22 Molarity (moldm-3 ) = Number of mole of solute (mol) Volume of solution(dm3 ) olution. b) What is the molarity of the solution. [RAM: Na,23; O, 16; H,1 ] A: 1) Find no of mole of solute, NaOH Mr m n n = No of mole of NaOH = 2) Molarity of solution = no of mole of solute Volume of solution Ans: 0.5 mol dm-3
Conversion of units of concentration from gdm-3 to moldm-3 oConcentration (Con) (gdm-3 ) Mr con M x MQ: Calculate the molarity of a sodium sulphate, Na2SO4 solution with a concentration of 28.4 gdm-3 . [RAM: O, 16; Na,23; S, 32] A: Molarity of Na2SO4 = Concentration Molar mass = 28.4 gdm-3 2Na + S + 4O = Ans: 0.2 moldm-3
23 or vice versa ÷ gmol-1 Molarity (M) (moldm-3 ) rnr x gmol-1 Q: The molarity of a bottle of nitric acid, HNO3 solution is 2.0 moldm-3 . What is the concentration of the solution in gdm-3 ? [RAM: H,1; N, 14; O,16] A: Concentration of HNO3 = molarity x molar mass of HNO3 = 2.0 moldm-3 x [H + N + 3O] = Ans: 126 g dm-3
NUMBER OF MOLE (INVOLVING SOLUTION) Q: Calculate the number of mole of 25.0 cm3 1.5 moldm-3 sodium hydroxide, NaOH solution. A: 1000 MV n = = Ans: 0.0375 mol AnsAns: 0.0375 mol n = number of mole of solute n (mol) Molarity/(moldm-3 )
24 1000 MV concentration )Volume of solution (cm3 ) Divide by 1000 to convert cm3 to dm3
DILUTION Add water M1V1 molarity of solution before dilution (moldm-3 ) volume of water before dilution (cm3 ) Q: Find the volume of 2.0 moldm-3 sulphuric acid, H2SO4 needed to prepare 100 cm3 1.0 moldm-3 sulphuric acid, H2SO4. A: Before dilution After dilution M1 2.0 moldm-3 M2 1.0 moldm-3 V1 ? V2 100 cm3 M1V1 = M2V2 (2.0)V1 = (1.0)(100) V1 = (1.0)(100) 2.0 Volume of H2SO4 before dilution = 50 cm3 Number of mole of solu
25 = M2V2 M2 = molarity of solution after dilution (moldm-3 ) total volume of solution after dilution (cm3 ) ute before dilution = Nombor of mole of solute after dilution Q: Calculate the volume of water needed to be added to 50 cm3 0.5 moldm-3 of hydrochloric acid, HCl to get 0.1 moldm-3 of hydrochloric acid, HCl. A: Before dilution After dilution M1 0.5 moldm-3 M2 0.1 moldm-3 V1 50 cm3 V2 ? M1V1 = M2V2 Ans: Volume of water needed to be added = 200 cm3
NEUTRALISATION Acid + Alkali salt + water ( Soluble base) MaVMbVvolume molarity of acid (moldm-3 ) volume of amolarity of soluble base (moldm-3 ) Regular formula us
26 Va = a Vb b of soluble base (cm3 ) Coefficient of acid from balanced equation Coefficient of alkali from balanced equation acid (cm3 ) ed for neutralisation Eg : 1 H2SO4 + 2 NaOH Na2SO4 + 2H2O
NEUTRALISATION Q: 50 cm3 1.0 moldm-3 sulphuric acid, H2SO4 solution is neutralized by 100 cm3 .sodium hydroxide solution, NaOH. Calculate the concentration of sodium hydroxide, NaOH in moldm-3 . Eq: H2SO4 + 2 NaOH Na2SO4 + 2 H2O Info given : Acid Acid H2SO4 Soluble base Bes terlarut NaOH Ma 1.0 moldm-3 Mb ? Va 50 cm3 Vb 100 cm3 a 1 b 2 MaVa = a MbVb b Ans: 1.0 moldm-3
27 NEUTRALISATION Q: Calculate the volume of 0.1 moldm-3 sulphuric acid, H2SO4 needed to completely react with 50 cm3 of 0.1 moldm-3 potassium hydroxide, KOH. Eq: Acid Acid H2SO4 Soluble base Bes terlarut NaOH Ma Mb Va Vb a b MaVa = a MbVb b Ans: 25 cm3
28 SALT/ GARAM Salt = A salt is an ionic compound formed when the H+ ion in an acid is replaced by a metal ion or an ammonium ion. Garam = Sebatian ion yang terbentuk apabila ion hidrogen, H+ daripada asid digantikan dengan ion logam atau ion ammonium. A) Solubility of salt/ Keterlarutan garam Soluble Insoluble Ni Na K Am Nitrate Sodium Potassium Ammonium NO3 - Na+ K+ NH4 + C Pb Carbonate lead CO3 2- Pb2+ Cl- AgCl PbCl2 SO4 2- BaSO4 PbSO4 CaSO4 Except NiNaKAm Angel Colin Peter Colin Naik Bas PbS Caspher
