UNIT 5
REDOX AND ELECTROCHEMISTRY
18.1 Oxidation and Reduction
*Scientists first described the reaction of metals with oxygen to form metal
oxides as oxidation, and described the removal of oxygen from metal oxides
to give pure metal as reduction.
*They later realized that electrons were transferred from one substance to
another.
Electron-transfer reactions = oxidation-reduction reaction (or redox)
Example of common redox reactions:
Battery: electrons transferred pass through some circuits that lights the
bulb of a flashlight.
Food metabolism: oxidation of carbohydrates to water and carbon
dioxide.
All combustion and corrosion processes
Bleaching clothes: bleach (NaOCl) is an oxidizing agent.
1. Oxidation-Reduction Reactions
Oxidation - refers to the loss of electrons by a reactant.
Reduction - refers to the gain of electrons by a reactant.
Always occur together. (LEO SAYS GER)
Example:
2Na + Cl2  2NaCl
2Na  2Na+ + 2e- (oxidation)
Cl2 + 2e-  2Cl- (reduction)
*the electron accepting substance is called the oxidizing agent and the
substance that supplies the electrons is called the reducing agent.
 Oxidizing agent and reducing Agents ARE FOUND ON THE
REACTANTS SIDE 
Therefore, Na is the reducing agent and Cl2 is the oxidizing agent.
Example 2:
Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)
Write the net ionic equation:
Cu + Ag+1 + NO3-1  Cu+2 + NO32- + Ag
Cu(s) + 2Ag+(aq)  Cu2+ + 2Ag(s)
Therefore, Copper metal is oxidized and silver ions are reduced.
Homework:
Pg. 715 # 1-9
18.2 Electronegativity and Oxidation Numbers
*Electronegativity: a measure of the ability of bonded atoms to attract
electrons that are shared with other atoms.
*Metals have low electronegativities and tend to lose electrons to other
atoms. Alkali metals have the lowest electronegativity values. Therefore
they lose electrons easily and are said to oxidize readily. Gold has the
highest electronegativity of any metal. Therefore, least reactive and difficult
to oxidized.
*Nonmetals have high electronegativities. Atoms of these elements accept
electrons readily. Therefore, they are easily reduced.
*The transfer of electrons is a redox reaction can be complete or incomplete.
In the case of the formation of ionic compounds the transfer of electrons is
complete. In the case of molecular compounds, the transfer is incomplete.
*We can keep track of the electrons being transferred by assigning oxidation
numbers to the various atoms.
*Oxidation number: The charge an element APPEARS to have when
electrons are counted by some arbitrary rules. A positive oxidation number
indicates a loss of electrons (oxidation) and a negative number indicates a
gain of electrons (reduction).
RULES:
1.
2.
3.
4.
5.
Each atom in free element has ox. no. = 0.
Zn O2 I2 S8
In simple ions, oxidation number = charge on ion.
-1 for Cl-, +2 for Mg2+
O has oxidation number = -2
(except in peroxides: in H2O2, O = -1)
oxidation number of H = +1
(except when H is associated with a metal as in NaH where it is -1)
Algebraic sum of oxidation numbers
= 0 for a compound
= overall charge for an ion
Assigning Oxidation Numbers
NH3
N = -3
H = +1
CaCl2
Ca = +2
Cl = -1
MoS2
Mo = +4
S = -2
Na2S2O3
S = +2
O = -2
ClO-
Cl = +1
O = -2
H3PO4
P = +5
H = +1
MnO4-
Mn = +7
O = -2
Cr2O72-
Cr = +6
O = -2
C3H8
C = (-8/3) H = +1
C3H6O
C = (-4/3)
H = +1
O = -2
Cr(NO3)3
Cr = +3
N = +5
O = -2
H2O2
H = +1
O = -1
(NH4)2CO3 N = -3
C=
Na = +1
O = -2
H = +1
O = -2
C=+4
P.S. Nonmetals can have several different oxidation numbers. However, the
oxidation number of an element never exceeds the group number of the
element.
*number of electrons loss = number of electrons gained
Example 1:
Which of the following reactions is a redox reaction?
