Easykemistry: May 2025

IONIC THEORY

Ionic theory as proposed by Arrhenius states that when an ionic compound is dissolved in water or melted, some or all its particles break up (dissociate) into free moving charged particles called ions. This dissociation into ions is called ionization.

These free ions move randomly in all direction inside the solution.

But as soon as an electric current is passed through the solution the ions will loss their randomness and become orderly aligning themselves around the opposite pole taking current in and out of the solution (electrodes).

Differences Between Electrolyte s and Non-electrolytes

ELECTROLYTE: This is a compound which allows the passage of electricity through it when in solution or in its molten state and is decomposed in the process.

examples of electrolytes include: dilute Acids and Alkalis and all electrovalent compounds like NaCl.

NON-ELECTROLYTES: These are compounds which do not conduct electricity at all whether in solution or when molten.

The non-electrolytes are mainly covalent compounds and only exist in the form of molecules. Examples includes vegetable oils, organic solvents like alcohols, benzene sugar solution, tetrachloromethane, etc.

Electrolytes are grouped in two

1. WEAK ELECTROLYTES: Weak electrolytes are compounds which ionize partially in solution

That is, there is slight dissociation of the ions in dilute solutions and they contain less ions in solution. Examples include organic acids, aqueous ammonia, etc.

CH₃COOH_(aq)→CH₃COO^—_(aq) + H⁺

2. STRONG ELECTROLYTES: Strong electrolytes are compounds which ionize completely in solutions.

They contain large amount of ions Example are salts, minerals acids, caustic alkalis.

NaCl_(aq) →Na⁺_(aq) + Cl^—_(aq)

Conductor and Non-conductor

CONDUCTORS: These are substances which allow the passage of electricity through them.

Examples include all metals in general. Silver is the best conductor followed by copper and ionic solutions

NON-CONDUCTORS: These are substance that do not conduct or allow electricity to pass through them.

They are also called Insulators. Examples include are wood, paper, air, rubber, plastic.

ELECTROLYSIS:- is the chemical decomposition of compound as a results of passing an electric current through a solution of the compound or its molten form.

Terminologies Used In Electrolysis

1. ELECTRODES: these are conductors in the form of plates rods or wires through which an electric current enters or leaves the electrolytes

2. ANODES: This is the positive electrode through which electrons leaves and current enters the electrolytes. It is also the electrode where oxidation occurs

3. CATHODE: This is the Cathode is a negative electrode through which electrons enter and current leaves the electrolytes. It is also the electrode where reduction takes place

4. Cations:- positively charged ions which migrate to the cathode ( negative electrode) during electrolysis.

5. Anions:_ negatively charged ions which migrate to the anode( positive electrode) during electrolysis.

5. ELECTROLYTIC CELL: This consists of a vessel containing an electrolyte with two electrodes connected to a suitable direct current supply.

During electrolysis of a solution of an ionic compound both cations and anions migrate to the cathode and anode respectively but only one of each ion will be preferentially discharged at the electrodes.

FACTORS AFFECTING THE PREFERENTIAL DISCHARGE OF IONS

1. Position of ions in the electrochemical series.

2. Concentration of ions

3. Nature of the electrodes.

1. POSITION OF IONS IN THE ELECTROCHEMICAL SERIES: The metals that are higher up the series are more electropositive and tend to remain as ions and hence the elements down the series are preferentially discharged. In the same manner, the more electronegative ions will have, the greater tendency to accept electrons.

2. CONCENTRATION: Increase in concentration of a given ion tends to promote its discharge from solution. The influence of concentration, however, is effective only when the two competing ions are closely positioned in the electrochemical series, while it is minimal if the ions are widely separated in their positions.

3. NATURE OF ELECTRODES: Some electrodes do not take part in electrolytic reaction. Hence. They are described as inert electrodes. Platinum and carbon (graphite) usually behave as inert electrode. Some electrode influence the discharge of ions because they have strong affinity for certain ions.

