Lesson 1. Electrochemistry
Outline
1.1. Oxidation-Reduction Reaction
1.2. Galvanic Cell
1.3. Cell Potential and Free Energy
1.4. Batteries and Fuel Cells
1.5. Electrolysis and Electroplating
1.6. Corrosion
Electrochemistry
Electrochemistry is the study of the relationship between electron flow and oxidation-reduction reactions or
termed as redox reactions. Applications of electrochemistry is numerous and important like in electrochemical
cells (commonly called batteries), electrolysis and electroplating. In electrochemical cells, electrons from
product-favored redox reactions are generated and transferred as an electrical current through an electrical
circuit. The voltage produced by an electrochemical cell depends on the oxidizing agent and reducing agents
used as reactants. However, not all product- favored redox reactions are beneficial, corrosion for example, is a
product- favored redox reaction but it results to damage to materials which is very costly.
On the other hand, electrolysis and electroplating are applications of redox reactions that are reactant-favored.
In an electrolysis cell, an external energy source causes an electric current to force a reactant-favored process
to occur. Electrolysis is important in the manufacture of many plated products like chrome plated objects.
This lesson discusses the basic principles of electrochemistry and its application.
1.1 Oxidation-Reduction Reaction
Oxidation-reduction reaction involves the transfer of electrons. Oxidation reaction is the loss of electrons from
a chemical species and reduction reaction is the gain of electrons. The overall redox reaction is a result of two
simultaneous half reactions: one half reaction for the oxidation and one half reaction for reduction.
Example 1. Take for example Cu wire in a solution of AgNO3
(Source: Brown, WH & Holme, TA. (2011). Chemistry for Engineering Students, 2nd edition)
In beaker a. Clean copper (Cu) wire is placed into a colorless solution of silver nitrate (AgNO3).
In beaker b. The color of the solution is changing to blue and silvery solids sticking on the wire appears.
In beaker c. As time passes, the solution turned blue and more silvery solids stick on the wire.
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The blue color is an indication of the presence of copper ions (Cu 2+). The copper solid had been oxidized
producing copper ions. In the oxidization of copper solid, copper losses two electrons. The oxidation reaction is:
Cu(s)
Cu 2+(aq) + 2e-
oxidation reaction
The silvery solid sticking in the copper wire is silver (Ag). The silver cation (Ag+) in the solution of silver nitrate
(AgNO3) had been reduced producing silver solid. In the reduction of silver cation, silver cation accepts only one
electron. The reduction reaction is:
Ag+(aq) + 1e-
Ag(s)
reduction reaction
The electrons can’t simply be lost. The electrons lost in the oxidation of a chemical species must always be
gained in the simultaneous reduction of another chemical species. We cannot have oxidation unless we also
have reduction. Hence, oxidation reaction and reduction reaction happens simultaneously.
The number of electrons lost or gained during oxidation and reduction reaction depends on the valence
electrons of the element. Valence electrons are the electrons in the highest energy level or outermost shell of
the atom’s electronic configuration. The valence electrons are similar to the group number where the element
belongs in the periodic table and this is true for the Group A elements.
The overall redox reaction for example 1 is:
Cu 2+(aq) + 2eAg(s) * [2]
Cu(s)
Ag+(aq) + 1eCu(s)
2Ag+(aq) + 2e-
(Ag is multiplied by 2 to balance the number of
electrons)
Cu 2+(aq) + 2e2Ag(s)
Cu(s) + 2Ag+(aq) + 2eCu(s) + 2Ag+(aq)
2Ag(s) + Cu2+(aq) + 2e- (2e- cancels)
2Ag(s) + Cu2+(aq) overall redox reaction
The overall reaction including the spectator ion (NO3– is the spectator ion) is:
2AgNO3 (aq) + Cu(s)
2Ag(s) + Cu(NO3)2 (aq)
Silver ion (Ag+) oxidizes copper solid (Cu) and copper solid reduces silver ion. The species undergoing oxidation
is referred to as a reducing agent and the species undergoing reduction is referred to as oxidizing agent. Ions
are charged particles, a negatively charge ion is called anion and the positively charge ion is called cation.
Example 2. For the reaction: Zn(s) + Cu2+ (aq)
Zn2+ (aq) + Cu(s)
a. Write the half reactions and label it as oxidation reaction or reduction reaction.
b. Determine the oxidizing agent and reducing agent.
