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Oxidation Reduction Reactions
Very few metals exist in pure form in nature (exceptions
are:
). Most exist as compounds with various other elements.
These are called
From metallurgy (the process of producing metals from their compounds or
refining) this process became known as reduction.
Fe2O3(s) + 3CO(g) 
2Fe(s) + 3CO2(g)
SnO2(s) + C(s) 
Sn(s) + CO2(g)
CuS(s) + H2(g) 
Cu(s) + H2S(g)
In each case the metal ore is ‘reduced’ to its elemental state. The substance used
to cause or promote the reduction of a metal compound to an elemental metal is
called the reducing agent. In the examples above the reducing agents are:
The reverse of this process is called oxidation, in which a metal is reacted with
another substance to form a compound. Oxidation reactions can be further broken
down into two sub-categories known as combustion (burning) and corrosion (rusting).
2Mg(s) + O2(g)  2MgO(s)
2Al(s) + 3Cl2(g)  2AlCl3(s)
Cu(s) + Br2(g) 
CuBr2(s)
The substance used to cause or promote the oxidation of a metal to produce a
metal compound is called the oxidizing agent. In these examples the oxidizing
agents are:
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Homework: Page 653 #2,3,4
Oxidation/reduction reactions can be explained by examining the transfer of
electrons between the atoms involved in the reaction.
Example:
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Write the ionic equation for the reaction:
The chlorine ions are not involved in the reaction (they are
).
Isolate the ‘real reactants’ and write half-reaction equations showing the transfer
of electrons:
For half-reactions the number of atoms on each side must be balanced (as always)
and the total charge must also be balanced on each side as well. In other words,
From these half-reactions it can be seen that Zn(s) loses electrons. The Zn(s) has
been oxidized.
The H+(aq) ions have gained electrons and have been reduced.
Homework: pg 656, #8  11
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Oxidation States
The oxidation state of an atom in a molecule/ion is
An oxidation number
Oxidation numbers are written with the +/- sign in front of the number. (i.e. +2, -1)
whereas valence numbers (the actual charge on an atom/molecule/ion) are written
with the +/- sign behind the number (i.e 2+, 1-).
Oxidation numbers and valence numbers are not necessarily the same!!!
Oxidation numbers are a contrived, arbitrary method to count electrons.
Common Oxidation Numbers
Atom or Ion
Oxidation
Number
Examples
All atoms in elements
Hydrogen in all compounds
Except hydrogen in hydrides
Oxygen in all compounds
Except oxygen in peroxides
All monatomic ions
Using these ‘known’ oxidation numbers, the oxidation numbers for any other
element in a compound/ion can now be determined.
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Examples
1. What is the oxidation number of carbon in methane (CH4)?
Hydrogen has an oxidation number of
There are 4 hydrogens, so the total oxidation number for the hydrogen in
this molecule is.
The molecule is electrically neutral (it does not have a net charge), so
therefore the carbon must have an oxidation number of
2. What is the oxidation number of manganese in the permanganate ion (MnO4-)?
Oxygen has an oxidation number of
There are 4 oxygen atoms, so the total oxidation number for the oxygen in
this molecule is
The ion has a net charge of 1-, so therefore the manganese must have an
oxidation number of
3. What is the oxidation number of sulfur in sodium sulfate?
Using the concept of oxidation numbers, the terms oxidation and reduction can be
further refined to mean:
Oxidation –
Reduction –
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Example
The determination of blood alcohol content from a sample of breath or blood
involves the reaction of the sample with acidic potassium dichromate solution. If
ethanol is present, chromium(III) ions, water, and acetic acid are produced.
Identify the oxidation and reduction components in the following reaction.
2Cr2O72-(aq) + 16H+(aq) + 3C2H5OH(aq)  4Cr3+(aq) + 11H2O(g) + 3CH3COOH(aq)
In biology, oxidation and reduction are often defined in terms of the addition or
removal of oxygen or hydrogen in relation to carbon.
The removal of oxygen results in the decrease of the oxidation number of carbon.
HCOOH  HCHO
C is +2
C is 0
carbon has been reduced
The removal of hydrogen results in the increase of the oxidation number of carbon.
