Redox Reactions
oxidation reduction reactions
Ch 22 sec 1
From the combustion of gasoline to the
metabolism of food - oxidation is
responsible
Oxidation - Loss of electrons
old definition - gain of oxygen
Reduction - Gain of electrons
old definition - loss of oxygen
ONE DOES NOT OCCUR WITHOUT THE
OTHER
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LEO the lion goes GER
Losing Electrons is Oxidation
Gaining Electrons is Reduction
Example
Mg + S ---> Mg2+ + S2-
Magnesium is oxidized (aka reducing
agent)
Sulfur is reduced (aka oxidizing agent)
Redox reactions are usually
presented as two components
Mg ----> Mg2+ + 2eS + 2e- ---> S2Identifying transfers of electrons is easy
for ionic reactions.
What about covalent where there is not
a transfer of electrons but a sharing?
Consider
2H2 + O2 ---> 2H2O
Which element is reduced and oxidized?
Explain
Oxygen is the electron hog. The partial
gain of electrons means it is reduced
and hydrogen’s partial loss means it is
oxidized.
Your Turn
4Fe + 3O2 ---> 2Fe2O3
Write the 2 redox reactions. Which is oxidized
and with is reduced.
answer
Fe --> Fe3+ + 3e- oxidized (reducing agent)
O2 + 2e- ---> O2- reduced (oxidizing agent)
Corrosion occurs more rapidly in the presence
of salts and acids. Why?
These are conducting solutions that make
electron transfer easier.
Sometimes Corrosion is good.
Aluminum oxidizes to form tightly
packed aluminum oxide particles. This is
a protective covering.
Iron needs to be coated to protect it
because the oxide is not tightly packed.
Ch 22 sec 2
Assigning Oxidation Numbers
Redox equations will be balanced using
oxidation numbers
Rules
1. The oxidation number of a monatomic ion
is equal to its ionic charge.
2. The oxidation number of hydrogen in a
compound is +1 except in a metal hydride.
Example - NaH H is -1
3. The oxidation number for oxygen is -2
unless in a peroxide H2O2 where it is -1
more rules
The oxidation number of an
uncombined atom (elemental form) is
0.
For any neutral compound, the sum of
the oxidation numbers of the atoms in
the compound must equal 0.
For a polyatomic ion, the sum of the
oxidation numbers must equal the ionic
charge of the ion.
What is the oxidation number of elements
in the following compounds?
1. SO2
S is +4 and O is -2
2. K2SO4
K is +1 S is +6 O is -2
Oxidation numbers change in redox
reaction. No change no redox.
An increase in the oxidation number
indicates oxidation
A decrease in the oxidation number
indicates reduction
Example - What is being oxidized and
what is being reduced?
2AgNO3 + Cu --> Cu(NO3)2 + 2Ag

+1
+5 -2
0
+2 +5 -2
0
Ch 22 sec 3
Classifying Reactions
Either electrons are transferred or they are
not.
Redox reactions include single-replacement,
combination, decomposition and combustion.
Others - double replacement and acid/base
reactions.
Color changes signify redox reactions
video example
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another redox clip
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Balancing Redox Reactions
Two Methods
Many reactions are too complex to be
balanced by trial and error.
All reactions should be balanced in this
manner. You were taught to balance atoms
because for simple reactions it works.
Balancing atoms may balance the equation
and may not be the correct balanced
equation. Atoms along with charges need to
be balanced.
oxidation-number-change method
balanced by comparing the increase
and decrease of oxidation numbers
Example:
Step 1 - Assign Oxidation Numbers
 +1 +6 -2
2
+1-2
0
+1-2+1 +3 -2
+4-
K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2
 +1 +6 -2
+1-2
0
+1-2+1 +3 -2
+4-2
 K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2
Step 2 - Identify which atoms are oxidized
and which are reduced.
Cr is reduced and S is oxidized
Onto step 3
Use a bracket line to connect the atoms that
undergo oxidation and another line to
connect those that undergo reduction. Then
write the oxidation-number change at the
midpoint of each line.
-3
 +1 +6 -2
+1-2
0
+1-2+1
+3 -2
+4-2
 K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2
+4
 Step 4 - Make the total increase in oxidation number
equal to the total decrease in oxidation number by
using appropriate coefficients.
(4)(-3)=-12
 K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2
(3)(+4)=+12
 The coefficient is the number of atoms needed.
 2 K2Cr2O7 + H2O + 3S --> KOH + 2Cr2O3 + 3SO2
Step 5 - Finish balancing by inspection
 2K2Cr2O7 + 2H2O + 3S --> 4KOH + 2Cr2O3 + 3SO2
 Your Turn  Balance using the oxidation-number-change method.
