Transcript File

Chemistry 30 Chapter 14
Electrochemical Cells pg. 610-669
14.1 Technology of Cells and
Batteries
Pg. 612-621
Introduction to Electrochemistry
 An electric cell converts chemical energy into electrical energy
 Alessandro Volta invented the first electric cell but got his inspiration from Luigi Galvani.
Galvani’s crucial observation was that two different metals could make the muscles of a frog’s legs
twitch. Unfortunately, Galvani thought this was due to some mysterious “animal electricity”. It was
Volta who recognized this experiment’s potential.
 An electric cell produces very little electricity, so Volta came up with a better design:
 A battery is defined as two or more electric cells connected in series to produce a
steady flow of current
 Volta’s first battery consisted of several bowls of brine (NaCl(aq))
connected by metals that dipped from one bowl to another
 His revised design, consisted of a sandwich of two metals
soaked in salt water (an electrolyte).
separated by paper
Cells and Batteries
 Alessandro Volta’s invention was an immediate technological success
because it produced electric current more simply and reliably than
methods that depended on static electricity.
 It also produced a steady electric current –something no other device
could do.
Cells and Batteries
 Electric cells are composed of two electrodes (solid
electrical conductors) and at least one electrolyte
(aqueous electrical conductor)
 In current cells, the electrolyte is often a moist paste (just
enough water is added so that the ions can move). Sometimes
one electrode is the cell container.
 In an electric cell or battery, the electrode that is defined
as the cathode is marked positive and the electrode that
is defined as the anode is marked negative.
 Electricity is the flow of elections.
 The electrons flow through the external circuit from the
anode to the cathode.
 To test the voltage of a battery, the red(+) lead is connected
to the cathode (+ electrode), and the black(-) lead is
connected to the anode (- electrode)
 Spontaneous: electrons flow from anode (-) to cathode (+)
 Nonspontaneous: electrons flow from anode (+) to cathode (-)
 Electrical energy is supplied to force this reaction to occur. (p.640)
Basic Cell Design and Properties
 A voltmeter is a device that measures the energy difference, per unit charge, between any
two points in an electric circuit (called electric potential difference or the voltage (V))
 I.e. A 9V battery releases 6* as much energy compared with the electrons from a 1.5V
battery.
 The voltage of a cell, depends mainly on the chemical composition of the reactants in the cell, as it
is a ratio of energy to charge and is not dependent on size. (eg. AA, B, C or D size
batteries)
 An ammeter is a device that measures the rate of flow of charge past a point in an electrical
circuit (called electric current)
 The larger the electric cell, the greater current (amperes (A)) that can be produced.
Electric potential difference is
like the potential energy
difference between 1kg of water
at the top of the dam and 1kg of
water at the bottom of the dam.
Electric current (or flow of
electrons) is like the flow of the
water. A larger drain would be like
a larger electric cell, allowing more
water (or electrons) to flow.
Basic Cell Design and Properties
 The power of a battery is the rate at which it produces electrical energy.
Power is measured in watts (W). Calculated using P = IV (Power =
current x potential difference)
 The energy density is a measure of the quantity of energy stored or
supplied per unit mass. Measured in J/kg.
 Table 1 summarizes some important electrical quantities and their units
Consumer, Commercial, and Industrial Cells
 Technology proceeded the understanding of cells and batteries.
Prediction-procedure-evidence-analysis cycle was used later on
to understand why it worked the way it did.
 Zinc chloride cell, dry cell, invented in 1865, Mercury and
alkaline dry cell were more advanced dry cells
 Problem is that the chemicals eventually become depleted and
irreversible reactions…
Primary cells cannot be
recharged, but are relatively
inexpensive
 See Table 2p. 616
Consumer, Commercial, and Industrial Cells
Secondary cells can be recharged using electricity, but are expensive
 E.g. Ni-Cd cells, lead-acid batteries, lithium-ion cell
o They discharge and are charged
Consumer, Commercial, and Industrial Cells
Fuel cells produce electricity by the reaction of a fuel that is continuously
supplied.
More efficient, and used for NASA vehicles, but still too expensive for general or
commercial applications
 More: Aluminum-Air cell – developed for
possible use in electric cars
Homework…
 Textbook:
 pg. 614 #1-5
 Pg. 619 # 8, 10, 13-15
 Pg. 621 # 1, 5, 8, 12
14.2a Voltaic Cells
Pg. 622-626
14.2 – Voltaic Cells
 Questions?