C2H6 
C2H4
C is -3
C is -2
carbon has been oxidized
Photosynthesis involves a series of reduction reactions where the oxidation number
of carbon is gradually changed from +4 in carbon dioxide to 0 in sugars. Cellular
respiration reverses this process by gradually oxidizing sugars until carbon is again
+4 in carbon dioxide.
Homework: Page 659 #12, 13, 14, 16
Page 662 #18, 19, 20
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Balancing Redox Reactions
There are two methods that can be used to balance redox reactions. One involves
using oxidation numbers and the other involves balancing the two half-reactions
involved in the reaction.
When balancing redox reactions
Method 1: Balancing Redox Reactions using Oxidation Numbers
Type 1: Simple Redox Reaction in a neutral solution
Balance the following redox reaction.
H2S(g) + O2(g)  SO2(g) + H2O(g)
Step 1: Assign each atom its oxidation number and identify the ones that change.
Step 2: Determine the change in electrons per molecule involved.
H2S has only one sulfur atom,
O2 has two oxygen atoms,
Step3: Find the lowest common factor to balance the gain and loss of electrons
for each reactant.
In this case the lowest common multiple is 12. So the H2S must be
multiplied by 2 and the O2 must be multiplied by 3. We now have the
reactants balanced as follows:
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Step 4: Balance the products using the coefficients from the reactants. (Don’t
change the coefficients for the reactants since they are already balanced!!)
Type 2: Redox Reaction in an acidic solution
Chlorate ions and iodine react in an acidic solution to produce chloride ions and
iodate ions. Balance the equation for this reaction.
Write the equation using the known reactants and products.
Step 1: Assign each atom its oxidation number and identify the ones that change.
Step 2: Determine the change in electrons per molecule involved.
ClO3 has only one chlorine atom,
I2 has two iodine atoms,
Step 3: Find the lowest common factor to balance the gain and loss of electrons
for each reactant.
In this case the lowest common factor is 30. So the ClO3- must be
multiplied by 5 and the I2 must be multiplied by 3. We now have the
reactants balanced as follows:
Step 4: Balance the products using the coefficients from the reactants. (Don’t
change the coefficients for the reactants since they are already balanced!!)
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Step 5: Balance the oxygen atoms by adding H2O(l) (since the reaction happens in
an acidic aqueous solution)
Step 6: Since the reaction occurs in an acidic solution, H+(aq) ions are added to
finish balancing the redox reaction.
Type 3: Redox Reaction in an basic solution
Methanol reacts with permanganate ions in a basic solution. Balance the redox
equation for this reaction.
Write the equation using the known reactants and products.
Step 1: Assign each atom its oxidation number and identify the ones that change.
Step 2: Determine the change in electrons per molecule involved.
CH3OH(aq) has only one carbon atom,
MnO4-(aq) has only one manganese atom,
Step 3: Find the lowest common factor to balance the gain and loss of electrons
for each reactant.
In this case the lowest common multiple is 6. So the CH3OH(aq) is not
changed at all and the MnO4-(aq) must be multiplied by 6. We now have the
reactants balanced as follows:
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Step 4: Balance the products using the coefficients from the reactants. (Don’t
change the coefficients for the reactants since they are already balanced!!)
Step 5: Balance the oxygen atoms by adding H2O(l) (since the reaction happens in
an basic aqueos solution)
Step 6: Balance the hydrogens by adding H+(aq) ions.
Step 7: Because this reaction happens in a basic solution, add an equal number of
hydroxide ions (OH-(aq)) to both sides of the equation.
Step 8: Combine any H+(aq) ions and OH-(aq) ions on the same side of the equation to
make H2O(l) and cancel any ‘excess’ water.
Method 2: Balancing Redox Reactions using Half-Reactions
This method is very similar to that of balancing redox reactions using oxidation
numbers. In this case you balance each part of the reaction separately (i.e. balance
the reduction part by itself and then the oxidation part by itself).
Example: Dichromate ions react with an acidic solution of iron(II) ions. The
products formed are iron(III) and chromium(III) ions as shown by the
following skeleton equation.
Fe2+(aq) + Cr2O72-(aq)  Fe3+(aq) + Cr3+(aq)
Step 1: Separate the equation into its two component parts (a reduction and an
oxidation).
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Step 2: Balance all the atoms other than oxygen and hydrogen in each halfreaction equation.
Step 3: Balance the oxygen atoms in each half-reaction by adding H2O(l) and then
balance the hydrogen atoms by adding H+(aq).
