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From your syllabus:
Core • Define electrolysis as the breakdown of an ionic compound, molten or in aqueous solution, by the passage of electricity

  • Describe the electrode products and the observations made during the electrolysis of:
    • – molten lead(II) bromide            
    • – concentrated hydrochloric acid           
    • – concentrated aqueous sodium chloride
    • – dilute sulfuric acid between inert electrodes (platinum or carbon)
  • State the general principle that metals or hydrogen are formed at the negative electrode (cathode), and that non-metals (other than hydrogen) are formed at the positive electrode (anode)
  • Predict the products of the electrolysis of a specified binary compound in the molten state
  • Describe the electroplating of metals
  • Outline the uses of electroplating
  • Describe the reasons for the use of copper and (steel-cored) aluminium in cables, and why plastics and ceramics are used as insulators


  • Relate the products of electrolysis to the electrolyte and electrodes used, exemplified by the specific examples in the Core together with aqueous copper(II) sulfate using carbon electrodes and using copper electrodes (as used in the refining of copper)
  • Describe electrolysis in terms of the ions present and reactions at the electrodes in the examples given
  • Predict the products of electrolysis of a specified halide in dilute or concentrated aqueous solution
  • Construct ionic half-equations for reactions at the cathode
    • – Describe the transfer of charge during electrolysis to include:the movement of electrons in the metallic conductor
    • – the removal or addition of electrons from the external circuit at the electrodes
    • – the movement of ions in the electrolyte
  • Describe the production of electrical energy from simple cells, i.e. two electrodes in an electrolyte. (This should be linked with the reactivity series in section 10.2 and redox in section 7.4.)
  • Describe, in outline, the manufacture of:
    • – aluminium from pure aluminium oxide in molten cryolite (refer to section 10.3)
    • – chlorine, hydrogen and sodium hydroxide from concentrated aqueous sodium chloride
(Starting materials and essential conditions should be given but not technical details or diagrams.)



Is the branch of chemistry that studies the processes that

    • produce a chemical reaction by using electricity (electrolysis)
    • produce electricity using chemical reactions (Battery, Electrochemical cells, etc)


Is the process by which ionic substances or molecular substances capable of release ions in solution are broken down into simpler substances using electricity. During electrolysis,gases or metals can be obtained at the electrodes.


Substances that can be electrolised:

  • Ionic substances soluble in water, salts, hydroxides, etc.
  • Covalent compounds that release ions in solutions (acids)

Ionic substances

Ionic substances are composed by a metal and a non-metal (NaCl, CuCl2, etc.) or a metal and a group of nonmetals (CuSO4, NaOH, etc.)

Ionic substances contain ions. If we analyze sodium chloride, Sodium ions (Na+) and Chloride ions (Cl-) are present. Hence, sodium chloride could be broken down by the use of electricity. 

We saw that ionic compounds only carry electricity if molten or dissolved in water, so in order for electrolysis to work, the ions need to be free, hence the substance has to be molten or dissolved in water. 

In our example, Sodium ions and Chloride ions will travel through the liquid and Sodium would be deposited in one of the electrodes and Chlorine gas will be bubbling in the opposite electrode. 

What happens during the electrolysis: 


    • Positive ions (cations) (Na+)move to the negative electrode (cathode) to receive electrons. This is called REDUCTION
      Na+ + 1e-  ==> Na      

This is the HALF EQUATION that shows the reduction of Na in the cathode

    • Negative ions (anions) move to the positive electrode (anode) and they lose electrons. This process is called OXIDATION
      Cl-   ==> Cl + 1e-     
Incorrect ↑ (Chlorine does not exist like that)

      2Cl-   ==> Cl2 + 2e-     
Correct (Chlorine is diatomic) 

This is the HALF EQUATION that shows the Oxidation of Cl- to Cl2 in the anode

Predicting which elements will obtained during electrolysis

Depending on the nature of the electrolyte and the ions present in the electrolyte, we will have different elements released next to the electrodes.


If we have a simple ionic substance that has been melted, there are only two ions present, for example: 

    • NaCl will release Na metal and Cl2 gas
    • KBr will release K metal and Br2 gas
    • PbS will release Pb metal and S solid

 In this case of the molten ionic compounds there are only 2 ions present. We can express the reactions that will take place in general as follows







Watch the following Video to understand how it works.

Electrolysis of Lead Bromide

In the particular case of our electrolysis of Lead Bromide, the ions present will be Pb2+ and Br-. The half equations will be as follows

At the CATHODE: Pb2+ + 2e-  ==> Pb
At the ANODE: 2Br-  ==> Br2  +  2e-

Electrolysis of Aluminum Oxide (Alumina)



When an ionic substance dissolves in water, its ions get separated. This process produces also a ionization of the water molecules, so besides the ions that correspond to the ionic substance we will also find protons (H+) and hydroxide ions (OH-) in solution. 

