Is the study of the relationship between electrical energy and chemical reactions.
It deals with the chemical action of electricity and the production of electrical energy by chemical reactions.

All redox reactions involve transfer of electrons from the oxidised substance to the reduced substance.
For example, zinc reacts with copper (II) ions as follows:

Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)

Explanation of Electrolysis

Electrolysis is the process of decomposing (breaking down) an ionicsubstance, called an electrolyte, into simpler substances using electricity. The chemical reaction of electrolysis occurs when an electric current is passed through a solution containing ions (ions are charged atoms, they have more or less electrons than protons which causes an imbalance, the overall charge can be negative or positive).

For the electrolyte to conduct electricity, it must be:

  • An ionic compound
  • In molten or aqueous state

The process of Electrolysis: Electrolysis of an ionic substance

The ionic substance is heated until it melts.

The ions are able to move freely after the ionic substance is melted.

The power supply is connected and the electrodes are charged.

The ions move to the oppositely charged electrode (the negative electrons move to the positive electrode, called the anode, and the positive electrons move to the negative electrode, called the cathode).

The electrodes give/take electrons from the ions which makes the ions neutral.

The ions become atoms (because they are neutral) and are deposited onto the electrode. 

 Electrolysis decomposition of compound using electricity
 Electrolyte an ionic compound which conducts electric current in molten or aqueous solution, being decomposed in the process.
 Electrode a rod or plate where electricity enters or leaves electrolyte during electrolysis.
Reactions occur at electrodes.
 Discharge  the removal of electrons from negative ions to form atoms or the gain of electrons of positive ions to become atoms.
  • positive electrode connected to positive terminal of d.c. source.
  • Oxidation occurs here.
  • Anode loses negative charge as electrons flow towards the battery, leaving anode positively charged.
  •  This causes anion to discharge its electrons here to replace lost electrons and also, negative charge are attracted to positive charge.
  •  negative electrode connected to negative terminal of d.c. source.
  • Reduction occurs here.
  • Cathode gains negative charge as electrons flow from the battery towards the cathode, making cathode negatively charged.
  • This causes cation to be attracted and gains electrons to be an atom.
  •  negative ion
  • attracted to anode.
  •  positive ion
  • attracted to cathode.


 Non-electrolytes Weak electrolytes  Strong electrolytes 
 Organic liquids or solutions Weak acids and alkalis Strong acids, alkalis and salt solutions
ethanol C2H5OH
tetrachloromethane CCl4
trichloromethane CHCl3
pure water H2O
sugar solution C12H22O11
molten sulphur S
limewater Ca(OH)2
ammonia solution NH3
aqueous ethanonoic acid CH3COOH
aqueous sulphurous acid H2SO3
aqueous carbonic acid H2CO3
aqueous sulphuric acid H2SO4
aquous nitric acid HNO3
aquous hydrochloric acid HCl
aqueous potassium hydroxide KOH
aqueous sodium hydroxide NAOH
copper(II) sulphate solution CuSO4

Electrolysis of Molten Compounds

  • Molten/aqueous ionic compounds conduct electricity because ions free to move.
  • In solid state, these ions are held in fixed position within the crystal lattice.
  • Hence solid ionic compounds do not conduct electricity.
  • When molten binary compound is electrolysed, metal is formed on cathode while non-metal is formed on anode.


Electrolysis of molten PbBr2

To make molten lead(II) bromide, PbBr2, we strongly heat the solid until it melts. To electrolyse it, pass current through the molten PbBr2.

What happens:

Ions present: Pb2+ and Br

Reaction at Anode

Br- loses electrons at anode to become Br atoms. Br atoms created form bond together to make Br2 gas.

2Br(aq) –> Br2(g)+ 2e

Reaction at Cathode

Pb2+ gains electrons at cathode to become Pb atoms becoming liquid lead (II).
Pb2+(aq) + 2e –> Pb(l)

Overall equation

PbBr2(l) –> Pb(l) + Br2(g)

Electrolysis of Aqueous Solution

Aqueous solutions contain additional H+ and OH ions of water, totaling 4 ions in the solution :

2 from electrolyte, 2 from water.

Only 2 of these are discharged.

Electrolysis of aqueous solutions use the theory of selective discharge.

