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Electrochemistry

Introduction to Redox Reactions

This is an introduction to oxidation-reduction reactions, also known as redox reactions.Redox reactions, or oxidation-reduction reactions, primarily involve the transfer of electrons between two chemical species. The compound that loses an electron is said to be oxidized, the one that gains an electron is said to be reduced.

In this lesson, we will discuss electrochemistry.This is a branch of chemistry that deals with the relationship between electricity (flow of electrons) and chemical reactions.
 

A redox reaction is a reaction in which there is simultaneous transfer of electrons from one substance to another. One substance loses electrons and those electrons are gained by another substance thus two processes take place in a redox reaction.
(i) Gain of electrons
(ii) Loss of electrons
The word redox is derived from two words. 'Red' is from reduction which describes the gain of electrons and 'ox' is from oxidation which describes the loss of electrons.
 

Zinc atom loses two electrons to get oxidized to zinc ion the process is referred to as oxidation. The copper ion gains two electrons to form copper atom. The process is known as reduction thw two processes combined are referred to as Redox.
The oxidizing agent is the species that is being reduced and the reducing agent is the species that is being oxidized.


 

The oxidation number is the number of electron(s) gained or lost by an element.
If an element has not lost or gained an electron , its oxidation number is zero. When an element gains one electron it acquires a negative charge of -1. The oxidation number is-1. If an elements gains two or three electrons then the oxidation number is -2 or -3.
When an element loses one electron, it acquires a positive charge of +1. The oxidation number is +1, if an element loses two or three electrons, then its oxidation number is +2 or +3.
Oxidation numbers help to keep an account of the electrons in a compound and also during the reaction.
 


 

Work out the oxidation number of the named element in each of the following compounds.
a) Sulphur in Potassium sulphate



 

Work out the oxidation number of Nitrogen is an ammonium ion (NH4+)


 

Determine the oxidation number of iron in the iron oxide (Fe2O3). Step I: The sum of oxidation number in Fe2O3 = )
Let the oxidation number of Fe be x.
Oxidation number O = -2
Step II: 2(x) + 3(-2) = 0
2x - 6 = 0
2x = 6
x = +3

In a redox reaction the reduced species is the oxidizing agent while the oxidized species is the reducing agent as illustrated in the redox equation shown below.

Zn (s) is the reducing agent, it gets oxidized to Zn2+(aq)

Cu 2+(aq) is the oxidising agent it gets reduced to Cu (s)
 

Given 2Fe2+ (aq) + H2O2(aq) + 2H+(aq) w 2Fe3+(aq) + 2H2O(l)
Identify (i) Reduced species
(ii) Oxidised species
Solution:Oxidation number of Fe2+ = =2
Oxidation number of (O) in H2O2
= +2+2x = O
2x = -2
X = -1
Oxidation number of Fe3+ = +3
Oxidation number of (O) in H2O
+2 + x = 0
x = -2
Reduced species H2O2 oxidized species Fe2+

Given 2Na (s) + 2H2O(l) 2NaOH(aq) + H2(g)
Identify the reduced species and the oxidized species
Step I :Water undergoes slight ionization to form hydrogen ions i.e.
H2O(l) H+(aq) + OH-(aq)
Step II: Sodium atoms are oxidized by losing electrons to form sodium ions
2Na(s) 2Na+(aq) + 2e oxidation
Step III :The electrons in released by sodium are picked up by H+ ions from the water, reducing them to hydrogen gas.
2H+(aq) + 2e H2(aq) reduction
 

In this lesson, we will discuss displacement reactions as redox reactions.

Displacement reactions involve reactions in which more reactive metals replace less reactive metals from the solution of their salt. This involves the transfer of electrons from the atoms or ions of one element to ions of the other. Therefore, displacement reactions are redox reactions.Click to play the video clip.

The following video clip shows the displacement reaction of iron and copper (II) Sulphate solution.

 

When iron is placed in blue solution of copper (II) sulphate, the blue colour fades to light green. Copper solid is deposited and is observed as brown solid deposit. The reaction is represented as follows.
Fe(s) + CuSO4 (aq) wFeSO4(aq) + Cu(s)
Ionically Fe(s) + Cu2+(aq)w Fe 2+ (aq) + Cu(s)
Iron is oxidized to Iron (II) while copper (II) ions is reduced to copper metal. This implies that the displacement reaction is also a redox reaction.
 

