Back to: O level chemistry notes full Uganda syllabus
Electricity is a form of energy produced as a result of flow of electrons through materials.
Investigation of electrical conductivity through solid materials
Examples of such materials include: copper wire, zinc wire, plastic, graphite, rubber.
Procedure
Connect the copper wire to the batteries through the bulb and the switch as shown below.
Results
The bulb produced light on complete connection with copper and zinc.
Conclusion
Copper and zinc wires conduct electricity and they are called conductors.
A conductor is a substance in solid form which can conduct electricity. Examples include; all metals and graphite (the only non metal that can conduct electricity.
When the above experiment was repeated using rubber and plastics, the bulb did not light indicating that they do not conduct electricity and are referred to as insulators or non conductors.
A non conductor is a substance in solid form that does not conduct electricity. Examples are all non metal except graphite.
Metals conduct electricity because they have delocalized, free or mobile electrons but non metals do not have these delocalized electrons as they are all locked up in bond formation
Investigation of electrical conductivity through liquid substances in solution
Examples: ethanol, urea, hydrochloric acid, copper (II) sulphate, ethanoic acid, water, ammonium hydroxide.
Procedure
- Put the liquid under investigation in an electrolytic cell.
- Dip two rods in the liquid which can either be a metal or carbon(graphite) called electrodes
- Connect the electrodes using a conductor to a bulb via a switch to the source of power (the batteries) as shown below.
- Close the switch.
- Repeat the experiment with hydrochloric acid, ethanol, ethanoic acid, water, urea, ammonium, ammonium hydroxide.
Results
When ethanol and urea were used there was no light produced indicating that they do not conduct electricity, they are there fore called non electrolytes.
When ammonium hydroxide and ethanoic acid were used, the bulb produced a dim light indicating that they weakly conduct and are there fore weak electrolytes.
When copper sulphate solution and hydrochloric acid were used, the bulb produced bright light indicating that they strongly conduct electricity and are strong electrolytes.
ELECTROLYSIS
This is the decomposition of a substance in solution form or molten form (electrolyte) as a result of passage of electric current. The decomposition of the electrolyte takes place at the electrodes.
Definitions of common terms used in electrolysis- Electrolytes
This is a substance in solution form or molten state that can conduct electricity. Electrolytes can be categorized as strong, weak or non electrolytes.
a) Strong electrolyte
This is a substance in solution form or molten state that ionizes completely and can easily conduct electricity. The electrolyte decomposes fully during electrolysis.
Examples include, all mineral acids, alkalis, ionic compounds.
b) Weak electrolytes
This is a substance in solution form which is only slightly ionized (partially ionized). The electrolyte is only partially decomposed by the electric current. Most of the ions of the electrolyte remain as un ionized ions or molecules.
Examples include:
c) Non electrolyte
Is a substance in solution form or aqueous state that doesn‘t conduct electricity. This substance is not decomposed at the electrodes. Example include: all covalent compounds, like pure water, benzene, methyl benzene, petrol and diesel.
- Electrodes
These are rods or plates or poles of conductors at which electrons enter and leave the electrolyte. The electrodes are either cathodes or anodes.
a) Anode
This is the positive electrode at which the electrons leave the electrolyte. Or is the positive electrode at which electrons enter the external circuit. It is normally connected to the positive terminal of the battery.
b) Cathode
This is the negative electrode at which the electrons enter the electrolyte or is the negative electrode at which the electrons leave the external circuit. It is connected to the negative end of the battery.
Simple electrolytic cell
In an electrolytic cell the electrolyte has to be in solution form or molten state as the ions have to be free to move so as to conduct electricity. Salts such as sodium chloride, and lead (II) bromide do not conduct electricity in solid state because the ions are held together by strong electrostatic forces of attraction and are not free and mobile. However, when the salts are melted or dissolved in water, the electrostatic forces are broken down and the ions become free and mobile and so conduct electricity.