29 C) Preparation of salt/ Penyediaan Garam Soluble salt Garam terlarut Insoluble salt Garam tak larut (For Na K Am salt) * Na+, K+, NH4 + Eg: Salt: NaNO3 , K2SO4, NH4Cl NaOH+ HNO3 NaNO3 + H2O 2 KOH + H2SO4 K2SO4 + H2O NH3 + HCl NH4Cl (For Non Na K Am salt) * Garam bukan Na+, K+, NH4 + Eg: Salt: CuCl2 (C) HCl + CuCO3 CuCl2 + H2O + CO2 (M) HCl + Cu x rxn (O) 2 HCl + CuO CuCl2 + H2O (H) 2 HCl + Cu(OH)2 CuCl2 + 2H2O * metal except Cu & Ag – unreactive metal Logam kecuali Cu& Ag – logam tidak reactif Double decomposition method/ Precipitation method Tindak balas penguraian ganda dua/ Pemendakan Eg: Salt: PbCl2 Chemical equation: Pb(NO3)2 + 2KCl PbCl2 +2KNO3 Ionic equation: Pb2+ + 2 Cl- PbCl2 Soluble + soluble Insoluble Larut + Larut Tak larut Acid + insoluble powder (CMOH – Carbonate, Metal. Oxide, Hydroxide) Asid + serbuk tak larut (KLOH - Karbonat logam, Logam reaktif, oksida logam, hidroksida logam) Acid + Alkali Asid + Alkali
30 Write the chemical equation to prepare: Tuliskan persamaan kimia untuk menyediakan: Insoluble Salt (Double decomposition) Soluble + Soluble PbSO4 Chem eq: Ionic eq : CaCO3 Chem eq: Ionic eq: PbI2 Chem eq: Ionic eq: ZnCO3 Chem eq: Ionic eq: Soluble Salt (Non Na K am) Acid + CMOH powder CuSO4 1. 2 3. MgCl2 1. 2 3. 4. Zn(NO3)2 1. 2. 3. 4. Soluble Salt (Na K am) Acid + Alkali Na2SO4 Chem eq: KCl Chem eq: Always use Ni Na Kam salt PREPARATION OF SALT PENYEDIAAN GARAM
Reagent Cation NaOH NH3 HCl/Dilute NaCl DiH2SDiNaNa+ - - Ca2+ White precipitate. Insoluble in excess. - Mg2+ White precipitate. Insoluble in excess. White precipitate. Insoluble in excess. Al3+ White precipitate. Soluble in excess to form colourless solution. White precipitate. Insoluble in excess. Zn2+ White precipitate. Soluble in excess to form colourless solution. White precipitate. Soluble in excess. Pb2+ White precipitate. Soluble in excess to form colourless solution. White precipitate. Insoluble in excess. White precipitate. Soluble in hot water. WpreciFe2+ Green precipitate. Insoluble in excess. Green precipitate. Insoluble in excess. Fe3+ Brown precipitate. Insoluble in excess. Brown precipitate. Insoluble in excess. Cu2+ Blue precipitate. Insoluble in excess. Blue precipitate. Soluble in excess and producing dark blue solution. NH4 + - - CATIOOnly Zn2+, Al3+, Pb2+ soluble in excess NaOH. ZAP Only Cu2+, Zn2+ soluble in excess NH3 CuZ CATION TEST
31 ilute SO4 / ilute a2SO4 KI Potassium Hexacyanoferrate (II) Potassium Hexacyanoferrate(III) Potassium thiocyanate Nessler’s Reagent White ipitate. Yellow precipitate. Soluble in hot water. Dark blue precipitate. Dark blue precipitate. Blood-red colouration Brown precipitate ON TEST Procedure: (For NaOH & NH3) 1. Add NaOH solution drop by drop into solution. 2. Shake and observe 3. Add NaOH solution until excess
32 Anion Sulphate, SO4 2- Chloride, Cl- Nitrate, NO3 - Carbonate, CO3 2- Code Su ha bac -wp C na – sini - wp Ni sa fes2-csa-br Diagram dilute HCl + BaCl2 solution dilute HNO3 + AgNO3 solution dilute H2SO4 + FeSO4 solution + Conc H2SO4 dilute HCl Procedure 1. Add dilute hydrochloric acid. 2. Add barium chloride solution. 1. Add dilute nitric acid 2. Add silver nitrate solution 1. Add dilute sulphuric acid 2. Add iron(II) sulphate solution 3. Shake the mixture 4. Add concentrated sulphuric acid drop by drop down the side of the test tube 1. Pour dilute hydrochloric acid in a test tube 2. Add calcium carbonate powder 3. Deliver gas into limewater Observation White precipitate formed White precipitate formed Brown ring formed Colourless gas bubble Lime water turns chalky Inference barium sulphate fomed/ SO4 - present silver chloride formed/ Cl- present NO3 - present CO2 gas released/ CO3 2- present Ionic eq Ba2+ + SO4 2- BaSO4 Ag+ + Cl- AgCl ANION TEST
