Electrochemistry

M e t a l l u r g i c a l E n g i n e e r i n g D e p a r t m e n t
0ctober 1st, 2013
Chapter 4
ASSIGNMENTNUMERICALSLECTURES
ELECTROCHEMISTRY
CY-105 APPLIED CHEMISTRY

of 50
Muhammad Umair Tariq (MY-065)
Syed Muhammad Tanveer (MY-017)
Anas Hanif (MY-037)
Ahmed Mazher (MY-011)
Raheel Nadeem (MY-059)
Ali Fakhar (MY-066)
Danish Khan (MY-069)
Course Instructor: SIR SAJJAD HAIDER.
SECTIONS:
1. THEORY SECTION.
2. CLASS AND HOMEWORK NUMERICALS.
3. ASSIGNMENT NUMERICALS.
4. IMPORTANT KEY POINTS AND QUESTIONS.
5. M.C.Q’s RELATING TO ELECROCHEMISTRY.

of 50
Electrochemistry
Electrolysis
Mechanism of
Electrolysis
Electrochemical
Cell
Half Cell ,
VoltaicCell
Corrosion and
Inhibation of
corrosion
Faraday's Law
Of Electrolysis

of 50
1. THEORY SECTION
Electrochemistry
It is the study of interconversion between chemical and electrical energy.
IMPORTANT TERMINOLOGIES USED IN ELECTROCHEMISTRY
Before we start the study of electrochemistry, we must be familiar with a few common terms:
i. Current is the flow of electrons through a wire or any conductor .
ii. Electrode is the material (a metallic rod/bar which conducts electrons into and out of a
solution.
iii. Anode is the electrode at which oxidation occurs. It sends electrons into the outer circuit. It
has negative charge and is shown a (-) in cell diagram.
iv. Cathode is the electrode at which electrons are received from the outer circuit. It has a
positive charge and is shown as (+) in cell diagrams.
v. Anode cell is the cell of the in which oxidation half reaction occurs. It contains the anode
vi. Cathode cell is the cell in which reduction half reaction occurs it contains the e cathode.
vii. Oxidation is the loss of electrons OR gain of O2 OR loss of H+ OR increase of oxidation
number.
viii. Reduction is the gain of electrons OR loss of O2 ORgain of H+ OR increase of reduction
number.
ix. Reducing agent are those which donates the electrons and is oxidized.
x. Oxidizing agent are those which accepts electrons and is reduced.
xi. Electrolyte is a compound in a solution or a molten compound which conducts electricity
and is decomposed by it in the process.
xii. Non-Electrolyte is a solution of a compound or a molten compound which does not conduct
electricity or become decomposed by it.
xiii. Salt-bridge in electrochemistry is a laboratory device used to connect the oxidation and reduction
half cells of a galvanic cell ( voltaic cell ) a type of electrochemical cell.
xiv. Electromotive force is the energy per unit charge that is converted reversibly from chemical,
mechanical or other forms of energy into electrical energy in a battery or dynamo.

of 50
ELECTROLYSIS
It is the chemical decomposition of a compound brought about by the passage of direct current
through solution of the compound or a molten compound.
MECHANISM OF ELECTROLYSIS
Before providing the current to the electrolytic cell the ions moves randomly in the solution. When
the current is applied to the cell. The an-ion of the electrolyte moves towards the anode and cat ion
towards the cathode.
Oxidation take place to the anode where electrons are lost and gain of electron (reduction) takes
place at cathode.
Note:
Anions are negatively charged that is why it moves towards anode which is positive and cat-ions are
positively charged hence move towards cathode.
EXAMPLE:
Consider electrolyte of HCl.
HCl → 𝐻+
+ 𝐶𝑙−
Here hydrogen ion moves towards cathode and chlorine ion moves toward anode.
At cathode:
𝐻+
+ 𝑒−
→ H
Each hydrogen ion picks up an electron from the cathode to become a hydrogen gas.
At anode:
𝐶𝑙−
→ Cl + 𝑒+
After the chloride ion loses its electron to the anode pair of chlorine atom unit to form chlorine gas.

of 50
DIFFERENCES
1. Electrolyte and Non-Electrolyte
2. Insulator and Non-Electrolyte
INSULATOR NON-ELECTROLYTE
They do not conduct electricity. They do not conduct electricity.
They are solid They are not solid
Example: Plastic Example: Covalent liquids or organic liquids.
3. Conductor and Electrolytes
CONDUTOR ELETROLYTES
They conduct electricity due to electrons. They conduct electricity due to ions.
They are solid substances. They are not solid substance either liquid
(solution) or molten.
It does not involve any chemical reaction. It involves chemical reaction.
They are not change in any way by supplying
electricity
They may change in any way by supplying
electricity.
EXAMPLE : metal, graphite. EXAMPLE: acid, alkalis, salt solution.
ELECTROLYTE NON-ELECTROLYTE
It is the ionic substances. It is the covalent substance.
It conducts electricity in a solution or in a molten
state.
It does not conduct electricity in a solution or
molten state.
They are inorganic compounds. They are mainly organic compounds.
For example:
Urea,benzene, trichloro methane etc.
For example:
Solution of sodium chloride, copper sulphates
etc.

of 50
STRONG ELECTROLYTE:
A strong electrolyte is an electrolyte that completely dissociates in solution. The solution will contain only
ions and no molecules of the electrolyte. Strong electrolytes are good conductors of electricity.
Examples:
HCl (hydrochloric acid), H2SO4 (sulfuric acid), NaOH (sodium hydroxide) and KOH (potassium hydroxide)
are all strong electrolytes.
Strong electrolytes conduct current very efficiently.
Completely ionized or dissociate when dissolved in water.
a. Soluble ionic compounds.
b. Strong acids (HNO3(aq) ,H2SO4 (aq) , HCl(aq))
HNO3 → H+ + NO3
- (100% ionization)
c. Strong bases (KOH and NaOH)
KOH → K+ + OH- (100% dissociation)
WEAK ELECTROLYTE :
A weak electrolyte is an electrolyte that does not completely dissociate in solution. The solution will contain
both ions and molecules of the electrolyte. Examples:
HC2H3O2 (acetic acid), H2CO3 (carbonic acid), NH3 (ammonia) are all weak electrolytes
a. Weak acids (organic acids → acetic, citric, butyric, malic etc.)
HC2H3O2 → H+ + C2H3O2
-
b. Weak bases (ammonia)
NH3 + H2O → NH4 + OH-

of 50
Laws of Electrolysis
Michael Faraday
Michael Faraday, was an English scientist who contributed to the fields of electromagnetism and
electrochemistry. His main discoveries include those of electromagnetic induction, diamagnetism and
electrolysis.
INTRODUCTION:
Faraday's laws of electrolysis are quantitative relationships based on the electrochemical
researches published by Michael Faraday in 1833.
Faraday’s Laws of Electrolysis give the relationship between the amount of material liberated at the
electrode and the amount of electric energy that is passed through the electrolyte. Faraday's work on the
chemical reaction produced when an electric current passes through a liquid resulted in the laws of
electrolysis.
There are 2 laws of Electrolysis:
1- Faraday's first law of electrolysis.
2- Faraday's second law of electrolysis.
Faraday's laws of electrolysis are quantitative relationships based on the electrochemical researches
published in 1833.
Faraday's first law of electrolysis:
Statement: “The amount of substance deposited or liberated at an electrode during electrolysis is directly
proportional to the quantity of current passing through the electrolyte”
Explanation:
It means that the mass of a substance produced by electrolysis is proportional to the quantity of
electricity used.
In order to produce more substance electrolytically we must use more electricity.
m α Q
m α It
m=zIt
where “Z” is known as electrochemical equivalent.

