From the above observations it can be deduced that


iin the ZnCu halfcell the


Zinc rodplate ionizes dissolves faster than the copper rodplate to form Zn2


Ionic equation Zns  Zn2aq  2e


blue copper ions in the Copper IIsulphate solution gains the donated electrons to form brown copper metalatoms


Ionic equation Cu2aq  2e  Cus


This reaction shows imply the Zinc rod has a higher tendency to ionize than copperThe Zinc rod has a higher net accumulation of electrons and is more negative compared to the copper rod which has lower accumulation of electrons The copper rod is therefore relatively more positive with respect to Zinc rod


When the two half cells are connected  electrons therefore flow from the negative Zinc rod through the external wire to be gained by copper ions This means a net accumulationincrease of Zn2 positive ions on the negative half cell and a net decrease in Cu2 positive ions on the positive half cell


The purpose of the salt bridge therefore is


icomplete the circuit


iimaintain balance of charges ions on both half cells


For the negative half cell the NO3 Cl from salt bridge decreaseneutralise the increased positiveZn2 ion


For the positive half cell the Na  K from salt bridge increase the decreased positiveCu2 ion


The voltmeter should theoretically registerread a 110Volts as a measure of the electromotive force emf of the cell Practically the voltage reading is lowered because the connecting wires have some resistance to be overcomed


A combination of two half cells that can generate an electric current from a redox reaction is called a voltaicelectrochemical cell


By convention a voltaicelectrochemical cell is represented


Ms  M2aq  N2 aq  Ns


metal rod of Msolution ofMsolution ofNmetal rod ofN


Note


aiMetal M must be the one higher in the reactivity series


iiIt forms the negative terminal of the cell


iiiIt must diagrammatically be drawn first on the left hand side when illustrating the voltaicelectrochemical cell


biMetal N must be the one lower in the reactivity series


iiIt forms the positive terminal of the cell


iiiIt must diagrammatically be drawn secondafter right hand side when illustrating the voltaicelectrochemical cell


Illustration of the voltaicelectrochemical cell


iZnCu cell


1 Zinc rod ionizes dissolves to form Zn2 ions at the negative terminal


Zns  Zn2aq  2e


2 Copper ions in solution gain the donated electrons to form copper atomsmetal


Cu2aq  2e  Cus


3Overall redox equation


Cu2aq  Zns  Zn2aq  Cus


4cell representation


Zns  1M Zn2aq  1MCu2aq  Cus E0  110 V


5cell diagram


iiMgCu cell


1 Magnesium rod ionizes dissolves to form Mg2 ions at the negative terminal


Mgs  Mg2aq  2e


Cu2aq  Mgs  Mg2aq  Cus


Mgs  1M Mg2aq  1MCu2aq  Cus E0  204 V


iiiFeCu cell


Fes  Fe2aq  2e


Cu2aq  Fes  Fe2aq  Cus


Fes  1M Fe2aq  1MCu2aq  Cus E0  078 V


ivAgCu cell


1 Copper rod ionizes dissolves to form Cu2 ions at the negative terminal


Cus  Cu2aq  2e


2 Silver ions in solution gain the donated electrons to form silver atomsmetal


2Agaq  2e  2Ags


3Overall redox equation


2Agaq  Cus  Cu2aq  2Ags


4cell representation


Cus  1M Cu2aq  1M2Agaq  2Ags E0  046 V


5cell diagram


Standard electrode potential E


The standard electrode potential E is obtained if the hydrogen half cell is used as reference The standard electrode potential E consist of inert platinum electrode immerseddipped in 1M solution of sulphuricVI acid H ions Hydrogen gas is bubbled on the platinum electrodes at


ia temperature of 25oC


iiatmospheric pressure of 101300Pa101300Nm21atm760mmHg76cmHg


iiia concentration of 1M1moledm3 of sulphuricVI acid H ions and 1M1moledm3 of the other half cell


Hydrogen is adsorbed onto the surface of the platinum An equilibriumbalance exist between the adsorbed layer of molecular hydrogen and H ions in solution to form a half cell


H2 g  H aq  e


The half cell representation is


Pt H2 g  H aq 1M


The standard electrode potential E is thus defined as the potential difference for a cell comprising of a particular element in contact with1M solution of its own ions and the standard hydrogen electrode


If the other electrode has a highergreater tendency to lose electrons than the hydrogen electrode the electrode is therefore negative with respect to hydrogen electrode and its electrode potential has negative E values


If the other electrode has a lowerlesser tendency to lose electrons than the hydrogen electrode the electrode is therefore positive with respect to hydrogen electrode and its electrode potential has positive E values


Table showing the standard electrode potential E of some reactions


Note


iE values generally show the possibilityfeasibility of a reduction processoxidizing strength


iiThe elementspecies in the half cell with the highest negative E value easily gain  acquire electrons


It is thus the strongest oxidizing agent and its reduction process is highly possiblefeasible The elementspecies in the half cell with the lowest positive E value easily donate  lose electrons


It is thus the strongest reducing agent and its reduction process is the least possiblefeasible


iiiThe overall redox reaction is possiblefeasible is it has a positive  E


If the overall redox reaction is not possible not feasible forced it has a negative  E


Sample standard electrochemical cell


Calculation examples on E


Calculate the E value of a cell made of


aZn and Cu


From the table above


Cu2 aq  2e  Cus E  034Vhigher E Right Hand Side diagram


Zn2 aq  2e Zns E  077Vlower E Left Hand Side diagram


Zns Zn2 aq  2e E  077reverse lower E to derive cell reaction  representation


Overall E  E higher E lower  E RHS  E LHS Eoxidized E reduced


Substituting


Overall E  034   077  110V


Overall redox equation


Cu2 aq  Zns  Zn2 aq  Cus E  110V


Overall conventional cell representation


Zns  Zn2 aq 1M  1MCu2 aq  Cus E  110V


Overall conventional cell diagram


Zinc and copper reaction has a positive overall E therefore is possiblefeasible and thus Zinc can displacereduce Copper solution


bMg and Cu


Mg2 aq  2e Mgs E  237Vlower E Left Hand Side diagram


Mgs Mg2 aq  2e E  237reverse lower E to derive cell reaction  representation


Overall E  E higher E lower  E RHS  E LHS E oxidized E reduced


Overall E  034   237  271V


Cu2 aq  Mgs  Mg2 aq  Cus E  271V


Mgs  Mg2 aq 1M  1MCu2 aq  Cus E  271V


cAg and Pb


From the table above


2Ag aq  2e  2Ags E  080Vhigher E Right Hand Side diagram


Pb2 aq  2e Pbs E  013Vlower E Left Hand Side diagram


Pbs Pb2 aq  2e E  013reverse lower E to derive cell reaction  representation


Substituting


Overall E  080   013  093V


Overall redox equation


2Ag aq  Pbs  Pb2 aq  2Ags E  093V


Pbs  Pb2 aq 1M  1M2Ag aq  Ags E  093V


dChlorine and Bromine


2e  Cl2g 2Cl aq E  136Vhigher E Right Hand Side diagram


2e  Br2aq 2Br aq E  013Vlower E Left Hand Side diagram


2Br aq  Br2aq  2e E  013reverse lower E to derive cell reaction  representation


Overall E   013   136  123V


2Br aq  Cl2g  2Cl aq  Br2aq E  123V


Cl2g  2Cl aq 1M  1M 2Br aq  Br2aq E  123V


Chlorine displaces bromine from bromine water When chlorine gas is thus bubbled in bromine water the pale green colour fades as displacement takes place and a brown solution containing dissolved bromine liquid is formed This reaction is feasible possible because the overall redox reaction has a positive E value


eStrongest oxidizing agent and the strongest reducing agent


From the table above


2e  F2g 2F aq E  287Vhighest E strongest oxidizing agent


2e  2K aq 2K aq E  292Vlowest E strongest reducing agent


2K aq  2K aq  2e E  292V reverse lower E to derive cell reaction  representation


Substituting


Overall E  287  292  579V


Overall redox equation


F2g  2Ks  2F aq  2K aq E  579V


Overall conventional cell representation


2Ks  2K aq1M  1M 2F aq  F2g E  579V


The redox reactions in an electrochemicalvoltaic is commercially applied to make the


aDry primaryLaclanche cell


bWet secondary accumulators


aDryprimaryLaclanche cell


Examine a used dry cell


Note the positive and the negative terminal of the cell Carefully using a knife cut a cross section from one terminal to the other


The dry cell consist of a Zinc can containing a graphite rod at the centre surrounded by a paste of


Ammonium chloride


Zinc chloride


powdered manganese IV oxide mixed with Carbon


Zinc actsserve as the negative terminal where it ionizesdissociates


Zns  Zn2aq  2e


Ammonium ions in ammonium chloride serve as the positive terminal where it is converted to ammonia gas and hydrogen gas


2NH4aq  2e  2NH3g  H2g


Ammonia forms a complex salt  compound ZnNH3 42 aq  tetramminezincII complex with the Zinc chloride in the paste


Manganese IV oxide oxidizes the hydrogen produced at the electrodes to water preventing any bubbles from coating the carbon terminal which would reduce the efficiency of the cell


Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fusednot mobile


Since the reactants are used up the dry primary Laclanche cell cannot provide continous supply of electricityThe process of restoring the reactants is called recharging


bWetSecondaryAccumulators


1 WetSecondaryAccumulators are rechargeable unlike dry primary Laclanche cellsWetSecondaryAccumulators are made up of


iLead plate that forms the negative terminal


iiLeadIV oxide that forms the positive terminal


2The two electrodes are dipped in concentrated sulphuricVI acid of a relative density 1213


3At the negative terminallead ionizes dissolves


Pbs  Pb2  2e


4At the positive terminal


i LeadIV oxide reacts with the hydrogen ions in sulphuricVIacid to form Pb2 aq ions


PbO2s  4Haq  2e  Pb2 aq  H2Ol


ii Pb2 aq ions formed instantly react with sulphate VI ions SO42 aq from sulphuric VIacid to form insoluble LeadII sulphate VI


Pb2 aq  SO42 aq  PbSO4s


5The overall cell reaction is called discharging


PbO2s Pbs  4Haq  2SO42 aq 2PbSO4s  2H2Ol E  20V


6The insoluble LeadII sulphate VI formed should not be left for long since fine LeadII sulphate VI will change to a course nonreversible and inactive form making the cell less efficient


As the battery discharges lead and leadIVoxide are depletedfinishedreduced and the concentration of sulphuricVIacid decreases


During recharging the electrode reaction is reversed as below


2PbSO4s  2H2Ol PbO2s Pbs  4Haq  2SO42 aq


8 A car battery has six Leadacid cells making a total of 12 volts


iiiELECTROLYSIS ELECTROLYTIC CELL


1Electrolysis is defined simply as the decomposition of a compound by an electric currentelectricity


A compound that is decomposed by an electric current is called an electrolyte Some electrolytes are weak while others are strong


2Strong electrolytes are those that are fully ionizeddissociated into many ions Common strong electrolytes include


iall mineral acids


iiall strong alkalissodium hydroxidepotassium hydroxide


iiiall soluble salts


3Weak electrolytes are those that are partiallypartly ionizeddissociated into few ions