 Why is it that electrodes and electrolytes determine a cell’s characteristics
such as voltage and current?
 What happens in different parts of the cell?
 To answer these questions, chemists use a cell that has two
electrodes and their electrolytes separated.
Voltaic Cells (aka Galvanic Cell)
 A device that spontaneously produces electricity by redox
 Uses chemical substances that will participate in a spontaneous redox reaction.
 The reduction half-reaction (SOA) will be above the oxidation half-reaction (SRA) in the activity series to ensure a
spontaneous reaction.
 Composed of two half-cells; which each consist of a metal rod or strip immersed in a solution of its
own ions or an inert electrolyte.
 Electrodes: solid conductors connecting the cell to an external circuit
 Anode: electrode where oxidation occurs (-)
 Cathode: electrode where reduction occurs (+)
 The electrons flow from the anode to the cathode (“a before c”) through an electrical circuit rather than
passing directly from one substance to another
 A porous boundary separates the two electrolytes while still allowing ions to flow to maintain cell
neutrality
 Often the porous boundary is a salt bridge, containing an inert aqueous electrolyte
(such as Na2SO4(aq) or KNO3(aq)),
 Or you can use a porous cup containing one
electrolyte which sits in a container of a second electrolyte.
 A Salt Bridge can be used to separate
the two electrolytes while still
permitting ions to move between the
two solutions
 A porous boundary (unglazed
ceramic) can be used as well to separates
the two electrolytes while still
permitting ions to move between the
two solutions
 Using this design, a cell can be split into two parts connected by
a porous boundary
 Each part, called a half-cell, consists of one electrode and one
electrolyte.
 A single line (I) indicates a phase boundary (such as the interface
of an electrode and an electrolyte in one half-cell)
 A double line (II) represents a physical boundary (such as a
porous boundary between half-cells)
Voltaic Cells (aka Galvanic Cells)
 Voltaic cells can be represented using cell notation:
The single line represents a phase
boundary (electrode to electrolyte)
and the double line represents a
physical boundary (porous
boundary)
 The SOA present in the cell always undergoes reduction at the cathode
 The SRA present in the cell always undergoes oxidation at the anode
Memory device:
“An ox ate a red cat”
Anode oxidation; reduction
cathode
Match the cell notation to the
descriptions
a)
Sn(s) Sn4+(aq) Cu2+(aq) Cu(s)
b) Mg(s) MgCl2(aq)
c)
Sn(s) SnCl2(aq)
SnCl4(aq) Sn(s)
Cu2+(aq) Cu(s)
Mg(s) Mg2+(aq)
Sn2+(aq) Sn(s)
f)
Copper placed in a solution of copper(II)
chloride and tin metal placed in a solution of
tin(II) ions
2.
A copper-magnesium cell
3.
Magnesium in a solution of magnesium
chloride and tin in a solution of tin(II)
chloride
4.
A tin(IV) ion solution containing tin and a
solution of magnesium ions containing
magnesium
5.
Two tin electrodes in solution of tin(II)
chloride and tin (IV) chloride respectively
6.
Copper place in a copper(II) solution and tin
place in a tin(IV) solution
CuCl2(aq) Cu(s)
d) Mg(s) Mg2+(aq)
e)
1.
Sn(s) SnCl2(aq) SnCl4(aq) Sn(s)
Note:
It is a common practice to write the cell notation with the anode on the left-hand side and the
cathode on the right-hand side. *Diagrams or photos of cells can have any arrangement
A Voltaic Cell Demonstration
 A voltaic cell is the arrangement of two half-cells separated
by a porous boundary.
Voltaic Cells – What is going on?
 Example: Silver-Copper Cell
1.
Cu(s) Cu2+(aq) Ag+(aq) Ag(s)
Use the activity series to determine which of the entities is the SOA.
The SOA present in the cell always undergoes a reduction at the cathode.
Write the reduction half reaction
2.
Use the activity series to determine which of the entities is the SRA.
The SRA present in the cell always undergoes an oxidation as the anode.
Write the oxidation half-reaction
3.
The cathode is the
electrode where the
strongest oxidizing agent
present in the cell reacts.
Balance the half-reactions and add together to create the net equation.
Memory device:
“An ox ate a red cat”
Anode oxidation; reduction
cathode
The anode is the electrode
where the strongest
reducing agent present in
the cell reacts.