33 A) Cation Test How to test for Cation Fe2+ Fe3+ Pb2+ Cu2+ Procedure Observation B) Anion test Anion Sulphate, SO4 2- Chloride, Cl- Nitrate, NO3 - Carbonate, CO3 2- Code Procedure 1) Add _______________ 2) Add _______________ 2) Add ______________ 2) Add _______________ 1) Add ________________ 2) Add ________________ Shake 3) Add _________________ _______________________ _______________________ Add ______________ __________________ __________________ Observation
34 How to convert ZnO to ZnCO3 ? How to convert CuCl2 to CuSO4 ? STEP 1 STEP 1 ZnO + H2SO4 ZnSO4 + H2O STEP 2 STEP 2 ZnSO4 + Na2CO3 ZnCO3 + Na2SO4 How to convert Cu(OH)2 to CuCO3 ? How to convert Mg(NO3)2 to MgSO4 ? STEP 1 STEP 1 STEP 2 STEP 2 Heat decomposition of salt CODE for observation on residue: * Zoro pa ku se put - ZnO panas kuning sejuk putih (solid is yellow when hot, white when cold) * Po pan pe se ku - PbO panas perang sejuk kuning (solid is brown when hot, yellow when cold) * CuCO3 – green , CuO - black How to test for the gases? CO2 Deliver the gas into lime water. Lime water turns chalky. NO2 Place a moist blue litmus paper near the mouth of the test tube Moist blue litmus paper turns red. O2 Place a glowing wooden splinter into the mouth of the test tube. Glowing wooden slinter ignites Try this! Challenge yourself! Heating of carbonate salt ZnCO3 ZnO + CO2 (zoropakuseput) PbCO3 PbO + CO2 (popanpeseku) CuCO3 CuO + CO2 (green) (black) Heating of nitrate salt 2 Zn(NO3)2 2ZnO + 4NO2 + O2 (zoropakuseput) 2 Pb(NO3)2 2 PbO + 4 NO2 + O2 (popanpeseku) 2 Cu(NO3)2 2 CuO + 4 NO2 + O2 (blue) (black) Nitrogen dioxide (Brown, acidic)
35 Heat decomposition of salt Chem eq PbCO3 2 Pb(NO3)2 2 PbO + 4 NO2 + O2 Observation: Observation: Solid Solid formed is brown when hot, yellow when cold Gas Brown gas formed which turns moist blue litmus paper red Colourless gas formed which ignites a glowing wooden splinter. Chem eq ZnCO3 Zn(NO3)2 Observation: Observation: Solid Gas
36 RATE OF REACTION COLLISION THEORY Factors Changes Effect Frequency of collision between particles Frequency of effective collision between particles Rate of reaction Size of reactant Decrease total surface area of reactant expose to collision increases Increase Increase Increase Concentration of solution Increase number of particles per unit volume increases Increase Temperature of solution Increases kinetic energy of particles increases Catalyst Present • catalyst will lower down the activation energy • More colliding particles able to overcome the lower activation energy Remain the same
37 Determine Particles Involved Chemical equation Mg + 2 HCl MgCl2 + H2 Ionic equation Mg + 2H+ + 2 Cl- Mg2+ + 2 Cl- + H2 Mg + 2H+ Mg2+ + H2 Particles involved Mg atom & H+ ion Chemical equation Zn + H2SO4 Ionic equation Particles involved Chemical equation CaCO3 + HCl Ionic equation Particles involved Chemical equation ZnCO3 + HNO3 Ionic equation Particles involved Chemical equation Na2S2O3 + H2SO4 Ionic equation Particles involved Chemical equation Na2S2O3 + HCl Ionic equation Particles involved
38 COLLISION THEORY 1. Temperature Eg: hydrochloric acid, HCl and sodium thiosulphate, Na2S2O3 Factor - When the temperature of solution increases, Explain (Effect) - H+ ion and S2O3 2- ion move faster / - the kinetic energy of particles increases FOC betw particle - frequency of collision between H+ ion and S2O3 2- ion increases FOEC - frequency of effective collision between particles increases Rate - rate of reaction increases 2. Concentration Eg: hydrochloric acid, HCl and zinc powder, Zn Factor - When the concentration of acid increases Explain (Effect) FOC betw particle FOEC Rate - rate of reaction increases 3. Size / Total surface area of reactant Eg: hydrochloric acid, HCl and calcium carbonate chip/ calcium carbonate powder Factor - When the size of calcium carbonate decreases, Explain (Effect) FOC betw particle FOEC Rate 4. Catalyst Eg: Decomposition of hydrogen peroxide, H2O2 with catalyst manganese(IV) oxide powder Factor Explain (Effect) 1) 2) FOC betw particle - frequency of collision between H2O2 molecule remain unchange FOEC -frequency of effective collision between particles increases Rate