Redox reactions involve reduction reaction and oxidation reaction taking place simultaneously.
From the displacement reaction, we can derive oxidation and reduction reactions.
Fe(s) + Cu2+ (aq) w Fe2+ (aq) + Cu (s
The oxidation reaction is Fe(s) w Fe2+ (aq) + 2e (s)
Oxidation is loss of electrons, iron loses two electrons to form iron (II) ions.
The reduction reaction is
Cu 2+ (aq) + 2e w Cu(s)
Reduction is gain of electrons. Copper (II) ions gain two electrons to form copper atoms.
 

Displacement reactions are also found in halogens. Halogens displace each other from their respective salt solutions according to their reactivity. The following video clip shows displacement reactions in halogens. (Courtesy of You Tube)

Pottasium bromide solution is colourless solution. On bubbling chlorine gas, the solution turned orange in colour. On the other hand, the colourless solution of potassium iodide turned black solid.
Explanation
Chlorine gas displaces bromide ions from solution to form orange solution.

Chlorine gas displaces iodide ions from its solution to form a black solid.

Considering the displacement reactions

and

Chlorine gas oxidizes bromide ions to bromine water and its also reduced to chloride ion. Chlorine gas also oxidizes iodine ions to iodine solid and is reduced to chloride ion.
Chlorine is therefore an oxidizing agent while the other ions (iodide and bromide) are reducing agents.
 

In this lesson we will discuss the electrochemical cell.

The following procedure is used in the setting up of an electrochemical cell. Click to play the video and note how this electrochemical cell is set up.

An excess of sulphate ions is left in the solution, which leads to excess negative charge in beaker B. This excess negative charge is removed either by potassium ions moving out of the salt bridge into the beaker containing Copper (II) sulphate solution or sulphate ions moving into the salt bridge.
 

 

The standards electrode potential (E o;) is the potential between an electrode and a solution of its ions at 250c when all gaseous reactants and products are at a partial pressure or one atmosphere and all other soluble substances in the reaction are at a concentration of one mole per litre each. For uniformity, a hydrogen half cell is chosen. It is arbitrarily assigned an electrode potential of zero (0). All other single electrode potentials are referred to as potentials on the hydrogen scale.
If the electrode reaction when connected to the standard hydrogen electrode involves reduction e.g. consumption of electrons, the electrode is given a positive sign. If it involves oxidation i.e. liberation of electrons, the electrode potential is given a negative sign. Thus the standard electrode potentials are sometimes referred to as standard reduction potentials.

The salt bridge has two main functions;It provides positive and negative ions to replace the ones consumed in the reduction half reaction and to balance the charges of ions from the oxidation half reaction and It completes the circuit by allowing ions to move from one half cell to the other.The following is the elctrochemical cell. (Courtesy of You Tube)

When the switch is closed, the bulb lights indicating that there is a flow of electrons through the wire. The ammeter and voltmeter are used to measure electric current and potential difference respectively. When the circuit is complete the concentration of Zn2+ was in beaker A (half-cell) increases as the zinc strip dissolves. Zinc gives up electrons in the process. This is an oxidation half reaction.
Zn(s) w Zn2+ (aq) + 2e
These electrons then flow in the external circuit. The reaction would stop if the net increase in the positive charge is not removed. The reaction does not stop because the positive charges in beaker A are removed either by zinc ions moving into the salt bridge or nitrate ions moving out of the salt bridge into the beaker with Zn2+ ions. The concentration of copper ions in beaker B decreases because they gain electrons given up by zinc and are converted to solid copper which is deposited on the copper strip.
Cu2+ (aq) + 2e w Cu(s)


 

The electrochemical cell described can be denoted as illustrated below

By convention the half cell of the more reactive metal is placed on the left hand side,
Hydrogen electrode if part of the cell is placed on the left.
A single line / represents a change in state or a phase between the metals and their ions e.g. Zn(s) / Zn2+(aq)
Double line // or double dotted indicates the salt bridge
By conversion, the more electro positive electrode (one that does not ionize easily) is put on the right hand side.