of 50
Electrochemical Equivalent “z”:
“If the mass is 1 kg and current of 1 ampere is passed through electrolyte in 1 second then the mass
deposited or liberated is called 1 electrochemical equivalent ”
OR
“The mass deposited by 1 coulomb charge is called 1 electrochemical equivalent”
z=m / It or z=m/Q
For example, when a charge of one coulomb is passed through silver nitrate solution, the amount of silver deposited
is 0.001118 g. this is the value of electrochemical equivalent of silver.
Importance of Faraday’s First law of Electrolysis:
With the help of faraday’s First law we are able to calculate:
1) The value of electrochemical equivalents of different substances.
2) The masses of substances produced by passing a known quantity of electricity through their solution.
Faraday’s Second Law of Electrolysis:
“It states that masses of different substances deposited or liberated when same quantity of current
is passed through different electrolytes connected in series are proportional to their chemical
equivalent masses”
Explanation:
Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass some quantity of
electricity through them for the same time. Now Ag, Cu and Al metals collect at the cathode. Their masses
are directly proportional to their equivalent masses.
According to Faraday, if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes,
we get which are the equivalent masses of Ag, Cu and Al respectively.
Ag+ 1e- Ag(S)
Cu2+ + 2e- Cu (S)
Al+3 +3e- Al(S)
Charge on 1 electron = 96500 coulomb = 1 faraday

of 50
Equivalent Mass:
“The mass deposited or liberated which is equal to 108 parts of silver, 35.5 parts of chlorine and 8
parts of oxygen”
.
Importance of Faraday’s Second Law:
By the help of 2nd law we can calculate:
 Equivalent weight of metal.
 Unit of electric charge (coulomb).
 Avogadro’s number.
Steps for doing numerical:
Step#1: Find Charge OR Coulomb
Step #2: Find Faraday
Step #3: Taking the calculated values of Charge and faraday, solve the Question further for given
equations or for given solutions.

of 50
2. Class Numericals
Q) A direct current of 0.01 ampere flows for 4 hours through 3 cells in series. They contain AgNO3, CuSO4
and AuNO3.
Calculate the mass deposited in each cell.
SOLUTION:
q=It
= (0.01)(4 x 60 x 60)
q= 144 coulomb
96500 coulomb = 1 Faraday
144 coulomb = 1/96500 x 144
F= 1.49 x 10-3 Farad.
Step #3: Taking the calculated values of Charge and faraday, solve the question further for given equations
Let X be the required amount of AgNO3, CuSO4 and AuNO3.
Ag+ 1e- → Ag(S)
108 g = 1 F
1.49 x 10-3 = X gm
X= 0.161 gm.
Cu2+ + 2e- → Cu(S)
63.5/2= 1 F
OR
2 F= 63.5 gm
1.49 x 10-3 = X
2X = 63.5 x (1.49 x 10-3)/2
X= 0.0473 gm.
Al+3 +3e- 𝐴𝑙(𝑠)
197 gm = 3 F
3 F= 197 gm
1.49 x 10-3 F= X
3X= 197 x (1.49 x 10-3)/3
X= 0.0983 gm

of 50
Q) A metal Ar=27 is deposited by electrolysis when 0.15 amp flows 3.5 hours, 0.176gm of metal is
deposited. What is the charge in the cation of this metal?
Solution:
q = It
=0.15 (3.5x60x60)
q= 1890 coulomb
96500 coulomb = 1 faraday
1890 coulomb = x
x= 1890/96500
x= 0.0196F
Step#3:
0.176gm= 0.0196F
27gm = x faraday
x = 0.0196×27/0.176
x = 3
charge is +3
Q) How many grams of oxygen gas liberated when 0.0565 ampere current passed during 185 seconds during
electrolysis
of water.
Solution:
q=It
= (0.0565)(185)
=10.452 coulomb

of 50
1F= 96500 Coulomb
10.452 C= 10.452/ 96500
= 1.083 x 10-4 F
Step#3:
1 F = 8 gm
1.083 x 10-4 F= X gm
X= 0.000864 gm
Q) Find the amount of current if 404mg Cu is deposited during 5 hours. Electrolyte is CuSO4.
q= It
I= q/t -------------(1)
Cu2+ + 2e- Cu(S)
63.5 gm = 2F
404 x 10-3 = X farad
X 63.5 gm = 8.08 x 10 -3
X = 1.272 x 10 -4 F
1 F= 96500 Coulomb
1.272 x 10 -4 F= 96500 x 1.272 x 10 -4 F
12.272 coulomb
I= q/t
= 12.272 / 5x60x60= 0.0680 Ampere.
HALF REACTION
Let us consider the reaction
2𝑁𝑎 + 𝐶𝑙2 → 2𝑁𝑎+
+ 𝐶𝑙−
It occur by the transfer of electron from Na to Cl. Na lose an electron and it said to be oxidized to Na+ ion at
the same time Cl gain a electron and is reduced to 𝐶𝑙−
ion. Such a reaction in which loss of electron
(oxidation) and gain of elctron (reduction) occurs simultaneously, is called an Oxidation Reduction reaction
or Redox Reaction.
The redox reaction can be considered as made up of two reactions. For examples the redox reaction
2𝑁𝑎 + 𝐶𝑙2 → 2𝑁𝑎+
+ 𝐶𝑙−
is composed of two reactions
2𝑁𝑎 → 2𝑁𝑎+
+ 2𝑒−
(Oxidation)
𝐶𝑙2 + 2𝑒−
→ 2𝐶𝑙−
(Reduction)

of 50
Each of the two reactions shows just oxidation or just the reduction portion of the overall redox reaction.
These two components of redox reactions are called Half reactions. The first half reaction that proceeds by
oxidation is often referred to as the Oxidation Half reaction. The second half reaction that occurs by
reduction is referred to as the Reduction Half reaction. When the two half reactions are added together the
sum is the net redox reaction.
ELECTROCHEMICAL CELLS
A device used for producing an electrical current from a chemical reaction (redox reaction) is called an
electrochemical cells.
Explanation:
When we read the above defination, the first question which came into our minds is how a redox reaction
can produce an electrical current? So we perform an experiment and check it.
Let consider a beaker containing 𝐶𝑢𝑆𝑂4 and a zinc rod. When bar of zinc is dipped in the solution of copper
sulphates redox reaction taking place on zinc bar and copper metal is deposited on the bar.
Now there is a burning question i.e. why Cu is reduced on zinc bar and why we use Zn bar?
We may use Cu bar dipped in 𝑍𝑛𝑆𝑂4 solution?
The answer is very simple. Cu is reduced because Cu has ability to reduced but this answer is not complete,
the real reason is Cu has greater redox potential value than Zn. Whenever reduction occurs deposition takes
place only and only at cathode.
If we dipped Cu bar in 𝑍𝑛𝑆𝑂4 no reaction occurs. Zn do not depositon Cu bar because it has less pedox
potential value than Cu.
The two half reactions are
𝐶𝑢+2
+ 2𝑒−
→ 𝐶𝑢 (reduction)
𝑍𝑛 → 𝑍𝑛+2
+ 2𝑒−
(oxidation)
The net reaction is

of 50
𝐶𝑢2+
+ 𝑍𝑛 → 𝐶𝑢( 𝑠) + 𝑍𝑛( 𝑎𝑞)
2+
In this change Zn is oxidized to give 𝑍𝑛2+
ions and 𝐶𝑢2+
are reduced to Cu atoms. The electrons released in
the first half reaction are used up by the second half reaction. Both the half reaction occurs on the zinc rod
itself and there is no net charge.
Now let the two half reaction occurs in separate compartments which are connected by a wire. The electrons
produced in the left compartment flow through the wire to the other compartment. However, the current will
flow for an instant then stop. The current stops flowing because of the charge built up in the two cells. The
electrons leave the left compartments and it would become positively charged. The right cell receives
electrons and become negatively charged both these factors oppose the flow of electrons which eventually
stops. This problem is solved by connecting a salt bridge (u shaped tube containing electrolyte) between the
two cells. It provides a passage of ion from one cell to the other cell. With this flow of ion, circuit is
complete and electrons pass freely through the wire to keep the net charge zero in the two compartments.
VOLTAIC CELL

of 50
A voltaic cell also known as a Galvonic Cell is a electrochemical cell in which electrical current is
generated by a spontaneous redox reaction.
Explanation:
Consider a electrochemical cell in which a rod of zinc (anode) is placed in 𝑍𝑛𝑆𝑂4 solution in the cell A. A
rod of Cu (cathode) is immersed in 𝐶𝑢𝑆𝑂4 solution in the cell B.
The Zn and Cu electrodes are joined by a copper wire .When the cell is set up, electrons flow from Zn
electrode through the wire to the Cu cathode. As a result,
IN CELL A
At anode half cell, redox reaction takes place. Zn dissolves in the solution to form more and more and more
𝑍𝑛+2
ions then at a certain time movement of 𝑒−
stops since current stops.
IN CELL B
At cathode half cell more and more 𝑒−
comes from cell A . The Cu ions in the cathode half cell pick up
electrons and are converted to Cu atoms on the cathode. Then at a certain time income of 𝑒−
stops since
current in cell B also stops.
In cell A 𝑍𝑛+2
stops the moment of 𝑒−
and in cell B 𝑆𝑂4
−
stops income of 𝑒−
.So we construct a Salt Bridge.
𝑍𝑛2+
+ 𝑆04
−
→ 𝑍𝑛𝑆𝑂4
Cell B Cell A