39 Sketch the graph Sketch the graph (Excess Zn added to HCl) Zn + 2 HCl ZnCl2 + H2 Height of graph (no of mole of limiting reactant - HCl) Steepness of graph (Factors affecting rate of reaction) 1) – 20 cm3 0.1 moldm-3 HCl 2) – 20 cm3 0.2 moldm-3 HCl 1) – no of mole = (20)(0.1) 1000 = 0.002 2) – no of mole = (20)(0.2) 1000 = 0.004 √ (higher graph) * Factor here is concentration 1) 0.1 moldm-3 HCl 2) 0.2 moldm-3 HCl √ (steeper graph) 1) – 40 cm3 0.1 moldm-3 HCl 2) – 20 cm3 0.2 moldm-3 HCl 1) – 50 cm3 0.1 moldm-3 HCl 2) – 25 cm3 0.15 moldm-3 HCl Volume of gas(cm3 ) time(s) 2 1
40 RATE OF REACTION 1. Calculation Excess Calcium carbonate is added into hydrochloric acid. Volume of gas(cm3 ) 22 39 50 57 60 60 Time (s) 30 60 90 120 150 180 Find a. Overall average rate of reaction b. Average rate of reaction at the first 2 minutes c. Average rate of reaction at the 2nd minute 2. Experiment Reactants I 25 cm3 0.5 moldm-3 sulphuric acid + excess magnesium strip II 25 cm3 1.0 moldm-3 sulphuric acid + excess magnesium strip III 25 cm3 1.0 moldm-3 sulphuric acid + excess magnesium powder a) Sketch the graph for experiment I, II and III of volume of gas versus time at the same axes. [3m]
41 b) Compare rate of reaction of Exp I and II Collision theory rate Rate of reaction in set II is higher than set I factor Concentration of sulpuric acid in set II is higher than Set I. effect Number of H+ ion per unit volume in set II is higher than Set I FOC between _ &_ Frequency of collision between H+ ion and Mg atom in set II is higher than set I FOEC Frequency of effective collision between H+ ion and Mg atom in set II is higher than set I b) Compare rate of reaction of Exp II and III Collision theory rate factor effect FOC bt _ &_ FOEC
BACK TO BASIC FORM 5 C A T A L Y S T C H E M I S T R Y M O D U L E J P N S A B A H
Chapter 1: Redox Reaction <Transfer of electron at a distance> Type of reaction Oxidation of Fe2+ by acidified KMnO4 Reactant Acidified potassium manganate(VII) Iron(II) sulphate Formula KMnO4 FeSO4 Particle involved in reaction MnO4 - ion _________________ _________ Iron(II) ion Observation Solution change from __________________ Solution change from _________________ Confirmation test - Add excess NaOH solution. Brown precipitate insoluble in excess NaOH solution Changes of particle MnO4 - ion Mn2+ ion Manganate(VII) ion to Manganese(II) ion ____________________ Iron(II) ion to iron(III) ion Half equation Overall ionic equation Terminal Positive ACEP Negative DEN Oxidation/ Reduction (Inference) ________accept electron and is reduced to _______ ACER Fe2+ ion donate electron and is ________ to Fe3+ ion DEO MnO4 - ion is reduced because the oxidation number of Mn decrease from _________ Fe2+ is oxidized because the oxidation number of ___increase from _______ Oxidizing and Reducing agent MnO4 - ion is the _________agent ________is the __________agent Transferring electron Electron transfer from Fe2+ ion to MnO4 - ion Conclusion Acidified potassium manganate(VII) solution oxidised Fe2+ ion to Fe3+ ion Diagram Before After Fe2+ Fe3+ 1