Common weak electrolytes include


iall organic acids


iiall bases except sodium hydroxidepotassium hydroxide


iiiWater


4 A compound that is not decomposed by an electric current is called nonelectrolyte Nonelectrolytes are those compounds substances that exist as molecules and thus cannot ionizedissociate intoany ions


Common nonelectrolytes include


i most organic solvents eg petrolparaffinbenzenemethylbenzeneethanol


iiall hydrocarbonsalkanes alkenesalkynes


iiiChemicals of lifeeg proteins carbohydrates lipids starch sugar


5 An electrolytes in solid state have fused joined ions and therefore do not conduct electricity but the ions cations and anions are free and mobile in molten and aqueous solution dissolved in water state


6During electrolysis the free ions are attracted to the electrodes An electrode is a rod through which current enter and leave the electrolyte during electrolysis An electrode that does not influencealter the products of electrolysis is called an inert electrode


Common inert electrodes include


iPlatinum


iiCarbon graphite


Platinum is not usually used in a school laboratory because it is very expensive Carbon graphite is easilyreadily and cheaply available from used dry cells


7The positive electrode is called AnodeThe anode is the electrode through which current enter the electrolyteelectrons leave the electrolyte


8The negative electrode is called Cathode The cathode is the electrode through which current leave the electrolyte  electrons enter the electrolyte


9 During the electrolysis free anions are attracted to the anode where they lose donate electrons to form neutral atomsmolecules ie


Ml  Ml  e for cations from molten electrolytes


Ms  Maq  e for cations from electrolytes in aqueous state  solution  dissolved in water


The neutral atoms molecules form the products of electrolysis at the anode This is called discharge at anode


10 During electrolysis free cations are attracted to the cathode where they gain acceptacquire electrons to form neutral atomsmolecules


X aq  2e  Xs for cations from electrolytes in aqueous state  solution  dissolved in water


2X l  2e  X l for cations from molten electrolytes


The neutral atoms molecules form the products of electrolysis at the cathode This is called discharge at cathode


11 The below set up shows an electrolytic cell


12 For a compound salt containing only two ionbinary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples


aTo determine the products of electrolysis of molten LeadIIchloride


iDecomposition of electrolyte into free ions


PbCl2 l  Pb 2l  2Cll


Compound decomposed into free cation and anion in liquid state


iiAt the cathodenegative electrode


Pb 2l  2e  Pb l


Cation  Pb 2 gains  accepts  acquires electrons to form free atom


iiiAt the anodepositive electrode


2Cll  Cl2 g  2e


Anion  Cl donatelose electrons to form free atom then a gas molecule


ivProducts of electrolysis therefore are


IAt the cathode grey beads solid lead metal


IIAt the anode pale green chlorine gas


bTo determine the products of electrolysis of molten Zinc bromide


ZnBr2 l  Zn 2l  2Brl


Zn 2l  2e  Znl


Cation  Zn2 gains  accepts  acquires electrons to form free atom


2Brl  Br2 g  2e


Anion  Br donatelose electrons to form free atom then a liquid molecule which change to gas on heating


IAt the cathode grey beads solid Zinc metal


IIAt the anode red bromine liquid  redbrown bromine gas


cTo determine the products of electrolysis of molten sodium chloride


NaCl l  Na l  Cll


iiAt the cathodenegative electrode


2Nal  2e  Na l


Cation  Na gains  accepts  acquires electrons to form free atom


iiiAt the anodepositive electrode


2Cll  Cl2 g  2e


IAt the cathode grey beads solid sodium metal


IIAt the anode pale green chlorine gas


dTo determine the products of electrolysis of molten Aluminium IIIoxide


Al2O3 l  2Al 3l  3O2l


4Al 3 l  12e  4Al l


Cation  Al 3 gains  accepts  acquires electrons to form free atom


6O2l  3O2 g  12e


Anion 6O2 donatelose 12 electrons to form free atom then three gas molecule


IAt the cathode grey beads solid aluminium metal


IIAt the anode colourless gas that relightsrekindles glowing splint


13 For a compound salt mixture containing many ions in an electrolytic cell the discharge of ions in the cell depend on the following factors


Position of cations and anions in the electrochemical series


1 Most electropositive cations require more energy to reduce gain electrons and thus not readily discharged The higher elements metals in the electrochemical series the less easilyreadily it is discharged at the cathode in the electrolytic cell


Table I showing the relative ease of discharge of cations in an electrolytic cell


Kaq  e  Ks least readilyeasily discharged


Naaq  e  Nas


Ca2aq  2e  Cas


Mg2aq  2e  Mgs


Al3aq  3e  Als


Zn2aq  2e  Zns


Fe2aq  2e  Fes


Pb2aq  2e  Pbs


2Haq  2e  H2g hydrogen is usually metallic


Cu2aq  2e  Cus


Hg2aq  2e  Hgs


Agaq  e  Ags most readilyeasily discharged


2The OH ion is the most readilyeasily discharged anion  All the other anionic radicalsSO42 SO32 CO32 HSO4 HCO3 NO3 PO43are notnever discharged The ease of discharge of halogen ions increase down the group


Table II showing the relative ease of discharge of anions in an electrolytic cell


4OH aq  2H2Ol  O2 g  4e most readilyeasily discharged


2 Iaq  I2aq  2e


2 Braq  Br2aq  2e


2 Claq  Cl2aq  2e


2 Faq  F2aq  2e


SO42 SO32 CO32 HSO4 HCO3 NO3 PO43 notneverrarely discharged


3aWhen two or more cations are attracted to the cathode the ion lower in the electrochemical series is discharged instead of that which is higher as per the table I above This is called selectivepreferential discharge at cathode


bWhen two or more anions are attracted to the anode the ion higher in the electrochemical series is discharged instead of that which is lower as per the table I above This is called selectivepreferential discharge at anode


4The following experiments shows the influence effect of selectivepreferential discharge on the products of electrolysis


iElectrolysis of acidified waterdilute sulphuricVI acid


Fill the Hoffmann voltameter with dilute sulphuricVI acid Connect the Hoffmann voltameter to a dc electric supply Note the observations at each electrode


Electrolytic cell set up during electrolysis of acidified waterdilute sulphuricVI acid


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


H2 SO4aq  SO42aq  2Haq


II Name the ions in acidified water that are attractedmove to


Cathode Haq from either sulphuricVI acid H2 SO4 or water H2O


Anode SO42aq from sulphuric VI acid H2 SO4 and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode 4Haq  4e  2H2g


Anode 4OH aq  2H2Ol  O2 g  4e


4OH ions selectively discharged instead of SO42 ions at the anode


IV Name the products of electrolysis of acidified water


CathodeHydrogen gas colourless gas that extinguishes burning splint with explosion pop sound


AnodeOxygen gas colourless gas that relights rekindles glowing splint


V Explain the difference in volume of products at the cathode and anode


The four4 electrons donatedlost by OH ions to form 1 molecule1volume1mole of oxygen O2gas at the anode are gainedacquiredaccepted by the four Haq ions to form 2 molecule2volume2mole of Hydrogen H2gas at the cathode


The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode


VI Why is electrolysis of dilute sulphuricVI acid called electrolysis of acidified water


The ratio of H2 g O2 g is 21 as they are combined in water This impliesmeans that water in the electrolyte is being decomposed into hydrogen and Oxygen gases The electrolysis of dilute sulphuric acid is therefore called electrolysis of acidified water


VI Explain the changes in concentration of the electrolyte during electrolysis of acidified water


The concentration of dilute sulphuric VI acid increases Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape The concentration mole of acid present in a given volume of solution thus continue increasingrising


iiElectrolysis of Magnesium sulphateVI solution


Fill the Hoffmann voltameter with dilute sulphuricVI acid Connect the Hoffmann voltameter to a dc electric supply Note the observations at each electrode


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


Mg SO4aq  SO42aq  Mg2aq


II Name the ions in Magnesium sulphateVI solution that are attractedmove to


Cathode Mg2aq from Magnesium sulphateVI solution Mg SO4 and Haq from water H2O


Anode SO42aq from Magnesium sulphateVI solution Mg SO4 and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode 4Haq  4e  2H2g


H ions selectively discharged instead of Mg2 ions at the cathode


Anode 4OH aq  2H2Ol  O2 g  4e


4OH ions selectively discharged instead of SO42 ions at the anode


IV Name the products of electrolysis of Magnesium sulphateVI solution


CathodeHydrogen gas colourless gas that extinguishes burning splint with explosion pop sound


AnodeOxygen gas colourless gas that relights rekindles glowing splint


V Explain the difference in volume of products at the cathode and anode


The four4 electrons donatedlost by OH ions to form 1 molecule1volume1mole of oxygen O2gas at the anode are gainedacquiredaccepted by the four Haq ions to form 2 molecule2volume2mole of Hydrogen H2gas at the cathode


The volume of Oxygen gas at the anode is thus a half the volume of Hydrogen produced at the cathode The volume of Hydrogen gas at the cathode is thus a twice the volume of Oxygen produced at the anode


VI Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphateVI solution


The concentration of dilute Magnesium sulphateVI solution increases


The ratio of H2 g O2 g is 21 as they are combined in water


Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products


The concentration mole of acid present in a given volume of Magnesium sulphateVI solution thus continue increasingrising


The set  up below was used during the electrolysis of aqueous magnesium sulphate using inert electrodes


Name a suitable pair of electrodes for this experiment


Identify the ions and cations in the solution


On the diagram label the cathode


Write ionic equations for the reactions that took place at the anode


Explain the change that occurred to the concentration of magnesium sulphate solution during the experience


During the electrolysis a current of 2 amperes was passed through the solution for 4 hours Calculate the volume of the gas produced at the anode1 faraday 96500 coulombs and volume of a gas at room temperature is 24000cm3


One of the uses of electrolysis is electroplating


What is meant by electroplating


Give tow reasons why electroplating is necessary


Concentration of the electrolytes


1High concentrations of cations andor anions at the electrodes block the ions that is likely to be discharged at the electrode This is called over voltage A concentrated solution therefore produces different products of electrolysis from a dilute one


2 The following experiments show the influenceeffect of concentration of electrolyte on the products of electrolysis


iElectrolysis of dilute and concentratedbrinesodium chloride solution


I Dissolve about 05 g of pure sodium chloride crystals in 100cm3 of water Place the solution in an electrolytic cell Note the observations at each electrode for 10 minutes Transfer the set up into a fume chamberopen and continue to make observations for a further 10 minute


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


NaClaq  Claq  Naaq


II Name the ions in sodium chloride solution that are attractedmove to


Cathode Naaq from Sodium chloride solution NaCl and Haq from water H2O


Anode Claq from sodiumchloride solution NaCl and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode 4Haq  4e  2H2g


H ions selectively discharged instead of Na ions at the cathode


Anode 4OH aq  2H2Ol  O2 g  4e


4OH ions selectively discharged instead of Cl ions at the anode


IV Name the products of electrolysis of dilute sodium chloride solution


CathodeHydrogen gas colourless gas that extinguishes burning splint with explosion pop sound


AnodeOxygen gas colourless gas that relights rekindles glowing splint


V Explain the difference in volume of products at the cathode and anode


Four4 electrons donatedlost by OH ions to form 1 molecule1volume1mole of oxygen O2gas at the anode are gainedacquiredaccepted by four Haq ions to form 2 molecule2volume2mole of Hydrogen H2gas at the cathode