of 50
ELECTROLYSIS OF BRINE
Brine is NaCl (concentrated) i.e. low H2O and high Na.
Brine is a concentrated or saturated solution of sodium chloride in water. The process of electrolysis uses
electrical energy to produce a chemical reaction in the sodium chloride solution which results in the
formation of sodium hydroxide, chlorine and hydrogen gas. The equation for this chemical reaction is as
follows.
Sodium chloride + Water → Sodium hydroxide + Hydrogen + Chlorine
NaCl + H2O → NaOH + H2 + Cl2
The manufacture of sodium hydroxide or caustic soda is an important industrial process.
The electrochemistry involved in the reaction involves an reduction oxidation reaction, which is sometimes
called a redox reaction.
In a solution of brine there are three chemical species: The sodium ion, Na+, the chloride ion, Cl- and water,
H2O. The easiest chemcial species to oxidise is the chloride ion. The easiest chemical species to reduce is
the water.
Reaction at anode:
The anode is the electrode where oxidation or the losses of electrons take place. In an electrolytic cell the
anode is positive in charge. The chloride ion is oxidized to chlorine gas.
2Cl-
(aq) → Cl + 2e-
OH- → ½H2 + O2(g).
Reaction at cathode:
Na+
(aq) + e- → Na(s)
In water the sodium ion is not reduced as it is much easier to reduce water. The sodium ion becomes a
spectator ion and is not involved in the reaction.
Overall reaction:
The overall chemical equation for the electrolysis of brine is
NaCl + H2O → H2 + Cl2 + NaOH
GOOD TO KNOW:
In industry the hydrogen gas must be separated from the chlorine gas or they will explode when exposed to
UV light producing hydrogen chloride.
𝐻2 + 𝐶𝑙2 → 2𝐻𝐶𝑙
Sodium hydroxide is caustic in nature and must be handled with care. It is used in industry in the
manufacture of soap by saponification. It is also used to make pulp and in the production of paper.

of 50
CORROSION
Corrosion is an electrochemical reaction composed of two half cell reactions, an anodic reaction and
cathodic reaction.
In anodic reactions electron releases and in cathodic reactions electron consumes.
An electrochemical reaction is defined as a chemical reaction involving the transfer of electrons. It is also a
chemical reaction which involves oxidation and reduction. Since metallic corrosion is almost always an
electrochemical process, it is important to understand the basic nature of electrochemical reactions. The
discoveries that gradually evolved in modern corrosion science have, in fact, played an important role in the
development of a multitude of technologies we are enjoying today.
An important achievement early in the history of electrochemistry was the production of power sources,
following the production of the first batteries by Alessandro Volta illustrates the principle of a Daniel cell in
which copper and zinc metals are immersed in solutions of their respective sulfates. The Daniel cell was the
first truly practical and reliable electric battery that supported many nineteenth-century electrical innovations
such as the telegraph. In the process of the reaction, electrons can be transferred from the corroding zinc to
the copper through an electrically conducting path as a useful electric current. Zinc more readily loses
electrons than copper, so placing zinc and copper metal in solutions of their salts can cause electrons to flow
through an external wire which leads from the zinc to the copper.
The difference in the susceptibility of two metals to corrode can often cause a situation that is called
galvanic corrosion named after Luigi Galvani, the discoverer of the effect. The purpose of the separator
shown in is to keep each metal in contact with its own soluble sulfates, a technical point that is critical in
order to keep the voltage of a Daniel cell relatively constant . The same goal can be achieved by using a salt
bridge between two different beakers. The salt bridge in that case, provides the electrolytic path that is
necessary to complete an electrochemical cell circuit. This situation is common in natural corrosion cells
where the environment serves as the electrolyte that completes the corrosion cell. The conductivity of an
aqueous environment such as soils, concrete, or natural waters has often been related to its corrosivity.
Zinc anode:
Zn(s)→ Zn+2+2e-
Copper cathode:
Cu2+ → 2e- + Cu(s)

of 50
The short-hand description is valid for both cells. Such a description is often used to simplify textual
reference to such cells.
Zn/Zn2, SO24 (Conc1)//Cu2, SO24 (Conc2))/Cu()
Conc1 andConc2 indicate respectively the concentration of zinc sulfate and copper sulfate that may differ
in the two half-cells while the two slanted bars (//) describe the presence of a separator. The same short-hand
description also identifies the zinc electrode as the anode that is negative in the case of a spontaneous
reaction and the copper cathode as positive.
The fact that corrosion consists of at least one oxidation and one reduction reaction is not always as
obvious as it is in chemical power cells and batteries. The two reactions are often combined on a single piece
of metal as it is illustrated schematically
A piece of zinc immersed in hydrochloric acid solution is undergoing corrosion. At some point on the
surface, zinc is transformed to zinc ions. This reaction produces electrons and these pass through the solid
conducting metal to other sites on the metal surface where hydrogen ions are reduced to hydrogen gas
Anodic reaction:
Zn(s)→2e- + Zn+2
Cathodic reaction:
2H+2+2e- → H2(g)
The nature of an electrochemical reaction typically illustrated for zinc .During such a reaction, electrons are
transferred, or, viewing it another way, an oxidation process occurs together with a reduction process. The
overall corrosion processes are summarized in
Overall corrosion reaction: Zn 2H  Zn2H2(g)
Briefly then, for corrosion to occur there must be a formation of ions and release of electrons at an anodic
surface where oxidation or deterioration of the metal occurs. There must be a simultaneous reaction at the
cathodic surface to consume the electrons generated at the anode. These electrons can serve to neutralize
must go on at the same time and at equivalent rates. However, what is usually recognized as the corrosion
process occurs only at the areas that serve as anodes.
Anodic Processes:
Let us consider in greater detail what takes place at the anode when corrosion occurs. For instance,
reconsider. This reaction involves the reduction of hydrogen ions to hydrogen gas,. This hydrogen
evolution reaction occurs with a wide variety of metals and acids, including hydrochloric, sulfuric,
perchloric, hydrofluoric, formic, and other strong acids. The individual anodic reactions for iron, nickel, and
aluminum are listed as follows:
Iron anodic reaction: Fe(s) Fe2  2e
Nickel anodic reaction: Ni(s) Ni2  2e
Aluminum anodic reaction: Al(s) Al3  3e
That is, the corrosion of metal M results in the oxidation of metal M to an ion with a valence charge of nand
the release of n electrons.
Some metals such as silver are univalent, while other metals such as iron, titanium, and uranium are
multivalent and possess positive charges as high as is general and applies to all corrosion reactions.

of 50
Cathodic Process
When hydrogen ions are reduced to their atomic form they often combine, as shown earlier, to produce
hydrogen gas through reaction with electrons at a cathodic surface. This reduction of hydrogen ions at
(OH ) ions and make the solution less acidic or more alkaline or basic at the corroding interface.
In neutral waters the anodic corrosion of some metals like aluminum, zinc, or magnesium develops enough
energy to split water directly.
Water splitting cathodic reaction: 2H2O(l) 2e- H2 2OH-
The change in the concentration of hydrogen ions or increase in hydroxyl ions can be shown by the use of
pH indicators, which change color and thus can serve to demonstrate and locate the existence of surfaces on
which the cathodic reactions in corrosion are taking place. There are several other cathodic reactions
encountered during the corrosion of metals. These are listed below:
Oxygen reduction:
(acid solutions) O2 4H4e 2H2O
(neutral or basic solutions) O2 2H2O 4e 4OH
Hydrogen evolution: 2H2e H2(g)
Metal ion reduction: Fe3e Fe2
Metal deposition: Cu22e Cu(s)
Hydrogen ion reduction, or hydrogen evolution, has already been discussed. This is the cathodic reaction
that occurs during corrosion in acids.
Inhibition of corrosion
Introduction:
Inhibition means to remove corrosion. There are a number of ways to remove corrosion.
Some methods are:
 Impermeable layer.
 Galvanizing
 Cathodic production.
 Sacrification method.
Impermeable layer:
Coating of impermeable layer on the material prevents it from corrosion. Sometimes this
measure fails because of porous paint.
Galvanizing:
Coating of zinc on a material can also prevent it from being corroded. This method is called
Galvanizing
Cathodic production:
This method was Invented in 1824 by the UK scientist Sir Humphrey Davy (1778-1829) to
protect the copper-cladded wooden ships from seawater corrosion. It is an electrical method of preventing
corrosion on metallic structures which are in electrolytes such as soil or water
Sacrification method:
This method is basically used in ships.Here Mg is used as a sacrificing agent. We attach Mg
to Fe. First Mg oxidizes, and then Fe will oxidize after a long time.