Type of reaction Oxidation of Fe2+ by acidified K2Cr2O7 Reactant Acidified Potassium dichromate(VI) Iron(II) sulphate Formula Particle involved in reaction ___________ Dichromate (VI) ion Fe2+ ion ___________ Observation Solution change from ____________________________ Solution change from __________________________ Confirmation test - Add excess NaOH solution. Brown precipitation insoluble in excess NaOH solution Changes of particle ______ion _____ion Dichromate (VI) ion to chromium(III) ion Half equation Overall ionic equation Cr2O7 2- + 14H+ + 6e- 2Cr3+ + 7H2O 6Fe2+ 6Fe3+ + 6e- Cr2O7 2- + 6Fe2+14H+ 2Cr3+ + 6Fe3++ 7H2O Terminal Positive ACEP Negative DEN Oxidation/ Reduction Cr2O7 2- ion _______electron and is reduced. ACER Fe2+ ion _______electron and is __________. DEO Cr2O7 2- ion is reduced because the oxidation number of Cr decrease from ___________ Fe2+ is oxidized because the oxidation number of ____ increase from ____________ Oxidizing and Reducing agent Cr2O7 2- ion is the ________agent Fe2+ ion is the ________agent Transferring electron Electron transfer from Fe2+ ion to Cr2O7 2- ion Conclusion Acidified Potassium dichromate(VI) ion oxidised iron(II) sulphate to iron(III) sulphate Diagram Before After 2
Try to do: Fe3+ with KI solution Type of reaction Reduction of Fe3+ by zinc Reactant Zinc Iron(III) sulphate Formula Zn Fe2(SO4)3 Particle involved in reaction Zn atom Zinc atom Observation Solution change from ________ _________ Confirmation test - Add excess _______________ into the test tube. ________________________________ _________________________________ Changes of particle Zn atom Zn2+ ion Zinc atom to zinc ion Half equation Overall ionic equation Terminal DEN ACER Oxidation/ Reduction (Inference) Zinc atom donate electron and is oxidized to _______ DEO Fe3+ ion accept electron and is reduced to ______ ACER Fe2+ is reduced because the oxidation number of Fe decrease from +3 to +2 Oxidizing and Reducing agent Fe3+ ion is the oxidizing agent Transferring electron Electron transfer from zinc to Fe3+ ion Conclusion Zinc reduced Fe3+ ion to Fe2+ ion Diagram Before After Fe3+ Fe2+ 3
Displacement of metal Type of reaction Displacement of metal Reactant Zinc Copper(II) sulphate Formula Particle involved in reaction Zn atom Zinc atom Observation Observation 1) 2) 3) Inference 1) 2) 3) Changes of particle Cu2+ ion Cu atom Copper(II) ion to copper atom Half equation Overall ionic equation Overall chemical equation Oxidation/ Reduction (Inference) Zinc atom donate electron and is oxidized to Zn2+ ion DEO ACER Cu2+ is reduced because the oxidation number of Cu decrease from +2 to 0 Oxidizing and Reducing agent Zinc atom is the reducing agent Transferring electron Electron transfer from zinc to Cu2+ ion Conclusion Zinc is more electropositive than copper. Zinc can displace copper from copper(II) sulphate solution. Diagram Before After Try to do: Cu with AgNO3 4
Displacement of halogen Do: KBr + Cl2 Type of reaction Displacement of halogen Reactant Chlorine Potassium iodide Formula Particle involved in reaction Cl2 molecule Chlorine molecule Observation Solution change from __________ to ___________ Solution change from _________to _________ Colour of aqueous layer in the test tube: Confirmation test - Add 2cm3 1,1,1-trichloroethane to the mixture and shake. Purple layer form at the bottom. Changes of particle Cl2 molecule Cl- ion chlorine molecule to chloride ion Half equation Overall ionic equation Overall chemical equation Oxidation/ Reduction (Inference) Chlorine molecule ___________and is ____________to Cl- ion ACER DEO Chlorine molecule is reduced because the oxidation number of ___decrease from _____ Oxidizing and Reducing agent _________is the oxidizing agent Transfering electron Concludion Chlorine is more electronegative than iodine. Chlorine can displace iodine from potassium iodide solution. Diagram Before After 5