Answers:
a. Half reactions
Zn(s)
Cu2+ (aq) + 2e-
Zn2+(aq) + 2eCu(s)
Oxidation reaction
Reduction reaction
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b. Zn(s) is the reducing agent
Cu2+(aq) is the oxidizing agent
1.2 Galvanic Cell
A galvanic cell is any electrochemical cell in which spontaneous chemical reaction can be used to generate an
electric current. The observation of electric current in an electrochemical cell led to the name electrochemistry,
this connects electricity and chemistry.
How could electricity be generated from a spontaneous chemical reaction? The oxidation and reduction
reaction presented in section 1.1 seems like an artificial idea because both oxidation and reduction reactions
takes place in one container; the copper and silver are directly in contact. What would happen if the half
reactions are in two separate containers? Say for example copper in one container and silver in another
container as shown in Figure 1.1 a.
Figure 1.1 A salt bridge is crucial in a galvanic cell
(Source: Brown, WH & Holme, TA. (2011). Chemistry for Engineering Students, 2nd edition)
In Figure 1.1a, copper solution is in one container and silver solution is in another container, no reaction occurs
because copper and silver are not in contact. Connecting the two solutions by a conducting wire, there will also
be no observable change because the wire cannot allow the flow of charge between the two solutions (copper
ion and silver ion cannot flow through the wire). Hence, there will only have a build-up of charges in the
electrodes. Electrodes are electrically conducting sites at which either oxidation and reduction takes place.
Oxidation occurs at the anode and reduction occurs at the cathode.
In Figure 1.1b, the two solutions are connected with a salt bridge instead of a wire. The salt bridge allows
migration of cations and anions where they are needed. A salt bridge contains strong electrolytes that allows
either cations or anions to migrate into the solutions. Moving charges through the salt bridge closes the circuit,
current can flow and a cell voltage can be measured. This set up in Figure 1.1b is called a galvanic cell.
A galvanic cell has the following characteristics: a) half reactions are in different sites; b) a salt bridge is inserted
between the two solutions allowing flow of charges/electrons; and c) a sustainable electric current is generated.
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The shorthand method representing the electrochemical cell is called cell notation. The cell notation is written
as follows:
Anode │ Electrolyte of Anode ║ Electrolyte of Cathode │ Cathode
Example 3. The cell notation for the oxidation-reduction reaction of copper and silver is written as:
Cu(s) │ Cu 2+ (aq) ║ Ag+ (aq)│ Ag(s)
Example 4. For the redox equation, Zn(s) + Cu2+ (aq)
Zn(s)│ Zn2+(aq)║ Cu2+ (aq)│Cu(s)
Zn2+(aq) + Cu(s), the cell notation is written as:
Clicker Exercise 1
1. For the following oxidation-reduction reactions, identify the half reactions and label them as oxidation
reaction or reduction reaction; and also write the cell notation.
a. Cu(s) + Ni 2+(aq)
Ni(s) + Cu 2+(aq)
3+
b. 2Fe (aq) + 3Ba (s)
3Ba 2+(aq) + 2Fe (s)
c. 2Br - (aq) + I2 (s)
Br2 (l) + 2I - (aq)
1.3 Cell Potential and Free Energy
Cell Potential
The buildup of charges on the electrodes is important because it means that there is a potential for electrical
work. This potential is called cell potential. Measurement of cell potential is done using a voltmeter. The cell
potential is associated with a particular half reaction. All half-cell potentials are tabulated as standard reduction
potentials. A standard reduction potential shows the potential of any half reaction when connected to a SHE
(standard hydrogen electrode). Below are examples of standard reduction potentials of some half reactions.
Half Reaction
Zn 2+(aq) + 2e
Fe 2+(aq) + 2e
2H+ (aq) + 2e
Cu 2+(aq) + 2e
Fe 3+(aq) + e
Ag + (aq) + e
Zn(s)
Fe(s)
H2(g)
Cu(s)
Fe 2+(aq)
Ag(s)
Standard Reduction Potential (V)
- 0.763
- 0.44
0.000
+ 0.337
+ 0.771
+ 0.7794
Note that in the above examples of standard reduction potentials, all half reactions are written in a form of
reduction reactions. A voltage that is positive means the SHE is the anode or the oxidation site, hence, positive
voltage means that the half reaction proceeds as written (reduction). If the voltage is a negative value, the SHE
is serving as cathode or the reduction site, hence, the half reaction proceeds as oxidation.