of 50
3. ASSIGNMENT NUMERICALS
QUESTION # 1
Electrolysis involves the chemical decomposition of a compound either when molten or in aqueous solution,
by the passage of an electric current.
(a)Explain why aqueous calcium nitrate can be electrolyzed but liquid pentane cannot.
ANSWER:
Aqueous calcium nitrate can be electrolyzed because in aqueous form it contain ions which are responsible
for electrolysis, but in liquid pentane ions are not formed that is why it cannot be electrolyzed.
(B) State the products of electrolysis of molten sodium chloride.
ANSWER:
Electrolysis of molten sodium chloride yields sodium metal and chlorine gas.Sodium ions are reduced at
cathode to sodium metal while chlorine gas is liberated at anode.
(C) State the products of electrolysis of concentrated aqueous sodium chloride.
ANSWER:
Electrolysis of concentrated aqueous sodium chloride yields hydrogen and chlorine ( with aqueous sodium
hydroxide remaining in solution.
The reason for the difference is that the reduction of Na (E=2.7V) is energetically more difficult than
reduction of water (-1.27V).
(D) A student investigates the electrolysis of aqueous copper (II) sulphate using the apparatus shown below.
.
Copper anode
(Positive electrode)
Copper cathode
(Negative
electrode)
Aqueous copper

of 50
The student weighs the copper cathode before and after the electrolysis.
Experiment Current used Time taken/s Mass of metal
Before starting/g After electrolysis/g
1 2.0 180 1.24 1.36
2 4.0 180 1.20 1.44
3 2.0 360 1.34 1.58
(i)Explain with the help of an equation, why the cathode increases in mass.
ANSWER:
Cathode increases in mass in this reaction because reduction occurs at cathode and copper is deposited at
cathode.
Cathode reaction:
𝐶𝑢2+
( 𝑎𝑞)+ 2𝑒−
→ 𝐶𝑢
(ii)In experiment 2 the student measures the mass of the anode both before and after the electrolysis. At the
start the anode has a mass of 1.45g.Determine the mass of the anode at the end of the electrolysis.
ANSWER:
A we can see in experiment 2 the mass of cathode at the start is 1.20g and at the end it has a mass of 1.44g.
Determine the mass of the anode at the end of the electrolysis, so taking the difference we get,
1.44-1.20=0.24g.
Adding the given mass we get the required mass at the anode,
0.24+1.45=1.69g.
(iii)The student does a fourth experiment, this time using a current of 8.0 A for 90 seconds. At the start the
anode has a mass of 1.51g. Predict the mass of cathode at the end of the electrolysis.
ANSWER:
First we calculate the charge by using current and time,
q=it
q=8×90
q=720 C.
For faraday,
96500C=1F
720C=
1
96500
× 720
=7.46× 10−3
F.

of 50
Since,
𝐶𝑢2+
+ 2𝑒 → 𝐶𝑢
2F = 63.5g
7.46× 10−3
𝐹 = x.
X =0.236g.
The total mass deposited on the cathode is calculated by adding the given mass in the calculated mass,
X=0.236+1.51
.
QUESTION # 2
Chlorine gas is manufactured by the electrolysis of brine using a diaphragm cell.
(a)write half equations ,including state symbols, for the reactions occurring at each of the electrodes of a
diaphragm cell.
ANSWER:
Reaction at cathode:
𝑁𝑎+( 𝑎𝑞) + 1e → Na
𝐻+
+ 2𝑒 → 𝐻2.
Reaction at anode:
2𝐶𝑙−( 𝑎𝑞) → 𝐶𝑙2(𝑔) + 2𝑒.
𝑂𝐻−
→ 𝑂2
(b)In the diaphragm cell the anode is made of titanium and the cathode is made of steel. Suggest why steel is
never used for anode.
ANSWER:
In electrolysis, the anodes are called as sacrificial anodes and they are usually made of zinc, aluminium and
magnesium which are more electronegative than steel so are able to supply electrons to the electropositive
steel because in electrolysis corrosion takes place so it is necessary to protect cathode so for cathodic
protection anodes are not made of steel.
(c)One important product made in the diaphragm cell is formed in the aqueous solution.
(i)What substance is produced in aqueous solution in the diaphragm cell?

X=1.746 gm

of 50
ANSWER:
Chlorine in water dissolves to produce HCL and HOCL being an oxidizing agent ,bleaches red litmus
(d) Explain with the help of equations how this compound is formed by electrolysis?
ANSWER:
𝐶𝑙2 + 𝐻2 𝑂 → 𝐻𝐶𝐿 + 𝐻𝑂𝐶𝑙.
HOCl + dye → 𝑑𝑦𝑒 + 𝑂( 𝑐𝑜𝑙𝑜𝑢𝑟𝑙𝑒𝑠𝑠) + 𝐻𝐶𝑙.
Question # 4
The electrolytic purification of copper can be carried out in an apparatus similar to the one shown below.
+ -
The impure copper anode contains small quantities of metallic nickel, zinc and silver, together with inert oxides
and carbon resulting from the initial reduction of the copper ore with coke. The copper goes into solution at the
anode, but the silver remains as the metal and falls to the bottom as part of the anode’sludge’. The zinc also
dissolves.
(A)(1)Write a half equation including state symbols for the reaction of copper at the anode and precipitate pure
copper onto the cathode. However, a small quantity of it is ‘wasted’ in dissolving the impurities at the anode which
then remain in solution. When a current of 20amp was passed through the cell for 10 hrs , it was found that 225
gm of pure copper was deposited on the cathode.
At the cathode, copper ions are deposited as copper.
At the anode, copper goes into solution as copper ions.
(1)Part 1: Calculate the following.
 Number of moles of copper produced at the cathode.
Pure copper
cathode
CuSO4
(aqu.)
Impure
copper
anode
Anode
‘sludge’

of 50
Answer :
No. of moles are represented by = mass of element / molecular mass of element.
So, no. of moles=225 / 63.5 = 3.54 mol.
No. of moles of copper produced at cathode is 3.54 mol.
 Number of moles of electrons that passed through the cell.
20 amp × 36000 secs = 720000 coulomb.
720000 C × 1F/96500 C = 7.4611 F.
7.46 F × 1mole e-
/1F = 7.46 mole e-
.
So, 7.46 moles of electrons passed through the cell.
 Number of moles of electrons needed to produce this copper.
7.46 mole of e-
× 1 mol. Of copper/2 mole e-
= 3.73 moles.
So , 3.73 moles are needed to produce this copper.
(1)Part 2: hence calculate the percentage of the current through the cell that has been ‘wasted’ in dissolving the
impurities at the anode.
(2) Explain why silver remains as the metal.
According to the electrochemical series copper (Cu) will deposit first and silver will remain as metal because
silver at anode which is below copper in the electrochemical series (reactivity series) doesn't go into
solution as ions. It stays as a metal and falls to the bottom of the cell as an "anode sludge" together with
any unreactive material left over from the ore.
(3) Predict what happens to the nickel at the anode.
Nickel according to the reactivity series will discharge into solution.
(4) Write a half equation including state symbols for the main reaction at the cathode.
AT CATHODE:
Cu+2
+ 2e-
Cu.
(5) Explain why zinc is not deposited on the cathode.
Metals above copper in the electrochemical series (like zinc) will form ions at the anode and go into
solution. However, they won't get discharged at the cathode provided their concentration doesn't get too
high.
The concentration of ions like zinc will increase with time, and the concentration of the copper ions in the
solution will fall. For every zinc ion going into solution there will obviously be one fewer copper ion
formed.
(6) Suggest why the blue colour of the electrolyte slowly fades as the electrolysis proceeds.
The blue colour of the electrolyte in cell fades when more and more Cu2+ ions are reduced to copper metal
and plated onto the cathode as a pink deposit.Since copper ions in solution are used up, the blue colour fades
Cu2+ + 2e- → Cu