Reactant Half Equation OXIDATION (Donates Electron) REDU(AcceptsFeSO4 + KMnO4 FeSO4 + K2Cr2O7 FeSO4 + Br2 FeSO4 + Cl2 Fe2+ Fe3+ + 2 e- Cl2 + 2Fe2(SO4)3 + Mg Fe2(SO4)3 + KI Cl2 + KI Br2 + KI Cl2 + KBr KMnO4 + KI K2Cr2O7 + KBr Cu + AgNO3 Zn + CuSO4 ACER DEO – accept electrons red
Observation UCTION s Electron) OXIDATION REDUCTION e- 2ClSolution changes from green to brown Solution change from yellow to colourless uction, donates electron oxidation 6
Electrochemistry Notes Electrolysis of molten lead(II) bromide, PbBr2 Electrode (carbon) Cathode - E (Cation) + Ion + E Anode - Ion (Anion) Ion in electrolyte Pb2+ Br - Changes of particle From Pb2+ ion to Pb atom From Br- ion to Br atom then to Br2 molecule Half Ion equation Pb2+ + 2e- Pb 1) Pb2+ ion moves to negative electrode, 2) Pb2+ ion accepts 2 electron to form lead atom. Br - Br + e2Br - Br2 + 2e1) Br– ion moves to positive electrode, 2) Br – ion donates 1 electron to form bromine atom. 3) 2 bromine atom combine to form a bromine molecule. Product Lead metal Bromine gas Observation Shiny grey solid found at the bottom of the crucible. Reddish-brown gas with a pungent smell is released. Oxidation/ Reduction (a) Pb2+ ion is reduced because the oxidation number of lead decreases from +2 to 0. (b) Pb2+ ion is reduced because Pb2+ ion accepts electron. (ACER) (a) Br- ion is oxidised because the oxidation number of bromine increases from -1 to 0. (b) Br- ion is oxidised because Br- ion donates electron. (DEO) Diagram Procedures: 1. Fill a crucible with PbBr2 solid. 2. Dip 2 carbon electrodes into PbBr2 solid. 3. Complete the circuit by connecting the electrodes with wires and batteries. 4. Heat the PbBr2 solid until completely melt. 7
Electrolysis of sodium chloride, NaCl solution (2 moldm-3) Electrode (carbon) Cathode (-) Anode (+) Ion moves to Na+ √ H+ EV √ Cl- OHCI Ions selectively discharge H+ ClHalf equation 2H+ + 2e- H2 2Cl - Cl2 + 2eExplanation 1) Na+ ion and H+ ion move to negative electrode, 2) H+ ion is selectively discharged because Eo value of H+ ion is more positive than Eo value of Na+ ion. 3) H+ ion accepts electron to form hydrogen molecule. 1) Cl– ion and OH- ion move to positive electrode, 2) Cl– ion is selectively discharged because the concentration of Cl- ion is higher than OH- ion 3) Cl- ion donates electron to form chlorine molecule. Product Hydrogen gas Chlorine gas Observation Colourless gas bubbles is released Greenish -yellow gas bubble with a pungent smell is released. Confirmation test Procedure: Insert a burning wooden splinter into the mouth of the test tube with collected gas. Procedure: Place a moist blue litmus paper to the mouth of test tube with collected gas. Observation: A ‘pop’ sound produced. Observation: Blue litmus paper change to red and then white. Diagram Before After Factors influence the choice of ions to be discharged: E.V = Eo value C.I = concentration of ion T.E = type of elctrode 8
Electrolysis of copper(II)sulphate, CuSO4 solution(carbon electrode) Electrode (carbon) Cathode (-) Anode (+) Ion moves to √ Cu2+ H+ EV SO4 2- √ OHEV Ion selectively discharged Cu2+ OHHalf equation Cu2+ + 2e- Cu 4OH- O2 + 2H2O + 4e- Explaination 1) Cu2+ ion and H+ ion move to negative electrode, 2)Cu2+ ion is selectively discharged because Eo value of Cu2+ ion is more positive than Eo value of H+ ion. 3) Cu2+ ion accepts 2 electron to form copper atom. 1) SO4 2- ion and OH- ion moves to positive electrode, 2) OH– ion is selectively discharged because Eo value of OH- ion is less positive than Eo value SO4 2- ion. Product Copper Oxygen Observation Brown solid deposited Colourless gas bubbles released Confirmation test - Insert a glowing wooden splinter into the mouth of the test tube with collected gas. The wooden splinter ignites. Electrolyte The intensity of the blue colour of the solution decreases because the concentration of Cu2+ ion decrease. Cu2+ ion accepts electron to become copper atom. Diagram Before After 9