Voltaic Cells – What is going on?
 Example: Silver-Copper Cell

Silver ions are the strongest oxidizing agents in the cell, so they undergo a reduction half-reaction at
the cathode, creating more Ag(s)

Copper atoms are the strongest reducing agents in the cell, so they give up electrons in an oxidation
half-reaction and enter the solution (Cu2+ = blue ions) at the anode.

Electrons released by the oxidation of copper atoms at the anode travel through the connecting
wire to the silver cathode. (Ag+(aq) win in the tug of war for e-’s over Cu2+(aq))

Since the positive silver ions are being removed from solution, you would assume that the solution
would become negatively charged. This does not happen. Why?

Cations (positively charged ions) move from the salt bridge into the solution in the cathode compartment to
maintain an electrically neutral solution.

Anions (negatively charged ions) move from the salt bridge into the solution in the anode compartment to
maintain an electrically neutral solution.
Voltaic Cell
Animation
Voltaic Cell Summary
 A voltaic cell consists of two-half cells separated
by a porous boundary with solid electrodes
connected by an external circuit
 SOA undergoes reduction at the cathode (+
electrode) – cathode increases in mass
 SRA undergoes oxidation at the anode (-
electrode) – anode decreases in mass
 Electrons always travel in the external circuit
from anode to cathode
 Internally, cations move toward the cathode,
anions move toward the anode, keeping the
solution neutral
Voltaic Cells with Inert Electrodes
 Inert electrodes are needed when the SOA or SRA involved in the reaction is not solid. If this is
the case, usually a graphite (C(s)) rod or platinum strip is used as the electrode.
 Inert (unreactive) electrodes provide a location to connect a wire and a surface on which a halfreaction can occur.
The copper electrode will decrease in mass and the blue colour of the
electrolyte increases (Cu2+), which indicates oxidation at the anode.
The carbon electrode remains unchanged, but the orange colour of the
dichromate solution becomes less intense and changes to greenish-yellow
(Cr3+), evidence that reduction is occurring in this half cell
Homework…
 Textbook pg. 626 #2-8
 Study for the 6 week exam over break!
 Enjoy the break!  
14.2b Standard Cells and Cell
Potential
Pg. 627-633
Standard Cells and Cell Potentials
 A standard cell is a voltaic cell in which each half-cell contains all entities shown in the half-
reaction equation at SATP conditions, with a concentration of 1.0 mol/L for the aqueous solutions.
 Standardizing makes comparisons and scientific study easier
 Standard Cell Potential, E0 cell = the electric potential difference of the cell (voltage) operating
under standard conditions. E°cell represents the energy difference (per unit charge) between the
cathode and anode.
E0 cell = E0r cathode – E0r anode
•
Where E0r is the standard reduction potential, and is a measure of a standard ½ cell’s
ability to attract electrons, thus undergoing reduction.
•
The higher the E0r , the stronger the OA
 The standard cell potential is the difference between the
reduction potentials of the two standard half-cells:
• All standard reduction potentials are based on the standard
hydrogen ½ cell being 0.00V. This means that all standard
reduction potentials that are positive are stronger OA’s than
hydrogen ions and all standard reduction potentials that are
negative are weaker.
• If the E0 cell is positive, the reaction occurring is
spontaneous.
• If the E0 cell is negative, the reaction occurring is
non-spontaneous
Standard Hydrogen Half-Cell
 The standard hydrogen half-cell is chosen as a reference and
is therefore assigned an electrode potential of exactly zero
volts. It is called a reference half-cell.
 Standard reduction potentials for all other half-cells are
measured relative to that of the standard hydrogen half cell.
 The relative potential of the hydrogen ion reduction half-
reaction is defined to be zero volts:
 A reduction potential that has a
positive value means that the
oxidizing agent is a stronger
oxidizing agent than hydrogen
ions.
 A negative reduction potential
means that the oxidizing agent is
a weaker oxidizing agent than
hydrogen ions.
Rules for Analyzing Standard Cells
1.
Determine which electrode is the cathode. The cathodes is the electrode where the strongest
oxidizing agent present in the cell reacts.
I.e. The OA that is closet to the top on the left side of the redox table = SOA
If required, copy the reduction half-reaction for the strongest oxidizing agent and its reduction potential
2.
Determine which electrode is the anode. The anode is the electrode where the strongest reducing
agent present in the cell reacts.