Step 4: Balance each half-reaction separately by adding electrons to balance the
charge.
Step 5: Balance the two equations to each other by multiplying by a common factor
to balance the number of electrons in each equation.
Step 6: Add the two half-reactions together to get the final balanced equation.
Cancel the electrons on both sides and any other common items on each
side of the equation.
For basic solutions:
Step 7: For reactions that occur in basic solutions, simply add OH-(aq) to both sides
of the equation to equal to the number of H+(aq) ions added.
Step 8: Combine the H+(aq) and OH-(aq) ions on the same side of the equation to
create H2O(l) and cancel any excess H2O(l) on either side of the equation.
Homework: Pg. 668 #1, 2, 3, 4 and Pg. 673 #6, 7, 8
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Predicting Redox Reactions
Fe2O3(s) + 3CO(g) 
2Fe(s) + 3CO2(g)
In this reaction the Fe2O3(s) is ________to Fe(s). The CO(g) used in this reaction is
called the_____________.
Reducing agent (RA)–
2Al(s) + 3Cl2(g)  2AlCl3(s)
In this reaction the Al(s) is _________to AlCl3(s). The Cl2(g) used in this reaction is
called the ____________.
Oxidizing agent (OA)–
Through a vast number of experiments a redox table (see Appendix C11) has been
constructed to outline the relative strengths of reducing agents and oxidizing
agents. Using this table reliable predictions can be made regarding whether or not
a redox reaction will occur when these two substances are mixed together.
Organization of Redox Tables
Redox Spontaneity Rule
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From the table in Appendix C11, fluorine gas (F2(g)) is the strongest oxidizing agent
(it is the least likely to give up electrons) and conversely, this makes it the weakest
reducing agent (the most likely to gain electrons).
Li+(aq) is the weakest oxidizing agent (it is the most likely to give up electrons) and
conversely, this makes it the strongest reducing agent (the least likely to gain
electrons).
In general,
Homework: Page 679 #14, 16, 18, 19
Predicting Redox Reactions in Solution
Step 1: List all of the possible entities that are present in the solution.
Step 2: Label all of the possible entities as either oxidizing or reducing agents.
Step 3: Use this information and the spontaneity rule to determine if the reaction
is spontaneous.
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Example: A solution of potassium permanganate is poured into an acidified iron(II)
sulfate solution. Will there be a spontaneous redox reaction?
Step 1: List all of the possible entities that are present in the solution.
Step 2: Label all of the possible entities as either oxidizing or reducing agents and
identify the strongest oxidizing agent and the strongest reducing agent.
Step 3: Use this information and the spontaneity rule to determine if the reaction
is spontaneous.
Homework: Page 683 #1  6, 11
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Technology of Cells and Batteries
When electrons move from one area to another they produce an electric current
(electricity!). Since redox reactions involve the transfer of electrons, these
chemical reactions can be used to produce electricity.
Electric Cell –
Battery –
The positive electrode is called the
The negative electrode is called the
The amount of electrical energy that a cell can produce is dependent on the
Voltage is a ratio of energy to charge. It is not dependent on the size of the cell
but rather the chemical composition of the reactants in the cell.
The total charge transferred by the movement of electrons in a cell/battery is
measured in coulombs (C).
Electric current (amperes (A) or coulombs/second)
The larger the cell, the greater the current that can be produced by the cell.
Example: The are many sizes of batteries (A, AA, AAA, C, D). All of these are 1.5V
batteries. However, the bigger batteries can produce more current.
Why do batteries have a ‘best before’ date?
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Inside the battery itself, a chemical reaction produces the electrons.
Electrons flow from the battery into a wire, and must travel from the
negative to the positive terminal (anode to cathode) for the chemical
reaction to take place. That is why a battery can sit on a shelf for a year and
still have plenty of power -- unless electrons are flowing from the negative
to the positive terminal, the chemical reaction does not take place. Once you
connect a wire, the reaction starts.
The power (watts (W)) of a cell or battery
The energy density or specific energy of a battery is
This has become an important measurement for technological reasons. For example
if the energy density is low (ie. lots of joules but a heavy mass as well) the battery
would not be very good for a car. If the energy density is high (i.e. lots of joules
but a low mass) if would be much more practical to be used in a car.