Examples of ELECTROLYSIS OF SPECIFIED ELECTROLYTES

1. ELECTROLYSIS OF DILUTE TETRAOXOSULPHATE (VI) ACID (ACIDIFIED WATER):

The electrolytic cell used is shown above. It is designed for collection of gaseous products at the two electrodes and it is known as Hofman Voltameter. The electrodes used are platinum foil. The ions present in the electrolyte are:

Cations Anions

H₂SO₄ →2H⁺_(g) + SO₄^2—_(aq)

H₂O → H⁺_(aq) + OH^—_(aq)

At the Cathode: H⁺ ion migrates to the cathode and takes up electrodes to form neutral hydrogen atoms.

H⁺_(aq) + e^—→ H_(g)

The hydrogen atoms then combine in pairs to form diatomic hydrogen gas molecule

H⁺_(g) + H⁺_(g)→ H_2(g).

At the anode: SO₄^2— and OH^—ions migrates to the anode. OH^—ions being lower in the electrochemical series is preferentially discharged and lose its electrons to the anode to become a neutral - OH group.

OH^—_(aq) →OH + e^—

The neutral –OH groups combine in pairs to form one molecule of water and one atom of oxygen.

OH + OH →H₂O_(l) + O_(g)

The oxygen atoms then react to form the diatomic oxygen gas molecule.

O_(g) + O_(g)→O_2(g)

Summary of Reactions

Cathodic half reaction – 4H⁺_(aq) + 4e^— →2H_2(g)

Anodic half reaction – 4OH^—_(aq)→2H₂O_(i) + O_2(g) + 4e^—

Over all reaction – 4H⁺_(aq) + 4OH^—_(aq)→2H_2(g) + 2H₂O_(l) + O_2(g)

2. ELECTROLYSIS OF CONCENTRATED SODIUM CHLORIDE SOLUTION (BRINE SOLUTION): The apparatus is the same as for dilute tetraoxosulphate (vi) acids. The cathode may be platinum or carbon but the anode must be carbon to resist attack by chlorine. The ion present are

Cations Anions

NaCl_(aq)→ Na⁺_(aq) Cl^—_(aq)

H₂O → H⁺_(aq) OH^—_(aq)

At the Cathode: Na⁺ and H⁺ ion migrates to the cathode. Although Na⁺ ion is higher concentration, H⁺ ion is preferentially discharged because of its lower position in the electrochemical series. H⁺ takes up electrodes to form neutral hydrogen atoms.

H⁺_(aq) + e^—→ H_(g)

The hydrogen atoms then combine in pairs to form diatomic hydrogen gas molecule

H⁺_(g) + H⁺_(g)→H_2(g).

Thus, hydrogen gas is obtained at the cathode.

At the anode: Cl^—and OH^—ions migrate to the anode. Cl^—ions is preferentially discharged because it is in higher concentration than OH^—ion. The effect of concentration is importance here because Cl^—and OH^—are close to each other in the series.

Cl^—_(aq)→Cl_(g) + e^--

The chlorine atoms then combine in pairs to give chlorine gas molecules.

Cl_(g) + Cl_(g)→Cl₂

Chlorine gas is obtained at the anode.

Summary of Reactions

Cathodic half reaction – 2H⁺_(aq) + 2e^—→ H_2(g)

Anodic half reaction – 2Cl^—_(aq)→ Cl_2(g) + 2e^—

Over all reaction – 2H⁺_(aq) + 2Cl^—_(aq) → H_2(g) + Cl_2(g)

3. ELECTROLYSIS OF COPPER (II) TETRAOXOSULPHATE (VI) SOLUTION USING DIFFERENT ANODES: With carbon or platinum electrode.

Cations Anions

CuSO₄ →Cu²⁺_(g) + SO₄^2—_(aq)

H₂O →H⁺_(aq) + OH^—_(aq)

At the Cathode: Cu²⁺ and H⁺ ion migrates to the cathode. Cu²⁺being lower than H⁺ in the electrochemical series is preferentially discharged to acquired two electrons to be deposited as metallic copper on the cathode.

Cu²⁺_(aq) + 2e^—→ Cu_(s)

OH^—_(aq) → OH + e^—

The neutral –OH groups combine in pairs to form one molecule of water and one atom of oxygen.

OH + OH → H₂O_(l) + O_(g)

The oxygen atoms then react to form the diatomic oxygen gas molecule.

