TOPIC 6: IONIC THEORY AND ELECTROLYSIS | CHEMISTRY FORM 3

IONIC THEORY AND ELECTROLYSIS

Ionic Theory

To account for the phenomena of electrolysis the Ionic Theory was put

forward by Arrhenius in 1880. The theory states that electrolytes are
made up of ions, which are built up in certain patterns called crystal
lattice. When these substances dissolve in water, the structure is
destroyed and the ions are set free to move.

Concentrated mineral acids such as sulphuric acid, hydrochloric acid and nitric acid do not contain ions but they consist of molecules. However, when they are diluted, the molecular structure is destroyed and ions are formed.

Electrolytes and Non-electrolytes

Distinguish electrolytes from non-electrolytes

The main purpose of this chapter is to investigate the effects which
electricity has on a range of substances, and to develop a thorough
explanation of those effects in terms of our present knowledge of atomic
structure.

Before we begin, it is important that we familiarize ourselves with different terms that we are going to use to explain different phenomena. It is crucial that the definitions and meanings of these terms be understood at the outset in order that concepts defined in this chapter are easily and clearly apprehended.

These terms are given hereunder:

Electrolysis: decomposition of a compound in solution or molten state by passing electricity through it.

Conductor: a solid substance that allows electricity to pass through it. All metals are included in this class.

Non-conductor or insulator: a solid substance that does not allow electricity to flow through it. All non-metals fall in this class.

Electrolyte: a substance which, when dissolved or molten, conducts electricity and is decomposed by it.

Non-electrolyte: a compound which cannot conduct electricity, be it in molten or solution state.

Electrode: a graphite or metal pole (rod) or plate through which the electric current enters or leaves the electrolyte.

Cathode: a negative electrode which leads electrons into the electrolyte.

Anode: a positive electrode which leads electrons out of the electrolyte.

Ion: a positively or negatively charged atom or radical (group of atoms).

Cation: a positive ion which moves to the cathode during electrolysis.

Anion: a negative ion which moves to the anode during electrolysis.

Electrolytes and non-electrolytes

Liquids

such as ethanol, paraffin, petrol and methylbenzene do not conduct
electricity. The bonding in these compounds is covalent. These
substances consist of molecules. There are no free electronsor charged
particles to flow through them. Solutions of covalent compounds, for
example sugar solution, do not conduct electricity.

These compounds are non-electrolytes. Non-electolytes exist only in the form of molecules and are incapable of ionization.

Ionic
compounds contain charged particles (ions), but in solid state, the
ions are firmly held in place and they are not free to move. An ionic
solid does not conduct electricity. However, the ions present can become
free to move if the solid is melted or dissolved in water. Then they
can conduct electricity. For example, solid sodium chloride cannot
conduct electricity but when melted or dissolved in water, the ions, Na⁺ and Cl^– are set free. Then these ions are free to move in solution and hence conduct electricity. These compounds are called electrolytes.

Weak and Strong Electrolytes

Categorize weak and strong electrolytes

Weak electrolytes
are compounds that are only partially or slightly ionized in aqueous
solutions. Some substances, for example, ethanoic acid solution ionize
partially.

CH₃COOH_(aq) ⇔CH₃COO^–_(aq) + H⁺_(aq)

Most
of the electrolytes exist in solution in the form of unionized
molecules. For example, in ordinary dilute (2M) ethanoic acid, out of
every 1000 molecules present, only 4 are ionized and 996 are unionized.

A solution of ammonia water is also a weak electrolyte, containing a relatively small proportion of ammonium and hydroxyl ions.

NH₄OH_(aq) ⇔NH₄⁺_(aq) + OH^–_(aq)

Most of the organic acids are weak electrolytes, e.g. tartaric, citric and carbonic acids.

However,
there is no sharp dividing line between weak and strong
electrolytes.Water is also a weak electrolyte. It ionizes only slightly.