Electrolysis of copper(II)sulphate, CuSO4 solution(copper electrode) Electrode (carbon) Cathode (-) Anode (+) Ion moves to √ Cu+ H+ EV SO4 2- OHTE Ions selectively discharged Cu2+ - Half Ion equation Cu2+ + 2e- Cu Cu Cu2+ + 2eExplaination 1) Cu2+ ion and H+ ion move to negative electrode, 2)Cu2+ ion is selectively discharged because Eo value of Cu2+ ion is more positive than Eo value of H+ ion. 3) Cu2+ ion accepts 2 electrons to form copper atom. 1) SO4 2- ion and OH- ion moves to positive electrode but no ions are selectively discharged. 2) Copper is an active electrode 2) Copper atom donates 2 electrons to form Cu2+ ion Product Copper Copper(II) ion Observation Brown solid deposited. Copper electrode become thicker. Copper electrode dissolves and become thinner. Electrolyte The intensity of the blue colour of the solution remain unchanged because the concentration of Cu2+ ion remain unchanged. Rate of Cu2+ ion discharged at cathode = Rate of Cu2+ ion formed at anode Diagram Before After 10
3 Application of electrolysis in industry: 1) Purification of metal Electrolyte Copper(II) sulphate solution, CuSO4 Electrode Anode (impure copper) Cathode (pure copper) Half equation Cu → Cu2+ + 2e- Cu2+ + 2e- → Cu Observation The impure copper electrode dissolves and becomes thinner. The pure copper electrode becomes thicker. Inference Copper atom donates electron to become copper(II) ion Copper(II) ion accept electron to become copper atom Product Copper(II ) ion Copper atom Diagram Before After 2) Electroplating Electrolyte Silver nitrate solution, AgNO3 Electrode Anode (silver) Cathode (iron spoon) Half equation Ag Ag+ + e- Ag+ + e- Ag Observation The silver electrode becomes thinner. Grey solid deposited. Inference Silver atom donates electron to become silver ion Silver ion accept electron to become silver atom Product Silver ion Silver atom Diagram Before After 3) Extraction of metal (Extract aluminium from aluminium ore) Electrolyte Aluminium oxide, Al2O3 Electrode Graphite (Carbon) Temperature 950˚C (cryolite lower the melting point) Cathode Anode Explanation Al3+ ions are discharged to form aluminium metal. O2- ions are discharged to form oxygen gas. Half equation Al3+ + 3e- → Al 2O2- → O2 + 4eChemical equation 2Al2O3 → 4Al + 3O2 Product Aluminium Oxygen gas 11
Exercise: 1) Electrolysis of sodium chloride, NaCl solution (2 moldm-3 ) <CONCENTRATED> 2) Electrolysis of sodium chloride, NaCl solution (0.0001 moldm-3 ) <DILUTE> Anode Cathode Ions move to Half equation Explanation 1) ______ and _______move to anode. 2) __________ is selectively discharged because ___________ ____________________________ ____________________________ Product Observation Confirmation test Procedure: Observation: Procedure: Observation: Anode Cathode Ions move to Half equation Explanation 1) ______ and _______move to anode. 2) __________ is selectively discharged because ___________ ____________________________ ____________________________ Product Observation Confirmation test Procedure: Observation: Procedure: Observation: 12
3) Electrolysis of CuSO4 solution using CARBON & COPPER Anode (Carbon) Anode (Copper) Half equation Explanation 1) ______ and _______move to anode. 2) __________ is selectively discharged because ___________ ____________________________ ____________________________ 1) Copper is an __________ electrode 2) ___________ donates electron to form ____________ . Product Observation Electrode: Electrode: Solution: Solution: Cathode (Carbon) Cathode (Copper) Half equation Observation 13