O_(g) + O_(g)→O_2(g)

Summary of Reactions

Cathodic half reaction – 2Cu²⁺_(aq) + 4e^— →2Cu_(s)

Anodic half reaction – 4OH^—_(aq) →2H₂O_(i) + O_2(g) + 4e^—

Over all reaction – 2Cu²⁺_(aq) + 4OH^—_(aq)→2Cu_(s) + 2H₂O_(l) + O_2(g)

USES OF ELECTROLYSIS

1. Extraction of highly electropositive metals such as Na, K, Mg, Al, Zn and highly electronegative non-metals such as F₂, Cl₂, O₂, etc.

2. Purification of less electropositive metals such as copper, mercury, silver, gold, etc.

3. Electroplating; that is, coating the surface of one metal with another metal to achieve improved appearance or prevent corrosion or give thickness to worn parts of machinery or combination of these.

4. Preparation of certain important substances such as sodium hydroxide, hydrogen and chlorine from electrolysis of brine using mercury cathode.

FARADAY’S LAWS OF ELECTROLYSIS

FARADAY’S FIRST LAW OF ELECTROLYSIS: State that the mass (m) of an element liberated at (or dissolved from) an electrode during electrolysis is directly proportional to the quantity of electricity (Q) that passed through it.

Mathematically

M α Q

Q = It

M α It

M = Zit

Where Z is a constant for any given substance, and is known as the electrochemical equivalent of the substance.

M = Mass of substance in gram

Q = Quantity of electricity in coulombs

I = Current in ampere

t = Time in seconds

VERIFICATION OF FARADAY’S FIRST LAW OF ELECTROLYSIS

Method

1. The beaker if filled to two – thirds of its volume with 0.1 mol per dm³ copper (II) tetraoxosulphate (VI) solution.

2. The copper plate (about 5cm by 3cm) are cleaned with emery paper and then weighed

3. The circuit is connected as shown above, and the variable resistor adjusted to maintain a current of 1A for 15 minutes.

4. The cathode is removed, washed with water and ethanol; dried and then reweighed

5. The cathode is replaced and the same current passed for another 15 minutes

6. The process of cleaning, drying and reweighing is repeated. In this way three more readings are obtained.

The results are tabulated as shown in the table below

Current (A)	Time(S)	Quantity of Electricity (c)	Mass of Copper
I₁	t₁	I₁ x t₁	M₁
I₂	t₂	I₂ x t₂	M₂
I₃	t₃	I₃ x t₃	M₃
I₄	t₄	I₄ x t₄	M₄
I₅	t₅	I₅ x t₅	M₅

A graph of mass of copper deposited (g) against the quanitity of electricity (C) is plotted

Conclusion: The mass of copper deposited at the cathode is directly proportional to the quantity of electricity passed.

FARADAY’S SECOND LAW OF ELECTROLYSIS: Faraday’s second law of electrolysis state that when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements discharged are inversely proportional to the charges on the ions of each of the elements respectively.

1 Faraday’s = 96500 coulombs = the minimum quantity of electricity required to liberate one mole of single – charge ions during electrolysis.

Mass = Quantity of Electricity

Molar mass Faraday

M Q

Mm F

VERIFICATION OF FARADAY’S SEOND LAW OF ELECTROLYSIS

Method

1. The two beakers are respectively filled to two – thirds of their volumes with 0.5 mol per dm³ copper (II) tetraoxosulphate (VI) solution and silver trioxonitrate(V) solution.

2. The copper and silver plate (about 5cm by 3cm) are cleaned with emery paper and then weighed

3. The circuit is connected as shown above, and the variable resistor adjusted to maintain a steady current of 0.5A passed through the solution for 20 minutes.

4. The cathode is removed, washed with water and ethanol; dried and then reweighed to obtain the masses of copper and silver deposited

5. The ratio of the number of moles of copper and silver deposited is calculated.

6. The process of cleaning, drying and reweighing is repeated. In this way three more readings are obtained.

Amounts n (Number of moles) = Mass of element deposited

Its relative atomic mass.