H₂O_(l)⇔H⁺_(aq) + OH^–_(aq)

Study shows that for every molecule of water ionized, there are 6 million molecules of water not ionized.Strong electrolytes
are compounds that are completely ionized in aqueous solutions. When
sodium chloride is dissolved in adequate water it ionizes completely
into Na⁺ and Cl^– ions. There are no NaCl solid
particles left unionized. All strong electrolytes (salts, the mineral
acids and caustic alkalis) ionize completely in solutions.

The Mechanisms of Electrolysis

Electrolytic Cells of Different Electrolytes in the Molten and Aqueous States

Set up electrolytic cells of different electrolytes in the molten and aqueous states

The
conductivity of ionic compounds is explained by the fact that ions move
in a particular direction in an electric field. This can be shown in
experiments with coloured salts. For example, copper (II) chromate (VI)
(CuCrO₄) dissolves in water to give a green
solution. This solution is placed in a U-tube. A colourless solution of
dilute hydrochloric acid (HCl) is then layered on top of the salt
solution in each arm. Graphite rods are fitted as shown in figure 13.3.
These rods (electrodes) carry the current into and out of the solution.

After passing the current for a short time, the solution around the cathode becomes blue. Around the anode, the solution becomes yellow. These colours are produced by the movement (migration) of the ions in the salt. The positive copper ions (Cu²⁺) are blue in solution. They are attracted to the cathode (negative electrode). The negative chromate ions (CrO₄^2-)
are yellow in solution. They are attracted to the anode (the positive
electrode). The use of coloured ions in solution has shown the direction
that positive and negative ions move in an electric field. Always
positive ions (cations) move to the cathode and negative ions (anions)
move to the anode.

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Ionic Migrations During Electrolysis and the Preferential Discharge of Ions at the Electrodes

Explain ionic migrations during electrolysis and the preferential discharge of ions at the electrodes

When
a salt such as sodium chloride is dissolved in water, its ions are set
free to move. So the solution can be electrolysed. Since the salts,
alkalis and acids are dissolved in water, most of the solutions are
aqueous. There is then a complication in electrolysis of such substances
in aqueous form. This is because the water used to dissolve them also
do ionize partially (it is a weak electrolyte)

Then
at each electrode, we get more than one ion for discharge, but only one
is supposed to be discharged. Take an example of electrolysis of copper
(II) sulphate solution using platinum electrodes. By ionic theory, the
solution ionizes thus:CuSO_{4 (aq)} → Cu²⁺_(aq) + SO₄^2-_(aq) (strong electrolyte)

During electrolysis, Cu²⁺and H⁺ions move to the cathode while SO₄^2-and OH^– ions move to the anode.

Cathode	Anode
Cu²⁺	SO₄^2-
H⁺	OH^–

In situations like this, the order of discharge of the ions at the electrode will depend on:

the position of the metal ion or radical in the electrochemical series;
the concentration or nature of the ions (or electrolyte) to be discharged; and
the nature of the electrodes used.

The position of ion or radical in the electrochemical series

*Cations*		*Anions*
K⁺	ease of dischargeincreases	SO₄^2-	ease of discharge increases
Ca²⁺		NO₃^–
Na⁺		Cl^–
Mg²⁺		Br^–
Al³⁺		I^–
Zn²⁺		OH^–
Fe²⁺
Pb²⁺
H⁺
Cu²⁺
Ag⁺

The
arrangement of ions above is the same as that of the electrochemical
series. If all other factors are constant, any ion will be discharged
from solution in preference to those above it.

Cathode		Anode
Cu²⁺		SO₄^2-
H⁺		OH^–
Cu²⁺+ 2e^–→ Cu_(s)copper is discharged (loses its charge)		4OH^–→2H₂O_(l)+O_2(g)+4e^–hydroxyl ion is discharged

In other words, we can say that the reaction that occurs at the cathode is reduction (electron gain) and that which occurs at the anode is oxidation (electron loss).