IONIC THEORY
Ionic theory was put forward to explain the phenomenon of electrolysis. According to ionic theory, electrolytes are believed to contain electrically charged particles called ions. The ions can be positively charged (cations) and are obtained from metals, hydrogen and ammonium or negatively charged (anions) and are obtained from non
Explanation of electrolysis by ionic theory
When current is not passed through an electrolyte, the ions are wandering randomly in solution.
When current is passed through the solution, the cathode attracts the to itself and the anode attracts to it self the anions.
Experiment to show that Lead (II) bromide only conducts electricity in molten form
Electrolysis of leads (II) bromide
The electrolysis is done using carbon electrodes (graphite) as shown below.
When carbon electrodes are dipped into solid Lead (II) bromide crystals and the circuit completed, the bulb did not light indicating that there was no conduction because the ions responsible for conduction were locked up in the solid crystal. When heated and molten liquid formed, there was conduction as the bulb lit because the ions were free to move and thus conducted.
When two or more ions of similar charges reach the electrode, one is preferentially selected for discharge and the selection depends on the following factors.
The position of the ion in the electrochemical series
The concentration of the electrolyte
The nature of electrodes
a) The position of metal or radical in the reactivity series
The ion that is lower in the electrochemical series is selected for discharge in preference to one above it.
Electrolysis of some electrolytes
- Electrolysis of dilute sulphuric acid (electrolysis of water)
The electrolysis is done by use of platinum or carbon rods for both electrodes. During the electrolysis, 2 volumes of hydrogen is produced at the cathode and one volume of oxygen is formed at the anode. Total acidity of the products remains the same as the products formed are elements of water.
At cathode
Hydrogen ions migrate to the cathode where they are discharged by receiving electrons from the cathode and form atoms. The hydrogen atoms pair up to form hydrogen gas (bubbles of a colorless gas that burns with a pop sound).
Equation at the cathode
At the anode
Both the sulphate ions and hydroxyl ions migrate to the anode. The hydroxyl ions are discharged in preference to the sulphate ions as it is below the sulphate ions in the electrochemical series.
Equation at the anode
- Electrolysis of dilute sodium hydroxide solution using carbon electrodes
Setup of the apparatus
At the cathode
Hydrogen ions are discharged in preference to sodium ions because it is below it in the electrochemical series. Therefore, at the cathode, bubbles of a colorless gas that burns with a pop sound is produced.
Equation
At the anode
At the anode, the hydroxyl ions are discharged giving off bubbles of a colorless gas that relights a glowing splint.
Equation
- Electrolysis of copper (II) Sulphate solution using copper cathode and platinum or graphite anode
Set up of apparatus
At the cathode
Both copper ions and hydrogen ions migrate to the cathode but the copper ions being lower than hydrogen in the electrochemical series, it‘s discharged in preference to hydrogen ions. There fore at the cathode, the copper ions gain electrons and are deposited as brown solids of copper.
Equation
At the anode
Both sulphate and hydroxyl ions migrate to the anode. The hydroxyl ions being lower than the sulphate ions in the electrochemical series are discharged forming water and hydrogen as final products.
Equation
Note
i. The blue color of the copper (II) Sulphate fades away with time as the copper ions which are responsible for the blue color are being discharged and deposited at the cathode as brown copper metal.
ii. The discharge of the hydroxyl ions at the anode disturbs the ionic equilibrium of water; therefore more water ionizes to restore this equilibrium. The excess
hydrogen ions produced combines with the undischarged sulphate ions forming sulphuric acid which makes the solution around the anode acidic.
b) The nature of electrode
Different electrodes for a given electrolyte may cause different products to be formed at the electrodes.
- Electrolysis of Copper(II) sulphate solution using copper electrodes
Set up of the apparatus
At the cathode
Both copper ions and hydrogen ions migrate to the cathode but copper ions are discharged in preference to hydrogen ions since it‘s below it in the electrochemical series.
Equation
At the anode
Both sulphate and hydroxide ions migrate to the anode but non is discharged. Instead, the copper anode goes into solution i.e. it dissolves to form copper ions. This process is called electrode ionization. Such a process in favored in this case as it requires less energy than the discharge of ions.