As2O3 + Cl2 + H2O --> H3AsO4 + HCl

(2)(-1)=-2
 +3 -2
1
0
+1 -2
+1 +5 -2
+1-
As2O3 + Cl2 + H2O --> H3AsO4 + HCl

(1)(+2)=+2
As2O3 + 2Cl2 +5H2O --> 2H3AsO4 + 4HCl
Using Half-Reactions
Good for ionic reactions
Two equations used - one shows
oxidation the other reduction.Then
combined together in the last step.
Balance using 1/2 reactions
Example:
 KMnO4 + HCl --> MnCl2 + Cl2 + H2O + KCl
Step 1 - Write the unbalanced equation in
ionic form. Place oxidation numbers above.
 +1
+7-2
+1
-1
+2
-1
0
+1 -2 +1
-1
 K1+ + MnO41- + H1+ + Cl1- --> Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-
 Step 2 - Write separate 1/2 reactions for the
oxidation reduction process.
 Cl1- --> Cl2
 MnO41- --> Mn2+
Step 3 - Balance the atoms in each 1/2
reaction. When reactions take place in an acid
solution you will need to use water and H+ to
balance hydrogen and oxygen atoms.
 2Cl1- --> Cl2
 MnO41- + 8H+ --> Mn2+ + 4H2O
 atoms are balanced but charges are not.
Step 4 - Add electrons to one side of each 1/2
reaction to balance the charges.
 2Cl1- --> Cl2 + 2e MnO41- + 8H+ + 5e- --> Mn2+ + 4H2O
Step 5 - numbers of electrons must be equal.
Multiply each reaction by a number to make
electrons equal. In this case make electrons
equal 10.
 10Cl1- --> 5Cl2 + 10e 2MnO41- + 16H+ + 10e- --> 2Mn2+ + 8H2O
Step 6 - Add the 1/2 reactions to show an
overall equation.
 10Cl1- + 2MnO41- + 16H+ + 10e- --> 5Cl2 + 10e- + 2Mn2+ + 8H2O
Remove terms that are the same on both
sides.
 10Cl- + 2MnO4 + 16H+ --> 5Cl2 + 2Mn2+ + 8H2O
 Step 7 - Add spectator ions and balance the
equation.
 From step 1
 K1+ + MnO41- + H1+ + Cl1- -->
Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-
 10Cl- + 2K+ + 2MnO4 + 16H+ + 6Cl- --> 5Cl2 + 2Mn2+
+ 4Cl-+ 8H2O + 2K+ + 2Cl Summarize spectator and nonspectator ions.
 16Cl- + 2K+ + 2MnO4 + 16H+ -->
5Cl2 + 2Mn2+ + 6Cl- + 8H2O + 2K+
The equation is now balanced for atoms and charge.
Now you can rewrite it into:
2KMnO4 + 16HCl --> 2MnCl2 + 5Cl2 + 8H2O + 2KCl
Balance using 1/2 reaction method
Your Turn:
S + HNO3 --> SO2 + NO + H2O
Step 1 - Write the unbalanced equation in
ionic form. Place oxidation numbers above.
 0
+1
+5 -2
+4-2
+2-2
+1 -2
S + H+ + NO3- --> SO2 + NO + H2O
Identify what is oxidized and what is reduced.
Step 2 - Write separate half-reactions for the
oxidation and reduction process.
S --> SO2
NO3- --> NO
Step 3 - Balance the atoms in each 1/2
reaction. When reactions take place in an acid
solution you will need to use water and H+ to
balance hydrogen and oxygen atoms.
2H2O + S --> SO2 + 4H+
4H+ + NO3- --> NO + 2H2O
note: charges are not balanced
Step 4 - Add electrons to one side of
each 1/2 reaction to balance the
charges.
2H2O + S --> SO2 + 4H+ + 4e4H+ + NO3- + 3e- --> NO + 2H2O
Step 5 - numbers of electrons must be
equal. Multiply each reaction by a
number to make electrons equal. In this
case make electrons equal 12.
6H2O + 3S --> 3SO2 + 12H+ + 12e16H+ + 4NO3- + 12e- --> 4NO + 8H2O
Step 6 - Add the 1/2 reactions to show
an overall equation.
6H2O + 3S + 16H+ + 4NO3- + 12e- -->
3SO2 + 12H+ + 12e- + 4NO + 8H2O
Remove terms that are the same on
both sides.