I.e. The RA that is closet to the bottom on the right side of the redox table = SRA
If required, copy the oxidation half-reaction (reverse the half-reaction)
3.
Determine the overall cell reaction. Balance the electrons for the two half reactions (but DO NOT
change the E0r) and add the half-reaction equations.
4.
Determine the standard cell potential, E0cell using the equation:
E0 cell = E0r cathode – E0r anode
Measuring Standard Reduction
Potentials
 Both voltage and the direction of current must be known.
 The magnitude of the voltage determines the numerical value of
the half-cell potential, and the direction of the current
determines the sign of the half-cell potential.
 Half-reaction equations can be multiplied by appropriate factors
to balance the electrons (see Text pg.629), but the reduction
potentials are not altered by the factors used to balance the
electrons.
E°cell = E°r (cathode) - E°r (anode)
 A positive cell potential (E°cell >0) indicates that the net
reaction is spontaneous – a requirement for all voltaic cells.
Standard Cells and Cell Potentials #1a
Standard Cells and Cell Potentials #1a
 Example: What is the standard potential of the cell represented below:
1.
Determine the cathode and anode
2.
Determine the overall cell reaction
3.
Determine the standard cell potential
Standard Cells and Cell Potentials #1b
Standard Cells and Cell Potentials #1b
Standard Cells and Cell Potentials #2
 Example: What is the standard potential of an electrochemical cell made of a cadmium electrode in
a 1.0 mol/L cadmium nitrate solution and chromium electrode in a 1.0 mol/L chromium(III) nitrate
solution?
SOA
Cd2+
SRA
(aq)
Cd(s)
Cr2+(aq) Cr(s)
cathode
H2O(l)
anode
E0 cell = E0r cathode – E0r anode
= (-0.40V) - (-0.91V)
= + 0.51V
The E0 cell is positive, therefore the reaction is spontaneous.
Standard Cells and Cell Potentials #3
 Example: A standard lead-dichromate cell is constructed. Write the cell notation, label the
electrodes, and calculate the standard cell potential.
SRA
Pb(s) Pb2+(aq)
Cr2O72-(aq)
anode
SOA
H+
(aq)
Cr3+(aq)
C(s)
cathode
E0 cell = E0r cathode – E0r anode
= (+1.23V) - (-0.13V)
= + 1.36V
The E0 cell is positive, therefore the reaction is spontaneous.
Standard Cells and Cell Potentials #4
 Example: A standard scandium-copper cell is constructed and the cell potential is measured. The
voltmeter indicates that copper the copper electrode is positive.
Sc(s) Sc3+(aq)
Cu2+(aq) Cu(s)
anode
E0 cell = +2.36V
cathode
Write and label the half-reaction and net equations, and calculate the standard reduction potential of
the scandium ion.
E0 cell = E0r cathode - E0r anode
2.36V = (+0.34V) - (x)
E0r anode = -2.02V
Cell Potentials Under Nonstandard
Conditions
 The electric potential difference or voltage of a cell decreases
slowly as the cell operates
 If electrons flow, oxidation and reduction occur, which in
turn, changes the concentration from the standard 1.0 mol/L
 If electrons are allowed to flow, eventually an equilibrium
will be reached when the flow ceases
 Nonstandard conditions for concentrations, temperature, and
pressure will cause differences between the cell potentials
predicted from a standard redox table and the ones measured in
the laboratory
 These differences are generally small if the conditions are
relatively close to standard values
 Other more important reason for discrepancies are the purity of
substances, the presence of oxide coatings on the metals, and the
type and size of porous boundaries
Homework…
 Textbook:
 Pg.631 #10,11
 Pg. 633 #12-16
14.2c Corrosion
Pg. 634-635
Human History…
 The Copper ages, Bronze
ages, Steel ages…
 Describe metals that were
widely refined from ores and
used for tools and weapons.
Corrosion
 An electrochemical process in which a metal reacts with
substances in the environment, returning the metal to an orelike state.
 Because we live in an oxidizing (oxygen) environment,
oxidation (corrosion) of some metals occurs spontaneously.
 As a metal is oxidized, metal atoms lose electrons to form
positive ions.
 Preventing corrosion and dealing with the effects of
corrosion are major economical and technological problems
for our society. Eg. A ship’s hull.
Corrosion…
 Metals vary in their ability to be oxidized (corroded).
 Some, such as silver and gold, are relatively weak reducing
agents and are therefore considered “noble” because of
resistance to corrosion.
