balance the atoms, beginning with the sulfur atoms.
S2O32− → 2SO32−
Step 3 Balance the oxygen and hydrogen atoms as if the solution
is acidic.
S2O32− + 3H2O → 2SO32−
S2O32− + 3H2O → 2SO32− + 6H+
Step 4 There are six hydrogen ions present, so adjust for basic
conditions by adding six hydroxide ions to each side.
S2O32− + 3H2O + 6OH− → 2SO32− + 6H+ + 6OH−
Step 5 Combine the hydrogen ions and hydroxide ions on the
right side into water molecules.
S2O32− + 3H2O + 6OH− → 2SO32− + 6H2O
Step 6 Remove three water molecules from each side.
S2O32− + 6OH− → 2SO32− + 3H2O
Step 7 The net charge on the left side is 8−, the net charge on the
right side is 4−. Add four electrons to the right side to balance
the charges.
S2O32− + 6OH− → 2SO32− + 3H2O + 4e−
Practice
Page 732 # 17-20
Page 734 # 21-24
The Half-Reaction Method for
Balancing Equations
• The half-reaction method for balancing equations
involves writing two balanced half-reactions for a
given redox reaction and combining them to form
one balanced net reaction.
• This method is especially useful for reactions that
take place under acidic or basic conditions.
Half-Reaction Method
1. Write the unbalanced net ionic equation.
2. Divide the unbalanced net ionic equation into an oxidation halfreaction and a reduction half-reaction.
 you may need to assign oxidation numbers to all of the
elements to determine what is oxidized and what is
reduced.
3. Balance the oxidation half-reaction and the reduction halfreaction independently.
4. Determine the least common multiple of the numbers of
electrons in the oxidation and reduction half-reactions
Half-Reaction Method
5. Use the coefficients to write each half-reaction so
that it includes the LCM of the number of electrons
6. Add the balanced half-reactions that include equal
numbers of electrons.
7. Remove the electrons from both sides of the
equation.
8. Remove any identical molecules or ions that are
present on both sides of the equation.
9. If you require a balanced chemical equation, include
any spectator ions in the chemical formulas.
Problem:
Write a balanced net ionic equation to show the reaction of
perchlorate ions, ClO4-, and nitrogen dioxide in acidic solution to
produce chloride ions and nitrate ions.
Step 1
ClO4− + NO2 → Cl− + NO3−
Step 2 Write two unbalanced half-reactions.
Oxidation: NO2 → NO3−
Reduction: ClO4− → Cl−
Step 3 Balance the two half-reactions for acidic conditions.
Step 4 The LCM of 1 and 8 is 8.
Step 5 Multiply the oxidation half-reaction by 8, so that equal numbers of
electrons are lost and gained.
8NO2 + 8H2O → 8NO3− + 16H+ + 8e−
Step 6 Add the half reactions.
8NO2 + 8H2O → 8NO3− + 16H+ + 8e−
ClO4− + 8H+ + 8e− → Cl− + 4H2O________
8NO2 +8H2O +ClO4− +8H+ +8e−→8NO3-+16H++8e−+Cl−+4H2O
Step 7 Simplify by removing 8 electrons from both sides.
8NO2 +8H2O +ClO4− +8H+ →8NO3- +16H+ +Cl− + 4H2O
Step 8 Simplify by removing 4 water molecules, and 8 hydrogen ions
from each side.
8NO2 + 4H2O + ClO4− → 8NO3- + 8H+ + Cl−
Practice
• Try questions 25-28 on page 738 & 739.
The Oxidation Number Method
for Balancing Equations
• The oxidation number method is a method of
balancing redox equations by ensuring that the total
increase in the oxidation numbers of the oxidized
elements equals the total decrease in the oxidation
numbers of the reduced elements.
Oxidation
Number
Method
1. Write an unbalanced equation, if it is not given.
2. Determine whether the reaction is a redox reaction by assigning an oxidation
3.
4.
5.
6.
7.
8.
number to each element wherever it appears in the equation.
If the reaction is a redox reaction, identify the element that undergoes an
increase in oxidation number and the elements the undergoes a decrease.
Find the numerical values of the increase and decrease.
Determine the smallest whole-number ratio of the oxidized and reduced
elements so that the total increase in oxidation numbers equals the total
decrease in oxidation numbers.
Use the smallest whole-number ratio to balance the numbers of atoms of
the element oxidized and the element reduced.
Balance the other elements by inspection, if possible.
For reactions that occur in acidic or basic solutions, include water molecules,
hydrogen ions, or hydroxide ions as needed to balance the equation.
Write a balanced net ionic equation to show the
formation of iodine by bubbling oxygen gas through a
basic solution that contains iodide ions.
Practice
• Try problems 33-36 on page 749.