The implications of the noted observations are: a) A large positive value for the standard reduction potential
implies that the substance is reduced readily and therefore a good oxidizing agent, reduction reaction occurs at
the cathode; b) A large negative value for the standard reduction potential implies that the substance is oxidized
readily and therefore a good reducing agent, oxidation reaction occurs at the anode.
The cell potential of a pair of half reactions is calculated as: E° cell = E° red – E° ox
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Example 5. For copper and iron: Fe 2+(aq) + 2eFe(s)
E° = -0.44V
Cu 2+(aq) + 2eCu(s)
E° = 0.337V
a. Identify the anode and the cathode.
b. Determine the cell potential of the galvanic cell of copper and iron
c. Write the overall redox reaction.
Answers:
a. Fe 2+ (aq) + 2eCu 2+(aq)+ 2e-
Fe(s) E° = -0.44V
Cu(s) E° = 0.337V
anode because the voltage is negative (oxidation)
cathode because the voltage is positive (reduction)
b. Calculation of the cell potential
E° cell = E° red – E° ox
E° cell = 0.337V – (- 0.44V) = 0.78V
c. The overall redox reaction
Fe(s)
Fe2+ (aq) + 2eCu2+ (aq) + 2eCu(s)
anode- oxidation reaction
cathode-reduction reaction
Adding the half reactions, gives:
Fe(s) + Cu 2+ (aq) + 2eFe(s) + Cu 2+ (aq)
Fe 2+(aq) + Cu(s) + 2eFe 2+(aq) + Cu(s)
(2e- cancel)
overall redox reaction
Example 6. For copper and silver, confirm that the potential of the galvanic cell is 0.462V.
Ag+(aq) + eAg(s)
E° = 0.7994V
Cu2+(aq) + 2eCu(s)
E° = 0.337V
Answers:
a. Ag+(aq) + eCu2+(aq) + 2e-
Ag(s)
E° = 0.7994V
Cu(s)
E° = 0.337V
cathode because the voltage is largely positive
(reduction)
anode because it is less positive (oxidation)
b. Calculation of the cell potential
E° cell = E° red – E° ox
E° cell = 0.7994V – (+0.337V) = 0.4624V
Free Energy
Free energy is the maximum work that can be done by a system. In a galvanic cell, this work done is the electrical
work. There is a relationship between free energy and cell potential and this relationship is defined in the
following equation: ΔG° = - nFE°
where: n = number of moles of e- (moles of electrons)
F = Faraday constant = 96,485 JV-1mol e-1
E° = cell potential
ΔG° = free energy or the Gibb’s free energy
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The ΔG° is negative, because a galvanic cell has a positive potential and spontaneously generates electrical
work. The negative sign is required for a spontaneous process.
Example 7. For the galvanic cell of zinc and chromium: Cr(s)│ Cr2+ (aq)║ Zn2+(aq) │ Zn (s)
a. Write the half reactions and determine the anode and cathode or oxidation reaction and reduction reaction
b. Determine the E° cell
c. Determine the ΔG°
Answers:
a. Half reactions
Cr(s)
Cr2+(aq) + 2eZn2+(aq) + 2eZn(s)
anode (oxidation reaction)
cathode (reduction reaction)
b. Calculation of the E° cell
From the table of standard reduction potentials
Cr2+((aq) + 2eCr(s)
E = -0.910V anode (oxidation reaction)
2+
Zn (aq) + 2eZn(s)
E = - 0.763 V
cathode (reduction reaction)
E° cell = E° red – E° ox
E° cell = -0.763 – (-0.910) = 0.147 V
c. Calculation of ΔG°
ΔG° = - nFE°
ΔG° = - 2 mol e- * 96,485 JV-1mol e-1 * 0.147 V
ΔG° = - 2.84 x 104 J
Clicker Exercise 2
1. Using the values from a table of standard reduction potential, identify the anode and cathode, calculate the
cell potential and write the overall redox reaction of the following cells:
a. Zn(s)│ Zn 2+(aq)║ Cr 3+ (aq) │ Cr (s)
b. Fe (s)│ Fe 2+ (aq)║ Hg 2+ (aq)│ Hg (l)
2. Calculate the cell potential and the free energy change for the following cells:
a. Ga(s)│ Ga3+ (aq) ║ Ag+(aq)│ Ag (s)
b. Zn (s)│ Zn 2+ (aq)║ Cr 3+ (aq)│ Cr (s)
1.4 Batteries and Fuel cells
The electrochemical reactions provide opportunities for using it constructively like for example in the
production of batteries. A battery is a cell or series of cells that generates an electric current. Batteries are
composed of many different materials and has many uses. Batteries can be classified as primary cell or
secondary cell.