of 50
(B) Most of the current passed through the cell is used to dissolve the copper at the anode and precipitate pure
copper onto the cathode. However,a small quantity of it is ‘wasted’ in dissolving the impurities at the anode which
then remain in solution. When a current of 20amp was passed through the cell for 10 hrs , it was found that 225
gm of pure copper was deposited on the cathode.
(B)Part 1: Calculate the following.
 Number of moles of copper produced at the cathode.
Answer :
No. of moles are represented by = mass of element / molecular mass of element.
So, no. of moles=225 / 63.5 = 3.54 mol.
No. of moles of copper produced at cathode is 3.54 mol.
 Number of moles of electrons needed to produce this copper.
7.46 ole of e-
× 1 mol. Of copper/2 mole e-
= 3.73 moles.
So , 3.73 moles are needed to produce this copper.
 Number of moles of electrons that passed through the cell.
20 amp × 36000 secs = 720000 coulomb.
720000 C × 1F/96500 C = 7.4611 F.
7.46 F × 1mole e-
/1F = 7.46 mole e-
.
So, 7.46 moles of electrons passed through the cell.
(B)Part 2: Hence calculate the percentage of the current through the cell that has been ‘wasted’ in dissolving the
impurities at the anode.
Question # 06
Define the following.
(1) Electrochemical equivalent (Z)
If the mass is 1kg and current of 1 amp pass through electrolyte in 1 second then the mass
deposited or liberated is called 1 electro chemical equivalent.
‘OR’
The mass deposited by 1 coulomb charge is called 1 electrochemical equivalent.
Z = m/It ; Z= m / Q.
(2) Standard electrode potential
In an electrochemical cell, an electrode potential is created between two dissimilar
metals. This potential is a measure of the energy per unit charge which is available from
the oxidation/reduction reactions to drive the reaction. Also, the standard electrode potential is the potential
difference between the electrode and an electrolyte (at 1 M), at standard conditions (1atm, 298K). It is
represented by Eo.
(3) Half reaction.

of 50
A half reaction is either the oxidation or reduction reaction component of a redox
reaction. A half reaction is obtained by considering the change in oxidation states of individual substances
involved in the redox reaction.
Redox
Reduction reaction Oxidation reaction.
½ oxidation reaction + ½ reduction reaction → redox reaction.
For Example
Ni(s) + 2𝐹𝑒3+
→ 𝑁𝑖2+
+ 2𝐹𝑒2+
Oxidation takes place at the anode and the electrode must be Ni↑ 𝑁𝑖2+
Ni(s)→ 𝑁𝑖2+
(𝑎𝑞) + 2𝐹𝑒
And the reduction occurs at the cathode
2𝐹𝑒2+
+2e→ 2𝐹𝑒2+
(4) Salt bridge.
A salt bridge in electrochemistry is a laboratory device used to connect the oxidation and
reduction half cells of a galvanic cell ( voltaic cell ) a type of electrochemical cell.
(5) Electromotive force.
The energy per unit charge that is converted reversibly from chemical ,mechanical or other
forms of energy into electrical energy in a battery or dynamo.
Question# 07
Define anti clockwise rule in electrolysis and if a direct current of 0.01amp flows for 4 hours through three cells in
a series. They contain aqueous AgNO3 , CuSO4 and Au(NO3)3. Calculate the mass of metal deposited in each.
(atomic masses. Ag=108, Cu=63.5, Au=197).
ANTI CLOCKWISE RULE:
It is a rule where positivity or negativity of an electrode is determined by considering the following steps.
(1) Arrange w.r.t to redox potential values i.e the reactions should first be written in descending
order of redox potential (Eo
) values.
Cu+2
+ 2e-
→ Cu. EO
= 0.34 volts.
Ag+
+ e-
→ Ag. EO
= 0.80 volts.
(2) Make anti clockwise heads.
Cu+2
+ 2e-
→ Cu. EO
= 0.34 volts.
Ag+
+ e-
→ Ag. EO
= 0.80 volts.
(3) The tail of arrow is the reactant and the head is product.

of 50
For the mass of metal :
Step#1:FindCharge OR Coulomb
q=It
= (0.01)(4 x 60 x 60)
q= 144 coulomb
Step#2: FindFaraday
96500 coulomb= 1 Faraday
144 coulomb= 1/96500 x 144
F= 1.49 x 10-3
Farad.
Step#3: Takingthe calculatedvaluesof Charge andfaraday,solve the Questionfurtherforgivenequations
Let X be the requiredamountof AgNO3,CuSO4 andAuNO3.
Ag+ 1e- → Ag(S)
108 g = 1 F
1.49 x 10-3
= X gm
X= 0.161 gm.
Cu2+
+ 2e- → Cu (S)
63.5/2= 1 F
OR
2 F= 63.5 gm
1.49 x 10-3
= X
2X = 63.5 x (1.49 x 10-3
)/2
X= 0.0473 gm.
Al+3
+3e- → Al(S)
197 gm = 3 F
3 F= 197 gm
1.49 x 10-3
F= X
3X= 197 x (1.49 x 10-3
)/3
X= 0.0983 gm

of 50
Question # 08
Contruct the cell in which reactions are taken place, which of the electrodes shall act as anode (-ve electrode) and
which one is cathode (+ve electrode).
(a) Zn + CuSO4 → ZnSO4 + Cu
(b) Cu + 2AgNO3 → Cu (NO3)2 + 2Ag
(c) Zn + H2SO4 → ZnSO4 + H2
(d) Fe + SnCl2 → FeCl2 + Sn
First we have to know that oxidation (loss of electrons) occurs at anode and reduction (gain of electrons) occurs at
cathode.
Oxidation: Zn  Zn2+
+ 2 e-
.
Reduction: Cu2+
+ 2 e  Cu
Steps :
1) First write in ionic form.
2) Then write half cell reactions and balance them.
3) Then determine whether the electrode is cathode or anode.
(a) Zn + CuSO4 → ZnSO4 + Cu.
1) Zn(s) + Cu+2
→ Zn+2
(aqu.) + Cu.
At cathode: Cu+2
+ 2e-
→ Cu.
At anode: Zn  Zn2+
+ 2 e-
.
2) So , since reduction occurs at cathode therefore Cu is –ve and oxidation occurs at anode therefore Zn is
+ve.
(b) Cu + 2AgNO3 → Cu (NO3)2 + 2Ag.
1) Cu+2
+Ag → Ag+
+ Cu.
At cathode: Cu+2
+ 2e-
→ Cu
At anode: Ag → Ag+1
+ e-
2) So , since reduction occurs at cathode therefore Cu is –ve and oxidation occurs at anode therefore Ag
is +ve.
(c) Zn + H2SO4 ZnSO4 + H2
1) Zn+2
+H2 2Zn + 2H+
.
2) At cathode: Zn2+
+ 2 e-
2 Zn.
At anode: H2 2H+
+ 2e-
.
3) So , since reduction occurs at cathode therefore Zn is –ve and oxidation occurs at anode therefore H
is +ve.

of 50
Question # 09
What is corrosion? Explain reactions associated with corrosion. (With equations and electrode potential values).
OR Define corrosion and explain the reaction involved in rusting of iron with equations and redox potential values.
CORROSION
Corrosion is an electrochemical reaction composed of two half cell reactions, an anodic reaction and
cathodic reaction.
In anodic reactions electron releases and in cathodic reactions electron consumes.
An electrochemical reaction is defined as a chemical reaction involving the transfer of electrons. It is also a
chemical reaction which involves oxidation and reduction. Since metallic corrosion is almost always an
electrochemical process, it is important to understand the basic nature of electrochemical reactions. The
discoveries that gradually evolved in modern corrosion science have, in fact, played an important role in the
development of a multitude of technologies we are enjoying today.
An important achievement early in the history of electrochemistry was the production of power sources,
following the production of the first batteries by Alessandro Volta illustrates the principle of a Daniel cell in
which copper and zinc metals are immersed in solutions of their respective sulfates. The Daniel cell was the
first truly practical and reliable electric battery that supported many nineteenth-century electrical innovations
such as the telegraph. In the process of the reaction, electrons can be transferred from the corroding zinc to
the copper through an electrically conducting path as a useful electric current. Zinc more readily loses
electrons than copper, so placing zinc and copper metal in solutions of their salts can cause electrons to flow
through an external wire which leads from the zinc to the copper.
The difference in the susceptibility of two metals to corrode can often cause a situation that is called
galvanic corrosion named after Luigi Galvani, the discoverer of the effect. The purpose of the separator
shown in is to keep each metal in contact with its own soluble sulfates, a technical point that is critical in
order to keep the voltage of a Daniel cell relatively constant . The same goal can be achieved by using a salt
bridge between two different beakers. The salt bridge in that case, provides the electrolytic path that is
necessary to complete an electrochemical cell circuit. This situation is common in natural corrosion cells
where the environment serves as the electrolyte that completes the corrosion cell. The conductivity of an
aqueous environment such as soils, concrete, or natural waters has often been related to its corrosivity.
Zinc anode: Zn(s) → Zn22e-
Copper cathode: Cu22e-
→ Cu(s)