The e.m.f of the cell is given by
E cell = E(right - hand side electrode) - E (left hand side electrode)
Or simply
E cell - E RHS - ELHS
The voltage of the cell is also known as the electro motive force (e.m.f)

The table below shows standard electrode potentials of some half cells.

The following is a worked out example on how to calculate the standard e.m.f of the given electrochemical cell.


 

By convention (i) the half cell of the more reactive metal is placed on the left hand side,
(iii) Hydrogen electrode if part of the cell is place on the left.
- A single lien/represents a change in state or a phase between the metals and their ions e.g. Zn(s)/Zn2+(aq)
- Double line// or double dotted indicates the salt bridge
- By conversion, the more electro positive electrode (one that does not ionize easily) is put on the right hand side.
The e.m.f of the cell is given by
E cell = E(rght ' hand side electrode) ' E (left hand side electrode)
Or simply
E cell ' E RHS - ELHS
The voltage of the cell is also known as the electro motive force (e.m.f)

The following is a worked out example showing how to Calculate the e.m.f of an electrochemical cell between Iron and copper(Fe(s)/Fe2+//Cu2+(aq)/Cu(s) )

The following is a worked out example showing how to Calculate the e.m.f of an electrochemical cell Zn(s)/Zn2+//Cu2+(aq)/Cu(s)

By the end of the lesson, the learners should be able to

:
Explain factors that affect preferential discharge of ions during electrolysis.

In this lesson we will discuss the process of electrolysis.

Electrolysis is the decomposition of a compound in molten or aqueous state by passing an electric current through it.Electrolysis in molten state involves binary electrolyte in which only two ions from molten compound are separated by passing electric current in them. The animation below shows electrolysis of molted lead (II) bromide solution. (Courtesy of You tube)

Molten lead (II) bromide solution contains two ions, Pb2+ ions and bromide (Br-) ions. During electrolysis, the molten compound is decomposed such that lead (II) ions migrates towards the negatively charged electrode (cathode) where it is deposited as grey solid. The equation at the cathode is
Pb2+ + 2e w Pb(s) grey solid
 

Bromide ions on the other hand, will migrate towards the positively charged electrode (anode) where it is evolved or released as red brown fumes of bromine gas. The equation at the anode is as follows.
Br-(aq) w Br + e-
Br- (aq) w Br + e-
2Br-(aq) w Br2 + 2e-
Red brown fumes
Note that the reaction an oxidation reaction.

Electrolysis of aqueous solution is different from that of molten solution because water as a solvent dissolves ionic compounds making the ions free to move in the solution as ions. In the process water dissociates into Hydrogen ions and hydroxide ions. Together with ions of the compounds dissolved the cations from the compound and hydrogen ions from water migrate to the cathode. While anions from the compounds and hydroxide ions from water migrate to the anode.
The ions from dissolved salt and those from dissociated water compete for discharge at the respective electrode, but at each electrode, only one type of ions get discharged. This is called preferential discharge.

Position of the metal in the electrochemical series. The case of discharge of ions during electrolysis depends on their position in the electrochemical series. Metals in this series are placed in an increasing order of their ability to lose electrons. Non metals are placed in an increasing order of their ability to gain electrons.The following illustration shows cations and Anions and their ease of discharge.

The video clip below shows the electrolysis of dilute Sodium chloride .Click to play the video and observe what happens carefully.

Na+ and H+(aq), H+ migrate to the cathode. H+ ions are preferentially discharged to Na+ because it is lower in the electrochemical series. It gains an electron to form H atom.
The hydrogen atoms combine to form H2 (g) molecules.

Cl - ion and OH- ions migrate to the anode. OH- ions are preferentially discharged because it is lower in electro chemical series.
Over all equation 4OH- (aq) w 2H2O(l) + O2(g)+ 4e-
Note that the volume of hydrogen gas produced is twice that of oxygen gas produced. This is shown in the mole ratio in the equations at the anode and cathode.