The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode


VI Explain the changes in concentration of the electrolyte during electrolysis of sodium chloride solution


The concentration of dilute sodium chloride solution increases


The ratio of H2 g O2 g is 21 as they are combined in water Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products The concentration moles of salt present in a given volume of sodium chloride solution continue increasingrising


II Dissolve about 20 g of pure sodium chloride crystals in 100cm3 of water Place the solution in an electrolytic cell Note the observations continuously at each electrode for 30 minutes in a fume chamberopen


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


NaClaq  Claq  Naaq


II Name the ions in sodium chloride solution that are attractedmove to


Cathode Naaq from Sodium chloride solution NaCl and Haq from water H2O


Anode Claq from sodium chloride solution NaCl and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode 2Haq  2e  H2g


H ions selectively discharged instead of Na ions at the cathode


Anode 2Cl aq  Cl2g  4e


Cl ions with a higher concentration block the discharge of OH ions at the anode


IV Name the products of electrolysis of concentrated sodium chloride solutionbrine


CathodeHydrogen gas colourless gas that extinguishes burning splint with explosion pop sound


AnodeChlorine gaspale green gas that bleaches dampmoistwet litmus papers


V Explain the difference in volume of products at the cathode and anode


Two 2 electrons donatedlost by Cl ions to form 1 molecule1volume1mole of Chlorine Cl2gas at the anode are gainedacquiredaccepted by two Haq ions to form 1 molecule1volume1mole of Hydrogen H2gas at the cathode


The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode The volume of Hydrogen gas at the cathode is equal to the volume of Chlorine produced at the anode


VI Explain the changes in concentration of the electrolyte during electrolysis of concentrated sodium chloride solutionbrine


The concentration of concentrated sodium chloride solutionbrine increases


The ratio of Cl2 g H2 g is 11 as they are combined in water


Water in the electrolyte is decomposed into only Hydrogen gas that escapes as products at cathode


The concentration moles of OH aq and Na ion as NaOH present in a given volume of electrolyte continue increasingrising


This makes the electrolyte strongly alkaline with high pH


As the electrolysis of brine continues the concentration of Cl ions decrease and oxygen gas start being liberated at anode


The electrolyte pH is thus lowered and the concentration of brine starts again increasing


iiElectrolysis of dilute and concentrated Hydrochloric acid solution


I Prepare about 50cm3 of 005 M of dilute Hydrochloric acid in 100cm3 solution Place the solution in an electrolytic cell Note the observations at each electrode for 10 minutes


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


HClaq  Claq  Haq


II Name the ions in dilute Hydrochloric acid solution that are attractedmove to


Cathode Haq from dilute Hydrochloric acid HCl and Haq from water H2O


Anode Claq from dilute Hydrochloric acid HCl and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode 4Haq  4e  2H2g


H ions selectively discharged instead of Na ions at the cathode


Anode 4OH aq  H2Ol O2 4e


4OH ions selectively discharged instead of Cl ions at the anode


IV Name the products of electrolysis of dilute Hydrochloric acid


CathodeHydrogen gas colourless gas that extinguishes burning splint with explosion pop sound


AnodeOxygen gas colourless gas that relights rekindles glowing splint


V Explain the difference in volume of products at the cathode and anode


Four4 electrons donatedlost by OH ions to form 1 molecule1volume1mole of oxygen O2gas at the anode are gainedacquiredaccepted by four Haq ions to form 2 molecule2volume2mole of Hydrogen H2gas at the cathode


The volume of Oxygen gas at the anode is half the volume of Hydrogen produced at the cathode The volume of Hydrogen gas at the cathode is twice the volume of Oxygen produced at the anode


VI Explain the changes in concentration of the electrolyte during electrolysis of dilute Hydrochloric acid


The concentration of dilute Hydrochloric acid increases


The ratio of H2 g O2 g is 21 as they are combined in water Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products The concentration moles of HCl present in a given volume of dilute Hydrochloric acid continue increasingrising


II Prepare about 50cm3 of 2M of Hydrochloric acid in 100cm3 solution Place the solution in an electrolytic cell Note the observations at each electrode for 30 minutes


CautionThis experiment should be done in the openfume chamber


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


HClaq  Claq  Haq


II Name the ions in 2M Hydrochloric acid solution that are attractedmove to


Cathode Haq from dilute Hydrochloric acid HCl and Haq from water H2O


Anode Claq from dilute Hydrochloric acid HCl and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode 4Haq  4e  2H2g


H ions selectively discharged instead of Na ions at the cathode


Anode 2Cl aq  Cl2 2e


OH ions concentration is lowCl ions concentration is higher at the anode thus cause over voltageblock discharge of OH ions


IV Name the products of electrolysis of 2M Hydrochloric acid


CathodeHydrogen gas colourless gas that extinguishes burning splint with explosion pop sound


AnodeChlorine gas Pale green gas that bleaches bluered moistwetdamp litmus papers


V Explain the difference in volume of products at the cathode and anode


Two2 electrons donatedlost by Cl ions to form 1 molecule1volume1mole of Chlorine Cl2gas at the anode are gainedacquiredaccepted by two Haq ions to form 1 molecule1volume1mole of Hydrogen H2gas at the cathode


The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the cathode The volume of Hydrogen gas at the cathode is twice the volume of Chlorine produced at the anode


VI Explain the changes in concentration of the electrolyte during electrolysis of 2M Hydrochloric acid


The concentration of Hydrochloric acid decreases


The ratio of H2 g Cl2 g is 11 as they are combined in Hydrochloric acid


Water in the electrolyte is decomposed only into Hydrogen gas that escapes as products at the cathode


There is a net accumulation of excess OH aq ions in solution


This makes the electrolyte strongly alkaline with high pH


Nature of electrodes used in the electrolytic cell


Inert electrodes carbongraphite and platinum do not alter the expected products of electrolysis in an electrolytic cell If anotherdifferent electrode is used in the electrolytic cell it altersinfluenceschanges the expected products of electrolysis


The examples below illustrate the influence of the nature of electrode on the products of electrolysis


iElectrolysis of copperII sulphateVI solution


I Using carbongraphite electrodes


Weigh Carbon graphite electrodes Record the masses of the electrodes in table I below Place the electrodes in 1M copperII sulphateVI solution in a beaker Set up an electrolytic cell


Close the switch and pass current for about 20 minutes Observe each electrode and any changes in electrolyte Remove the electrodes from the electrolyte Wash with acetonepropanone and allow them to dry Reweigh each electrode


Sample results


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


CuSO4aq  SO42aq  Cu2aq


II Name the ions in 1M copperII sulphateVI solution that are attractedmove to


Cathode Cu2 aq from copperII sulphateVI solution and Haq from water H2O


Anode SO42aq from copperII sulphateVI solution and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode 2Cu2 aq  4e  2Cug


Cu2 ions are lower than H ions in the electrochemical series therefore selectively discharged at the cathode


Anode 4OH aq  H2Ol  O2 4e


OH ions ions are higher than SO42 ions in the electrochemical series therefore selectively discharged at the cathode


IV Name the products of electrolysis of 1M copperII sulphateVI solution


Cathode2 moles of copper metal as brown solid coat


AnodeOxygen gas Colourless gas that relights rekindles glowing splint


V Explain the changes that take place at the cathode and anode


Four4 electrons donatedlost by OH ions to form 1 molecule1volume1mole of Oxygen O2gas at the anode are gainedacquiredaccepted by two Cu2aq ions to form 2 moles of brown copper solid that deposit itself at the cathode


The moles of oxygen gas at the anode is equal to the moles of copper produced at the cathode


VI Explain the changes in electrolyte during electrolysis of 1M copper II sulphateVI solution


iThe pH of copperII sulphateVI solution lowersdecreases The salt becomes more acidic Water in the electrolyte is decomposed only into Oxygen gas from the OH ions that escapes as products at the anode There is a net accumulation of excess H aq ions in solution This makes the electrolyte strongly acidic with low pH


ii Cu2 aq ions are responsible for the blue colour of the electrolyte copperII sulphate VI solution As electrolysis continues blue Cu2 aq ions gain electrons to form brown Copper The blue colour of electrolyte therefore fadesbecome less blue


iiiCopper is deposited at the cathode This increases the mass of the cathodeOH ions that produce Oxygen gas at anode come from water Oxygen escapes outaway without increasing the mass of anode


II Using copper electrodes


Weigh clean copper plates electrodes Record the masses of the electrodes in table I below Place the electrodes in 1M copperII sulphateVI solution in a beaker Set up an electrolytic cell


Close the switch and pass current for about 20 minutes Observe each electrode and any changes in electrolyte Remove the electrodes from the electrolyte Wash with acetonepropanone and allow them to dry Reweigh each electrode


Sample results


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


CuSO4aq  SO42aq  Cu2aq


II Name the ions in 1M copperII sulphateVI solution that are attractedmove to


Cathode Cu2 aq from copperII sulphateVI solution and Haq from water H2O


Anode SO42aq from copperII sulphateVI solution and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode Cu2 aq  2e  Cus


Cu2 ions are lower than H ions in the electrochemical series therefore selectively discharged at the cathode


Anode Cu s  Cu2aq  2e


Both OH ions and SO42 ions move to the anode but none is discharged The copper anode itself ionizesdissolvesdissociate because less energy is used to remove an electronionize dissociate copper atoms than OH ions


IV Name the products of electrolysis of 1M copperII sulphateVI solution using copper electrodes


Cathode1 moles of copper metal as brown solid coat Cathode increasedeposits


AnodeAnode erodesdecrease in size


V Explain the changes that take place during the electrolytic process


iCathode


Cu2 ions are lower than H ions in the electrochemical series therefore selectively discharged at the cathode Cu2 ions have greater tendency to acceptgainacquire electrons to form brown copper atomssolid that deposit itself and increase the masssize of the cathodeThe copper deposited at the cathode is pure


H ions accumulate around the cathode Electrolyte thus becomes strongly acidic around the cathode


Cu2 ions in solution are responsible for the blue colour of electrolyte Blue colour of electrolyte fade around the cathode


iiAnode


Copper atom at the anode easily ionizes to release electrons The anode therefore keeps decreasing in masseroding The amount of copper that dissolveerode is equal to the mass of copper deposited This is called electrode ionization


Electrode ionization is where the anode erodesdecrease and the cathode depositsincrease during electrolysis The overall concentration of the electrolyte remains constant


14In industries electrolysis has the following usesapplications


aExtraction of reactive metals from their ores


Potassium sodium magnesium and aluminium are extracted from their ores using electrolytic methods


bPurifying copper after exraction from copper pyrites ores


Copper obtained from copper pyrites ores is not pure After extraction the copper is refined by electrolysing copperIIsulphateVI solution using the impure copper as anode and a thin strip of pure copper as cathode Electrode ionization take place there


iAt the cathode Cu2 aq  2e  Cus Pure copper deposits on the strip


iiAt the anode Cus Cu2 aq  2e impure copper erodesdissolves


cElectroplating


The label EPNSElectro Plated Nickel Silver on some steelmetallic utensils mean they are platedcoated with silver andor Nickel to improve their appearanceadd their aesthetic valueand preventslow corrosionrusting of iron Electroplating is the process of coating a metal with another metal using an electric current During electroplatingthe cathode is made of the metal to be coatedimpure