Primary Cell
A primary cell is a single use battery that cannot be recharged. The most prevalent type of primary cell is
alkaline battery. The anode in the alkaline battery is zinc electrode and the cathode is derived from manganese
(IV) oxide. Figure 1.2 shows the essential design features of alkaline battery. The half reactions are as follows:
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Anode:
Cathode:
Overall reaction:
Zn(s) + 2OH- (aq)
2MnO2(s) +H2O(l) + 2eZn(s) + 2MnO2(s) + 2H2O (l)
Zn(OH)2 (s) + 2eMn2O3(s) + 2OH- (aq)
Zn(OH)2(s) + Mn2O3 (s)
Figure 1.2 Essential design features of a primary cell
(Source: Brown, WH & Holme, TA. (2011). Chemistry for Engineering Students, 2nd edition)
Primary cells have various applications. In some instances, the battery should be small and long lasting like for
example batteries for medical devices. Mercury batteries fill this role. In a mercury battery, zinc is the anode and
mercury (II) oxide is the cathode. Figure 1.3 shows the design features of zinc-mercuric oxide battery. The half
reactions are as follows:
Anode:
Cathode:
Overall reaction:
Zn(s) + 2OH- (aq)
HgO(s) + H2O(l) + 2eZn(s) + HgO(s) + H2O(l)
Zn(OH)2(s) + 2eHg(l) + 2OH- (aq)
Zn(OH)2(s) + Hg(l)
Figure 1.3 Design features of zinc-mercuric oxide battery
(Source: Brown, WH & Holme, TA. (2011). Chemistry for Engineering Students, 2nd edition)
Secondary Cell
Secondary cell is a rechargeable battery. Secondary cells are very popular because of cellphones, digital
cameras, and computer devices like palmtop and laptop. An example of a secondary cell is nickel-cadmium
battery. In the nickel-cadmium battery, the anode is cadmium and the cathode is nickel. The half reactions are
as follows:
Anode:
Cathode:
Overall reaction:
Cd(s) + 2OH- (aq)
NiO(OH(s) + H2O (l) + eCd(s) + 2NiO(OH) (s) + 2H2O (l)
Cd(OH)2(s) + 2eNi(OH)2 (s)+ OH- (aq)
Cd(OH)2 (s) + 2Ni(OH)2 (s)
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The Ni-Cad batteries can be expended and recharged many times, but sometimes they are susceptible to a
performance decreasing memory effect. This memory effect involves the inability to use all of the possible
chemical energy of the battery unless it is completely discharged. Successive recharging results in shorter times
before the battery appears to die. The chemistry that gives rise to this effect involves the formation of a thin
layer of material on the electrodes inside the battery, thus, limiting the oxidation-reduction reactions needed
to generate electrical work. Figure 1.4 shows the design features of Ni-Cad battery.
Figure 1.4 Design features of nickel-cadmium battery
(Source: Brown, WH & Holme, TA. (2011). Chemistry for Engineering Students, 2nd edition)
Fuel Cells
A fuel cell is a voltaic cell in which the reactants can be supplied continuously and the products of the cell
reaction are continuously removed. It uses a chemical reaction to produce electrical energy. It can also be
refueled on an on-going basis. The most common fuel cell is based on the reaction of hydrogen and oxygen to
produce water. Hydrogen flows into the anode compartment and oxygen gas flows into the cathode
compartment. The electrodes in each compartment are usually porous carbon which is impregnated with
platinum catalyst. Oxygen is reduced in the cathode and hydrogen is oxidized at the anode. The half reactions
are as follows:
Anode:
Cathode:
Overall reaction:
H2
2H+ + 2e
+
O2 + 4H + 4e2H2O
2H2 + O2
2H2O
Just like in any other voltaic cells, the half reactions are physically separated. Electrons flow from the anode to
the cathode through an external circuit, while protons pass through a special proton exchange membrane
separating the two electrode compartments. Fuel cells are used in a variety of specialized applications, including
powering instrumentation aboard spacecraft.