3S + 4H+ + 4NO3- --> 3SO2 + 4NO + 2H2O
Step 7 - Add spectator ions and balance the
equation. This reaction does not have
spectator ions. Therefore final answer is:
3S + 4HNO3 --> 3SO2 + 4NO + 2H2O
Electrochem Intro
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Ch 23 sec 1
Electrochemical Cells
When a strip of zinc metal is placed in CuSO4.
Electrons are transferred from Zn to Cu.
Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)
The flow of electrons is an electric current.
Any conversion between electrical and
chemical energy is an electrochemical
process. The device that converts between
the two is an electrochemical cell.
Voltaic Cells
Convert chemical energy into electrical
energy
Half-cell is part of the cell where
oxidation or reduction takes place.
A half-cell consists of a metal rod or
strip immersed in a solution of its ions.
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zinc-copper reaction
One 1/2 cell has a zinc rod immersed in zinc
sulfate
One 1/2 cell has a copper rod immersed in
copper (II) sulfate
Cells are separated by a salt bridge
a tube containing a strong electrolyte often K2SO4
A wire carries the electrons in the external
circuit from the zinc rod to the copper rod.
The driving force is the spontaneous redox
reaction
Electrode - a conductor in a circuit that
carries electrons to or from a substance.
Anode - electrode at which oxidation occurs.
Electrons are produced at the anode so it it
labeled the negative electrode.
Cathode - electrode at which reduction takes
place. Electrons are consumed at the
electrode. The cathode is labeled the positive
electrode.
Neither electrode is really charged. The
moving electrons balance any charge that
might build up as oxidation and reduction
occur.
The electrochemical process in a zinc-copper
voltaic cell. These steps occur at the same time.
1. Electrons are produced at the zinc rod.
Zn --> Zn2+ + 2e- (anode)
2. The electrons leave the the zinc anode and
travel through the external circuit to the
copper rod. If a bulb is in the circuit it will
light.
3. Electrons enter the copper rod and interact
with copper ions in solution.
Cu2+ + 2e- --> Cu
To complete the circuit, both positive
and negative ions move through the
aqueous solutions via the salt bridge.
Summary:
Zn --> Zn2+ + 2eCu2+ + 2e- --> Cu
Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)
Dry Cells (batteries)
A voltaic cell in which the electrolyte is a
paste.
A zinc container is filled with a thick , moist
electrolyte paste of manganese (IV) oxide,
(MnO2), ZnCl2, NH4Cl and water.
A graphite rod is embedded in the paste.
The zinc container is the anode and the
graphite rod is the cathode.
The thick paste and its surrounding paper
liner prevent the contents of the cell from
freely mixing.
Dry Cell Reactions:
Zn --> Zn2+ + 2e2MnO2 + 2NH4+ + 2e- --> Mn2O3 + 2NH3 + H2O
The graphite serves as a conductor even
though it is the cathode. The Mn is actually
reduced.
Alkaline battery - same as dry cell but the
paste is KOH to prevent buildup of ammonia
gas.
Lead Storage Batteries
A battery is actually a group of cells
connected together.
A 12 volt car battery consists of 6 cells.
One set of grids packed with spongy lead
(anode).
The other grid is (cathode) is packed with
PbO2.
The grids are immersed in concentrated
sulfuric acid.
The reactions:
Pb + SO42- --> PbSO4 + 2e- (oxidation)
PbO2 + 4H+ + SO42- + 2e- --> PbSO4 +
2H2O (reduction)
Overall
Pb + PbO2 + 2H2SO4 --> 2PbSO4 + 2H2O
The sulfate slowly builds up and
concentration of acid decreases.
The car’s generator reverses the
reaction.
Fuel Cells
Cells with renewable electrodes.
A voltaic cell in which a fuel substance
undergoes oxidation and from which electrical
energy is continuously obtained.
Do not have to be recharged
emits no air pollutants, operates quieter,
more cost effective than an electrical
generator.
The hydrogen-oxygen fuel cell
3 compartments separated by 2 carbon
electrodes.
Oxygen (the oxidizer) is fed into the
cathode compartment.
Hydrogen (the fuel) is fed into the
anode compartment.
The gases diffuse slowly.
The electrolyte in the center is a hot,
concentrated solution of KOH.
Electrons enter from the oxidation 1/2
reaction at the anode then pass through
an external circuit to enter the
reduction 1/2 reaction at the cathode.
Summary:
2H2 + 4OH- --> 4H2O + 4e- (anode)
O2 + 2H2O + 4e- -->4OH- (cathode)
The overall reaction is:
2H2 + O2 --> 2H2O
Ch 23 sec 2
Cell Potentials
Electrical Potential - cells ability to
produce an electric current. Measured
in Volts.