2NaBr(aq) + Cl2(aq)  2NaCl(aq) + Br2(aq)
+1 -1
0
+1 -1
0
2HCl(aq) + Na2S(aq)  2NaCl(aq) + H2S(g)
+1 -1
+1 -2
+1 -1
+1 -2
Example 2:
Redox reactions that you are familiar with:
Mg(s) + O2(g)  MgO(s)
0
0
+2 -2
Mg is oxidized and O is reduced
Mg(s) + HCl(aq)  MgCl2(aq) + H2(g)
0 +1 -1
+2 -1
0
+
Mg is oxidized and H is reduced
Zn + H2SO4  ZnSO4 + H2
0
+1 -2
+1 -1
0
+
Zn is oxidized and H is reduced
Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O
0
+1+5 -2
+2 +5-2
+4 -2 +1 -2
Cu is oxidized and N is reduced
Homework
Pg. 726 # 9-16
18.3 Half-Reaction method to balance equations
(Balancing Redox Reactions)
STEP 1
-Calculate the oxidation numbers of all elements in the equation
and identify the substance oxidized and the substance reduced.
STEP 2
-Separate the total reaction into an oxidation half-reaction and a
reduction half-reaction.
-Balance the number of atoms of the element reduced or
oxidized.
-To each half-reaction, add the number of electrons lost or
gained. Electrons lost are added to the product side and
electrons gained are added to the reactant side.
STEP 3
-Balance the total charges by adding H+ (in acidic and neutral
solutions) or OH- (in basic solutions).
STEP 4
-Add H2O to balance the hydrogens and oxygens
STEP 5
-Combine the two half reactions so that the number of electrons
lost equals the number of electrons gained.
-Cancel any substances appearing on both sides of the total
equation.
Example 1: Balance the following half reaction:
O2-  O2
STEP
Write the oxidation number and
identify the substance oxidized
or reduced.
Write the substance oxidized and
the oxidation product
O2-  O2
-2
0
Oxidation (loss 2e each)
O2-  O2
Balance the number of atoms
Of the element oxidized.
2O2-  O2
Add the total number of
Electrons lost to the product side
2O2-  O2 + 4e
Example 2: Balance the following redox equation occuring in acidic solution
Cr2O7-2  Cr3+
STEP
Write the oxidation number and
identify the substance oxidized
or reduced.
Write the substance reduced and
the reaction product
Cr2O7-2  Cr3+
+6 -2
+3
Reduction (gain 3e each)
Cr2O7-2  Cr3+
Balance the number of atoms
of the element reduced.
Cr2O7-2  2Cr3+
Add the total number of
electrons gained to the reactant
side
Cr2O7-2 + 6e 2Cr3+
Since this reaction is done in
acidic solution, use H+ to
balance the charges.
14H+ + Cr2O7-2 + 6e 2Cr3+
Now add H2O to balance the 14H+ + Cr2O7-2 + 6e 2Cr3+ + 7H2O
hydrogens and oxygens.
Example 3: Balance the following redox equation occuring in acidic solution
MnO4- + SO32-  Mn2+ + SO42STEP
Write the oxidation number and
identify the substance oxidized
or reduced.
Write the substance oxidized and
the oxidation product
MnO4- + SO32-  Mn2+ SO42+7 -2 +4 -2
+2
+6 -2
Gains 5e (reduction)
loses 2e (oxidation)
SO32-  SO42-
Balance the number of atoms
of the element reduced.
SO32-  SO42-
Add the total number of
electrons lost to the product
side
SO32-  SO42- + 2e
Since this reaction is done in
acidic solution, use H+ to
balance the charges.
SO32-  SO42- + 2e + 2H+
Now add H2O to balance the
hydrogens and oxygens.
SO32- + H2O  SO42- + 2e + 2H+
Now repeat steps 2-4 for the
Substance reduced
MnO4-  Mn2+
MnO4- + 5e  Mn2+
MnO4- + 5e + 8H+  Mn2+
MnO4- + 5e + 8H+  Mn2+ +
4H2O
Combine the two half reactions
SO32- + H2O  SO42- + 2e +
2H+
So that the number of electrons
Lost equals the number of
(multiply by 5)
MnO4- + 5e + 8H+  Mn2+ +
4H2O
Electrons gained.