At cathode

In CONCENTRATED solutions of nickel/lead compound, nickel/lead will be discharged instead of hydrogen ions of water which is less reactive than nickel/lead.
In VERY DILUTE solutions, hydrogen, copper and silver ions are preferable to be discharged, according to its ease to be discharged.
Reactive ions (potassium, sodium, calcium, magnesium, aluminium) will NEVER BE DISCHARGEDin either concentrated or dilute condition. Instead, hydrogen ions from water will be discharged at cathode.
At anode
In CONCENTRATED solutions, iodine/chlorine/bromine ions are preferable to be discharged, although it’s harder to discharged compared to hydroxide ions.
In VERY DILUTE solutions containing iodide/chloride/bromide ions, hydroxide ions of water will be discharged instead of iodide/chloride/bromide, according to ease of discharge.
Sulphate and nitrate are NEVER DISCHARGED in concentrated/dilute solutions.


A. Concentration Solutions

Electrolysis of Concentrated NaCl

What happens:

Ions Present: Na+, H+, OH and Cl

Reaction at Anode

Cl loses electrons at anode to become Cl atoms, although OH is easier to discharge.
Cl atoms created form covalent bond together to make Clgas.
2Cl– (aq) –> Cl(g) + 2e
Reaction at Cathode
H+ gains electrons at cathode to become H atoms becoming hydrogen gas
2H(aq) + 2e –> H2 (l)

Overall Equation

2HCl(l) –> H2(l) + Cl2(g)

Note: any cation and anion left undischarged in solution forms new bonds between them.
E.g. in above, leftovers Na+ and OH combine to form NaOH.

B. Very Dilute Solutions

Electrolysis of Dilute H2SO4

What happens:

Ions Present: 
H+, OH and SO42-

Reaction at Anode

OH- loses electrons at anode to become O2 and H2O.
4OH (aq) –> O2 (g) + 2H2O (l) + 4e
Reaction at Cathode
H+ gains electrons at cathode to become H atoms becoming hydrogen gas.
2H+(aq) + 2e –> H2 (g)
Overall Equation
Both equations must be balanced first.
The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.
(2H+(aq) + 2e –> H2 (g)) x 2 = 4H +(aq) + 4e– –> 2H(g)
Now we can combine the equations, forming:
4H+ (aq) + 4OH(aq) –> 2H(g) + O(g) + 2H2O (l)
4H+ and 4OH+ ions, however, combine to form 4H2O molecules.
Hence: 4H2O (l) –> 2H2 (g) + O(g)+ 2H2O (l)
H2O molecules are formed on both sides.
Therefore, they cancel the coefficients: 2H2O (l) –> 2H(g) + O(g)

Since only water is electrolysed, the sulfuric acid now only becomes concentrated.

Electrolysis using different types of electrodes

Inert Electrodes are electrodes which do not react with electrolyte or products during electrolysis.

Eg. platinum and graphite.

Active Electrodes are electrodes which react with products of electrolysis, affecting the course of electrolysis.

Eg. copper.

A. Electrolysis of CuSO4 Using Inert Electrodes (e.g. carbon)

What happens:

Ions Present
: Cu2+, H+, OH and SO42-

Reaction at Anode

OH loses electrons at anode to become O2 and H2O.
4OH (aq) –> O(g) + 2H2O (l) +4e
Reaction at Cathode
Cu2+ gains electrons at cathode to become Cu atoms becoming liquid copper.
Hydrogen ions are not discharged because copper is easier to discharge.
Cu2+ (aq) + 2e –> Cu (s)
Overall Equation
Both equations must be balanced first.
The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.
(Cu2+ (aq) + 2e –> Cu (s)) x 2 = 2Cu2+ (aq) + 4e –> 2Cu (s)
Now we can combine the equations, forming:
2Cu(OH)(aq) –> 2Cu (s) + O(g) + 2H2O (l)
Since copper ions in solution are used up, the blue colour fades.
Hydrogen and sulphate ions left forms sulphuric acid.
B. Electrolysis of CuSO4 Using Active Electrodes (e.g. copper)

Ions Present
: Cu2+, H+, OH and SO42-

Reaction at Anode
Both SO42- and OH gets attracted here but not discharged. Instead, the copper anode discharged by losing electrons to form Cu2+. So, the electrode size decreases.
Cu (s) –> Cu2+ (aq) + 2e
Reaction at Cathode
Cu2+ produced from anode gains electrons at cathode to become Cu atoms becoming copper. Hence, the copper is deposited here and the electrode grows.
Cu2+ (aq) + 2e –> Cu (s)
Overall Change
There is no change in solution contents as for every lost of Cu2+ ions at cathode is replaced by Cu2+ions released by dissolving anode.
Only the cathode increases size by gaining copper and anode decreases size by losing copper.
We can use this method to create pure copper on cathode by using pure copper on cathode and impure copper on anode.