Result

At cathode, copper is deposited

At anode, oxygen is given out (liberated)

The solution at the end of electrolysis is colourless and acidic because in the electrolyte there are left H⁺ and SO₄^2- ions, which remain moving freely in the solution as ionic sulphuric acid.

The concentration or nature of electrolyte

Take an example of electrolysis of sodium chloride using platinum electrodes when in:

aqueous solution;
concentrated aqueous solution; or
fused or molten state.

aqueous solution

NaCl → Na⁺+ Cl^–

H₂O⇔H⁺ + OH^–

During electrolysis:

Cathode	Anode
Na⁺and H⁺	Cl^–and OH^–
H⁺ions are discharged in preference to Na⁺ ions: 2H⁺+ 2e^–→ H_2(g)	OH^–ions are discharged in preference to Cl^–ions:4OH^–→2H₂O_(l)+O_2(g)+4e^–

Result

At cathode, hydrogen is given out

At anode, oxygen is out

The concentration of the solution remains constant since all Na⁺ and Cl^– ions remain undisturbed in the solution. This means that the Na⁺ and Cl^– ions that are left in the solution are equivalent to sodium chloride solution.

<!–[endif]–>Concentrated aqueous solution

NaCl → Na⁺+ Cl^–

H₂O⇔H⁺ + OH^–

Cathode	Anode
Na⁺and H⁺H⁺ ions are discharged in preference to Na⁺ ions since Na⁺ and H⁺ are very far from each other in e. c.s2H⁺+ 2e^–→ H_2(g)	Cl^–and OH^–Cl^–ions are discharged in preference to OH^– ions since Cl^– and OH^– ions are very close to each other in the e. c. s (and because there are more Cl^– ions in the solution) 2Cl^–→ Cl_2(g) + 2e^–

Results

At cathode H_2(g) is given off
At anode Cl_2(g) is given off
The solution becomes progressively more alkaline as the electrolysis goes on because Na⁺ and OH^–ions
remain in solution as caustic soda (sodium hydroxide) solution. This is
the main method used in the manufacture of sodium hydroxide in
industry. Na⁺_(aq)+ OH^–_(aq) → NaOH_(aq)

Fused or molten sodium chloride

NaCl → Na⁺+ Cl^–

During electrolysis:

Cathode	Anode
Na⁺ + e^– → Na_(l)	2Cl^– → Cl_2(g) + 2e^–

Result

At cathode sodium is liberated (deposited)
At anode chlorine is given off

The nature of electrodes (inert vs active electrodes)

<!–[endif]–>If dilute sulphuric acid iselectrolysed using platinum electrodes:

H₂SO₄ → 2H⁺ + SO₄^2-

H₂O⇔H⁺ + OH^–

Cathode	Anode
2H⁺+ 2e^–→ H_2(g)	SO₄^2- and OH^– 4OH^–→2H₂O_(l)+O_2(g)+4e^–

Result

At cathode H_2(g) is given out
At anode O_2(g) is given out
The solution becomes acidic at the end of electrolysis because of the acidic ions (SO₄^2-) left in the finala solution.

Charge flow during electrolysis

The
coulomb is the electrolytic unit of charge. A current of one ampere is
the rate of flow of charge equal to one coulomb per second.The charge is
calculated from the knowledge of the number of seconds for which a
steady current is passed.

Current in circuit	Time taken	Total charge
1 ampere	1 second	1 coulomb
1 ampere	10 seconds	10 coulombs
20 amperes	10 seconds	200 coulombs
A amperes	t seconds	At coulombs

Therefore charge = quantity of electricity (Q) = I×t

Flow of charge required to liberate 1 mole of element during electrolysis

Electrolysis always produces chemical reactions. Consider a reaction (at cathode) in which one mole of silver Ag⁺ ions is discharged and deposited.Ag⁺ + e^–→ Ag_(s)

In this case, 1 mole of electrons (e^–) is required to discharge 1 mole of Ag⁺ ions to produce 1 mole of silver atoms (Ag_(s)).
1 mole of electrons is a large charge and experiments show that it is
equal to 96500 coulombs. Therefore 1 mole of electrons = 96500 coulombs.
This is called the faraday.