 Others, such as the metals found in Group 1 and 2 on the
periodic table, are very strong reducing agents and are,
therefore, easily oxidized.
 In general, any metal appearing below the various oxygen
half-reactions in a redox table will be oxidized in our
environment.
Corrosion…
 Why is the corrosion or rusting of iron such a major
problem, but the corrosion of aluminium, which is a much
stronger reducing agent, is not?
 Because the nature of the oxide that forms on the surface of the
metal… A freshly cleaned surface of aluminium rapidly oxidizes
in air to form aluminium oxide.
 The aluminium oxide adheres tightly to the surface of the metal. This
prevents further corrosion by sealing any exposed surfaces.
 Iron does not adhere very well, flaky. Also will corrode with other
things than just oxygen.
Rusting of Iron
 Studies of the corrosion of iron have shown that the presence
of both oxygen and water is required and the iron is
converted into iron hydroxides and oxides.
 Step 1: is thought to be the oxidation of iron at a wet
exposed surface.
 Iron(II) ions diffuse through the water on the iron surface
while the electrons easily travel through the iron metal,
which is an electrical conductor.
 The electrons are picked up by oxygen molecules dissolved in
water on the surface at a point away from the original oxidation
site.
 The combination of iron(II) ions and hydroxide ions forms a low-
solubility precipitate of iron(II) hydroxide which is further
oxidized by oxygen and water to form iron(III) hydroxide to
form a mixture of iron(III) hydroxide and hydrated iron(III)
oxide, a yellow-brown solid.
 This simplified description helps to explain why certain conditions
promote rusting:
 A dry environment allows little or no corrosion.
 Eliminating either water or oxygen in the water makes the reduction of aqueous
oxygen impossible.
 If oxidizing agents other than oxygen are present, the iron can still be corroded
through spontaneous redox reactions. (This helps to explain the corrosion of iron
in acidic environments, eg. Acid rain SOA compared to normal rain.)
 In general, electrolytes accelerate rusting.
 Ships in sea water rust faster, in winter when salt is on the roads… poor
cars. 
 They conduct electricity and accelerate rusting.
 Summary: The rusting of iron requires the presence of oxygen and water and is
accelerated by the presence of acidic solutions, electrolytes, mechanical stresses,
and contact with less active metals.
Corrosion Prevention
 Corrosion protection:
 Paint and other similar coatings (no scratches)
 Tin and zinc used as metallic coatings (zinc galvanizing of steel/iron)
 Oxidation is the loss of electrons and occurs at the anode of a voltaic
cell… so use….
 Cathodic protection in which iron is forced to become the
cathode by supplying the iron with electrons, either using an
impressed current or a sacrificial anode.
An impressed current is an electric current forced to flow toward
an iron object by an external potential difference. (eg. is used for
pipelines)
 A sacrificial anode is a metal more easily oxidized than iron and
connected to the iron object to be protected.

Homework…
 Textbook:
 pg.673 "Practice" # 18-24
 pg. 637-638 # 2,4, 7-11
Electrolytic Cells 14.3a
Pg. 639-645
The SOA undergoes reduction at the cathode (- electrode)
The SRA undergoes oxidation at the anode (+ electrode)
Eocell less than zero= nonspontaneous
Memory device:
“An ox ate a red cat”
Anode oxidation (SRA); reduction (SOA) cathode
Electrolytic Cells
 The term “electrochemical cell” is often used to refer to a:
 Voltaic Cell – one with a spontaneous reaction
SOA over SRA on the activity series
Eocell (standard cell potential) greater than zero = spontaneous
 Electrolytic cell – one with a nonspontaneous reaction
SOA below SRA – i.e. zinc sulfate and lead solid cell
Eocell less than zero= nonspontaneous
 Why would anyone be interested in a cell that is not spontaneous?
 This would certainly not a good battery choice, but by supplying electrical energy to a
nonspontaneous cell, we can force this reaction to occur.
 This is especially useful for producing substances, particularly elements. I.e. the zinc sulfate cell
discussed above is similar to the cell used in the industrial production of zinc metal.
Electrolytic Cells
 Electrolytic Cell – a cell in which a nonspontaneous redox reaction is
forced to occur; a combination of two electrodes, an electrolyte and an
external power source.