Impurities of anode falls under it.


Electroplating is coating an object with thin layer of metal by electrolysis. This makes the object protected and more attractive.

Object to be plated is made to be cathode and the plating metal is made as anode.

The electrolyte MUST contain plating metal cation.

Plating Iron object with Nickel

Reaction at Anode

Ni2+ discharged from anode into solution. So, the electrode size decreases.
Ni (s) –> Ni2+ (aq) + 2e
Reaction at Cathode
Ni2+ produced from anode gains electrons at cathode to become Ni atoms becoming nickel. Hence, the nickel is deposited here and the electrode grows.
Ni2+ (aq) + 2e– –> Ni (s)
Overall Change
There is no change in solution contents while iron object receives nickel deposit.

Uses of Electroplating.


Creation of Electric Cells by Electrolysis

A Simple cell or an Electric cell is a device that converts chemical energy into electrical energy, and it consists of 2 electrodes made of 2 metals of different reactivity.

In a simple cell, the MORE REACTIVE metal/electrode is ALWAYS designated the NEGATIVE electrode

The anode (negative electrode) is made of more reactive metal. This is because they have more tendency of losing electrons.

The cathode (positive electrode) is made of less reactive metal.

The further apart the metals in the reactivity series, the higher the voltage created.

The electrons in a simple cell will ALWAYS flow from the NEGATIVE electrode (made of the MORE reactive metal) to the POSITIVE electrode.

A simple electric cell using zinc and copper

Observation: bubbles of hydrogen gas appear at the copper rod.
Explanation: Zinc is more reactive than copper. Thus, it is more electropositive than copper, meaning that zinc loses electrons more easily than copper. As a result, oxidation occurs at the zinc rod (the anode) and zinc metal loses electrons to become zinc ions, that is, Zn (s) – 2e –> Zn2+(aq)

The electrons then flow from the zinc rod to the copper rod through the external circuit. At the copper rod, reduction occurs – the hydrogen ions in solution accept these electrons to form hydrogen gas; 
2H+(aq) + 2e –> H2 (g) 
This explains why bubbles of gas are produced at the copper rod when the two rods are connected by a wire.

The magnitude of the voltage (potential difference) is related to the positions of the two metals in the reactivity series. The further apart the two metals, the larger will be the potential difference (voltage) produced.

Electrolytic cell vs Electrochemical cell
Electrolytic cell Electrochemical cell 
Converts chemical energy into electrical energy Converts electrical energy into chemical energy
Redox reaction is spontaneous and is responsible for the production of electrical energy Redox reaction is not spontaneous and electrical energy has to be supplied to initiate the reaction
The two half-cells are set up in different containers which are connected through a salt bridge or porous partition Both the electrodes are placed in the same container in a solution of molten electrolyte.
The anode is the negative electrode while the cathode is the positive electrode.
The reaction at the anode is oxidation and the reaction at the cathode is reduction.
The anode is the positive electrode while the cathode is the negative electrode.
The reaction at the anode is oxidation and the reaction at the cathode is reduction.
Electrons are supplied by the species getting oxidized.
They move from anode to the cathode in the external circuit.
An external battery supplies the electrons, which enter through the cathode and come out through theanode.

Factors affecting electrolysis


Type of electrode

If the concentration of a particular ion is high, then this can alter the preferential discharge
If dilute hydrochloric acid is electrolysed, hydrogen gas is given off at the cathode and oxygen gas at the anode.
However, when concentrated hydrochloric acid is electrolysed, hydrogen gas is still given off at the cathode, but chlorine gas is given off at the anode.
This is because although the chloride ion is harder to discharge than the hydroxide ion, its high concentration makes it more likely to be discharged.
Type of electrode
Eg. electrolysis of aqueous copper(II) sulphate solution
Use Carbon Electrodes:
Carbon electrodes are inert and so do not affect the electrolysis
At the anode, we have a choice of sulphate or hydroxide ions, and hydroxide ions are easier to discharge so oxygen gas is given at the anode
4OH (aq) + O2 (g) —> O2 (g) + 2H2O (l) + 4e
At the cathode, we have a choice of copper or hydrogen ions. Copper ions are easier to discharge so we will see a pink deposit of copper metal on the carbon electrode
Cu2+ (aq) + 2e —> Cu (s)
Use Copper Electrodes
Copper electrodes are active and so will affect electrolysis
At the anode, the copper electrode dissolves into solution
Cu (s) —> Cu2+ (aq) + 2e
At the cathode, the copper ions are deposited as pink copper metal
Cu2+ (aq) + 2e —> Cu (s)

Terms used in electrolysis

  1. Conductors

These are substances which allow an electric current to pass through them e.g. metals and graphite.