Example


During the electroplating of a spoon with silver


ithe spoonimpure is placed as the cathodenegative terminal of battery


iithe pure silver is placed as the anodepositive terminal of battery


iiithe pure silver erodesionizesdissociates to release electrons


Ags Ag aq  e impure silver erodesdissolves


iv silver Agions from electrolyte gain electrons to form pure silver deposits  coat cover the spoonimpure


Ag aq  e Ags pure silver deposits coatcover on spoon


15The quantitative amount of products of electrolysis can be determined by applying Faradays 1st law of electrolysis


Faradays 1st law of electrolysis states that the massamount of substance liberatedproducedused during electrolysis is directly proportional to the quantity of of electricity passedused


aThe SI unit of quantity of electricity is the coulombC The coulomb may be defined as the quantity of electricity passedused when a current of one ampere flow for one secondie


1Coulomb  1 Ampere x 1Second


The Ampere is the SI unit of currentI


The Second is the SI unit of timet therefore


Quantity of electricityin Coulombs  CurrentI x timet


Practice examples


1 A current of 2 amperes was passed through an electrolytic cell for 20 minutes Calculate the quantity of electric charge produced


Working


Substituting converting time to second  2 x 20 x 60


2400 C


2 A current of 2 amperes was passed through an electrolytic96500 coulombs of charge were produced Calculate the time taken


Timet in seconds  Quantity of electricityin Coulombs


CurrentI in amperes


Substituting  96500


48250 seconds


3 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell  Calculate the amount of current used


CurrentI in amperes  Quantity of electricityin Coulombs Timet in seconds


Substitutingconverting time to second 96500


10 x 60


1608333 Amperes


bThe quantity of electricity required for one mole of electrons at the anodecathode is called the Faraday constantF It is about 96500 Coulombsie


The number of Faradays used required is equal to the number of electrons used at cathodeanode during the electrolytic process eg


Cu2 require to gain 2 moles of electrons2 Faradays 2 x 96500 coulombs of electricity at the cathode


Al3 require to gain 3 moles of electrons3 Faradays 3 x 96500 coulombs of electricity at the cathode


Na require to gain 1 moles of electrons1 Faradays 1 x 96500 coulombs of electricity at the cathode


2H require to gain 2 moles of electrons2 Faradays 2 x 96500 coulombs of electricity at the cathode to form 1molecule of hydrogen gas


2O2 require to losedonate 4 moles of electrons4 Faradays 4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen O2 gas


4OH require to losedonate 4 moles of electrons4 Faradays 4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water


cThe massamount of products at the cathodeanode is related to the molar mass of the substance andor the volume of gases at standardroom temperature and pressure as in the below examples


Practice examples


1Calculate the mass of copper deposited at the cathode when a steady current of 40 amperes is passed through copperIIsulphateVI for 30 minutes in an electrolytic cell Cu635 1F  96500C


Working


Quantity of electricityin Coulombs  CurrentI x timet


Substituting converting time to second  4 x 30 x 60


7200 C


Equation at the cathode Cu2 aq  2e  Cus


2 mole of electrons  2 Faradays  2 x 96500 C produce a mass molar mass of copper thus


2 x 96500C  635 g


72000C  7200 x 635  23689 g of copper


2 x 96500


2aIf 32 g of Lead were deposited when a current of 25 amperes was passed through an electrolytic cell of molten LeadIIbromide for 20 minutes determine the Faraday constantPb  207


Quantity of electricity in Coulombs  CurrentI x timet


Substituting converting time to second  25 x 20 x 60


3000 C


If 32g of Lead  3000C


Then 207 g of Lead  207 x 3000  1940625 C


Equation at the cathode Pb2 l  2e  Pbl


From the equation 2 moles of electrons  2 Faradays  1940625 C


1mole of electrons  1 Faraday  1940625  9703125 C


bWhat is the volume of bromine vapour produced at the anode at room temperature1mole of gas at room temperature and pressure  24000cm3


Method 1


Equation at the anode Br l  Br2g  2e


From the equation 2 moles of electrons  2 Faradays  1940625 C  24000cm3


3000 C  3000 x 24000


Method 2


Mole ratio of products at Cathode anode  11


Moles of Lead at cathode  32  00155moles  moles of Bromine


1 moles of bromine vapour  24000cm3


00155moles of Bromine  00155 x 24000  372 cm3


Method 3


Ratio of Faradays used to form products at Cathode anode  22


2 x 9703125 C produce 24000cm3 of bromine vapour


Then 3000 C  3000 x 24000cm3  3710145cm3


2 x 9703125


3What mass of copper remain from 20 at the anode if a solution of copperIIsulphateVI is electrolysed using a current of 1 ampere flowing through an electrolytic cell for 20 minutesCu 635 1Faraday  96487 coulombs


Working


Substituting converting time to second  1 x 20 x 60


1200 C


Equation at the cathode Cu2 aq  2e  Cus


2 mole of electrons  2 Faradays  2 x 96500 C erodedissolve a mass molar mass of copper thus


2 x 96500C  635 g


1200C  1200 x 635  03948g of copper deposited


2 x 96500


Mass of copper remaining  Original mass  mass dissolvederoded


20 03948  16052 g of copper remain


4 Calculate the current passed if a mass of 0234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper IIsulphateVI


Cu 635 1F  96500C


Equation at the cathode Cus  Cu2 aq  2e


2 mole of electrons  2 Faradays  2 x 96500 C produce a mass molar mass of copper thus


635 g  2 x 96500C


0234 g  0234 x 2 x 96500  7112126 C


CurrentI in amperes  Quantity of electricityin Coulombs Timet in seconds


Substitutingconverting time to second 7112126 C


4x 60


29634 Amperes


5 aWhat quantity of electricity will deposit a mass of 243 g of Zinc during electrolysis of a solution of Zinc IIsulphateVI


Zn 65 1F  96500C


Working


Equation at the cathode Zn2 aq  2e  Zns


2 mole of electrons  2 Faradays  2 x 96500 C erodedissolve a mass molar mass of Zinc thus


65 g  2 x 96500


243 g  243 x 2 x 96500  72152308 C


bCalculate the time in minutes it would take during electrolysis of the solution of Zinc IIsulphateVI above if a current of 40 Amperes is used


Timet in seconds  Quantity of electricityin Coulombs


CurrentI in amperes


Substituting  72152308  18038077 seconds  300635 minutes


4 60


6When a current of 15 amperes was passed through a cell containing M3 ions of metal M for 15 minutes the mass at cathode increased by 026 gFaraday constant  96500C


a Calculate the quantity of electricity used


Quantity of electricity in Coulombs  CurrentI x timet


Substituting converting time to second  15 x 15 x 60


1350 C


Determine the relative atomic mass of metal M


Equation at the cathode M3 aq  3e  Ms


1350 C of electricity  026 g of metal M


3 mole of electrons  3 Faradays  3 x 96500 C produce a mass molar mass of M thus


RAM of M  026 g x 3 x 96500  557556No units


7An element P has a relative atomic mass 88When a current of 05 amperes was passed through fused chloride of P for 32 minutes and 10seconds 044 g of P was deposited at the cathode Determine the charge on an ion of PFaraday constant  96500C


Working


Substituting converting time to second  05 x 32 x 60  10


044 g of metal P are deposited by 965C


88g of of metal P are deposited by 88 x 965 193000 C


96500 C  1 mole of electrons  1 Faradays  single charge


193000 C  193000  2 molesFaradayscharges  symbol of ion  P2


8 During purification of copper by electrolysis 148 g of copper was deposited when a current was passed through aqueous copper IIsulphateVI for 2  hours Calculate the amount of current that was passed Cu 635 1F  96500C


Equation at the cathode Cu2 aq  2e Cus


2 mole of electrons  2 Faradays  2 x 96500 C produce a mass molar mass of copper thus


635 g  2 x 96500C


148 g  148 x 2 x 96500  42551181 C


CurrentI in amperes  Quantity of electricityin Coulombs Timet in seconds


Substitutingconverting time to second 42551181C


2 x 60  30 x60


04728 Amperes


17 Practically Faraday 1st law of electrolysis can be verified as below


Verifying Faraday 1st law of electrolysis


Procedure


Weigh clean copper plates electrodes Record the masses of the electrodes in table I below Place the electrodes in 1M copperII sulphateVI solution in a beaker Set up an electrolytic cell


Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutesRemove the electrodes from the electrolyte Wash with acetone propanone and allow them to dry Reweigh each electrode


Sample results


Answer the following questions


I Write the equation for the decomposition of the electrolytes during the electrolytic process


H2Ol  OH aq  Haq


CuSO4aq  SO42aq  Cu2aq


II Name the ions in 1M copperII sulphateVI solution that are attractedmove to


Cathode Cu2 aq from copperII sulphateVI solution and Haq from water H2O


Anode SO42aq from copperII sulphateVI solution and OH aq from water H2O


III Write the equation for the reaction during the electrolytic process at the


Cathode Cu2 aq  2e  Cus


Cu2 ions are lower than H ions in the electrochemical series therefore selectively discharged at the cathode


Anode Cu s  Cu2aq  2e


Both OH ions and SO42 ions move to the anode but none is discharged The copper anode itself ionizesdissolvesdissociate as less energy is used to remove an electronionize dissociate copper atoms than OH ions


IV Name the products of electrolysis of 1M copperII sulphateVI solution using copper electrodes


Cathode125 g of copper metal as brown solid coatdeposits


Anode125 g of copper metal erodesdecrease in size


V iHow many moles of electrons are used to depositerode one mole of copper metal at the cathodeanode


From the equation at anodecathode 2 moles


iiHow many Faradays are used to depositerode one mole of copper metal at the cathodeanode


From the equation at anodecathode  2 moles  2 Faradays


iiiCalculate the quantity of electric charge used


Working


Quantity of electricity in Coulombs  CurrentI x timet


Substituting converting time to second  2 x 20 x 60


VI i Calculate the quantity of electricity required to depositerode one mole of copper at the cathodeanodeCu635


Since 125 g of copper  2400C


Then 635 g 1mole of copper  635 x 2400  121920 C


iiDetermine the Faraday constant from the results in Vi above


From the equation at


Cathode Cu2 aq  2e  Cus


Anode Cu s  Cu2aq  2e


2 moles  2 Faradays  121920 C


1 moles  1 Faradays  121920  60960 C


iii The faraday constant obtained above is far lower than theoreticalExplain


high resistance of the wires used


temperatures at 25oC were not kept constant


plateselectrodes used were not made of pure copper


plateselectrodes used were not thoroughly clean copper


Further practice


1An element P has a relative atomic mass of 88 When a current of 05 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds 044g of P were deposited at the cathode Determine the charge on an ion of P 1 faraday  96500 Coulombs


2During electrolysis of aqueous copper II sulphate 144750 coulombs of electricity were used Calculate the mass of copper metal that was obtained


Cu  64 1 Faraday  96500 coulombs  3 mks


3A nitrate of a metal M was electrolysed 118 g of metal was deposited when a current of 4 ampheres flow for 16 minutesDetermine the formula of the sulphateVIsalt of the metal