 Electrolysis – the process of supplying electrical energy to force a nonspontaneous
redox reaction to occur
 The external power source acts as an “electron pump”; the electric energy is used to
do work on the electrons to cause an electron transfer
Electrons (go the same direction as in voltaic
cells) are pulled from the anode and pushed to
the cathode by the battery or power supply
Comparing Electrochemical Cells:
Voltaic and Electrolytic
It is best to think of “positive” and “negative” for electrodes as labels, not charges!!
It is best to think of “positive” and “negative” for electrodes as labels, not charges!!
We say in table 1
 Note: The theoretical definitions of cathode and anode are the
same for both voltaic and electrolytic cells.
 Note: A negative sign for a cell potential indicates that the
chemical process is nonspontaneous. The more negative the cell
potential, the more energy is required.
 Note: Theoretically, the SOA present in a particular mixture has
the greatest attraction for electrons and gains electrons at the
cathode.
A secondary cell
 A secondary cell can be used to illustrate the difference
between an electric cell and an electrolytic cell.
 As the cell discharges, electrical energy is spontaneously
produced and the cell functions as an electric cell.
 When the cell is recharged, the electrical energy forces the
products to react and re-form the original reactants. During
recharging, the secondary cell is functioning as an electrolytic
cell.
Procedure for Analyzing Electrolytic Cells
 Use the redox table to identify the SOA and SRA
 Don’t forget to consider water for aqueous electrolytes.
 Write equations for the reduction (cathode) and oxidation (anode) half-reactions.
Include the reduction potentials if required.
 Balance the electrons and write the net cell reaction including the cell potential.
E0 cell = E0r cathode - E0r anode
 If required, state the minimum electric potential (voltage)
to force the reaction to occur. (The minimum voltage is
the absolute value of E0 cell)
 If a diagram is requested, use the general outline in
Figure 6, and add specific labels for chemical entities.
Analyzing Electrolytic Cells #1
 Example: What are the cell reactions and the cell
potential of
the aqueous potassium iodide electrolytic cell?
 Identify major entities and identify the SOA and SRA.
 Write the half-reaction equations and calculate the cell potential.
 State the minimum electric potential (voltage) to force the reaction to occur.
Electrons must by supplied with a
minimum of +1.37 V from an external
battery or other power supply to force the
cell reactions.
Potassium-Iodide Electrolytic Cell
 In the potassium iodide electrolytic cell, litmus paper
does not change colour in the initial solution and turns
blue only near the electrode from which gas bubbles.
Why?
 At the other electrode, a yellow-brown colour and a dark
precipitate forms. The yellow brown substance produces
a purplish-red colour in the halogen test (pg. 805). Why?
Analyzing Electrolytic Cells #2
 Example: An electrolytic cell containing cobalt(II) chloride solution and lead electrodes is
assembled. The notation for the cell is as follows:
a)
Predict the reactions at the cathode and anode, and in the overall cell.
b) Draw and label a cell diagram for this electrolytic cell, including the power supply.
c)
What minimum voltage must be applied to make this cell work?
Analyzing Electrolytic Cells #3
 Example: An electrolytic cell is set up with a power supply connected to two nickel electrodes
immersed in an aqueous solution containing cadmium nitrate and zinc nitrate.
 Predict the equations for the initial reaction at each electrode and the net cell reaction. Calculate
the minimum voltage that must be applied to make the reaction occur.
The Chloride Anomaly
 Some redox reactions predicted using the SOA and SRA
from a redox table do not always occur in an electrolytic
cell.
 The actual reduction potential required for a particular
half-reaction and the reported half-reaction reduction
potential may be quite different (depending on the
conditions or half-reactions)
 This difference is known as the half-cell overvoltage, it is
generally much greater for the production of oxygen than for
chlorine.
 “As an empirical rule, you should recognize that chlorine gas is
produced instead of oxygen gas in situations where chloride and
water are the only reducing agents present.”
Electrolytic Cells
 Summary:
 An electrolytic cell is based upon a reaction that is nonspontaneous; the Eocell for
the reaction is negative.
An applied voltage of at least the absolute value of Eocell is required to force the
reactions to occur.
 The SOA undergoes reduction at the cathode (- electrode)
 The SRA undergoes oxidation at the anode (+ electrode)
 Electrons are forced by a power supply to travel from the anode to the cathode
through the external circuit.