of 50
The short-hand description is valid for both cells. Such a description is often used to simplify textual
reference to such cells.
Zn/Zn2, SO24 (Conc1)//Cu2, SO24 (Conc2))/Cu()
Conc1 andConc2 indicate respectively the concentration of zinc sulfate and copper sulfate that may differ
in the two half-cells while the two slanted bars (//) describe the presence of a separator. The same short-hand
description also identifies the zinc electrode as the anode that is negative in the case of a spontaneous
reaction and the copper cathode as positive.
The fact that corrosion consists of at least one oxidation and one reduction reaction is not always as
obvious as it is in chemical power cells and batteries. The two reactions are often combined on a single piece
of metal as it is illustrated schematically
A piece of zinc immersed in hydrochloric acid solution is undergoing corrosion. At some point on the
surface, zinc is transformed to zinc ions. This reaction produces electrons and these pass through the solid
conducting metal to other sites on the metal surface where hydrogen ions are reduced to hydrogen gas
Anodic reaction:
Zn(s)→2e- + Zn+2
Cathodic reaction:
2H+2+2e- → H2(g)
The nature of an electrochemical reaction typically illustrated for zinc .During such a reaction, electrons are
transferred, or, viewing it another way, an oxidation process occurs together with a reduction process. The
overall corrosion processes are summarized in
Overall corrosion reaction: Zn 2H  Zn2H2(g)
Briefly then, for corrosion to occur there must be a formation of ions and release of electrons at an anodic
surface where oxidation or deterioration of the metal occurs. There must be a simultaneous reaction at the
cathodic surface to consume the electrons generated at the anode. These electrons can serve to neutralize
must go on at the same time and at equivalent rates. However, what is usually recognized as the corrosion
process occurs only at the areas that serve as anodes.
Anodic Processes:
Let us consider in greater detail what takes place at the anode when corrosion occurs. For instance,
reconsider. This reaction involves the reduction of hydrogen ions to hydrogen gas,. This hydrogen
evolution reaction occurs with a wide variety of metals and acids, including hydrochloric, sulfuric,

of 50
perchloric, hydrofluoric, formic, and other strong acids. The individual anodic reactions for iron, nickel, and
aluminum are listed as follows:
Iron anodic reaction: Fe(s) Fe2  2e
Nickel anodic reaction: Ni(s) Ni2  2e
Aluminum anodic reaction: Al(s) Al3  3e
That is, the corrosion of metal M results in the oxidation of metal M to an ion with a valence charge of nand
the release of n electrons.
Some metals such as silver are univalent, while other metals such as iron, titanium, and uranium are
multivalent and possess positive charges as high as is general and applies to all corrosion reactions.
Cathodic Process
When hydrogen ions are reduced to their atomic form they often combine, as shown earlier, to produce
hydrogen gas through reaction with electrons at a cathodic surface. This reduction of hydrogen ions at
(OH ) ions and make the solution less acidic or more alkaline or basic at the corroding interface.
In neutral waters the anodic corrosion of some metals like aluminum, zinc, or magnesium develops enough
energy to split water directly.
Water splitting cathodic reaction: 2H2O(l) 2e- H2 2OH-
The change in the concentration of hydrogen ions or increase in hydroxyl ions can be shown by the use of
pH indicators, which change color and thus can serve to demonstrate and locate the existence of surfaces on
which the cathodic reactions in corrosion are taking place. There are several other cathodic reactions
encountered during the corrosion of metals. These are listed below:
Oxygen reduction:
(acid solutions) O2 4H4e 2H2O
(neutral or basic solutions) O2 2H2O 4e 4OH
Hydrogen evolution: 2H2e H2(g)
Metal ion reduction: Fe3e Fe2
Metal deposition: Cu22e Cu(s)
Hydrogen ion reduction, or hydrogen evolution, has already been discussed. This is the cathodic reaction
that occurs during corrosion in acids.
Inhibition of corrosion
Introduction:
Inhibition means to remove corrosion. There are a number of ways to remove corrosion.
Some methods are:
 Impermeable layer.
 Galvanizing
 Cathodic production.
 Sacrification method.

of 50
Impermeable layer:
Coating of impermeable layer on the material prevents it from corrosion. Sometimes this measure
fails because of porous paint.
Galvanizing:
Coating of zinc on a material can also prevent it from being corroded. This method is called
Galvanizing
Cathodic production:
This method was Invented in 1824 by the UK scientist Sir Humphrey Davy (1778-1829) to protect the
copper-cladded wooden ships from seawater corrosion. It is an electrical method of preventing corrosion on metallic
structures which are in electrolytes such as soil or water
Sacrification method:
This method is basically used in ships. Here Mg is used as a sacrificing agent. We attach Mg to
Fe.First Mg oxidizes, and then Fe will oxidize after a long time.
Question # 10
Write three definitions of oxidation and reduction reactions with examples ; also explain second law of electrolysis.
Definitions of oxidation and reduction reactions:
Reduction reaction:
(1) Loss of O2
(2) Increase of H+
ion.
(3) Increase of e-
.
(4) Increase of oxidation no.
Examples: Zn  Zn2+
+ 2 e-
.
Zn is the reducing agent, and Zn2+
the oxidizing agent.
Oxidation reaction:
(1) Gain of O2.
(2) Decrease of H+
ion.
(3) Decrease of e-
.
(4) Decrease of oxidation no.
Examples: Cu2+
+ 2 e  Cu
Cu2+
is the oxidizing agent and Cu the reducing agent.

of 50
Faraday’s Second Law of Electrolysis:
“It states that masses of different substances deposited or liberated when same quantity of current
is passed through different electrolytes connected in series are proportional to their chemical equivalent masses”
Explanation:
Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass
some quantity of electricity through them for the same time. Now Ag Cu and Al metals collect at the cathode. Their
masses are directly proportional to their equivalent masses.
According to Faraday,if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes,
we get which are the equivalent masses of
Ag, Cu and Al respectively.
Ag+ 1e-
→ Ag(S)
Cu2+
+ 2e-
Cu (S)
Al+3
+3e-
Al(S)
Charge on 1 electron=96500 coulomb= 1 faraday

of 50
Question # 11
State the methods of prevention of corrosion by sacrificial anode method with diagram.
Corrosion can be prevented by application of electrochemistry principles. This basically falls into two distinct
areas,sacrificial anodes and cathodic protection by impressed currents.
Sacrificial Anodes.
In this preventative technique, corrosion is allowed to occur on a piece of metal that is extraneous to the
structure, for example, a zinc attached to a steelboat hull. The zinc corrodes in place of the steel hull. The principles
behind this process were discussed previously and will not be repeated except to show the relevant Evans diagram.
Cathodic Reaction 1
Anodic Reaction 2
icorr 1+2
Ecorr 1+2
E
(V)
log Current Density
A/cm
2
Cathode Reaction 3
Anode Reaction 3
Total Cathode 1+3
In this case the anode#3 was protected from corrosion by anode reaction #2. One initial principle is that the
sacrificial material must have a potential lower than the material it is trying to protect.
Simple examples of the application of this protection technique include:-
Galvanized bolts, automobile steel and mail boxes, zincs placed outboard engines and steelboat hulls, aluminum
blocks on oil rigs, etc.
Typical coatings which work on this principle are:-
Zinc on steel, aluminum on steel, cadmium on steel.
It should be remembered that cadmium is only slightly below steel in the galvanic series. As such it does not
have much "Throwing power" which is the ability to protect over large distances.

of 50
Question # 12
Consider the reaction
2Ag+
+ Cd 2Ag + Cd+2
Standard electrode potentials are Eo
Ag
+
/Ag =0.80 volts and E0
Cd
+2
/Cd = - 0.40 volts.
(1) What is the standard electrode potential (Eo
) for this reaction.
Eo
= E - E
= EREDUCTION - EOXIDATION
= 0.80 – (-0.4)
EO
= 1.20 volts.
(2) Which electrode is negative electrode?
 The electrode which has low redox potential value is –ve electrode.
 In this case Cd has low value of redox potential i.e -0.40 volts. So , Cd is –ve.