Faraday constant  96500  RAM of X  590


Working


Q  It  4 x 16 x 60  3840 C


118 g of X  3840 C


590 g  590 x 3840  192000 C


96500 C  1Faraday


192000 C 192000 C x1  2F thus charge of M  M2


96500 C


Valency of M is 2 thus formula of sulphateVIsalt MSO4


4 Below is the results obtained when a current of 20ampheres is passed through copperIIsulphateVIsolution for 15 minutes during electrolysis using copper electrode


Initial mass of cathode  10 g


Final mass of cathode  16 g


Change in mass of cathode  060 g


iDetermine the change in mass at the anode Explain your answer


Mass decrease  06g


Electrode ionization take place where the cathode increase in mass form the erosion of the anode


iiCalculate the quantity of electricity required to deposit one mole of copperCu 635


Q It  2 x 15 x 60  1800 coulombs


Method 1


060 g of copper 1800 coulombs


635 g  635 x 1800  190500 Coulombs


Method 2


Moles of Copper  Mass  060  94488 x10 3 moles


Molar mass 635


94488 x10 3 moles  1800 coulombs


1 Mole  1 x 1800 coulombs  190500381 coulombs


94488 x10 3 moles


iiiDetermine the oxidation number of copper produced at the cathode and hence the formula of its nitrate Vsalt 1 Faraday  96500 Coulombs


96500 Coulombs  1 Faraday


190500381 coulombs  190500381 coulombs x 1


96500 Coulombs


19741 Faradays  2Fwhole number


Charge of copper  2  Oxidation number


Valency of copper  2 hence chemical formula of nitrate Vsalt  Cu NO32


RADIOACTIVITY


A INTRODUCTION  CAUSES OF RADIOCTIVITY


Radioactivity is the spontaneous disintegrationdecay of an unstable nuclide


A nuclide is an atom with defined mass number number of protons and neutrons atomic number and definite energy


Radioactivity takes place in the nucleus of an atom unlike chemical reactions that take place in the energy levels involving electrons


A nuclide is said to be stable if its neutron proton ratio is equal to one np  1


All nuclide therefore try to attain np  1 by undergoing radioactivity


Examples


iOxygen nuclide with 168 O has 8 neutrons and 8 protons in the nucleus therefore an np  1 thus stable and do not decaydisintegrate


iiChlorine nuclide with 3517 Cl has 18 neutrons and 17 protons in the nucleus therefore an np  10588 thus unstable and decaysdisintegrates to try to attain np  1


iiUranium nuclide with 23792 U has 206 neutrons and 92 protons in the nucleus therefore an np  22391 thus more unstable than 23592 U and thus more readily decays  disintegrates to try to attain np  1


iii Chlorine nuclide with 3717 Cl has 20 neutrons and 17 protons in the nucleus therefore an np  11765 thus more unstable than 3517 Cl and thus more readily decays  disintegrates to try to attain np  1


ivUranium nuclide with 23592 U has 143 neutrons and 92 protons in the nucleus therefore an np  15543 thus more stable than 237 92U but also readily decays  disintegrates to try to attain np  1


All unstable nuclides naturally try to attain nuclear stability with the production of


ialpha particle decay


The alpha  particle has the following main characteristic


iis positively chargedlike protons


ii has mass number 4 and atomic number 2 therefore equal to a charged Helium atom  42He2


iii have very low penetrating power and thus can be stopped blockedshielded by a thin sheet of paper


iv have high ionizing power thus cause a lot of damage to living cells


v a nuclide undergoing decay has its mass number reduced by 4 and its atomic number reduced by 2


Examples of alpha decay


210 84 Pb  x 82 Pb  42He 2


210 84 Pb  206 82 Pb  42He 2


226 88 Ra  222 y Rn  42He 2


226 88 Ra  222 86 Rn  42He 2


x y U  23490 Th  42He 2


238 92 U  23490 Th  42He 2


x y U  23088 Ra  2 42He 2


238 92 U  23088 Ra  2 42He 2


210 84 U  xy W  10


210 84 U  17064 W  10


210 92U  xy W  6


210 92U  18680W  6


iiBeta  particle decay


The Beta  particle has the following main characteristic


iis negatively chargedlike electrons


iihas no mass number and atomic number negative one1 therefore equal to a fast moving electron 0 1e


iii have medium penetrating power and thus can be stopped blockedshielded by a thin sheet of aluminium foil


iv have medium ionizing power thus cause less damage to living cells than the  particle


v a nuclide undergoing  decay has its mass number remain the same and its atomic number increase by 1


Examples of beta  decay


123 x Na  2312Mg  0 1e


23 11 Na  2312Mg  0 1e


2 234 x Th  y91 Pa  0 1e


234 90 Th  y91 Pa  0 1e


3 20770Y  x y Pb  30 1e


20770Y  207 73Pb  30 1e


4 x y C  147N  0 1e


14 6 C  147N  0 1e


5 1 x n  y1H  0 1e


1 0 n  11H  0 1e


6 42He  411H  x 0 1e


42He  411H  2 0 1e


7 22888Ra  22890Th  x


22888Ra  22892Th  4


8 23290Th  21282Pb  2   x


23290Th  21282Pb  2   5


9 23892U  22688 Ra  x   3


23892U  22688 Ra  2   3


10 21884Po  20682Pb  x   3


21884Po  20682Pb  4  3


iiiGamma y particle decay


The gamma y particle has the following main characteristic


iis neither negatively chargedlike electronsbeta nor positively chargedlike protonsalpha therefore neutral


iihas no mass number and atomic number therefore equal to electromagnetic waves


iii have very high penetrating power and thus can be stopped blockedshielded by a thick block of lead


iv have very low ionizing power thus cause less damage to living cells unless on prolonged exposure


v a nuclide undergoing y decay has its mass number and its atomic number remain the same


Examples of gamma y decay


3717Cl  3717Cl  y


146C  146C  y


The sketch diagram below shows the penetrating power of the radiations from a radioactive nuclide


radioactive nuclide sheet of paper aluminium foil thick block of lead


radiation source block rays block rays block yrays


rays rays yrays


The sketch diagram below illustrates the effect of electric magnetic field on the three radiations from a radioactive nuclide


Radioactive disintegrationdecay naturally produces the stable 20682Pb nuclide isotope of leadBelow is the 238 92 U natural decay series Identify the particle emitted in each case


Write the nuclear equation for the disintegration from


i238 92 U to 23490 T


238 92 U  23490 T  4 2 He 2


238 92 U  23490 T


ii238 92 U to 222 84 Rn


238 92 U  22284 Rn  4 4 2 He 2


238 92 U  22284 Rn  4


230 90 Th undergoes alpha decay to 222 86 Rn Find the number of  particles emitted Write the nuclear equation for the disintegration


Working


230 90 Th  222 86 Rn  x 4 2 He


Method 1


Using mass numbers


230  222  4 x  4 x  230  222  8


x  8  4  2


Using atomic numbers


90  86  2 x  2 x  90  86  4


x  4  2  2


Nuclear equation


230 90 Th  222 86 Rn  2 4 2 He


214 82 Pb undergoes beta decay to 214 84 Rn Find the number of  particles emitted Write the nuclear equation for the disintegration


214 82 Pb  214 84 Rn  x 0 1 e


Using atomic numbers only


82  84  x  x  82  84  2


x  2


214 82 Pb  214 84 Rn  2 0 1 e


238 92 U undergoes beta and alpha decay to 206 82 Pb Find the number of  and  particles emitted Write the nuclear equation for the disintegration


Working


238 92 U  206 82 Pb  x 0 1 e  y 4 2 He


Using Mass numbers only


238  206  4y  4y  238  206  32


y  32  8


Using atomic numbers only and substituting the 8 above


238 92 U  206 82 Pb  8 4 2 He  x 0 1 e


92  82  16   x


x  6


Nuclear equation


238 92 U  206 82 Pb  6 0 1 e  8 4 2 He


298 92 U undergoes alpha and beta decay to 214 83 Bi Find the number of  and  particles emitted Write the nuclear equation for the disintegration


298 92 U  210 83 Bi  x 4 2 He  y 0 1 e


298  214  4x  4x  298  214  84


y  84  21


Using atomic numbers only and substituting the 21  above


238 92 U  214 83Bi  21 4 2 He  y 0 1 e


92  83  42   y


92  83  42   x


x  33


298 92 U  210 83 Bi  21 4 2 He  33 0 1 e


BNUCLEAR FISSION AND NUCLEAR FUSION


Radioactive disintegrationdecay can be initiated in an industrial laboratory through two chemical methods


a nuclear fission


b nuclear fusion


aNuclear fission


Nuclear fission is the process which a fast moving neutron bombards hits knocks a heavy unstable nuclide releasing lighter nuclide three daughter neutrons and a large quantity of energy


Nuclear fission is the basic chemistry behind nuclear bombs made in the nuclear reactors


The three daughter neutrons becomes again fast moving neutron bombarding  hitting knocking a heavy unstable nuclide releasing lighter nuclides three more daughter neutrons each and a larger quantity of energy setting of a chain reaction


Examples of nuclear equations showing nuclear fission


10 n  235 b U  9038 Sr  c 54Xe  310 n  a


10 n  2713 Al  2813 Al  y  a


10 n  28a Al  b11 Na  42 He


a0 n  147 N  14b C  11 H


10 n  11 H  21 H  a


10 n  235 92 U  95 42 Mo  139 57 La  210 n  7 a


b Nuclear fusion


Nuclear fusion is the process which smaller nuclides join together to form larger  heavier nuclides and releasing a large quantity of energy


Very high temperatures and pressure is required to overcome the repulsion between the atoms


Nuclear fusion is the basic chemistry behind solarsun radiation


Two daughter atomsnuclides of Hydrogen fusejoin to form Helium atomnuclide on the surface of the sun releasing large quantity of energy in form of heat and light


21H  21H  abHe  10 n


21H  a  32He


21H  21H  a  11 H


4 11H  42He  a


147H  a  178O  11 H


C HALF LIFE PERIOD t12


The halflife period is the time taken for a radioactive nuclide to spontaneously decay disintegrate to half its original mass amount


It is usually denoted t 12


The rate of radioactive nuclide disintegrationdecay is constant for each nuclide


The table below shows the halflife period of some elements


The less the half life the more unstable the nuclide element


The halflife period is determined by using a GeigerMuller counter GM tube


A GM tube is connected to ratemeter that records the countrates per unit time


This is the rate of decay disintegration of the nuclide


If the countrates per unit time fall by half then the time taken for this fall is the halflife period


Examples


aA radioactive substance gave a count of 240 counts per minute but after 6 hours the count rate were 30 counts per minute Calculate the halflife period of the substance


If t 12  x


then 240 x120 x60 x30


From 240 to 30 3x 6 hours


x  t 12   6  3


2 hours


b The count rate of a nuclide fell from 200 counts per second to 125 counts per second in 120 minutes


Calculate the halflife period of the nuclide


If t 12 x


then


200 x100 x50 x25 x125


From 200 to 125 4x 120 minutes


x  t 12   120  4


30 minutes


c After 6 hours the count rate of a nuclide fell from 240 counts per second to 15 counts per second on the GM tube Calculate the halflife period of the nuclide


then 240 x120 x60 x30 x15


From 240 to 15 4x 6 hours


x  t 12   6  4  15 hours


d Calculate the mass of nitrogen13 that remain from 2 grams after 6 halflifes if the halflife period of nitrogen13 is 10 minutes