Limitations of Batteries
Battery manufacturing is a major industry and considerable researches are conducted to produce longer lasting,
better performing or lighter batteries. The most common cause for the loss of performance of batteries is
corrosion. When you forgot to remove batteries from radio or flashlight that you did not use for months, it is
more likely that you will found the batteries to be corroded and possibly even allowing fluids to leak out of the
cells. To limit the performance-diminishing effects of corrosion on batteries, one method is the protective
plating of materials used in the battery by the process of electrolysis.
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1.5 Electrolysis and Electroplating
Electrolysis
Electrolysis is a process of passing an electric current through an ionic solution or molten salt to produce a
chemical reaction. Electrolytic cells are divided into two categories based on the nature of the electrodes. If the
electrodes are chemically inert materials that simply provide a path for electrons, the process is called passive
electrolysis. Passive electrolysis is used in industry to purify metals that corrode easily. When the electrodes are
part of the electrolytic reaction, the process is called active electrolysis. Active electrolysis is used to plate
materials to provide resistance to corrosion.
Electroplating
Electroplating is a process of depositing a thin coat of metal by using electricity. Electroplating prevent
corrosion and is also vital to some functionality of coated piece. In electroplating, the important parameters to
look into are: current; time; and amount of metal to coat.
Current: The base unit of current is Ampere (A); 1A = 1Cs-1
Charge: If a current pass through a circuit for a certain time the charge can be calculated as
Charge = current x time
Q=I*t
where: Q= charge (unit is Coulomb; C)
I = current (unit is ampere or Coulombs/second; C/s)
t = time (unit is seconds; s)
Example 8. In a process called flash electroplating, a current of 2.50 x10 3 A passes through an electrolytic cell
for 5 minutes. How many moles of electrons are driven through the cell?
Given:
Required:
I = 2.50 x 103 A or 2.50 x 103 C/s
t = 5 min
moles of e- driven to the cell.
Solution:
a. Solve for charge (Q)
Q=I*t
Q =2.50 x 103 A * (5 min * 60 s/1min)
Q = 2.50 x 103 C/s * 300 s
Q = 7.50 x 105 C
b. Solve for the moles e- driven to the cell using the Faraday constant (F = 96,485 JV-1mol e-1)
Moles e- = Q / F
Moles e- = 7.50 x 105 C / 96, 485 JV-1 mol e-1
Moles e- = 7.50 x 105 C / 96,485 C mol e-1
Moles e- = 7.77 mol e1J = 1CV
1J/V = 1C
Example 9. Suppose that a batch of parts is plated with copper in an electrolytic bath running at 0.15 V and 15
A at exactly 2 hours. What is the energy cost of this process if the electric utility charges the company P0.050
per KWh?
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Given:
Required:
Voltage = 0.15V
Current = 15 A or 15 C/s
Energy cost for the process
Cost = P0.050/kWh
time = 2 hours
Solution:
a. Solve for charge (Q)
Q=I*t
Q = 15 C/s * (2 hr* 60 min/h * 60 s/min)
Q = 1.08 x 105 C
b. Solve for energy
E = Charge * Voltage
E = 1.08 x 105 C * 0.15 V
E = 1.08 x 105 J/V * 0.15 V
E = 1.62 x 104 J
c. Convert energy to KWh and determine cost
1kWh = 3.60 x 106 J
1 watt = 1 Js-1
Energy cost = 1.62 x 104 J * (1 kWh/ 3.60 x 106 J) * P0.050/kWh
Energy cost = P 0.00023
Example 10. An electrolysis cell deposits gold from Au+(aq) and operates for 15 minutes at a current of 2.30 A.
What mass of gold is deposited?