The electrical potential results from a
competition for electrons.
E0 = reduction potential
Cell Potential
E0 = E0red - E0Oxid
Standard Cell Potential
measured potential when the ion
concentrations in the half-cells are 1M,
gases are at 101 kPa and temperature
is 25oC
Half-cell potentials cannot be measured.
So, the 1/2 cells are compared to
assigning a hydrogen cell 0.00 V where
reduction of Hydrogen occurs.
Reduction takes place at the cathode
Oxidation takes place at the anode
Calculating 1/2 cell potential examples:
2 examples
Zinc and Hydrogen half cells.
A voltmeter reads +0.76 V.
The zinc is oxidized (anode)
Hydrogen Ions are reduced (cathode)
oxidation Zn --> Zn2+ + 2ereduction 2H+ + 2e- --> H2
cell reaction Zn + 2H+ --> Zn2+ + H2
therefore,
E0 = E0red - E0Oxid
E0 = E0H+ - E0Zn2+
+0.76V = 0.00V - E0Zn2+
E0Zn2+ = -0.76V
The value is negative because the tendency
for zinc ions to be reduced to zinc metal is
less than the tendency of hydrogen ions to be
reduced to hydrogen gas. Therefore it is
negative because
 Zinc is oxidized.
Second example this time for copper.
Reduction Cu2+ + 2e- --> Cu
Oxidation H2 --> 2H+ + 2eCell reaction Cu2+ + H2 --> Cu + 2H+
Cell potential measured at +0.76V.
So, E0 = E0red - E0Oxid
+0.34V = E0Cu2+ - 0.00H+
E0Cu2+ = +0.34V
The potential for Cu ions to be reduced is
higher than the potential for H ions to be
reduced. Therefore it is positive because
Copper is reduced.
Table 23.2 in your book on page 688 has a
list of all the 1/2 cell potentials.
Calculating Standard Cell Potentials
To function a cell must be constructed
of 2 half-cells.
If the cell potential for a given redox
reaction is positive then the reaction is
spontaneous.
Calculating Standard Cell Potentials
Example
Determine the cell reaction, the
standard cell potential, and the half-cell
that acts as the cathode for a voltaic
cell composed of the following halfcells.
Fe3+ + e- --> Fe2+ E0Fe3+ = +0.77V
Ni2+ + 2e- --> Ni
E0Ni2+ = -0.25V
Answer:
Reduction takes place in the Fe3+ half cell.
So, this cell is the Cathode
The 1/2 reactions are:
Ni --> Ni2+ + 2e2[Fe3+ + e- --> Fe2+]
The cell reaction is:
Ni + 2Fe3+ --> Ni2+ + 2Fe2+
Cell potential is:
E0 = E0red - E0Oxid
+0.77V - (-0.25V) = +1.02V
The reaction is spontaneous.
Your Turn:
 Calculate E0cell to determine whether
the following redox reaction is
spontaneous as written:
Ni + Fe2+ --> Ni2+ + Fe
From table 23.2 in your text
E0Ni2+ = -0.25V
E0Fe2+ = -0.44V
answer
Nickel is oxidized Iron is reduced
E0= -0.44V - (-0.25V) = -0.19
Reaction will proceed in the reverse
Electrolytic Cells
23.3
Electrolysis - making a nonspontaneous
reaction go.
ex - silverplated dishes and utensils, goldplated jewelry and chrome-plated auto parts
The electrolytic cell is an electrochemical cell
used to cause a chemical change through the
application of electrical energy (DC Current).
Similarities between voltaic and electric cells:
electrons flow from the anode to the cathode
reduction occurs at cathode
oxidation anode
Differences
Voltaic is spontaneous
Electrolytic is result of outside push of
electrons
Cathode is negative electrode
(connected to negative electrode of
battery) and vice versa for anode.
Electrolysis of water produces hydrogen and
oxygen gas.
An electrolyte is often used to help conduct
an electric current. The electrolyte is usually
redoxed.
The electrolysis of Brine (salt water) produces
Cl2 + H2 + NaOH
Electrolysis of molten NaCl produces:
liquid Na used in sodium vapor lamps and as
the coolant in some nuclear reactors.
chlorine gas used to sterilize drinking water
and the manufacturing of PVC and pesticides.
Which one comes off of which electrode?
sodium gains and electron - cathode
Electroplating
The depositing of a thin layer of a metal
on an object
Common metal used to plate:
Ag, Au,Cu, Ni, Cr
An object to be silver plated is made
the cathode. Why?
Anode is the silver to be deposited
Electrolyte is a silver salt.
Also used to purify metals and various
other processes you will read about.