If t 12  x then


2 x1 2x05 3x025 4x01255x006256x003125


After the 6th half life 003125 g of nitrogen13 remain


e What fraction of a gas remains after 1hour if its halflife period is 20 minutes


then 60 20  3x


1 x 12 2x 14 3x 18


After the 3rd halflife 18 of the gas remain


f 348 grams of a nuclide A was reduced to 435 grams after 270daysDetermine the halflife period of the nuclide


348 x174 2x87 3x435


From 348 to 4353x 270days


x  t 12   270  3


90 days


g How old is an Egyptian Pharaoh in a tomb with 2grams of 14C if the normal 14C in a present tomb is 16gramsThe halflife period of 14C is 5600years


If t 12  x  5600 years then


16 x8 2x4 3x2


3x   3 x 5600


h 100 grams of a radioactive isotope was reduced 125 grams after 81daysDetermine the halflife period of the isotope


If t 12  x then


100 x50 2x25 3x125


From 100 to 1253x 81days


x  t 12


81  3


27 days


A graph of activity against time is called decay curve


A decay curve can be used to determine the halflife period of an isotope since activity decrease at equal time interval to half the original


iFrom the graph show and determine the halflife period of the isotope


From the graph t 12 changes in activity from


100  50    20  0   20 minutes


50  25    40  20   20 minutes


Thus t   20 minutes


iiWhy does the graph tend to O


Smaller particles will disintegrate decay to half its original


There can never be Ozero particles


D CHEMICAL vs NUCLEAR REACTIONS


Nuclear and chemical reaction has the following similarities


iboth involve the subatomic particles electrons protons and neutrons in an atom


iiboth involve the subatomic particles trying to make the atom more stable


iiiSome for of energy transferreleaseabsorb fromto the environment take place


Nuclear and chemical reaction has the following differences


i Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom


Chemical reactions mainly involve outer electrons in the energy levels an atom


ii Nuclear reactions form a new element


Chemical reactions do not form new elements


iii Nuclear reactions mainly involve evolutionproduction of large quantity of heatenergy


Chemical reactions produce or absorb small quantity of heatenergy


ivNuclear reactions are accompanied by a loss in massmass defectDo not obey the law of conservation of matter


Chemical reactions are not accompanied by a loss in mass mass defect hence obey the law of conservation of matter


vThe rate of decay disintegration of the nuclide is independent of physical conditions temperaturepressure purityparticle size


The rate of a chemical reaction is dependent on physical conditions temperaturepressurepurityparticle size surface area


E APPLICATION AND USES OF RADIOCTIVITY


The following are some of the fields that apply and use radioisotopes


aMedicine Treatment of cancer to kill malignant tumors through radiotherapy


Sterilizing hospital surgical instruments equipments by exposing them to gamma radiation


b Agriculture


If a plant or animal is fed with radioisotope the metabolic processes of the plantanimal is better understood by tracing the route of the radioisotope


c Food preservation


Xrays are used to kill bacteria in tinned food to last for a long time


d Chemistry


To study mechanisms of a chemical reaction one reactant is replaced in its structure by a radioisotope eg


During esterification the O joining the ester was discovered comes from the alkanol and not alkanoic acid


During photosynthesis the O released was discovered comes from water


e Dating rocksfossils


The quantity of 14C in living things plantsanimals is constant


When they die the fixed mass of 14C is trapped in the cells and continues to decaydisintegrate


The halflife period of 14C is 5600 years


Comparing the mass of 14C in living and dead cells the age of the dead can be determined


F DANGERS OF RADIOCTIVITY


All rays emitted by radioactive isotopes have ionizing effect of changing the genetic make up of living cells


Exposure to theses radiations causes chromosomal and or genetic mutation in living cells


Living things should therefore not be exposed for a long time to radioactive substances


One of the main uses of radioactive isotopes is in generation of large cheap electricity in nuclear reactors


Those who work in these reactors must wear protective devises made of thick glass or lead sheet


Accidental leakages of radiations usually occur


In 1986 the Nuclear reactor at Chernobyl in Russia had a major explosion that emitted poisonous nuclear material that caused immediate environmental disaster


In 2011 an earthquake in Japan caused a nuclear reactor to leak and release poisonous radioactive waste into the Indian Ocean


The immediate and long term effects of exposure to these poisonous radioactive waste on human being is of major concern to all environmentalists


G SAMPLE REVISION QUESTIONS


The figure below shows the behaviour of emissions by a radioactive isotope x Use it to answer the question follow


a Explain why isotope X emits radiations 1mk


is unstable has np ratio greaterless than one


b Name the radiation labeled T 1mk


alpha particle


c Arrange the radiations labeled P and T in the increasing order of ability to be deflected by an electric filed 1mk


T  P


a Calculate the mass and atomic numbers of element B formed after 21280 X has emitted three beta particles one gamma ray and two alpha particles


Mass number


212  0 beta o gamma  2 x 4  alpha  204


Atomic number


80  1 x3 beta  0 gamma  2 x 2  alpha 79


bWrite a balanced nuclear equations for the decay of 21280 X to B using the information in a above


21280 X  20479B  242He  3 01 e  y


Identify the type of radiation emitted from the following nuclear equations


i 146 C  147N


Beta


11 H  10 n  21H


y gamma


iii 23592 U  9542Mo  13957La  10 n


7   seven beta particles


23892 U  23490Th


alpha


146 C  11 H  157N


ygamma


X grams of a radioactive isotope takes 100 days to disintegrate to 20 grams If the halflife period isotope is 25 days calculate the initial mass X of the radio isotope


Number of halflifes   100  25   4


20g  40g  80g 160g  320g


Original mass X  320g


Radium has a halflife of 1620 years


iWhat is halflife


The halflife period is the time taken for a radioactive nuclide to spontaneously decay disintegrate to half its original mass amount


bIf one milligram of radium contains 268 x 10 18 atoms how many atoms disintegrate during 3240 years


Number of halflifes   3240  1620   2


1 mg 1620 05mg 1620 025mg


If 1mg  268 x 1018 atoms


Then 025 mg   025 x 268 x 1018   67 x 1017


Number of atoms remaining  67 x 1017


Number of atoms disintegrated


268 x 1018  67 x 1017


201 x 1018


The graph below shows the mass of a radioactive isotope plotted against time


Using the graph determine the half  life of the isotope


From graph 10 g to 5 g takes 8 days


From graph 5 g to 25 g takes 16  8  8 days


Calculate the mass of the isotope dacayed after 32 days


Number of half lifes 328  4


Original mass  10g


10g1st 5g2nd253rd 1254th 0625 g


Mass remaining  0625 g


Mass decayed after 32 days  10g  0625 g  9375g


A radioactive isotope X2 decays by emitting two alpha a particles and one beta  to form 214 83Bi


aWrite the nuclear equation for the radioactive decay


21286 X  214 83Bi  242He  01 e


bWhat is the atomic number of X2


c After 112 days 116 of the mass of X2 remained Determine the half life of X2


1x 1 2 x 1 4 x 1 8x 1 16


Number of t 1 2 in 112 days  4


t 1 2  112  28 days


1Study the nuclear reaction given below and answer the questions that follow


126 C step 1127 N step 2 1211Na


a126 C and 146 C are isotopes What does the term isotope mean


Atoms of the same element with different mass number number of neutrons


bWrite an equation for the nuclear reaction in step II


127 N  1211Na  0 1e


cGive one use of 146 C


Dating rocksfossils


Study of metabolic pathwaysmechanisms on plantsanimals


Study the graph of a radioactive decay series for isotope H below


Name the type of radiation emitted when isotope


i H changes to isotope J


AlphaMass number decrease by 4 from 214 to 210yaxis


atomic number decrease by 2 from 83 to 81xaxis


ii J changes to isotope K


BetaMass number remains 210yaxis


atomic number increase by 1 from 81 to 82xaxis


b Write an equation for the nuclear reaction that occur when isotope


iJ changes to isotope L


21081 J  21084L  3 0 1e


iH changes to isotope M


21483 H  20682M  3 0 1e  2 4 2He


Identify a pair of isotope of an element in the decay series


K and M


Have same atomic number 82 but different mass number K210 and M206


aA radioactive substance emits three different particles


Identify the particle


iwith the highest mass


Alpha


ii almost equal to an electron


Beta


1aState two differences between chemical and nuclear reactions2mks


i Nuclear reactions mainly involve protons and neutrons in the nucleus of an atomChemical reactions mainly involve outer electrons in the energy levels an atom


ii Nuclear reactions form a new element Chemical reactions do not form new elements


iii Nuclear reactions mainly involve evolutionproduction of large quantity of heatenergyChemical reactions produce or absorb smaller quantity of heatenergy


ivNuclear reactions are accompanied by a loss in mass mass defect


Chemical reactions are not accompanied by a loss in mass


vRate of decay disintegration of nuclide is independent of physical conditionsThe rate of a chemical reaction is dependent on physical conditions of temperaturepressurepurityparticle size surface area


bBelow is a radioactive decay series starting from 21483 Bi and ending at 20682 Pb Study it and answer the question that follows


Identify the particles emitted in steps I and III 2mks


I  particle


III  ray


iiWrite the nuclear equation for the reaction which takes place in a step I


21483Bi  21081Bi  4 2 He


b step 1 to 3


21483Bi  21081Bi  4 2 He  2 0 1 e


c step 3 to 5


21082Pb  20682Pb  4 2 He  2 0 1 e


c step 1 to 5


21483Bi  20682Pb  2 4 2 He  3 0 1 e


The table below give the percentages of a radioactive isotope of Bismuth that remains after decaying at different times


iOn the grid below  plot a graph of the percentage of Bismuth remainingVertical axis against time


iiUsing the graph determine the


I Half  life of the Bismuth isotope


II Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 016g 2mks


d Give one use of radioactive isotopes in medicine 1mk


14aDistinguish between nuclear fission and nuclear fusion 2mks


Describe how solid wastes containing radioactive substances should be disposed of 1mk


biFind the values of Z1 and Z2 in the nuclear equation below


Z1 1 94 140 1


U  n  Sr  Xe  2 n


92 0 38 Z2 0


iiiWhat type of nuclear reaction is represented in b i above


A radioactive cobalt 6128Co undergoes decay by emitting a beta particle and forming Nickel atom


Write a balanced decay equation for the above change 1 mark


If a sample of the cobalt has an activity of 1000 counts per minute determine the time it would take for its activity to decrease to 6250 if the halflife of the element is 30years 2 marks


Define the term halflife


The diagram below shows the rays emitted by a radioactive sample


Identify the rays SR and Q


S Beta   particleray


R Alpha  particleray


Q Gamma y particleray


b State what would happen if an aluminium plate is placed in the path of ray RS and Q