H+ ion is easily discharged at the cathode compared to Na+ ions while hydroxide ions OH- are preferentially discharged at the anode compared to the chloride ion.The ease of discharge of ions during electrolysis of various solutions are summarized in the table below.


 

We have observed that ions lower in the electrochemical series will be preferentially discharges under normal conditions. However, if the ion that requires more energy to discharge is present in greater concentration than the other one lower in the electro chemical series, it will be discharged preferentially.

Click to play the following video clip to observe Electrolysis of concentrated sodium chloride (brine) solution.(Courtesy of You Tube)


-

Click to play the following video clip to observe Electrolysis of concentrated sodium chloride (brine) solution
- Half fill the beaker with concentrated sodium chloride solution as shown in the illustration below
- Switch on current
- Note the colour of gas collected at the anode and cathode
- Test the gas produced at each electrode with blue and red litmus paper
Observe what happens at the anode and cathode.
 

At the cathode, Na+ ion and H+ ions migrate to the cathode. H+ ions are preferentially discharged to Na+ ions because it is lower in electro chemical series . H+ ions gain an electron to form H atom.
The equation at the cathode is as follows
2H+ (aq) + 2e w H2(g)
A colourless gas is produced which extinguishes a burning splint with a pop sound. The gas is Hydrogen.
Cl - and OH-(aq) ions migrate to the anode. OH- ions are preferentially discharged because it is lower in the elecro chemical series
2Cl- (aq w Cl2(g) ) + 2e-

The gas produced bleaches most litmus paper and is green-yellow in color and is therefore chlorine.

The concentration of ions in solution determines the products at the respective electrodes. A cation or anion whose concentration is high is preferentially discharged if the ions are close in the electrochemical series. During electrolysis of concentrated sodium chloride, H+ and Cl- ions are discharged while Na+ and OH- are left in solution. This makes the resulting solution alkaline.

The choice of electrodes during electrolysis can alter the order of discharge of the ions. To discuss this, consider electrolysis of copper (II) sulphate solution using graphite electrodes and copper electrodes. Click to play the video to observe what happens

 

At the cathode, both Cu2+ and H+ ions migrate towards it. Cu2+ ions are lower in the electrochemical series and are therefore preferentially discharged by gaining electrons to form a brown solid deposit of copper. The equation for the reaction is
Cu2+(aq) + 2e w Cu(s)
At the anode, SO42- and OH- ions migrate there. We would expect OH- ions to be oxidized. But because of the nature (type) of electrodes, i.e. copper, none of the ions are discharged. Instead copper electrode dissolves. The equation for the reaction is
Cu(s) w Cu2+(aq) + 2e-

Copper (II) ions make copper (II) sulphate solution blue. For each ion removed at the cathode, it is replaced when copper dissolves at the anode. Therefore the solution remains blue and the concentration of the solution remains constant. Note that the mass lost at the anode is equal to the mass gained at the cathode.

The following video clip shows the electrolysis of Copper (II) sulphate using copper electrodes. Click to play the video and observe what happens carefully

Cu2+ ions and H+ ions migrate to the cathode. Copper ions are preferentially discharged because they are lower in the electrochemical series. They gain electrons to form a brown solid deposit of copper metal. The equation for the reaction is:

Cu2+ (aq) + 2e w Cu(s)

At the cathode SO42+ ions and OH - ions migrate. However OH - ions

preferentially lose electrons being lower in the electrochemical series.

They combine to form water and Oxygen gas as shown in the following equation

40H-(aq) w 2H20(l) + O2 + 4e-

Copper ions give copper (II) Sulphate solution the blue colour. As these ions are discharged and deposited at the cathode as Copper solid, the colour of the solution fades and will eventually become colourless. Having removed the Copper and Hydroxide ions from solution Hydrogen ions and Sulphate ions remain forming sulphuric acid.