The
number of faradays (moles of electrons) required to liberate 1 mole of
an element during electrolysis is deduced from the equation for the
electrode reaction.

Example 1

Element	Electrode reaction	Faradays
Sodium	Na⁺ + e^– → Na_(s)	1
Copper	Cu²⁺ + 2e^– → Cu_(s)	2
Aluminium	Al³⁺ + 3e^– → Al_(s)	3
Chlorine	2Cl^–→ Cl_2(g) + 2e^–	2

That
is, 1 faraday is needed to deposit 1 mole of sodium atoms (23g), 3
faradays to deposit 1 mole of aluminium atoms (27g) and 2 faradays to
liberate 1 mole of chlorine gas (71g).

The mass of an element liberated by 1 coulomb of electricity during electrolysis is called electrochemical equivalent of that element.
The mass of an element deposited or liberated by 1 faraday during electrolysis is called chemical equivalent of that element.

Experiments to Identify the Products of Electrolysis when Different Electrolytes are Used

Perform experiments to identify the products of electrolysis when different electrolytes are used

Activity 1

Perform experiments to identify the products of electrolysis when different electrolytes are used

Experiments to Identify the Products of Electrolysis when Different Electrodes are Used

Perform experiments to identify the products of electrolysis when different electrodes are used

Activity 2

Perform experiments to identify the products of electrolysis when different electrodes are used

Laws of Electrolysis

Experiments to Relate Masses Liberated and Quantity of Electricity Passed

Carry out experiments to relate masses liberated and quantity of electricity passed

The
laws expressing the quantitative results of electrolysis were first
stated by a British chemist called Michael Faraday. The laws assert that
the amount (expressed in moles) of an element liberated during
electrolysis depends on:

the time of passing the steady current;
the magnitude of the steady current passed; and
the charge on the ion of the element.

An Experiment to Verify Faraday’s First Law of Electrolysis

Carry out an experiment to verify Faraday’s First Law of Electrolysis

The
fact that the amount of a substance liberated during electrolysis
depends upon these factors can be proved by conducting experiments. The
product of time (measured in seconds) and the current passed (measured
in amperes) gives a measure of electricity known as the quantity of
electricity.

Quantity of electricity (Q) [coulombs]= current (I) [amperes]×time (t) [seconds]

Q = I × t

Because
of this relationship, factors (1) and (2) may be included in the same
experiment. The experiment to determine the effect of time on the amount
of element deposited or liberated is carried out by passing a steady
current through a solution of the compound of that element for different
lengths of time.

The
following table summarises the specimen results obtained by passing a
steady current (0.21 amps) through a solution of copper (II) sulphate
for 15, 30, 45 and 60 minutes. The last column shows the mass of copper
deposited.

Specimen results for electrolysis of copper (II) sulphate

Current (amps)	Time(s)	Quantity of electricity (coulombs)	Mass of copper deposited (grams)
0.21	15 × 60=900	900 ×0.21=189	0.063
0.21	30 × 60=1800	1800 ×0.21=378	0.129
0.21	45 × 60=2700	2700 ×0.21=576	0.187
0.21	60 × 60=3600	3600 × 0.21=756	0.250

The
relationship between the amount of copper deposited and the quantity of
electricity passed can be assessed by considering the values in the
last two columns in the table. This data may be represented in a graph
illustrated in figure 6.4. The shape of the graph shows a straight line
passing through the origin. From this fact, it is clear that the mass of
copper deposited is directly proportional to the quantity of
electricity passed. This is in accordance to what Faraday formulated in
his First Law.

Graph of mass deposited versus amount of electricity passed

Faraday’s First Law of Electrolysis states that the
mass of a substance liberated at (or dissolved from) an electrode
during electrolysis is directly proportional to the quantity of
electricity passing through the electrolyte.