Question # 13:
What is redox potential? How will you obtain redox potential of a complete cell?
Redox potentials are used to infer the direction and free energy cost of reactions involving electron transfer,one of
the most important types of biochemical reactions. Such reduction-oxidation reactions are characterized by a free
energy change.
If we are given two half cell reactions with there respective redox potential values then e.g
 CO+2
(aqu.) + 2e-
CO -0.28 volts.
 Br2(aqu.) + 2e-
2Br-
(aqu.) 1.07 volts.
- - - -
1.35 volts.
Then, redox potential values add up to 1.35 volts.this is how we obtain redox potential value of a complete cell.
Question # 14
Consider equations with reduction potentials and answer the followings.
E/V
O2 (g) +4H+
(aqu.) + 4e-
2H2O (l) +1.23
Cl2 (g) + 2e-
2Cl-
(aq) +1.36
F2 (g) +2e-
2F-
(aqu.) +2.87

of 50
1-State and explain what occurs when Flourine is bubbled into water.
When fluoride is bubbled in into the water then after these standard reactions,
By applying anticlockwise rule.
O2 (g) +4H+
(aqu.) + 4e-
2H2O (l) +1.23
2F2 (g) +4e-
4F-
(aqu.) +2.87
- - - -
O2 + 4H+
(aqu.) + 4F(aqu.)
-
2H2O + 2F2 -1.64 volts.
So,
 O2 is formed when F2 is bubbled in water H2O.
 Water is oxidized and fluorine is reduced.
(1) Write an equation for the expected reaction of chlorine with water using the above data.
O2 (g) +4H+
(aqu.) + 4e-
2H2O (l) +1.23
2Cl2 (g) + 4e-
4Cl-
(aq) +1.36
- - - -
O2 + 4H+
(aqu.) + 4Cl-
(aqu.) 2H2O + 2F2 -0.13 volts.
So,
 O2 is formed.
 Water is oxidized and chlorine is reduced.
(2) In fact the reaction between chlorine and water is usually represented by the equation.
Cl2(g) + H2O(l) H+
(aqu.) + Cl-
(aqu.) + HClO(aqu.)
Suggest why this reaction occurs rather than that in (2).
 This is because of little difference between redox potential values.
 Cl2 mix with water to form bleaching agent.
Question # 15
A direct current of 0.01 amp flows for 4 hours through three cells in series. They contain aqueous AgNO3 , CuSO4
and Au(NO3)3. Calculate the mass of metal deposited in each. (atomic masses. Ag=108, Cu=63.5, Au=197
respectively).
q=It
= (0.01)(4 x 60 x 60)
q= 144 coulomb
96500 coulomb = 1 Faraday
144 coulomb = 1/96500 x 144
F= 1.49 x 10-3
Farad.

of 50
Step #3: Taking the calculated values of Charge and faraday,solve the Question further for given equations
Let X be the required amount of AgNO3,CuSO4 and AuNO3.
Ag+ 1e- → Ag(S)
108 g = 1 F
1.49 x 10-3 = X gm
X= 0.161 gm.
Cu2+ + 2e- → Cu(S)
63.5/2= 1 F
OR
2 F= 63.5 gm
1.49 x 10-3 = X
2X = 63.5 x (1.49 x 10-3)/2
X= 0.0473 gm.
Al+3 +3e- 𝐴𝑙(𝑠)
197 gm = 3 F
3 F= 197 gm
1.49 x 10-3 F= X
3X= 197 x (1.49 x 10-3)/3
X= 0.0983 gm
Question # 16
A metal of Ar = 27 is deposited by electrolysis when 0.15amp flows for 3.5 hrs,0.176gm of metal is deposited.
What is the charge on the cation of these metals?
q=It
=0.15 (3.5x60x60)
q= 1890 coulomb
96500 coulomb = 1 faraday
1890 coulomb = x
x= 1890/96500
x= 0.0196F
Step#3:
0.176gm= 0.0196F
27gm = x faraday
x = 0.0196x27/0.176
x = 3
charge is +3.

of 50
1. IMPORTANT KEY POINTS AND QUESTIONS.
Questions:
(1) Pak has crisis of electricity. U.P.S is used. State two laws of cell used in U.P.S.?
Answer:
(1) Faraday's first law of electrolysis:
Statement: “The amount of substance deposited or liberated at an electrode during electrolysis is directly
proportional to the quantity of current passing through the electrolyte”
Explanation:
It means that the mass of a substance produced by electrolysis is proportional to the quantity of electricity
used.
In order to produce more substance electrolytically we must use more electricity.
m α Q
m α It
m=zIt
where “Z” is known as electrochemical equivalent.
Definition of Electrochemical Equivalent “z”:
“If the mass is 1 kg and current of 1 ampere is passed through electrolyte in 1 second then the mass
deposited or liberated is called 1 electrochemical equivalent ”
OR
“The mass deposited by 1 coulomb charge is called 1 electrochemical equivalent”
z=m / It or z=m/Q
For example, when a charge of one coulomb is passed through silver nitrate solution, the amount of silver deposited
is 0.001118 g. this is the value of electrochemical equivalent of silver.
Importance offaraday’s First law ofElectrolysis:
With the help of faraday’s First law we are able to calculate:
1) The value of electrochemical equivalents of different substances.
2) The masses of substances produced by passing a known quantity of electricity through their solution.
(2) Faraday’s Second Lawof Electrolysis:
“It states that masses of different substances
deposited or liberated when same quantity
of current
is passed through different electrolytes
connected in series are proportional to their
chemical equivalent masses”

of 50
Explanation:
Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass
some quantity of electricity through them for the same time. Now Ag Cu and Al metals collect at the cathode. Their
masses are directly proportional to their equivalent masses.
According to Faraday,if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes,
we get which are the equivalent masses of
Ag, Cu and Al respectively.
Ag+ 1e- Ag(S)
Cu2+
+ 2e- Cu (S)
Al+3
+3e- Al(S)
Charge on 1 electron = 96500 coulomb = 1 faraday
Equivalent Mass:
“The mass deposited or liberated which is equal to 108 parts of silver, 35.5 parts of chlorine and 8 parts of oxygen”
.
Importance ofFaraday’s Second Law:
By the help of 2nd
law we can calculate:
 Equivalent weight of metal.
 Unit of electric charge (coulomb).
 Avogadro’s number.
(2) Rate of corrosion depends on surface exposure?
Answer:please paste ‘inhibition of corrosion’ here.
(3) Define cathodic reaction , corrosion and equation with redox potential values?
Answer:please paste from ‘corrosion to inhibition’ here.
Important points:
Carbon atom has no charge. Charge on carbon will depned upon with what species it is attached to, if it is attached
to more electronegative element than itself, then the charge should be +ve and if it is attached to more electropositive
element then the charge should be -ve on carbon. Although in general carbon forms covalent bonds most of the time
so it does not form ionic species like 4+ (by loss of 4 electrons) or 4- (by gain of 4 electrons). It shares its 4 electron
with 4 others to form covalent bonds.

of 50
(1) If metal is reduced then redox potential value is more.
(2) If metal is oxidized then redox potential value is small.
When there is reduction there will be deposition always on cathode.
When battery is attached then anode = +ve and cathode = -ve and if battery is not attached then anode = -ve and
cathode = +ve.
The electrode which has low redox potential value is –ve electrode.
In cathodic reaction electron consumes whereas in anodic reaction electron releases.
Whenever you have ions always write the state. E.g :-
Cl2(g) + H2O(l) H+(aqu.) + Cl-(aqu.) + HClO(aqu.)
Gold has high redox potential value so it does not corrode.