The following video clip shows the electrolysis of Copper (II) sulphate using graphite electrodes. Click to play the video and observe what happens carefully

At the cathode, both CU2+ and H+ ions migrate towards IT. CU2+ ions are lower in the electrochemical series and are therefore preferentially discharged by gaining electrons to form a brown solid deposit of copper. The equation for the reaction is
CU2+(aq) + 2e CU(s)
At the anode, SO42- and OH- ions migrate there. We would expect OH- ions to be oxidized. But because of the nature (type) of electrodes, i.e. copper, none of the ions are discharged. Instead copper electrode dissolves. The equation for the reaction is
CU(s) CU2+(aq) + 2e-

Copper (II) ions, CU2+ ions make copper (II) sulphate solution blue. For each ion removed at the cathode, it is replaced when copper dissolves at the anode. Therefore the solution remains blue and the concentration of the solution remains constant. Note that the ions in mass at the anode is equal to the mass gained at the cathode.

In this lesson we will discuss quantity of electricity.

When electricity is passed through an ionic compound in molten or solution form, it is decomposed to obtain products at the electrodes. Experiments have shown that the mass or volume of elements of the products formed depends on the following factors:
(i) Amount of electricity passed
(ii) Length of time taken to pass the steady current
(iii) Change on the ions of the element making electrolyte.

The coulomb is the practical unit used to measure the quantity of electricity or charge. One coulomb (1C) of charge is transferred if a current of one amphere (1A) flows through a circuit for one second (1s).
That is
1C=1amp/sec
Generally, if Q coulombs of electricity flow along a wire for t seconds, the electric current (I), produced is given by,
I=Q/t
Rearranging the equation, we obtain,
Q=It where, Q is the charge in coulombs
I is the current in ampheres
t is the time in seconds

The following is a worked out example showing how to Calculate the quantity of electricity passed when a current of 0.6 ampere flow for 50minutes.

The following is a worked out example of the electrolysis of molten binary compound. Follow the example carefully.

Michael Faraday investigated the quantities of substances deposited during electrolysis. He found that the mass of the substance deposited or liberated at the electrode depends on the quantity of electricity passed.To find the mass of copper deposited from aqueous copper (II)sulphate by a known quantity of electricity the current was recorded in amperes and the time in seconds.

A sample of the results were recorded in the table below.

Table of results obtained

The information on the table can be presented in a graph as shown in the illustration below.

Graph showing quantity of electricity passed and mass of copper deposited

The graph shows a straight line passing through the origin. This indicates that the mass of copper deposited is directly proportional to the quantity of electricity passed.
 

The relationship between the mass of a substance produced and the quantity of electricity passed is the basis of faradays law of electrolysis.

Faraday's first law of electrolysis states that the amount of substance consumed or produced at one of the electrodes during electrolysis is directly proportional to the amount of electricity that is passed through the electrolytic cell.

The following is a worked out example for the electrolysis of Molten Sodium Chloride. Follow the example carefully.

The following is a worked out example for the electrolysis of copper (II)sulphate solution. Follow the example carefully.

Through many experiments, it has been established that one mole of electrons is equivalent to 96500 C. This quantity of electricity is the Faraday's constant and is denoted as F.

1F = 1e = 96500C

Consider the following equations:
1 mole of Ag+ ion requires 1 mole of electrons or 1F to deposit 1 mole of Ag atoms

Ag(aq) + (aq) + e- w Ag (s)

1 mole of Cu2+ ion requires 2 moles of electrons or 2F to deposit 1 mole of Cu atoms

Cu2+(aq) + 2e w Cu (s)

1 mole of Al3+ ion requires 3 moles of electrons or 3F to deposit 1 mole of Al atoms

Al3+ + 3e w Al(s)

Therefore, 1F=1mole of electrons=96,500C.
 

The following is a worked out example showing the electrolysis of copper (II)sulphate solution. Follow the example carefully.

The following is a worked out example showing the electrolysis of an aqueous solution of a salt of metal M. Carefully follow up the example.


 

The process of electrolysis is also used in the refining of metals. The following video clip shows how Silver metal is refined using electrolysis. (Courtesy of You Tube).

Electrolysis is also used in electroplating metals. The following video clip shows how Copper metal is used to electroplate a silver coin. (Courtesy of You Tube)

The process of electrolysis has many applications.Electrolysis is used in the extraction of metals. Click to play the following video to observe how electrolysis is used to extract a metal.(Courtesy of You Tube).

 

 




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