 Internally, anions move toward the anode and cations move toward the cathode
Homework…

- Textbook:
- pg. 640 #3
- pg. 644 #5,6
- pg. 645 # 7,8,10,12,13,
- Diploma:
-EX #49-54
Science and Technology of
Electrolysis 14.3b
Pg. 646-651
Science and Technology of Electrolysis
 Electrolysis – the process of supplying electrical energy to
force a nonspontaneous redox reaction to occur.
 Production of Elements
 Most elements occur naturally combined with other elements in
compounds.
 Potassium, sodium, calcium are abundant but reactive so only found
combined in nature.
 This is due to very negative reduction potentials for these
elemental metals.
 Consequently, these metals are easily oxidized by practically
all other substances.
 Many metals can be produced by electrolysis of solutions of their ionic
compounds, but two difficulties arise:
 1. Many naturally occurring ionic compounds have a low solubility in
water,
 2. Water is a stronger oxidizing agent than many active metal cations.
 To overcome these difficulties a technological design in used in which
water is not present.
 Fortunately, ionic compounds can be melted. These molten ionic
compounds are good electrical conductors and can functions as the
electrolyte in the cell.
 Fig. 7: Sir. Humphry Davy
 Eg. pg. 647 example 3: Lithium chloride make for an excellent battery to
produce lithium and chloride gas.
Production of Aluminum 3rd most abundant
element on Earth
 Did ya know?! (pg. 647)
 Watch How Its Made! 
 Initial efforts to produce aluminum by electrolysis were




unproductive because it common ore, Al2O3(S) has high melting
point (2072oC)
Scientists found the Al dissolves in the molten mineral cryolite,
Na3AlF6
Forms a molten conducting mixture with a melting point around
1000oC
1855 $45,000 Kg
A few years later  90 cents Kg
 Because of the Hall-Heroult invention to produce Aluminium…
The Chlor-Alkali Process
 The most important nonmetal produced by electrolysis is chlorine
 The chlor-alkali process is the electrolysis of aqueous sodium
chloride (brine) to produce chlorine, hydrogen, and sodium
hydroxide
 This process is done in Dow Chemical in Fort Saskatchewan, AB:
 Sodium chloride is extracted from large underground deposits using hot water
and pumped to electrolytic cells at the surface.
 Hydrogen gas: ammonia, hydrogen peroxide, margarine, and to crack
petroleum, used as a fuel to produce electricity
 Chlorine: disinfectant for drinking water, to manufacture bleach, plastics,
pesticides, and solvents.
 Sodium hydroxide: used in industry to make cellophane, pulp and paper,
aluminium and detergents.
 The anode reaction in the electrolysis is unexpected since water
is a slightly stronger reducing agent then chloride ions
 This is the chlorine anomaly
Refining of Metals
 In the production of metals, the initial product is usually an
impure metal.
 Electrorefining uses an electrolytic cell to obtain high-grade
metals at the cathode from an impure metal at the anode.
 ex: electrorefining of copper, (if impurities in copper it doesn’t
conduct electricity as well)
 (Similarly, Electrowinning is another method of purifying
metals is to reduce metal cations from a molten or aqueous
electrolyte at the cathode of and electrolytic cell.)
Electroplating
- Plating of a metal at the cathode of an electrolytic cell
- Vapour deposition is another method used to cover the surface of
an object with a metal.
- Dipping is another method, an object is dipped into a molten
metal that solidifies on the surface. Eg. Zinc plated nails for
exterior use.
- Stainless steal has a chromium oxide coating to prevent
corrosion—> How It’s Made?!
Homework…
 Textbook:
 pg. 649 #20,21
 pg. 651 #4, 7,10, 11
 Diploma:
 EX # 55-62
 2013 # 13,14, 16 NR #6,7,10, 11
Cell Stoichiometry 14.4
Pg. 652-657
Stoichiometry Calculations
(Measured (known) quantity)
solids/liquids
m
gases
V, T, P
solutions
c, V
electrochemical cells Q
solutions
c, V
gases
V,T,P
solids/liquids
m
(Required (unknown) quantity)

n
mole
ratio

n
Cell Stoichiometry
 First, pg 653 Figure 1 Mr. Faraday
 The quantity of electricity that passes through a cell
determines the masses of substances that react or are
produced at the electrodes.
 As in all stoichiometry, this relationship establishes a mole
ration of electrons to some other substance in the halfreaction equation.
 The number of electrons is determined indirectly.
Background
 Charge (Q) is determined by multiplying the electric current (I),
(measured in C/s) by the time (t) measured is seconds.