The
quantity of electricity is measured in coulombs where a coulomb is the
passage of an electric current of one ampere for one second. Let

m be the mass of the substance liberated;
I be the current passed in amperes; and
t be the time in seconds.

We can therefore represent the first law mathematically as:m α I × t or m = Z × I × t where Z
is the proportionality constant referred to as electrochemical
equivalent of the substance liberated. Electrochemical equivalent is the
mass of a substance (element) liberated by 1 coulomb of electricity
during electrolysis.

An Experiment to Verify Faraday’s Second Law of Electrolysis

Carry out an experiment to verify Faraday’s Second Law of Electrolysis

The
third (3) factor mentioned previously as affecting the amount of
substance liberated during electrolysis may also be investigated
experimentally. Because our interest is the effect of the charge on the
ions present in solution, we need to keep the quantity of electricity
fixed whilst varying the types of the ions in solution. This may be
achieved by passing the same quantity of electricity through two cells,
with ions of different charges in each cell.

Apparatus for confirmation of Faraday’s Second Law

The
experiment is conducted using two voltameters. The two voltameters are
connected in series as shown. The first one is a copper voltameter and
the second is a silver voltameter.The copper voltameter has copper electrodes in a solution of copper (II) sulphate. Hence, in this voltameter, copper ions are discharged and deposited at the cathode.

The silver voltammeter has silver electrodes in a solution of silver nitrate. The discharged silver ions are deposited at the cathode.

Before
the experiment is started, each cathode electrode in the voltameters is
cleaned, dried and weighed. The electrodes are then connected to the
circuit, after which a suitable current is passed for a measured period
of time.

After
this, the cathode are removed from the voltameters, cleaned, dried and
reweighed. The increase in mass of the two cathode electrodes represents
the respective amounts of copper and silver deposited at the cathodes.
The quantity of electricity required to deposit one mole of each element
is calculated using Faraday’s Second Law of Electrolysis.

Specimen results

Current flowing	=	0.45A
Duration of current flow	=	25 minutes
Mass of copper deposited	=	0.221g
Mass of silver deposited	=	0.755g

The
results show that the masses of silver and copper deposited are
different. A comparison of the amounts of each of the elements deposited
can be made simple by calculating the number of moles of atoms of each
of the element deposited.

Thus:

Amount of copper deposited = 0.221/63.5mole = 0.0035 mole

Amount of silver deposited = 0.755/107.8mole = 0.0070 mole

It
is seen that twice as many atoms of silver are deposited as atoms of
copper. The difference in amount of each element deposited arises from
the difference in charges on ions of the element concerned.

The
change on the copper ion is twice that on the silver ionand therefore
twice the quantity of electricity will be required to liberate one mole
of copper as for the liberation of one mole of silver.

This
relationship is in accordance to Faraday’s Second Law of Electrolysis,
which describes the relationship between the amount of element deposited
and the charge on the ions of that element. Faraday’s Second Law of
Electrolysis states that when the same quantity of electricity is
passed through solutions of different electrolytes the relative numbers
of moles of the elements deposited are inversely proportional to the
charges on the ions of each of the elements respectively.

In
order to discharge one mole of monovalent ions such as hydrogen ion,
Sodium ion, Silver ion and Chlorine ion,96500 C of electricity are
required. This quantity of electricity has been experimentally
determined and is known as the Faraday constant.

It
represents one mole of electrons, which is the same as the quantity of
electrons required to discharge one mole of Siliverions to give one mole
of silver atom. The validity of Faraday’s Second Law of Electrolysis is
evident from the following observations:

One faraday (IF) discharges one mole of H⁺, Na⁺, Ag⁺, Cl^– and OH^–ions.
Two Faradays (2F) discharges one mole of Cu²⁺, Pb²⁺, Mg²⁺, Ca²⁺, Fe²⁺, etc ions.
Three Faradays (3F) discharge one mole of Al³⁺, Fe³⁺, etc. ions.