Ris blockedstoppeddo not pass through


Qis not blockedpass through


Sis blockedstoppeddo not pass through


cThe diagram bellow is the radioactive decay series of nuclide A which is 24194PuUse it to answer the questions that follow The letters are not the actual symbols of the elements


aWhich letter represent the  Explain


ishortest lived nuclide


Lhas the shortest half life


iilongest lived nuclide


PIs stable


iii nuclide with highest np ratio


Lhas the shortest half life thus most unstable thus easilyquickly decaydisintegrate


iv nuclide with lowest np ratio


Pis stable thus do not decaydisintegrate


bHow long would it take for the following


iNuclide A to change to B


10 years half life of A


ii Nuclide D to change to H


27days 162000years70000years16days


232000 years and 43 days


iii Nuclide A to change to P


Study


ATHE RATE OF CHEMICAL REACTION


CHEMICAL KINETICS


The rate of a chemical reaction is the time taken for a given massamount of products to be formed The rate of a chemical reaction is also the time taken for a given massamount of reactant to be consumed used up


Some reactions are too slow to be determined eg rusting decomposition of hydrogen peroxide and weathering


Some reactions are too fast and instantaneous eg neutralization of acid and basesalkalis in aqueous solution and double decompositionprecipitation


Other reactions are explosive and very risky to carry out safely eg reaction of potassium with water and sodium with dilute acids


The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries


Theories of rates of reaction


The rate of a chemical reaction is defined as the rate of change of concentrationamount of reactants in unit time It is also the rate of formation of given concentration of products in unit time ie


Rate of reaction  Change in concentrationamount of reactants


Time taken for the change to occur


Rate of reaction  Change in concentrationamount of products formed


Time taken for the products to form


For the above therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product


The SI unit of time is seconds but minutes and hours are also used


aThe collision theory


The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of smalltinyminute particles like ions atoms and molecules


The collision theory proposes that


ifor a reaction to occur reacting particles must collide


iinot all collisions between reacting particles are successful in a reaction Collisions that initiate a chemical reaction are called successful  fruitful effective collisions


iiithe speed at which particles collide is called collision frequency


The higher the collision frequency the higher the chances of successful  fruitful effective collisions to form products


ivthe higher the chances of successful collisions the faster the reaction


vthe average distance between solid particles from one another is too big for them to meet and collide successfully


vidissolving substances in a solvent make the solvent a medium for the reaction to take place


The solute particle distance is reduced as the particle ions are free to move in the solvent medium


viisuccessful collisions take place if the particles colliding have the required energy and right orientation which increases their vibration and intensity of successful  fruitful effective collisions to form products


bThe Activation EnergyEa theory


The Enthalpy of activationHa Activation EnergyEa is the minimum amount of energy which the reactants must overcome before they react Activation EnergyEa is usually required needed in bond breaking of the reacting particles


Bond breaking is an endothermic process that require an energy input


The higher the bond energy the slower the reaction to start of


Activation energy does not influence whether a reaction is exothermic or endothermic


The energy level diagrams below shows the activation energy for exothermic and endothermic processesreactions


Energy level diagram showing the activation energy for exothermic processes reactions


Activated complex


Energy level diagram showing the activation energy for endothermic processes reactions


The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions but can theoretically exist


Exothermic reaction proceeds without further heating external energy because it generates its own energyheat to overcome activation energy


Endothermic reaction cannot proceed without further heating external energy because it does not generates its own energyheat to overcome activation energy It generally therefore requires continuous supply of more energyheat to sustain it to completion


3 Measuring the rate of a chemical reaction


The rate of a chemical reaction can be measure as


iVolume of a gas in unit time


if reaction is producing a gas as one of the products


if reaction is using a gas as one reactants


iiChange in mass of reactantsproducts for solid productsreactants in unit time


iiiformation of a given mass of precipitate in unit time


iva certain mass of reactants to completely form productsdiminish


Reactants may be homogenous or heterogenous


Homogenous reactions involve reactants in the same phasestate eg solidsolidgasgasliquidliquid


Heterogenous reactions involve reactants in the different phasestate eg solidliquidgasliquidsolidgas


4 Factors influencingalteringaffectingdetermining rate of reaction


The following factors alterinfluenceaffectdetermine the rate of a chemical reaction


aConcentration


bPressure


c Temperature


dSurface area


eCatalyst


Influence of concentration on rate of reaction


The higher the concentration the higher the rate of a chemical reaction An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products


Practically an increase in concentration reduces the time taken for the reaction to take place


Practical determination of effect of concentration on reaction rate


Method 1a


Reaction of sodium thisulphate with dilute hydrochloric acid


Procedure


Measure 20cm3 of 005M sodium thisulphate into a 50cm3 glass beaker Place the beaker on a white piece of filter paper with ink mark X on it Measure 20cm3 of 01M hydrochloric acid solution using a 50cm3 measuring cylinder Put the acid into the beaker containing sodium thisulphate Immediately start off the stop watchclock Determine the time taken for the ink mark X to become invisible obscured when viewed from above Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1 Sample resultsTable 1


For most examining bodiescouncilsboards the above results score for


a complete table as evidence for all the practical work done and completed


b iConsistent use of a decimal point on time as evidence of understandingknowledge of the degree of accuracy of stop watchesclock


iiConsistent use of a minimum of four decimal points on inversereciprocal of time as evidence of understandingknowledge of the degree of accuracy of scientific calculator


c accuracy against a school value based on candidates teachersresults submitted


d correct trend time increase as more water is addedacid is diluted in conformity with expected theoretical results


Sample questions


1 On separate graph papers plot a graph of


ivolume of acid usedxaxis against time Label this graph I


ii volume of acid usedxaxis against 1t Label this graph II


2 Explain the shape of graph I


Dilutingadding water is causes a decrease in concentration


Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products


Sketch sample Graph I


Sketch sample Graph II


3From graph II determine the time taken for the cross to be obscuredinvisible when the volume of the acid is


i 13cm3


From a correctly plotted graph


1t at 13cm3 on the graph  275 x 102


t  1  275 x 102  363636 seconds


ii 15cm3


1t at 15cm3 on the graph  335 x 102


t  1  335 x 102  298507 seconds


iii 15cm3


1t at 17cm3 on the graph  40 x 102


t  1  40 x 102  250 seconds


iv 19cm3


1t at 19cm3 on the graph  465 x 102


t  1  465 x 102  215054 seconds


4From graph II determine the volume of the acid used if the time taken for the cross to be obscuredinvisible is


i25 seconds


1t  125  40 x 102


Reading from a correctly plotted graph


40 x 102 correspond to 170 cm3


ii30 seconds


1t  130  333 x 102


333 x 102 correspond to 147 cm3


iii40 seconds


1t  140  25 x 102


25 x 102 correspond to 123 cm3


4 Write the equation for the reaction taking place


Na2S2O3 aq  2HClaq  2NaCl aq SO2 g  Ss  H2Ol


Ionically


S2O32 aq  2H aq  SO2 g  Ss  H2Ol


5Name the yellow precipitate


Colloidal sulphur


Method 1b


Reaction of sodium thisulphate with dilute hydrochloric acid


You are provided with


20M Hydrochloric acid


04M sodium thiosulphate solution


Procedure


Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker Place the beaker on a white piece of filter paper with ink mark X on it


Add 50cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate


Immediately start off the stop watchclock Determine the time taken for the ink mark X to become invisible obscured when viewed from above


Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1


Sample resultsTable 1


Note concentration of diluted solution is got


C1V1C2V2  04 x 25  C2x 25 04M


C1V1C2V2  04 x 20  C2x 25 032M


C1V1C2V2  04 x 15  C2x 25 024M


C1V1C2V2  04 x 10  C2x 25 016M


Sample questions


iConcentration of sodium thiosulphate against time Label this graph I


iiConcentration of sodium thiosulphate against against T1Label this graph II


2 Explain the shape of graph I


Dilutingadding water causes a decrease in concentration


From graph II


Determine the time taken if


i12cm3 of sodium thisulphate is diluted with 13cm3 of water


At 12cm3 concentration of sodium thisulphate


C1V1C2V2  04 x 1 2  C2x 25 0192M


From correct graph at concentration 0192M  24 x102


It  24 x102 t  416667seconds


ii22cm3 of sodium thisulphate is diluted with 3cm3 of water


At 22cm3 concentration of sodium thisulphate


C1V1C2V2  04 x 22  C2x 25 0352M


From correct graph at concentration 0352M  36 x102


It  36 x102 t  277778seconds


Determine the volume of water and sodium thiosulphate if T1 is 30 x101


From correct graph at T1  30 x101  concentration  065 M


C1V1C2V2  04 x 25  065 M x V2  153846cm3


Volume of water  25  153846cm3  96154cm3


Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used


Mole ratio Na2S2 O3 HCl 12


Moles of Na2S2 O3  0192M x 12  2304 x 103 moles


Mole ratio HCl 2304 x 101 moles  1152 x 103 moles


Molarity o f HCl  1152 x 103 moles x 1000  02304M


Method 2


Reaction of Magnesium with dilute hydrochloric acid


Procedure


Scub 10centimeter length of magnesium ribbon with sand papersteel wool Measure 40cm3 of 05M dilute hydrochloric acid into a flask Fill a graduated gas jar with water and invert it into a trough Stopper the flask and set up the apparatus to collect the gas produced as in the set up below


Carefully remove the stopper carefully put the magnesium ribbon into the flask  cork tightly Add the acid into the flask Connect the delivery tube into the gas jar Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below


Sample results Table II


Sample practice questions


1Plot a graph of volume of gas produced yaxis against time


2Explain the shape of the graph


The rate of reaction is faster when the concentration of the acid is high


As time goes on the concentration of the acid decreases and therefore less gas is produced


When all the acid has reacted no more gas is produced after 210 seconds and the graph flattens


3Calculate the rate of reaction at 120 seconds


From a tangent at 120 seconds rate of reaction  Change in volume of gas


Change in time


From the tangent at 120seconds V2  V1  9684  12  02cm3sec1


T2  T1 15090 60


4 Write an ionic equation for the reaction taking place


Mg2s  2Haq  Mg2aq  H2 g


5 On the same axis sketch then explain the curve that would be obtained if


i 01 M hydrochloric acid is used Label this curve I


ii10 M hydrochloric acid is used Label this curve II


Observation


Curve I is to the right


Curve II is to the left


Explanation


A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction


An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction


Both graphs flatten after some time indicating the completion of the reaction


bInfluence of pressure on rate of reaction


Pressure affects only gaseous reactants


An increase in pressure reduces the volumeBoyles law in which the particles are contained


Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effectivesuccessfulfruitful collision to form products


An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react


At industrial level the following are some reactions that are affected by pressure


aHaber process for manufacture of ammonia


N2g  3H2g  2NH3g


bContact process for manufacture of sulphuricVIacid


2SO2g  O2g  2SO3g


cOstwalds process for the manufacture of nitricVacid


4NH3g  5O2g  4NO g  6H2O l


The influence of pressure on reaction rate is not felt in solids and liquids


This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom Kinetic Theory of matter


cInfluence of temperature on rate of reaction


An increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency


Increase in temperature increases the particles which can overcome the activation energy Ea


A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half


Practical determination of effect of Temperature on reaction rate


Method 1


Reaction of sodium thisulphate with dilute hydrochloric acid


Procedure


Measure 20cm3 of 005M sodium thisulphate into a 50cm3 glass beaker


Place the beaker on a white piece of filter paper with ink mark X on it


Determine and record its temperature as room temperature in table 2 below


Measure 20cm3 of 01M hydrochloric acid solution using a 50cm3 measuring cylinder


Put the acid into the beaker containing sodium thisulphate


Immediately start off the stop watchclock


Determine the time taken for the ink mark X to become invisible obscured when viewed from above