(multiply by 2)
5SO32- + 5H2O  5SO42- + 10e +
10H+
2MnO4- + 10e + 16H+  2Mn2+ +
8H2O
Combine the two half reactions
5SO32- + 2MnO4- + 6H+  5SO42- +
2Mn2+ + 3H2O
Homework
Pg. 732 #17-28
9.4
Technology of Cells and Batteries
1. Electric Cells
18th century: Luigi Galvanic
Electric cells call galvanic
Allessandro Volta
cells or voltaic cells
*electric cell: device that continuously converts chemical energy into
electrical energy.
*battery: group of two or more electric cells connected to each other in
series (p. 685, figure 1 and 2)
2. Basic Cell Design and Properties
Example: place zinc metal into CuSO45H2O(aq)
*redox reaction takes place
*electrons transfer from zinc to copper ion
*zinc oxidizes more readily then copper (see activity
series)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
*half-reactions: Zn(s)  Zn2+(aq) + 2e
Cu2+(aq) + 2e  Cu(s)
(oxi)
(red)
*this reaction could not be used as a source of current if the reaction is done
in a single beaker. The electrons provided by the zinc would move directly
to the aqueous Cu2+ ions on contact.
Zinc-Copper Galvanic Cell
*galvanic cell: electric cells adapted for scientific study
*electrons can pass from the zinc electrode to the solution of Cu2+ ions
through an external wire (and through any device to be powered)
voltmeter
e
Zn
Cu
salt bridge
ZnSO4(aq)
Zn becomes Zn2+
CuSO4(aq)
Cu2+ becomes Cu
Anode
(-)
Oxidation
SO42Cathode
(+)
reduction
Zinc and copper rod are electrodes
Anode - the site of oxidation (zinc)
Cathode - the site of reduction (copper)
*electrons travel from the anode to the cathode through a metal wire
(external circuit)
*the force that pushes the electrons is the potential difference (or
electromotive force, emf) measured in volts. This force is due to the
difference in energy of an electron at the two electrodes.
*the voltage is a measure of the joules of energy that can be delivered per
coulomb of charge as the current moves through the circuit.
*the half cells are connected by a salt bridge (internal circuit)
*allows negative/positive ions to move to the other container in order to
maintain electrical neutrality of the two solutions
*as the reaction continues the voltage decreases and finally becomes zero
when the reaction reaches equilibrium - the cell is then dead.
Notation for an electrochemical cell
Zn(s)|Zn2+(aq) (1.0 M)||Cu2+(aq) (1.0M)|Cu(s)
*by convention the anode (oxidation) appears on the left and the cathode
(reduction) on the right
*single vertical line indicate the contact boundaries between phases in each
half cell
*double vertical line represents the salt bridge
Example Problem
When the metal electrode of the electrochemical cell shown below ar
connected by a voltmeter, the meter indicates a potential difference of
0.46V.
Using the activity series (p. 383) identify the half reactions taking place at
each electrode.
Write the shorthand notation for the cell, identifying the anode and cathode
and indicate the direction of movement of electrons and ions
voltmeter
Cu
Ag
salt bridge
Cu(NO3)2(aq)
(1.0M)
AgNO3(aq) (1.0M)
Answer:
Lower in the series is oxidized (more active). Therefore, Cu is oxidized.
Cu(s)  Cu2+(aq) + 2e
(oxidation) Anode
Ag+(aq) + 1e  Ag(s)
(reduction) Cathode
Cu(s)|Cu2+(aq)(1.0M)||Ag+(aq)(1.0M)|Ag(s)
Electrochemical cells can be made from half cells in which either the
reactant or the product of the half reaction is a gas
Example:
voltmeter
e
Zn
H2(g)
salt bridge
ZnSO4(aq)
Platinum electrode
1.0 M HCl
*a hydrogen half cell is linked by a salt bridge to a zinc half cell
*hydrogen gas is bubbled into a solution of 1.0 M HCl over a platinum
electrode
*the platinum does not react but serves as a surface where exchange of
electrons can take place
*from activity series Zn(s) is oxidized, H+(aq) are reduced.