Application of Electrolysis in Industry 1) Purification of metal Electrolyte Copper(II)nitrate solution, Cu(NO3)2 Electrode Anode (impure copper) Cathode (pure copper) Half equation Observation Inference Diagram 2) Electroplating Electrolyte Silver nitrate solution, AgNO3 Electrode Anode (silver) Cathode (iron spoon) Half equation Observation Inference Diagram 3) Extraction of metal (Extract aluminium from aluminium ore, Al2O3) Electrolyte: Molten Aluminium oxide, Al2O3 Electrode: Carbon Add Cryolite : To ____________________________________ Anode Cathode Half equation Product Observation 14
Voltaic Cell Cell P Cell Q Anode Cathode Anode Cathode Eq: Obs: Suggest electrode X : _______________, Solution Y: __________________________ Metal Displacement Half eq : 1) 2) Ionic eq : Chem eq: Anode (- terminal) Cathode (+ terminal) Half equation Inference Fe(NO Observation 3)2 Fe Zn Zn(NO3)2 Observation Inference 1) 2) 3) 4) Copper Silver nitrate solution 15
Electrolyte Anode Electro de - Ion Product Half equation Observation PbBr2 (molten) C Al2O3(molten) C HBr(aq) 1moldm-3 C HCl(aq) 0.0001moldm-3 C NaCl (aq)2 moldm-3 C Na2SO4 (aq) C H2SO4 (aq) Pt AgNO3 (aq) Ag KI(aq)1 moldm-3 C CuCl2(aq) 1moldm-3 C CuSO4 (aq) C SO4 2- √ OHO2 4OH- O2 + 2H2O + 4eColourless gas bubble CuSO4 (aq) Cu ZnSO4 (aq) C Cu(NO3)2 (aq) Cu AgNO3 (aq) Ag EleFactors influence tC.I =T.E
Cathode Factor Electrode + Ion Product Half equation Observation Factor C C C C C C Pt Ag C C E.V C √Cu2+ H+ Cu Cu2+ + 2e- Cu Brown solid E.V Cu C Fe Fe ectrolysis Table the choice of ions to be discharged: E.V = Eo value = concentration of ion E= type of elctrode 16
Reactant Reaction React with Condition Chemical equpropene Hydrogenation Hydrogen gas, H2 -Nickel, Ni - 180o C Chlorine/Bromine water, Cl2/ Br2 Or bromine in tetrachloromethane - Steam, H2O - Phosphoric acid, H3PO4 - 300o C - 60 atm Hydrogen halide, HCl - Acidified potassium manganate(VII) solution, KMnO4/H+ - Polymerisation Trace of oxygen 200o C 1200 atm * Combustion Oxygen, O2 - ChapteChemica
uation Product (Structural formula & name) Procedure (Testing for carboncarbon double bond) Observation propane - - * Bubble propene gas into bromine water and shake Bromine water change from ______ ____________ __________ - - - - * Bubble propene gas into acidified potassium manganate(VII) solution and shake KMnO4 change from ______ ____________ __________ - - Carbon dioxide, water - - er 2: Carbon Compound al Properties of Alkene 17
Reactant Code Reaction name Reacts with Chebutan-1-ol C Combustion Oxygen O Acidified postassium dichromate(VI) solution D Unglazed porcelain chip E Glacial Ethanoic acid & concentrated sulphuric acid (as catalyst) Chemical P
emical equation Product (structural formula & name) Observation Carbon dioxide, water 1) Colour of flame: ____________________ 2) Combustibility: ____________________ 3) Presence of soot: ____________________ butanoic acid 1) Colour change on acidified K2Cr2O7 solution : __________ __________________ 2) Colour of distillate: ____________________ 3) Smell of distillate: ____________________ 4) Effect on blue litmus paper: _______________ 1) Colour of gas collected: ____________ 2) Colour change of Bromine water after testing with product: ____________________ 2) Colour change of acidified KMnO4 solution after testing with product: ____________________ 1) Smell of ester: ______________ 2) Colour of ester: _______________ 3) Density of ester: ____________________ Properties of Alcohol (CODE) 18
Chemical Properties of Carboxylic Acid (CMOH E) Reactant React with Chemical equation ethanoic acid C Metal carbonate CuCO3 2 C3H7COOH + CuCO3 (C3H7COO)2Cu + CO2 + H2O Butanoic acid + Copper(II) carbonate copper(II)butanoate + carbon dioxide + water M Reactive metal Zn O Metal oxide MgO H Metal hydroxide Pb(OH)2 E Esterification C3H7OH 19