Measure another 20cm3 separate portion of the thisulphate into a beaker heat the solution to 30oC


Add the acid into the beaker and repeat the procedure above Complete table 2 below using different temperatures of the thiosulphate


Sample resultsTable 2


Sample practice questions


Plot a graph of temperaturexaxis against 1t


2aFrom your graph determine the temperature at which


i1t is


I 003


Reading directly from a correctly plotted graph  3225 oC


II 007


Reading directly from a correctly plotted graph  480 oC


ii t is


I 30 seconds


30 seconds  1t 130 0033


Reading directly from a correctly plotted graph 0033  335 oC


II 45 seconds


45 seconds  1t 145 0022


Reading directly from a correctly plotted graph 0022  290 oC


III 25 seconds


25 seconds  1t 125 004


Reading directly from a correctly plotted graph 004  360 oC


b From your graph determine the time taken for the cross to become invisible at


i 575 oC


Reading directly from a correctly plotted graph at 575 oC 0094


1t  0094


t 10094  106383 seconds


ii 45 oC


Reading directly from a correctly plotted graph at 45 oC  0062


1t  0062


t 10094  161290 seconds


iii 35 oC


Reading directly from a correctly plotted graph at 35 oC  0047


1t  0047


t 10047  212766 seconds


Method 2


Reaction of Magnesium with dilute hydrochloric acid


Procedure


Scub 5centimeter length of magnesium ribbon with sand papersteel wool


Cut the piece into five equal one centimeter smaller pieces


Measure 20cm3 of 10M dilute hydrochloric acid into a glass beaker


Put one piece of the magnesium ribbon into the acid swirl


Immediately start off the stop watchclock


Determine the time taken for the effervescencefizzingbubbling to stop when viewed from above


Record the time in table 2 at room temperature


Measure another 20cm3 portions of 10M dilute hydrochloric acid into a clean beaker


Heat separately one portion to 30oC 40oC  50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence fizzing bubbling to stop when viewed from above


Record each time to complete table 2 below using different temperatures of the acid


Sample resultsTable 1


Sample practice questions


Plot a graph of temperaturexaxis against 1t


2aCalculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1gMg 240


Moles  Mass of magnesium  10  4167 x 10 2 moles


Molar mass of Mg 24


bCalculate the number of moles of hydrochloric acid used


Moles of acid  molarity x volume of acid


10 x 20  20 x 10 2 moles


cCalculate the mass of magnesium that remain unreacted


Mole ratio Mg HCl  12


Moles Mg   moles HCl


x 20 x 10 2 moles  10 x 10 2 moles


Mass of reacted Mg  moles x molar mass


10 x 10 2 moles x 24  024 g


Mass of unreacted Mg  Original total mass  Mass of reacted Mg


10 g  024  076 g


bCalculate the total volume of hydrogen gas produced during the above reactions


Mole ratio Mg  H2  11


Moles of Mg that reacted per experiment  moles H2 10 x 10 2 moles


Volume of Hydrogen at stp produced per experiment  moles x 24 dm3


10 x 10 2 moles x 24 dm3  024dm3


Volume of Hydrogen at stp produced in 5 experiments 024 dm3 x 5


12 dm3


3aAt what temperature was the time taken for magnesium to react equal to


70 seconds  1t 170 001429


Reading directly from a correctly plotted graph 001429  280 oC


40 seconds  1t 140 0025


Reading directly from a correctly plotted graph 0025  320 oC


bWhat is the time taken for magnesium to react if the reaction was done at


i 550 oC


Reading directly from a correctly plotted graph at 550 oC 1t  80 x 102


t  180 x 102  125 seconds


ii 470 oC


Reading directly from a correctly plotted graph at 470 oC 1t  60 x 102


t  160 x 102  166667 seconds


iii 330 oC


Reading directly from a correctly plotted graph at 330 oC 1t  27 x 102


t  127 x 102  37037 seconds


4 Explain the shape of the graph


Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products


dInfluence of surface area on rate of reaction


Surface area is the area of contact An increase in surface area is a decrease in particle size Practically an increase in surface area involves chopping cutting solid lumps into smaller pieceschips then crushing the chips into powder Chips thus have a higher surface area than solid lumps but powder has a highest surface area


An increase in surface area of solids increases the area of contact with a liquid solution increasing the chances of successfuleffectivefruitful collision to form products The influence of surface area on rate of reaction is mainly in heterogeneous reactions


Reaction of chalkcalcium carbonate on dilute hydrochloric acid


Procedure


Measure 20cm3 of 10 M hydrochloric acid into three separate conical flasks labeled C1 C2 and C3


Using a watch glass weigh three separate 25g a piece of white chalk Place the conical flask C1 on an electronic balance


Reset the balance scale to 00


Put one weighed sample of the chalk into the acid in the conical flask Determine the scale reading and record it at time 00


Simultaneously start of the stop watch


Determine and record the scale reading after every 30 seconds to complete Table I Repeat all the above procedure separately with C2 and C3 to complete Table II and Table III by cutting the chalk into small pieceschips for C2 and crushing the chalk to powder for C3


Sample resultsTable 1


Sample resultsTable 1I


Sample resultsTable III


Sample questions


1Calculate the loss in mass made at the end of each time from the original to complete table III and III


2On the same axes plot a graph of total loss in mass against time xaxes and label them curve I II and III from Table I II and III


3Explain why there is a loss in mass in all experiments


Calcium carbonate react with the acid to form carbonIVoxide gas that escape to the atmosphere


4Write an ionic equation for the reaction that take place


CaCO3s  2Haq  Ca2aq  H2Ol  CO2g


5SulphuricVIacid cannot be used in the above reaction On the same axes sketch the curve which would be obtained if the reaction was attempted by reacting a piece of a lump of chalk with 05M sulphuricVIacid Label it curve IV Explain the shape of curve IV


Calcium carbonate would react with dilute 05M sulphuricVIacid to form insoluble calcium sulphateVI that coat cover unreacted Calcium carbonate stopping the reaction from reaching completion


6Calculate the volume of carbonIVoxide evolvedmolar gas volume at room temperature  24 dm3 C 120 O 16O Ca400


Method I


Mole ratio CaCO3s  CO2g  11


Moles CaCO3s used  Mass CaCO3s  0025 moles


Molar mass CaCO3s


Moles CO2g  0025 moles


Volume of CO2g  moles x molar gas volume


0025 moles x 24 dm3  0600 dm3600cm3


Method II


Molar mass of CaCO3s  100g produce 24 dm3 of CO2g


Mass of CaCO3s 25 g produce 25 x 24  0600dm3


7From curve I determine the rate of reaction loss in mass per secondat time 180 seconds on the curve


From tangent at 180 seconds on curve I


Rate  M2M1  208  1375  0625  0006944g sec1


T2 T1 222132 90


8What is the effect of particle size on the rate of reaction


A larger surface area is a reduction in particle size which increases the area of contact between reacting particles increasing their collision frequency


Theoretical examples


1 Excess marble chips were put in a beaker containing 100cm3 of 02M hydrochloric acid The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below


aWhy was there a loss in mass


Carbon IV oxide gas was produced that escape to the surrounding


bCalculate the average rate of loss in mass between


i 0 to 2 minutes


Average rate M2M1  180  00  18  900g min1


T2 T1 20  00 2


i 6 to 8 minutes


Average rate M2M1  320  295  025  0125g min1


T2 T1 80  60 2


iii Explain the difference between the average rates of reaction in i andii above


Between 0 and 2 minutes  the concentration of marble chips and hydrochloric acid is high therefore there is a higher collision frequency between the reacting particles leading to high successful rate of formation of products


Between 6 and 8 minutes  the concentration of marble chips and hydrochloric acid is low therefore there is low collision frequency between the reacting particles leading to less successful rate of formation of products


cWrite the equation for the reaction that takes place


CaCO3s  2HCl aq  CaCO3 aq  H2Ol  CO2g


dState and explain three ways in which the rate of reaction could be increased


iHeating the acid increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency


iiIncreasing the concentration of the acidincreasing in concentration reduces the distances between the reacting particles increasing their chances of effectivefruitfulsuccessful collision to form products faster


iiiCrushing the marble chips to powderthis reduces the particle sizeincrease surface area increasing the area of contact between reacting particles


eIf the solution in the beaker was evaporated to dryness then left overnight in the open explain what would happen


It becomes wet because calcium II chloride absorbs water from the atmosphere and form solutionis deliquescent


fWhen sodium sulphate VI was added to a portion of the contents in the beaker after the reaction  a white precipitate was formed


iName the white precipitate


CalciumIIsulphateVI


iiWrite an ionic equation for the formation of the white precipitate


Ca2aq  SO42aqCaSO4s


iiiState one use of the white precipitate


Making plaster for building


Manufacture of plaster of Paris


Making sulphuricVIacid


gi Plot a graph of total loss in massyaxes against time


iiFrom the graph determine the rate of reaction at time 2 minutes


From a tangentslope at 2 minutes


Rate of reaction  Average rate M2M1  225  130  095  03958g min1


T2 T1 320  08 24


iiiSketch on the same axes the graph that would be obtained if 002M hydrochloric acid was used Label it curve II


e Influence of catalyst on rate of reaction


Catalyst is a substance that alter the rate speed of a chemical reaction but remain chemically unchanged at the end of a reaction Biological catalysts are called enzymes A catalyst does not alter the amount of products formed but itself may be altered physically eg from solid to powder to fine powder Like biological enzymes a catalyst only catalyse specific type of reactions


Most industrial catalysts are transition metals or their compounds Catalyst works by lowering the Enthalpy of activationHaactivation energy Ea of the reactants The catalyst lowers the Enthalpy of activationHaactivation energy Ea by


i forming short lived intermediate compounds called activated complex that break up to form the final products


ii being absorbed by the reactants thus providing the surface area on which reaction occurs


A catalyst has no effect on the enthalpy of reaction Hr but only lowers the Enthalpy of activationHaactivation energy EaIt thus do not affectinfluence whether the reaction is exothermic or endothermic as shown in the energy level diagrams below


Energy level diagram showing the activation energy for exothermic processes reactions


Activated complex


Ea Catalysed


Energy level diagram showing the activation energy for endothermic processes reactions


The following are some catalysed reaction processes


aThe contact process


VanadiumV OxideV2O5 or platinumPt catalyses the oxidation of sulphurIVoxide during the manufacture of sulphuricVI acid from contact process


SO2g  O2g V2O5 SO3g


To reduce industrial cost of manufacture of sulphuric VI acid from contact process VanadiumV OxideV2O5 is used because it is cheaper though it is easily poisoned by impurities


bOstwalds process


Platinum promoted with Rhodium catalyses the oxidation of ammonia to nitrogenIIoxide and water during the manufacture of nitricVacid


4NH3g  5O2g PtRh 4NO g  6H2Ol


cHaber process


Platinum or iron catalyses the combination of nitrogen and hydrogen to form ammonia gas


N2g  3H2g Pt or Fe 2NH3g