Zn(s)  Zn2+(aq) + 2e
2H+(aq) + 2e  H2(g)
Overall reaction: Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
Shorthand representation: Zn(s)|Zn2+(aq) (1.0M)||H+(aq) (1.0M)|H2(g)(100
kPa),Pt(s)
Standard Electrode Potentials (p. 804)
*reduction potential - the tendency of a half cells ions and molecules to gain
electrons
*cannot measure the electrical potential of one half cell - we can only
measure the voltage difference or emf between 2 half cells by combining
them into an electrochemical cell.
Example: Zinc-Copper cell - cell voltage is 1.10V
Ie. reduction potential of metal ions is 1.10 V greater for copper ions
than for zinc ions (copper is above zinc on activity series)
Or oxidation potential (the tendency to lose electrons) is 1.10 V
greater for zinc than for copper.
Standard Half Cell Potentials
*Standard hydrogen electrode has been assigned a standard potential (o) of
0.00V.
2H+ + 2e  H2(g)
(1.0 M)
100 kPa
o = 0.00 V,
T = 25oC
Standard conditions: 1.0 M solution, pressure 100 kPa, temperature 25oC
*the measured potential difference of a standard electrochemical cell
containing a hydrogen half cell is the standard potential of the second hal
cell.
Standard reduction Potential
*is a measure of the tendency to gain electrons from the hydrogen half cell
(list on page 804)
Standard oxidation potential
*is a measure of the tendency to lose electrons form the hydrogen half cell
Example
Zn2+ + 2e  Zn(s)
Zn(s)  Zn2+ + 2e
oreduction = -0.76V (not easily red)
ooxidation = 0.76V
Calculating Standard Cell Potential
Example: Calculate the potential for the following cell:
Fe(s)|Fe2+(aq)(1.0M)||Cu2+(aq)(1.0M)|Cu(s)
Anode (Oxi)
Cathode (red)
Fe(s)  Fe2+(aq) + 2e
Cl2(g) + 2e  2Cl-
o = 0.76V
o = 1.36V
ocell= 2.12V
Reduction potential depends on the concentration. Therefore when
coefficients in an equation change, o stays the same.
Example:
Ag+(aq) + e  Ag(s)
2Ag+(aq) + 2e  2Ag(s)
o = 0.80V
o = 0.80V
Consumer, Commercial, and Industrial Cells
Textbook: p.688-694
a) Describe a typical lead-acid cell.
b) How can a car battery produce an electromotive force of 12V if the cell reaction
has a standard potential of only 2V?
c) What are the anode and cathode reactions during the discharging of a lead-acid
cell?
d) What is the net reaction taking place?
e) How can the charge be checked? How can it be recharged?
f) What are the disadvantages of this type of battery?
a) Draw and describe the construction of a zinc chloride dry cell.
b) What are some of the advantages and disadvantages of these cells?
c) What are the half reaction of an alkaline dry cell?
d) How does an alkaline dry cell differ from a zinc chloride dry cell?
a) Describe a nickel-cadmium cell
b) What are the half-reactions?
c) What are the advantages and disadvantages of this type of cell?
a) Describe a mercury cell. Where are these cells used?
b) What are the anode, cathode and net cell reaction for this cell?
c) What are the advantages of this type of battery?
a) Describe a fuel cell.
b) What are the anode, cathode and net cell reaction?
c) What are the advantages and disadvantages of this type of cells?
Electrolysis and Corrosion
Textbook: p. 710-713, 730-744
What is electrolysis?
What is an electrolytic cell?
a) Draw and label the electrolytic cell that decomposes molten sodium chloride.
b) Write the oxidation and reduction half reactions for this cell.
c) Write the overall (cell) reaction taking place in this electrolytic cell.
a) Write the oxidation and reduction half-reactions for the electrolysis of water.
b) Write the cell reaction.
a) Describe the process of electroplating.
b) How is silver electroplated?
a) What difficulty had to be overcome befor aluminum could be purified by the
electrolysis of Al2O3?
b) Draw a labelled diagram of the apparatus used to produce aluminum by the
Hall-Heroult method.
c) Write the half-reactions and net cell reaction for this process.
d) Describe problems associated with this method of aluminum production.