Observation: On passing the same quantity of electricity through the solutions, the ratio of the number of moles of copper and silver deposited is 1:2.

This ratio is inversely proportional to the ratio of the charges on the ions, Cu²⁺ and Ag⁺ , or the number of moles of electrons required to liberate 1 mole each of the ions.

Cu²⁺_(aq) + 2e^— Cu_(S) and Ag⁺_(aq) + e^— Ag_(s)

2 moles 63.5g 1 mole 108g

Conclusion: When the same quantity of electricity is passed through solution of different electrolytes the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of each of the elements respectively.

CALCULATION BASED ON THE FIRST LAW OF ELECTROLYSIS

1. In an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited in 30mins is 0.66g. Calculate the error in the ammeter reading.[electrochemical equivalent of copper =0.00033gC^-1]

Solution

M = ZIt

M = 0.66g, Z = 0.00033gC^-1, t = 30mins = 1800 seconds

I = M/Zt

I = 0.66/0.00033 x 1800

I = 0.66/0.594

I = 1.11A

The error in the ammeter reading is 1.11 – 1 = 0.11A

2. Calculated the time in minutes, required to plate a substance of total surface area 300cm², a layer of copper 0.6mm thick, if a constant current of 2A is maintained. Assuming the density of copper is 8.8g/cm³ and one coulombs liberates 0.00033g copper.

Solution

Given that area = 300cm² , thickness = 0.6mm = 0.06cm

Mass = 0.00033g, density = 8.8g/cm³

Density = mass/volume

Mass = density x volume

Mass = 8.8 x 300 x 0.06

Mass = 158.4g

From M = Zit

t = m/ZI

t = 158.4/2 x 0.00033

t = 240000secs

t= 4000mins

3. At what time must a current of 5Amp pass through a solution of zinc sulphate to deposited 1g of zinc. Electrochemical equivalent (e.c.e) = 0.0003387

Solution

Given current = 5Amp

e.c.e = 0.0003387g/c

Mass = 1g

M = ZIt

t = M
ZI

time = 1
0.0003387 x 25

time = 1,180secs

time = 19mins 40 seconds

4. In an electrolysis experiment, a cathode of mass 5g is found to weigh 5.01g, after a current of 5A flows for 50 seconds. What is the electrochemical equivalent for the deposited substance?

Solution

M = 5.01 – 5 = 0.01g, t = 50sec, I = 5A

M = ZIt

Z = m/It

Z = 0.01/50 x 5

Z = 0.01/250

Z = 4 x 10^-5g/c

5. The electrochemical equivalent of silver is 0.0012g/c. if 0.36g of silver is to be deposited by electrolysis on a surface by passing a steady current for 5.0 minutes. Calculate the value of the current.

Solution

Z= 0.0012g/c, m = 0.36g, t = 5mins = 300secs,

M = ZIt

I = m/Zt

I = 0.36/0.0012 x 300

I = 0.36/0.36

I = 1A.

CALCULATION BASED ON THE SECOND LAW OF ELECTROLYSIS

6. A current of 4.5A is passed through a solution of gold salt for 1 hour 45 minutes. Calculate

(i) The mass of gold deposited

(ii) The number of moles of gold deposited

(iii) If the same current is used, find the time taken for 5.5g of gold to be deposited (Au = 197, 1 Faraday = 96500c)

Solution

(i) Au⁺ + e^—→ Au

197g 1F 197g

Mass = Quantity of Electricity
Molar mass Faraday

M Q
Mm F

Mass = ?

Molar mass = 197g

Quantity of electricity = I x t =

I = 4.5A

t = 1 hour 45minutes = 105 minutes = 105 x 60 = 6300 seconds

Quantity of electricity = I x t = 4.5 x 6300 = 28,350C

Faraday = 96500F

Mass = 28,350
197 96500

Mass = 0.29378
197

Mass of Gold deposited = 57.88g

(ii) Number of mole = Mass
Molar mass

Number of mole = 57.88
197

The number of mole of Gold deposited = 0.30mol

(iii) Mass = Quantity of Electricity
Molar mass Faraday

M Q
Mm F

5.5 = Quantity of Electricity
197 96500

0.02792 = Quantity of Electricity

96500

Quantity of Electricity = 2694.28C. This is the quantity of electricity (Q) required for 5.5g of Au to be deposited.