Any substance which allows the electric current to pass through it, is called an electrical conductor. The substance can be solid metallic in nature or can be in fused state as well but the flow of energy in form of electrical charge takes place only when there is less resistance to the flow of this charge. The lesser the resistance the higher the flow electrical charge through a solid body and more the number of free ions in fused state, easier the flow of charge in it.

In a conductor, electric current can flow freely, in an insulator it cannot. Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.
“Conductor” implies that the outer electrons of the atoms are loosely bound and free to move through the material.
Most atoms hold on to their electrons tightly and are insulators. In copper, the valence electrons are essentially free and strongly repel each other.
Any external influence which moves one of them will cause a repulsion of other electrons which propagates,
“domino fashion” through the conductor.Simply stated, most metals are good electrical conductors, most nonmetals are not. Metals are also generally good heat conductors while nonmetals are not.


In a metal or graphite’s, flow of current is movement of electrons

  1. Non conductor /insulators

Are materials which do not allow an electric current to flow through them

Using a burette

12.90                        12.40

15.80                         28.65

21.55                         49.75

6.50                           40.75

33.80                         39.70

Calculations involving moles in solution

What is the morality of solution prepared by dissolving 56gms of potassium hydroxide with water and solution made up of 1 litre?

K             o               H

R.F.M of KOH   = (1×39) + (1×16) +   (1×1)



Morality = 1

Procedures in titration apparatus (burette), pipette 25 or 20cm3, 2 conical flasks, white tile, 2 beakers.

Wash apparatus with water.

Remove as much water as possible from the beakers and collect solution. (100 mls of each solution –acid in a glass beaker and alkali or carbonate solution in plastic.

Rinse burette with a little of the solution, it will be used to measure

Do the same with a pipette.

Fill the burette up to above the 0 mark (about full) if it doesn’t start from zero.

In case you use a bead burette remember to remove the air bubble

Set your burette so the lower meniscus is between (somewhere) 0 and 1

Clamp burette in vertical position and record the initial burette reading.

Note: All burette reading must be recorded to 2 decimal places.

Pipette solution and transfer it into clean conical flasks and add two drops of indicator.

Titrate the first solution by running solution into it from the burette until the colour of the indicator changes.

Record the final burette reading.

Find the volume of solution you have used by subtracting the initial reading from final reading.

Before you start the 2nd experiment make sure you have enough solution to take you through that experience.


Exp 2

You are provided with solution BA, and BA2


BA1 = 0.1M (solution of acid HNX

BA2 = 0.1M sodium hydroxide

You are required to determine the stoichiometry (reaction mole ratio) of the acid. Base


Pipette 20/25cm3 of BA2 into a clean conical flask and add to it 2 drops of pthenothalene indicator.

Titrate the solution with BA, from the Burette.

The end point is when the colour of the indicator just turns colourless

Report titration until you got consist result and record them in the table below


Litre values to be averaged =

= 23.80cm3


  1. Calculate
  2. The number of moles of the acid used.

100cm 3of acid contains 0.1 M

1cm3 of acid contains

1050cm3 of acid contains  x 10.50

= 0.00105 moles

The number of moles of sodium hydroxide used.

100cm 3of NaoH contains 0.1 Moles

1cm3 of NaoH contains

1050cm3 of NaoH contains  x 25.0

= 0.0025 moles

Hence determine the stoichiometry for the reaction between acid and base.

HnX:  NaoH

:: 2


State the basicity of the acid n

n=2        : Basicity = 2



Welcome to FAWE

STEM Elearning

We at FAWE have built this platform to aid learners, trainers and mentors get practical help with content, an interactive platform and tools to power their teaching and learning of STEM subjects, more

How to find your voice as a woman in Africa

© FAWE, Powered by: Yaaka DN.