Describe the production of magnesium and sodium
How is copper purified?
Describe how iron corrodes, including the factors involved, and the different
reactions in the process. List ways that the corrosion process can be slowed.
10.1 Electrolysis
*electrolytic cell is a type of cell that uses energy to move electrons from
lower potential energy to a position of higher potential energy.
*converts electrical energy to chemical energy
*process is called electrolysis
*non-spontaneous reaction
*electrolytic cell includes electrodes, at least one electrolyte, an external
source of electricity, and external circuit
*possible for electrolytic cells to have the two half-reactions take place in
the same container
Galvanic CellElectrolytic CellSpontaneous reactionNon-spontaneous
reactionConverts chemical energy to electrical energyConverts electrical
energy to chemical energyAnode (negative): ZincAnode (positive):
CopperCathode (positive): CopperCathode (negative): ZincOxidation at
anode: Zn  Zn2+ + 2eOxidation at anode: Cu  Cu2+ +
2eReduction at cathode: Cu2+ + 2e  CuReduction at cathode: Zn2+ +
2e  ZnCell reaction: Zn + Cu2+  Zn2+ + CuCell reaction: Cu +
Zn2+  Cu2+ + Zn
Predicting the products of electrolysis for an aqueous solution
*examine all possible half-reactions and their reduction potentials
*find the overall reaction that requires the lowest external voltage
(negative cell potential closest to zero)
EXAMPLE:
Predict the products of the electrolysis of 1 mol/L LiBr(aq)
Step 1: List the half-reactions and their reduction potentials
Br2 + 2e  2BrEo = 1.066
V
O2 + 4H+ + 4e  2H2O
E = 0.815 V
2H2O + 2e  H2 + 2OHE = -0.414 V
Li+ + e  Li
Eo = -3.040 V
*There are two possible oxidation half-reactions at the anode: the
oxidation of bromide ion in the electrolyte, or the oxidation of water.
*There are two possible reduction half-reactions at the cathode: the
reduction of lithium ions in the electrolyte, or the reduction of water.
Step 2: Combine pairs of half-reactions to produce four possible overall
reactions.
Reaction1: the production of lithium and bromine
2Li+ + 2Br-  2Li + Br2
E o cell  E o cathode  E o anode
 3.040 V  1.066 V
 4.106 V
Reaction 2: the production of hydrogen and oxygen
2H2O  2H2 + O2
E o cell  E o cathode  E o anode
 0.414 V  0.815 V
 1.229 V
Reaction 3: the production of lithium and oxygen
4Li+ + 2H2O  4Li + O2 + 4H+
E o cell  E o cathode  E o anode
 3.040 V  0.815 V
 3.855 V
Reaction 4: the production of hydrogen and bromine
2H2O + 2Br-  H2 + 2OH- + Br2
E o cell  E o cathode  E o anode
 0.414 V  1.066 V
 1.480 V
The electrolysis of water requires the lowest external voltage. Therefore,
the predicted products of this electrolysis are hydrogen and oxygen.
10.3 Stoichiometry of Cell Reactions
*connecting the concepts of stoichiometry and electrochemistry
*current: flow of electrons through an external circuit. Units are ampere (A)
*charge: product of the current flowing through a circuit and the time for
which it flows. Units are coulomb (C)
*one coulomb is the quantity of electricity that flows through a circuit in one
second if the current is one ampere.
q  It
where q is the charge in coulombs, I is the electric current in amperes, and t
is the time in seconds
*charge on a mole of electrons is known as one faraday (1 F)
Charge on one mole of electrons  F
1.602  10-19 C 6.022  10 23 e


1e
1 mol
 9.647  10 4 C / mol
*Faraday’s law states that the amount of a substance produced or consumed
in an electrolysis reaction is directly proportional to the quantity of
electricity that flows through the circuit.
It
ne 
F
Look at sample problems 1 and 2 on page 748.