Given:
Required:
t = 15 minutes
I = 2.30 A
amount of gold deposited (grams Au(s))
Solution:
a. Write a balanced half reaction for Au+(aq)
Au+(aq) + eAu(s)
b. Solve for charge (Q)
Q= I * t
Q = 2.30 C/s * (15 min*60 s/min)
Q = 2.070 x 103 C
c. Calculate moles of e- using Faraday constant
Moles e- = Q / F
Moles e- = 2.070 x 103 C / 96, 485 C mol e-1
Moles e- = 2.15 x 10-2 mol ed. Calculate moles Au(s) using mole ratio
From the balanced half reaction in letter a, 1 mole of e- produces 1 mole Au(s)
Moles Au(s) = moles e- * mole ratio
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Moles Au(s) = 2.15 x 10-2 mol e- * 1 mol Au(s)/ 1 mol eMoles Au(s) = 2.15 x 10-2 mol Au(s)
e. Convert mole Au(s) to grams using the formula weight
Grams Au(s) = moles Au(s) * Formula weight of Au(s)
Grams Au(s) = 2.15 x 10-2 mol Au(s) * 197 g/mol
Grams Au(s) = 4.23 g Au(s)
1.6 Corrosion
Corrosion is the degradation of metals by chemical reaction with the environment. It generally involves a slow
combination of oxygen and water with metals to form compounds called oxides or hydrated oxides. Corrosion
is an example of an electrochemical process, but corrosion alone does not define electrochemistry. Rusting or
corrosion is seen to be disastrous. It causes accidents in industry, on highways, and in homes. It is wasteful
financially costing industrialized nations 4-5% of their gross domestic products annually.
Metals differ in its behavior towards corrosion, how fast a metal corrodes depends on its reactivity. More
reactive metals corrode faster than less reactive metals. But corrosion behavior between aluminum and iron is
different. Aluminum has greater tendency to corrode because it is more reactive than iron, but corrosion of
aluminum is less problematic than iron. This difference in corrosion behavior of aluminum and iron lies in the
nature of the products of corrosion reaction. Aluminum oxide (Al 2O3), the product of corrosion of aluminum is
less problematic because it is very unreactive, it sticks tightly to the surface of the metal serving as a protective
layer such that no cracks are formed preventing further corrosion.
Corrosion occurs in a variety of forms. The most common visible form of corrosion is uniform corrosion which
rusting of automobile bodies is an example. Another important form of corrosion is galvanic corrosion, which
occurs when two different metals are in contact to each other in the presence of an appropriate electrolyte.
Another form is crevice corrosion which is a major problem in many large machines, this happens when two
pieces of metal touch each other but tend to leave a small gap. At this gap or crevice, the metals are more likely
to corrode unless the joint is covered with a coating. Other forms of corrosion require specific conditions and
many of these situations are common in designs of machinery.
Corrosion results to great economic loss, hence, corrosion control is very important. The most common means
of protecting a material from corrosion is coating the more reactive metal with less reactive metal. This process
is called galvanizing or the metal is plated, it undergoes a process called electroplating. Another widely used
coating practice is painting. Paint coatings protect the material from exposure to water and oxygen.
Cathodic protection is also a means of protecting a metal from corrosion. Cathodic protection is a process of
creating a galvanic corrosion condition where the material that easily rusts is made as the cathode by
connecting it electrically to a more reactive metal that serves as the sacrificial anode (Figure 1.5). Corrosion of
iron, for example, can be prevented by choosing magnesium as the sacrificial anode that is connected
electrically to the iron. Since, magnesium has a reduction potential that is more negative than iron, the
magnesium is oxidized and the iron is reduced. The iron is made as the cathode and magnesium is the anode.
Using sacrificial anode is one effective method of corrosion prevention.
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Iron pipe (cathode)
Magnesium (sacrificial anode)
Figure 1.5 Cathodic protection
Clicker Exercise 3
1.Suppose that you have a part that requires a tin coating. You calculated that you need to deposit 3.60 g of tin
to achieve an adequate coating. If your electrolysis cell (using Sn 2+) runs at 2 A, how long must you operate the
cell to obtain the desired coating?
2. In a copper plating experiment in which copper metal is deposited from copper (II) ion solution, the system is
run for 2.6 hours at a current of 12A. What mass of copper is deposited?
References
1. Brown, WH & Holme, TA. (2011). Chemistry for Engineering Students. 2nd edition, Brooks/Cole, Centage
Learning, USA.
2. Gaffney J & Marley N. (2017). General Chemistry for Engineers. 1st edition, Elsevier Publishing.
3. Moore, JW, Stanitski, CL & Jurs, PC. (2005). Chemistry: The Molecular Science. 2nd edition, Brooks/Cole,
Thomson learning, Canada.
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