Batteries that are made up of cells that can be used only once are called
Batteries that are made up of cells that can be reused (rechargeable) are called
Fuel cells are a special type of primary cell,
Homework: Read pages 691 – 693 and make a table with the following headings:
 Fuel cell name
 Reactant at cathode
 Reactant at anode
 Electrolyte
 Potential uses
 Advantages
 Disadvantages
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Galvanic Cells
Galvanic Cell –
A half-cell consists of
There are two basic designs of galvanic cells. One uses a ‘salt bridge’ which is a Ushaped tube filled with a non-reactive electrolyte (ie. positive and negative ions
that are not part of the net reaction). Cotton plugs help to prevent the mixture of
the two reacting solutions, but allow ions to pass.
The other uses a porous porcelain cup that contains one of the electrodes and
electrolytes. This is immersed in a beaker that has the other electrode and its
corresponding electrolyte. (see page 695 in the textbook)
Galvanic cells can be represented using a shorthand cell notation:
The single line (|) indicates a phase boundary (between the electrode and its
electrolyte) and the double line (||) indicates a physical boundary (the salt bridge
or a porous porcelain cup).
Example 1: For the following galvanic cell, determine which metal is the cathode
and which metal is the anode.
Cu(s) | Cu(NO3)2(aq) || Zn(NO3)2(aq) | Zn(s)
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Example 2: For the following galvanic cell, determine which metal is the cathode
and which metal is the anode.
Cu(s) | Cu(NO3)2(aq) || AgNO3(aq) | Ag(s)
Explaining Observations in Galvanic Cells
Evidence and Interpretation of the Copper-Zinc Cell
Evidence
Interpretation
Copper electrode increases in mass;
intensity of the blue colour of the
electrolyte decreases
The zinc electrode decreases in mass
A voltmeter indicates that the copper
electrode is the cathode (positive) and
the zinc electrode is the anode
(negative)
The salt bridge (or the porous porcelain cup) is important to maintain electrical
neutrality in each half-cell. This allows ions to pass to compensate for the
electrons being lost from the anode or being gained by the cathode. Without this
the galvanic cell would stop after a very short period of time.
Galvanic cells with inert electrodes
________ and __________are commonly used as inert electrodes. These inert
electrodes simply provide a surface for the half-reaction to occur.
Homework: Page 700 #3,5
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Standard Cells and Cell Potentials
Standard Cell –
Standard Reduction Potential (Ero) –
Standard Cell Potential (ΔEo) –
Reference half-cell –
Using Appendix C11 (Relative Strengths of Oxidizing and Reducing Agents), the
standard cell potential for any two half-cells can be determined. In addition, the
cathode and anode can be identified as well as the direction of the spontaneity of
the reaction.
Example 1: For the following galvanic cell, determine the identity of the cathode
and anode, calculate the standard cell potential and write the net
equation for the reaction.
C(s) | Cr2O72-(aq) H+(aq) Cr3+(aq) || Pb2+(aq) | Pb(s)
Step 1: find the half-cell reactions in Appendix 11
Cr2O72-(aq) + 14H+(aq) + 6e- 
Pb2+(aq) + 2e-  Pb(s)
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Cr3+(aq) + 7H2O(l)
Ero = 1.23
Ero = -0.13
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Step 2: Determine the identity of the cathode and anode.
Since the chromate ion reaction has the higher reduction potential, it is the
________ in this reaction.
Since the lead reaction has a lower reduction potential, it is the ______ in this
reaction.
NOTE:
Step 3: balance the equations by finding the lowest common multiple for the
number of electrons transferred (in this case 6)
Example 2: A voltmeter indicates that in a standard copper-scandium galvanic cell
(using Cu2+), copper is the cathode and the standard cell potential is
2.36V. Write the half-reactions for the two cells, the net equation,
the shorthand cell notation for the cell, and determine the standard
reduction potential for the scandium ion (Sc3+).
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Step 1: Write the half-cell reactions for the copper and scandium.
Step 2: Find the lowest common multiple to balance the electrons in the two
equations (in this case 6).
Multiply the copper half-reaction by 3 (do not change the standard reduction
potential value).
Multiply the scandium half-reaction by 2.
Add the two reactions together for the net reaction.
The shorthand cell notation for this galvanic cell is:
The standard reduction potential for the scandium ion (Sc3+) is:
Homework: Pg 708 #
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