Note
i. During the experiment, the anode loses mass and the cathode gains mass. The loss in mass at the anode equals the gain in mass at the cathode. The change in mass at either electrode is proportional to the quantity of electricity passed through the electrolyte.
ii. The intensity of the blue color of copper (II) sulphate remains constant as the process is a mere transfer of the copper ions from anode to cathode. i.e. the copper from the anode goes into solution as ions to replace the lost copper ions at the cathode.
iii. Overall concentration of the electrolyte remains constant.
Electrolysis of dilute sodium chloride using a mercury cathode and graphite/platinum anode
At the anode
Both the chloride and hydroxyl ions migrated to the anode but the hydroxyl ions being lower than the chloride ions in the reactivity series, the hydroxyl ions are discharged by losing electrons in preference to chloride ions. Therefore bubbles of colorless gas that relights a glowing splint is observed at the anode.
Equation
At the cathode
Both sodium and hydrogen ions move to the cathode. In this case, sodium ions are discharged despite the fact that hydrogen ions are lower than it in the electrochemical series. This is because the process requires less energy and the sodium atoms produced form an amalgam with mercury.
c) Concentration
Increase in concentration of ions tends to promote its chance of being discharged. E.g. if concentrated Hydrochloric acid is used.
- Electrolysis of concentrated hydrochloric acid using graphite electrodes
The electrolysis is done in the set up of apparatus below.
At the cathode
Hydrogen ions migrate to the cathode where they are discharged forming bubbles of a colorless gas that burns with a pop sound (hydrogen gas).
Equation
At the anode
Both the chloride and the hydroxyl ions migrate to the anode but the hydroxyl ions despite them being lower than chloride ions in the electrochemical series are not discharged. Instead the chloride ions are discharged since they are present in a much higher concentration. Therefore, a green-yellow gas that bleaches damp litmus paper is observed indicating that the gas is chlorine gas.
Equation
- Electrolysis of concentrated sodium chloride solution using graphite electrodes
Set up of the apparatus
At the anode
Both the chloride and hydroxyl ions migrate to the anode but chloride ions being present in a much higher concentration are discharged in preference to the hydroxyl ions. There fore, bubble of a green yellow gas that bleaches damp litmus paper (chlorine) is observed.
Equation
At the cathode
Both sodium and hydrogen ions migrate to the cathode. Hydrogen being lower in the electrochemical series are discharged in preference to sodium ions. Therefore, bubbles of a colorless gas that burns with a pop sound is produced at the cathode.
Equation
LAWS OF ELECTROLYSIS
The laws of electrolysis express the quantitative results of electrolysis and were first stated by faraday. These laws assert that, the amount of substance (in moles) liberated during electrolysis depends on,
i) The time of passing the steady current
ii) The magnitude of the steady current passed
iii) The charge on the ion of the element
Faraday’s first law of electrolysis
It states that, the amount (mass) of substance liberated/deposited at electrodes during electrolysis is directly proportional to the quantity of electricity passed.
The quantity of electricity passed is the product of time (seconds) and current (ampere).
=0.156Faradays.
Faraday’s second law of electrolysis
It states that, the mass or number of moles of a substance deposited at the electrode is inversely proportional to the charge on its ion. i.e. when the same quantity of electricity is passed through solutions 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.
For monovalent elements like sodium, silver and potassium. 1 mole of electron is needed to liberate 1 mole of the element.
For divalent elements such as copper, iron (II), calcium and zinc. 2 moles of electrons are needed to deposit 1 mole of the substance.
Examples
- A current of 12A is passed through a solution of aluminium chloride for 3 minutes. Calculate the quantity of electricity required to deposit one mole of aluminium.
3 moles of electrons deposit 1 mole of aluminium. But 1 Faraday=1 mole of electrons
Therefore According to equation. 1 mole of Al is deposited by 3F i.e.
(3×96500)C= 289500C.
- What is the amount of copper deposited when 96500C of electricity is passed through a solution of copper(II)sulphate
2 moles of electrons deposit 1 mole of copper. Therefore 2 F deposits 1 mole of copper
(2×96500)C deposits 1 mole of copper
1C deposits (1×1/193000) moles of copper
96500C deposits (96500x1x1/193000) moles of copper
=0.5 moles of copper.