The electrolyte will be formed as follows:

    • Positive ions in this case will be the metal ion (M+) and Hydrogen ion (H+)
    • Negative ions will be the non metal ion from the salt (NM-)and the hydroxide ions (OH-)

Let's analyze what occurs in each electrode individually. 

At the negative electrode (cathode), where the reduction occurs. 


    • The more reactive  an element, the better it will stay as an ion.
    • If the metal is more reactive than hydrogen, the metal will stay in the solution and hydrogen gas will be collected in the cathode. 
    • if the metal is less reactive than hydrogen, Hydrogen will stay as an ion (H+) and the metal will be forming in the electrode

To know which element is more reactive, refer to the table below

ELEMENT          Reactivity     Symbol     How to remember    
Potassium         reactivity           K   Please
Sodium    Na   Send 
Calcium    Ca   Cats
Magnesium    Mg   Monkeys
Aluminum    Al   And
Carbon    C   Cute
Zinc    Zn   Zebras
Iron    Fe   Into
Tin    Sn   Two
Lead    Pb   Lovely
Hydrogen    H   Hot
Copper    Cu   Countries.
Silver    Ag   Sincerely
Gold    Au General
Platinum    Pt Pluto

In general, you can think the processes the same as the following examples:


    • If Copper  (Cu2+ and Hydrogen  (H+are in the solution, copper will be formed because it is less reactive than Hydrogen 
    • If Magnesium (Mg2+ are in the solution with Hydrogen, the metal will stay in the solution and H2 bubbles will be seen. 


At the positive electrode (anode), where the oxidation occurs.

In the solutions to be electrolyzed, you will find only two type of anions present:

      • The anion from the ionic compound
      • The hydroxide ions from the water.

Luckily, in your IGCSE LEVEL EXAMS you will find only two types of anions

      • Halogen ions (halides) Cl, Br, etc. 
      • Sulfate ions (SO42-) or Nitrate ions (NO3-)

We know that Oxidation is the process of losing electrons but also to gain oxygen. If you calculate the oxidation number of sulfur in sulfate ions or nitrogen in nitrate ions, you will understand that they are already in their maximum oxidation number. In other words, Sulfate or Nitrate ions cannot be oxidized further. That's why in the case of having sulfate ions or nitrate ions, oxygen gas will be formed.

You do not need to remember all that.

      • If you just have an halide ion, the halogen will  be formed,
      • if you have a sulfate or nitrate ion, oxygen will be formed.


  Ions present in  solution             Element collected in the anode
  chloride, Cl     and hydroxide, OH–   chlorine, Cl2
  bromide, Br     and hydroxide, OH–   bromine, Br2
  iodide, I–           and hydroxide, OH–   iodine, I2
  sulfate, SO42-   and hydroxide, OH–   oxygen, O2
  nitrate, NO3-    and hydroxide, OH–  oxygen, O2


Electrolysis of diluted Hydrochloric acid

electrolysis hcl before electrolysis-HCl
CATHODE HALF EQUATION=  2 H+  +  2 e-  ==>  H2
ANODE HALF EQUATION=  2 Cl-  ==>  Cl2  +  2 e- 
OVERALL EQUATION=  2 Cl-+2 H==>  H2  +   Cl2


Electrolysis of Brine (Concentrated solution of NaCl)

 The following is an explanation of the electrolysis of brine in the lab

electrolysis brine before brine-lab-after
CATHODE HALF EQUATION=  2 H+  +  2 e-  ==> H2
ANODE HALF EQUATION=  2 Cl-==>  Cl2+  2 e- 
Resulting solution=  Concentrated sodium hydroxide solution. 

INDUSTRIAL Electrolysis of Brine 

Electrolysis of brine is one of the most important processes you need to remember.  

The industrial process has a chamber that is separated in two sections, so the chlorine gas and the hydrogen gas do not mix with each other. 


All products of electrolysis of brine have important applications in the industry


    • killing bacteria in drinking water
    • making bleach
    • making disinfectants
    • making hydrochloric acid
    • making PVC


    • making ammonia
    • making margarine
    • making HCl
    • Rocket fuel

Sodium hydroxide

    • making soap
    • making paper
    • making ceramics 


Electrolysis of copper

Copper is a good conductor of electricity, and is used extensively to make electrical wiring and components. The extraction of copper from copper ore is done by reduction with carbon. However, the copper produced is not pure enough for use as a conductor, so it is purified using electrolysis.

In this process, the positive electrode (the anode) is made of the impure copper which is to be purified. The negative electrode (the cathode) is a bar of pure copper. The two electrodes are placed in a solution of copper(II) sulfate.

The animation shows what happens when electrolysis begins. Copper ions leave the anode and are attracted to the cathode, where they are deposited as copper atoms. The pure copper cathode increases greatly in size, while the anode dwindles away. The impurities left behind at the anode form a sludge beneath it.






© Analia Sanchez