Analía Bellizzi – Chemistry Classes

Ronald Reagan Senior High School

Electrochemistry

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)

Electrolysis

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     

This is the HALF EQUATION that shows the Oxidation of Cl to Cl2 in the anodeCorrect ↑(Chlorine is diatomic)

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.

ELECTROLYSIS - DIFFERENT TYPES

1.MOLTEN IONIC SUBSTANCE

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 BrThe half equations will be as follows

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

Electrolysis of Molten Sodium Chloride

Industrial electrolysis of molten Sodium Chloride requires extreme temperatures.

  • Calcium Chloride is added to the mixture to lower the melting point of Sodium Chloride.
  • Molten sodium metal floats in the mixture because of its low density.
  • The anode is carbon
  • The cathode is iron
  • Chlorine gas is collected separately
NaCl electrolysis

Electrolysis of Aluminum Oxide (Bauxite) to obtain pure aluminum metal 

2.DISSOLVED IONIC SUBSTANCE OR ACID SOLUTION

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. 

FACTS:

    • 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

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.

Sumarizing: 

EXAMPLES YOU NEED TO KNOW

A. Electrolysis of diluted Hydrochloric acid

BEFORE

AFTER

ELECTROLYTE= HCl Solution
CATIONS PRESENT= H+
ANIONS PRESENT= Cl– and OH
CATHODE HALF EQUATION= 2 H+  +  2 e  ==>  H2
ANODE HALF EQUATION= 2 Cl–  ==>  Cl2  +  2 e 
OVERALL EQUATION= 2 Cl+2 H==>  H2  +   Cl2

B. IN THE LAB= Electrolysis of Brine (Concentrated solution of NaCl)

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

BEFORE

AFTER

ELECTROLYTE=NaCL Solution
CATIONS PRESENT=H+ and Na+
ANIONS PRESENT=Cl– and OH
CATHODE HALF EQUATION=2 H+  +  2 e  ==>  H2
ANODE HALF EQUATION=2 Cl–  ==>  Cl2  +  2 e 
OVERALL EQUATION=2 Cl+2 H==>  H2  +   Cl2
RESULTING SOLUTION=Concentrated NaOH 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. 

EXPLANATION

VIDEO

ELECTROLYTE=NaCL Solution
CATIONS PRESENT=H+ and Na+
ANIONS PRESENT=Cl– and OH
CATHODE HALF EQUATION=2 H+  +  2 e  ==>  H2
ANODE HALF EQUATION=2 Cl–  ==>  Cl2  +  2 e 
OVERALL EQUATION=2 Cl+2 H==>  H2  +   Cl2
RESULTING SOLUTION=Concentrated NaOH solution
All products of electrolysis of brine have important applications in the industry

Chlorine

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

Hydrogen

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

Sodium hydroxide

    • making soap
    • making paper
    • making ceramics 

C. Electrolysis of water

BEFORE

AFTER

ELECTROLYTE=H2SO4, Solution
CATIONS PRESENT=H+
ANIONS PRESENT=SO42- and OH
CATHODE HALF EQUATION=2 H+  +  2 e  ==>  H2
ANODE HALF EQUATION=4 OH  ==>  O2  + 2H2O + 4 e 
OVERALL EQUATION=4 OH+4 H==> 2 H2  + O2  + 2H2O

Electrolysis of water using Hofmann apparatus

The hofmann apparatus is used to collect the gases  in their inverted burettes. Gases can be extracted from the top. 

Hoffman apparatus

Electrolysis of Copper (II) sulfate solutions

With carbon electrodes:

Using inert electrodes (carbon or platinum)

At the cathode Copper ions are discharged:
2Cu2+ + 4e-  → 2Cu (s)
The copper coats the electrode.
At the anode Oxygen bubbles off:
4OH-  → 2H2O+ O2  +4e-
So copper and oxygen are produced.
The blue color of the solution fades as the copper ions are discharged, since the copper in the solution is not replenished

With copper electrodes:

At the cathode Again, copper is formed, and coats the electrode:
2Cu2+ + 4e-  → 2Cu (s)
At the anode The anode dissolves, giving copper ions in solution:
2Cu (s) → 2Cu2+ + 4e-
Both electrodes take part in the reactions (they are not inert)

The anode dissolves, giving copper ions. These move to the cathode, to form copper. So copper moves from the anode to the cathode.

The colour of the solution does not fade.

Electrolysis of copper (copper purification)

 

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.

BEFORE

AFTER

ELECTROLYTE=CuSO4 SOLUTION
CATIONS PRESENT=Cu2+ and  H+
ANIONS PRESENT=SO42-   and  OH
CATHODE HALF EQUATION= Cu2+  +  2 e  ==>  Cu
ANODE HALF EQUATION=Cu  ==>  Cu2+ +  2 e- 
RESULTING SOLUTION=Same Molarity of CuSO4 SOLUTION

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.

ELECTROPLATING

In the electroplating process, a pure metal is used as anode and a soluble salt of the same metal is the electrolyte. the substance to be plated is placed as cathode. 

The metal atoms in the anode oxidize (lose electrons) forming ions, which will migrate to the solution. The electrons lost in this process will travel through the wire and battery towards the cathode. 
On the cathode, the reduction takes place (RED-CAT) , the metal ions in the solution will deposit on the cathode, gaining the extra electrons and forming a layer of metal on the surface. 

Note that the anode metal is the same as the metallic ion in the solution. 

the concentration of ions in the solution do not change due to the same amount of electrons gained and lost in the two processes. 

The example in the video above is used to electroplate with copper. 

the same process can be done to electroplate with other metals like silver. 

 

BEFORE

AFTER

ELECTROLYTE=AgNO3 SOLUTION
CATIONS PRESENT=Ag+ and  H+
ANIONS PRESENT=NO3–   and  OH
CATHODE HALF EQUATION= Ag+   e  ==>  Ag
ANODE HALF EQUATION=Ag  ==>  Ag+ +   e- 
RESULTING SOLUTION=Same Molarity of AgNO3  SOLUTION