General applications of Faraday’s laws of electrolysis
Faraday‘s laws of electrolysis can be used to determine,
- Quantity of electricity passed through an electrolyte for a specific time.
- The mass (moles) or volume of substances liberated/ deposited during electrolysis.
- The relative atomic mass of substances
- The charge on the ions discharged.
Examples of calculations
Determination of quantity of electricity passed and mass /volume of substances liberated during electrolysis
- A current of 0.5 A flows for 1 hour, 40 minutes and 20 seconds through copper(II) sulphate solution. Determine
i) The quantity of electricity passed
ii) The mass of copper deposited
iii) The volume of oxygen liberated
(1 mole of gas occupies 24000cm3 r.t.p, Cu=64, O=16, 1F=96500C)
2 moles of electrons deposit 1 mole of copper. Therefore 2 Faradays deposit 1mole of copper
(2×96500)C deposits 64g of copper
1C deposits (1×64/193000) g of copper
3010C deposits (3010x1x64/193000) g of copper
=0.998g of copper
From the equation,
4 moles of electrons liberate 1 mole of oxygen. Therefore, 4 Faradays liberate 1mole of oxygen
(4×96500)C liberates 24000cm3 of oxygen
1C liberates (1×24000/386000) cm3 of oxygen
3010C liberates (3010x1x24000/386000)cm3 of oxygen
=187.2cm3
- A current of 0.5 A flows through nickel (II) sulphate solution for 6 minutes and 30 seconds. Determine,
i) The mass of nickel deposited
ii) The volume of gas produced at the anode.
(Ni=59, 1 mole of gas occupies 22.4 dm3 at s.t.p., 1 f =96500C)
From the equation,
4 moles of electrons liberate 1 mole of oxygen. Therefore, 4 Faradays liberate 1mole of oxygen
(4×96500)C liberates 22.4dm3 of oxygen
1C liberates (1×22.4/386000) dm3 of oxygen
195C liberates (195x1x195/386000)dm3 of oxygen
=0.0113dm3 of oxygen was evolved.
Exercise
- Calculate the volume of oxygen and hydrogen produced when a current of 2A is passed through dilute sulphuric acid for 20 minutes.(Answer Oxygen=0.139 litres, hydrogen=0.278 litres)
- Calculate the mass of copper deposited on the cathode when current of 3A is passed through copper(II) sulphate solution for 25 minutes.(Answer is 1.49g)
- 3A was passed through a cell for 20minutes containing dilute sulphuric acid. Find the volume of hydrogen at 25˚C and 120kp.(Answer is 385cm3 NOT 417.8cm3)
- The same amount of current was passed through molten sodium chloride and through cryolite containing aluminiun oxide. If 4.60 g of sodium were liberated in one cell, what is the mass of aluminium liberated in the other cell? (Answer=1.8g
Determination of relative atomic mass of substances
The RAM of a substance discharged during electrolysis can be estimated if a steady current passed through its solution, time taken and mass deposited are known.
Example
- A current of 0.25 A flowing for 13 minutes through a nickel (II) Chloride solution deposits 0.059 g of nickel. Determine the RAM of nickel.
Solution
Q=It
Given that I=0.25 A
t= (13×60)s =780s.
Q=(0.25×780) =195C
195C deposits 0.059 g of nickel
1C deposits (0.059/195) g of nickel
But from the equation
0.1=0.2/Charge on X
Charge on X=0.2/0.1 =2.