Q = It
(C) = (Ampere)(second)
(Coulomb) = (Coulombs per second) x (second)
One coulomb is the quantity of charge transferred by a current of 1 Ampere
during 1second.
Example: Calculate the charge that passes through one 300kA cell in a 24 hour period.
Q = It
= (300kA x 1000A/kA)(24 h x 3600s/h)
= (300000C/s )(86400s) = 2.6 x 1010C
Practice: Calculating Charge
 Pg. 653 #1,2-4
Faraday’s Law
 “The mass of an element produced or consumed at an electrode is directly proportional to the
time the cell operated, as long as the current was constant.”
 He also found that “9.65 x 104C of charge is transferred for every mole of electrons that flows in
the cell. This value is the molar charge of electrons, also called Faraday’s constant”
F = 9.65 x 104 C _.
mol e This constant can by used as a conversion factor in converting electric charge to moles, very
similarly to the way that molar mass is used to convert mass to a chemical amount.
 I.e. 2.5 mol x 2.02 g = 5.05 g
15000 C x 1 mol e- = 0.16 mol emol
9.65 x 104 C
Practice: Faraday’s Law #1
 Convert a current of 1.74 A for 10.0 min into an amount of
electrons:
We do not have a direct way of measuring amount of electrons, so we need to calculate
charge first (Q = It), and then use Faraday’s constant to calculate the amount of
electrons.
Q = It = (1.74 C ) (10.0min x 60 s ) = 1044 C x 1 mol = 0.0108 mol
s
min
9.65 x 104 C
You can also write this as a single equation (using cancellation of units):
Practice: Faraday’s Law #2
 How long, in minutes, will it take a current of 3.50 A to transfer
0.100 mol of electrons?
Remember: Amperes = Coulombs/second
Practice: Calculating Charge
 Pg. 654 #5-7
Half Cell Calculations
 Since the mass of an element produced at an electrode depends on the
amount of transferred electrons (Faraday’s Law), a half-reaction equation
showing the number of electrons involved is necessary to do stoichiometric
calculations.
 This applies to all electrochemical cells, whether voltaic or electrolytic.
 Separate calculations are carried out for each electrode, although the same
charge and therefore the same amount of electrons passes through each electrode in
a cell.
 Remember: The only new part of the stoichiometry is the calculation involving
the amount of electrons based on Faraday’s constant
Practice: Half-Cell Calculations #1
 What is the mass of copper deposited at the cathode of a copper
electrorefining cell operated at 12.0 A for 40.0 min?
 First identify and write the appropriate half-cell.
 Copper is being deposited at the cathode, copper(II) ions must be gaining
electrons to form copper metal.
 Write the equation for this reduction and list all the information given.
 Do we have enough information to solve for the amount of electrons (moles of
electrons)?
Practice: Half-Cell Calculations #1
 What is the mass of copper deposited at the cathode of a copper electrorefining
cell operated at 12.0 A for 40.0 min?
 Yes, we can solve for the number of moles, and then use the mole ratio to convert
from a chemical amount of one substance to another.
 The last step is to convert to the quantity requested in the question, in this case the
mass of the copper metal
 Could we do this as one equation instead?
Practice: Half-Cell Calculations #1
 What is the mass of copper deposited at the cathode of a copper electrorefining
cell operated at 12.0 A for 40.0 min?
Practice: Half-Cell Calculations #2
 Silver is deposited on objects in a silver electroplating cell. If 0.175 g of silver is to be
deposited from a silver cyanide solution in a time of 10.0 min, predict the current
required.
 Write the balanced equation for the half-cell reaction, list the measurements and
conversion factors.
 Convert to moles, use the mole ratio, convert to the current (C/s)
Cell Stoichiometry Summary
1.
Write the balanced equation for the half-cell reaction of the substance produced
or consumed. List the measurements and conversion factors for the given and
required entities.
2.
Convert the given measurements to an amount in moles by using the appropriate
conversion factor. (M, c, F)
3.
Calculate the amount of the required substance by using the mole ratio from the
half-reaction equation.
4.
Convert the calculated amount to the final quantity
appropriate conversion factor. (M, c, F)
by using the
Homework…
 Textbook:
 Pg. 657 #1,3,5,7,9a
 Diploma:
 EX # 63,64,66
 2012: MC#15,16,17, 19, 20
 2012 NR# 6-10