of 50
5. M.C.Q’s RELATING TO ELECROCHEMISTRY
(1). Faraday’s laws of electrolysis are related to the
a. atomic number of the cation.
b. atomic number of the anion.
c. equivalent weight of the electrolyte
d. speed of the cation
Ans: (c)
(2). A solution containing one mole for litre each of Cu(NO3)2, AgNO3, Hg2(NO3)2 and Mg(NO3)2 is electrolysed
using inert electrodes.
Standard electrode potentials in volts (reduction potentials) are:
Ag+| Ag = 0.80
Hg22+ |Hg = 0.79
Cu2+|Cu = 0.34
Mg2+|Mg = -2.37
With increasing voltage, the sequence of deposition of metals on the cathode will be:
a. Ag,Hg,Cu, Mg
b. Mg,Cu,Hg,Ag
c. Ag,Hg, Cu
d. Cu,Hg, Ag
Ans: ( c )
(3). The electric charge or electrode deposition of one gram equivalent of a substance is:
a. one ampere for one second
b. 96,500 coulombs per second
c. one ampere for one hour
d. charge on one mole of electrons
Ans: (d)
(4). The reaction ½ H2(g) + AgCl(s) = H+(aq) + Cl-
(aq) +Ag(s) occurs in the galvanic cell:
a. Ag|AgCl(s) |KCl(soln) |AgNO3(soln) |Ag
b.Pt|H2(g) |HCl(soln) |AgNO3(soln) |Ag
c. Pt|H2(g) |HCl(soln) |AgCl(s) |Ag
d. Pt|H2(g) |KCl(soln) |AgCl(s) |Ag

of 50
Ans: (c)
In cell (c) reactions occurring are
AgCl(s) + e- -> Ag(s) + Cl-
(aq)
½ H2(g) -> H+
(aq) + e-
adding ½ H2(g) + AgCl(s) = H+(aq) + Cl-(aq) +Ag(s)
(5) Choose the single correct or the single incorrect statement
A. At the anode, oxidation takes place only when used as a battery
B. At the anode, oxidation takes place in a battery and in a electrolysis operation
C. At the cathode, oxidation takes place only when used as a battery
D. At the cathode, oxidation takes place in a battery and in a electrolysis operation
Answer B.
Consider...
It is a convention to call the reduction electrode a cathode in a battery or in a electrolysis operation. The oxidation
reaction occurs at the anode.
(6)All chemical reactions that supply the power to a battery are oxidation reduction reactions. True or false?
Answer True!
Consider...
Only RedOx reactions involve electron transfer. Even concentration cells involve oxidation and reduction of the same
material. Note that Ag+
+ Cl-
AgCl(s) is an ionic reaction, not a redox reaction.
(7)The half-cell using the reaction:
2 H+
(aq, 1.00 F) + 2 e ® H2(g, 1atm)
has a half cell potential of zero because
E. it is so defined,
F. hydrogen is not very reactive,
G. its potential is absolutely zero,
H. it is not a useful electrode.
Answer A
Consider...
No cell potential is ABSOLUTELY zero. H2 is reactive. This is not the reason at all.
(8)The notation to indicate a boundary between two phases in a electrochemical cell is
I. |
J. /
K. ||
L.
Answer A.
Consider...
The vertical bar | is used to indicate boundary between two phases.
Pt | H2 | H+
(1.0 M) represents the hydrogen half cell.
(9)The notation to indicate a salt bridge between two half electrochemical cells is
M. |

of 50
N. /
O. ||
P.
Answer C.
Consider...
Only | and || are used among the four notations. Two vertical bars, ||, represent a salt bridge.
10. Electricity can pass through molten lead(II) bromide because of the presence of
a. free electrons
b. moveable ions
c. moveable atoms
d. lead metal
11. When a dilute salt water is electrolysed, a colorless gas is given off at the anode. The gas is
a. hydrogen
b. steam
c. oxygen
d. chlorine
12. A solution of copper(II) sulphate is electrolysed, using carbon electrodes. The pinkish deposit which forms on one
of the electrodes is
a. copper
b. copper(I) oxide
c. copper(II) oxide
d. copper(III) sulphide
13. A solution of copper(II) sulphate is electrolysed, using copper electrodes. Which of the following would happen?
a. the anode loses weight
b. the cathode loses weight
c. the solution darkens in color
d. the solution lightens in color
14. An electrolyte is always
a. an acid or alkali
b. an aqueous solution
c. a liquid
d. a molten solid
15. Anions are formed by
a. metals gaining electrons
b. metals losing electrons
c. non-metals gaining electrons
d. non-metals losing electrons
16. Which of these anions is never discharged at the positive electrode during electrolysis?
a. NO3-
b. OH-
c. I-
d. O2-

of 50
17. In the electrolytic manufacture of aluminium, what is the anode made of?
a. copper
b. graphite
c. platinum
d. steel
18. In which electrolyte would a carbon cathode increase in mass during electrolysis?
a. aqueous copper(II) sulphate
b. concentrated hydrochloric acid
c. concentrated aqueous sodium chloride
d. dilute sulphuric acid
19. Chlorine is manufactured commercially by the electrolysis of aqueous sodium chloride (brine). Which other
important products are made in the process?
a. hydrochloric acid and hydrogen
b. hydrogen and sodium
c. hydrogen and sodium hydroxide
d. sodium and sodium hydroxide
20. An electric current is passed through aqueous potassium sulphate, K2SO4.
What is formed at the cathode (negative electrode)?
a. hydrogen
b. oxygen
c. potassium
d. sulphur
21. What happens when molten lead(II) chloride is electrolysed?
a. chloride ions gain electrons at the cathode
b chloride ions lose electrons at the anode
c. lead(II) ions lose electrons at the cathode
d. lead(II) ions move towards the anode
22. Which element is liberated at a carbon cathode when aqueous sodium chloride is electrolysed?
a. chlorine
b. hydrogen
c. oxygen
d. sodium
23. Which change always takes place when aqueous copper(II) sulphate is electrolysed?

of 50
a. copper is deposited at the negative electrode
b. oxygen is evolved at the positive electrode
c. sulphate ions move towards the negative electrode
d. the color of the solution fades
24. Which element is liberated at the cathode by the electrolysis of an aqueous solution containing its ions?
a. bromine
b. chlorine
c. hydrogen
d. oxygen
25. Aqueous copper(II) sulphate is electrolysed using copper electrodes. Which observations will be made?
at anode (positive) at cathode (negative)
a anode dissolves pink solid forms
b anode dissolves pink solid forms
c color gas forms color gas forms
d color gas forms pink solid forms
26. Why is cryolite, Na3AlF6, used in the extraction of aluminium from aluminium oxide?
a. to dissolve aluminium oxide
b. to prevent the anodes from burning away
c. to prevent the oxidation of the aluminium
d. to remove impurities from the aluminium oxide
27. When sodium chloride was electrolysed, sodium was produced at the negative electrode. In which form was the
sodium chloride during the electrolysis?
a. concentrated aqueous solution
b. dilute aqueous solution
c. molten
d. solid
28. In which instance is there no change in the concentration of the solution during electrolysis?
a. concentrated sodium chloride solution between carbon electrodes
b. copper(II) sulfate solution between copper electrodes
c. copper(II) sulfate solution between platinum electrodes
d. dilute sodium chloride solution between platinum electrodes

of 50
29. An example of a weak electrolyte is
a. alcohol
b. salt solution
c. sugar solution
d. ammonia solution
30. Electroplating iron with zinc is called galvanising. The reaction at the cathode is shown by the equation
a. Fe (s) ---> Fe2+ (aq) + 2e-
b. Fe2+ (aq) + 2e- ---> Fe (s)
c. Zn (s) ---> Zn2+ (aq) + 2e-
d. Zn2+ (aq) + 2e- ---> Zn (s)
31. The circuit shown below was set up, with brass as the anode.
Which electrode reactions will occur on closing the switch?
Anode reaction Cathode reaction
a. Copper dissolves preferentially. Copper is deposited.
b. Copper dissolves preferentially. Hydrogen is evolved.
c. Zinc dissolves preferentially. Hydrogen is evolved.
d. Zinc and copper both dissolve. Copper is deposited.
32. During the electrolysis of concentrated sodium chloride in a cell, chlorine, hydrogen, and sodium hydroxide are
produced. What is the molar ratio of these products?
Chlorine Hydrogen Sodium hydroxide
a. 1 1 1
b. 2 1 2

of 50
c. 2 1 1
d. 2 2 1
Answers
10. b
11. c
12. a
13. a
14. c
15. c
16. a
17. b
18. a
19. c
20. a (H+ and K+ ions in the electrolyte migrate to the cathode. H+ are preferentially discharged to form hydrogen gas
because it is lower down in the electrochemical series than K+ ions)
21. b (the negative chloride ions will migrate to the anode and become oxidised at the anode to form chlorine gas)
22. b (the ions attracted to the cathode are H+ and Na+ ions. H+ is preferentially discharged to form hydrogen gas)
23. b
24. c
25. b
26. a
27. c
28. b
29. d
30. d
31. c
32. b

Electrochemistry

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Andere mochten auch

Andere mochten auch (6)

Ähnlich wie Electrochemistry

Ähnlich wie Electrochemistry (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Electrochemistry