Q = It

t = Q/I

t = 2694
4.5

t = 598.7

The time taken is 9.98 minutes

7. Calculate the current that must be passed into a solution of aluminium salt for 1hr.30minutes in order to deposited 1.5g of Aluminium (Al = 27)

Solution

Al³⁺ + 3e^— →Al_(s)

27g 3F 27g

Mass = Quantity of Electricity
Molar mass Faraday

M Q
Mm F

Mass = 1.5g, Molar mass = 27g/mol

Quantity of electricity = ?

Faraday = 3 x 96500 = 289500

1.5 = Quantity of electricity
27 289500

0.0556 = Quantity of electricity
289500

Quantity of electricity = 16096.2C

Quantity of electricity = I x t

I = Q/t

t = 1 hr.30mins = 90mins = 90 x 60 = 5400seconds

I = 16096.2
5400

The current (I) = 2.98Amperes

8. 0.222g of a divalent metals is deposited when a current of 0.45A is passed thought a solution of its salt for 25 minutes. Calculate the relative atomic mass of the metal. (1 Faraday = 96500 coulombs)

Solution

M²⁺ + 2e^—→M

1mole 2F 1 mole of atoms.

Mass = Quantity of Electricity
Molar mass Faraday

M Q
Mm F

Mass = 0.222g, molar mass = ?

Quantity of electricity = It = 0.45 x 25 x 60 = 675C

Faraday = 2 x 96500 = 193000F

0.222 = 675
Mm 193000

0.222 = 0.003497
Mm

Molar mass = 0.222
0.0003497

The relative atomic mass of metal M is 63.5g/mol

Therefore metal M is copper.

9. A given quantity of electricity was passed through three electrolytic cells connected in series containing solutions of Silver trioxonitrate (V), Copper(II) tetraoxosulphate (VI) and Sodium Chloride respectively. If 10.5g of Copper are deposited in the second electrolytic cell. Calculate

(a) The mass of Silver deposited in the first cell.

(b) The Volume of Chloride liberated in the third cell at 18^oC and 760mmHg pressure. (Ag=108, Cu=63.5, 1Faraday=96500C, molar volume of gases at s.t.p =22.4dm³.)

Solution

In the second cell

Cu²⁺_(aq) + 2e^— →Cu_(s)

63.5g 2F 63.5g

Mass = Quantity of Electricity
Molar mass Faraday

M = Q
Mm F

mass of copper = 10.5g, molar mass of copper 63.5g/mol, Faraday = 2 x 96500 = 193000C, Quantity of electricity = ?

10.5 = Quantity of electricity
63.5 193000

0.16535 = Quantity of electricity
193000

Quantity of electricity = 31913.4 Coulombs

(a) The mass of silver deposited in the first cell

Ag⁺_(aq) + e^— →Ag_(s)

108g 1F 108g

Mass = Quantity of Electricity
Molar mass Faraday

M Q
Mm F

M = 31913.4
108 96500

M = 0.3307
108

Mass = 108 x 0.3307

Mass of silver deposited = 35.7g

(b) Volume of chlorine liberated in the third cell

2Cl^—→Cl₂ + 2e^—
2F

Mass = Quantity of Electricity
Molar mass Faraday

M 31913.4
Mm 2 x 96500

Mole = 31913.4
193000

Mole = 0.1654mol

1 molar volume of gas = 22.4dm³ x mole

Volume of chlorine = 22400cm³ x 0.1654

Volume of chlorine at s.t.p = 3704cm³

Convert the volume at s.t.p to its volume at 18⁰C and 740mmHg

P₁V₁ = P₂V₂
T₁ T₂

Where P₁ = 760mmHg, P₂ = 740mmHg

T₁ = 273 K T₂ = 18 + 273 = 291K

V₁ = 3704cm³ V₂ = ?

V₂ = P₁V₁T₂ 760 x 3704 x 291
P₂T₁ 740 x 273

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Friday, 16 May 2025

ELECTROLYSIS at a glance