Relationship between the Chemical Equivalents of Elements and Quantity of Electricity Passed

Relate the chemical equivalents of elements and quantity of electricity passed

A
steady current of 4 amperes is passed through aqueous copper (II)
sulphate solution for 1800 seconds using platinum electrodes. Calculate:

mass of copper deposited
mass of oxygen liberated

Given:

Atomic weight of copper = 63.5
Atomic weight of oxygen = 16
1 Faraday = 96500 C

Solution

<!–[endif]–>Cathode reaction: Cu²⁺ + 2e^– → Cu_(s)

In this case, 2 Faradays of electricity are required to deposit one mole of copper atom 63.5g.

This means 2 × 96500 C liberates 63.5g of copper.

Quantity of electricity passed = I×t = 4×1800C. So, if 2×96500 C liberates 63.5g, thenI×t = 4×1800Cwill liberate2.4g of copper.

Therefore, mass of copper deposited = 2.4g

Anode reaction: 4OH^–→2H₂O_(l)+ O_2(g)+ 4e^–

The reaction shows that 4 moles of electrons (4 Faradays) are lost during the reaction process.

Therefore, 4 × 96500 C are needed to liberate one mole (32g) of oxygen.

<!–[endif]–>Quantity of electricity flowing = 4 × 1800 C

Therefore, mass of oxygen liberated = 0.6g

Application of Electrolysis

Electrolysis
has several uses in industry. Its main application has been in the
fields of manufacture of chemicals and in the purification of metals for
which other purification methods prove either too difficult or highly
expensive to apply. Some applications of electrolysis are as discussed
below:

The Industrial Purification of Copper by Electrolysis

Outline the industrial purification of copper by electrolysis

Some
metals can be purified by means of electrolysis. This process is used
in industry to purify copper, which must be very pure 99.9% for
electrical wiring. Copper made by roasting the sulphide ore is about
99.5% pure (so it has an impurity level of 0.5%). This level of impurity
cuts down electrical conductivity significantly.

This
is how the electrolytic purification (refining) process is carried
out:The anode is made of a large block of impure copper. The cathode is a
thin sheet of pure copper. The electrolyte is copper (II) sulphate
solution.During the refining process, the copper atoms of the impure
block become ions (the anode dissolves).Cu → Cu²⁺ + 2e^–

The ions from the solution become atoms.

Cu²⁺ + 2e^– → Cu_(s)

They
stick onto the cathode. A layer of pure copper builds up on the
cathode. As electrolysis takes place, the cathode gains mass as copper
is deposited on it. As a result, the cathode gets smaller while the
cathode gets bigger as electrolysis proceeds. Eventually the whole
cathode dissolves.

Purification of copper by electrolysis

Only
pure copper sticks to the cathode. Most impurities fall to the bottom
of the electrolytic cell. They form a solid material (anode sludge or slime)
which contains small quantities of precious metals such as silver, gold
and platinum. The precious metals recovered from the slime are purified
and sold.

An Experiment on Electroplating of Metallic Materials

Carry out an experiment on electroplating of metallic materials

Electroplating
is the coating of a metal with a layer of another metal by means of
electrolysis. Electrolysis can be used to coat a thin layer of a less
reactive metal onto a more reactive metal. The thin layer of less
reactive metal will provide protection from corrosion for the more
reactive metal underneath. It may also make the product more attractive.

The
object to be coated should be made the cathode and the coating material
should be the electrolyte. The most commonly used metals for
electroplating are copper, chromium, silver and tin.

Steel
can be electroplated with chromium or tin. This prevents the steel from
rusting and gives it a shiny, silver finish. This is also the idea
behind chromium-plating articles such as car bumpers, kettles, bath
taps, etc. Chromium does not corrode, it is a hard metal that resists
scratching and wear, and can also be polished to give an attractive
finish.

Nickel
can be electroplated with silver. This will make nickel more
attractive.The diagram below shows how a steel jug is electroplated with
silver. The jug becomes the cathode of an electrolytic cell. The anode
is made of silver. The electrolyte is a solution of a silver compound,
for example silver nitrate.