2 When a metal of relative atomic mass 207 is deposited by electrolysis, a current of 0.0600A flowing for 66 minutes increases the mass of the cathode by 0.254g.
a) Find,
i) The number of moles deposited
ii) The number of moles of electrons that have passed
b) Deduce the number of units of charge on the cation of the metal
Solution
Given that: RAM=207, I=0.060A, t= (66×60) s =3960s, mass of metal deposited=0.254g
i) Moles=Mass deposited /relative atomic mass =0.253/207 =0.00123 moles.
ii) Quantity of electricity =It =(0.0060×3960)C =237.6C
96500C liberates 1 mole of electron
1 C liberates (1/96500) moles of electrons
237.6 liberates (237.6×1/96500) moles of electrons
=0.00246 moles
Therefore, moles of electrons that have passed is 0.00246 moles.
iii) From
Application of electrolysis
a) Electroplating
This is the coating of one metal by metal using electricity. In electroplating, the metal to be coated is made the cathode and the plating metal is made the anode. The electrolyte must contain ions of the metal used for coating e.g. solution of a salt of the metal used as a coating on the object. Electroplating prevents rusting and improves on the appearance of metals. For an effective electroplating; a clean cathode must be used; a steady current must flow; a steady temperature must be maintained and the concentration of the electrolytes must also be steady.
Setup used to electroplate iron nail with zinc.
b) Purification of metals.
Metals such as copper and zinc may be purified by electrolysis.
Example: purification of copper
Set up of apparatus
In this case, the impure metal is made the anode and the pure metal the cathode. The electrolyte is usually an acidified solution of a salt containing the metal to be purified. In this case acidified copper (II) sulphate solution is used. During electrolysis, the impure copper anode dissolves and their fore losses mass as copper (II) ions (cu2+ )are formed.
c) Extraction of reactive metals like sodium, and aluminium.
d) Manufacture of chemicals such as sodium hydroxide and chlorine
Manufacture of sodium hydroxide by electrolysis
Sodium hydroxide is manufactured industrially by the electrolysis of concentrated sodium chloride (Brine) using carbon anode and a layer of mercury as the cathode. The cell is called a mercury cathode cell.
At the anode
Both chloride and hydroxyl ions move to the anode. Due to the high concentration of the chloride ions, it is discharged in preference to hydroxyl ions, therefore forming chlorine gas at the anode.
e) Anodization and dyeing of aluminium
Aluminium objects may be coated with a thin layer of aluminium oxide using electricity, a process known as anodization. During the process, the alluminium object is made the anode and the electrolyte is dilute suphuric acid. The aluminium oxide coating is important in preventing corrosion.
Sample questions on electrolysis
- Explain briefly what is meant by the following terms: conductor, non conductor, electrolyte, non electrolyte, electrolysis, electrodes, strong electrolyte and weak electrolyte, giving an example of each. Explain the ionic theory in relation to electrical conduction. How possible is it that Solid lead(II) bromide does not conduct electricity yet molten lead(II) bromide conducts.
- Describe experiments to demonstrate the products formed in the electrolysis of solutions of (a) sulphuric acid (b) sodium sulphate (c) copper(II) sulphate.
- State faradays laws of electrolysis. Describe carefully what happens when copper(II) sulphate solution is electrolysed between (a) platinum and (b) copper electrodes and when sodium chloride solution is electrolysed between (a) platinum and (b) carbon electrodes. Mention two areas where electrolysis is applied.
- What mass of silver and what volume of oxygen at s.t.p. would be liberated in electrolysis by 9650C of electricity? (Ans. 10.8g ; 560cm3)
- How many moles of electrons are required to produce by electrolysis: (a) 27 grams of aluminium (b) 8 grams of oxygen. (Ans. (a) 3; (b) 1)
- An element X has a relative atomic mass of 88. When a current of 0.5 amp was passed through the fused chloride of X for 32 minutes 10 seconds, 0.44g of X was deposited at the cathode. (1 faraday = 96500C). (a) calculate the number of faradays needed to liberate 1 mole of X. (b) write the formula for the X ion (c) write the formula for the hydroxide of X. (Ans.(a)=2 (b)= (c) X(OH)2).
- Explain how you can prepare by electrolysis a sample of pure copper from impure copper.
- Describe carefully what happens when copper(II)sulphate solution is electrolysed between platinum electrodes and write equations for the reaction at both electrodes. Calculate the mass of each product of electrolysis if the current was stopped after the passage of 0.01 Faraday.
- Describe five applications of electrolysis.