JK-TK/Newdata huggingface dataset:
Set up an electrolytic cell. Close the switch and pass a steady current of 2 amperes by adjusting the rheostat for exactly 20 minutes.Remove the electrodes from the electrolyte. Wash with acetone/ propanone and allow them to dry. Reweigh each electrode. Sample results Mass of cathode before electrolysis 7.00 g Mass of anode before electrolysis 7.75 g Mass of cathode after 8.25 g Mass of anode after 6.50 g
45 electrolysis electrolysis Change in mass at cathode after electrolysis 1.25 g Change in mass at anode after electrolysis 1.25 g Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) CuSO4(aq) -> SO42-(aq) + Cu2+(aq) II. Name the ions in 1M copper(II) sulphate(VI) solution that are attracted/move to: Cathode- Cu2+ (aq) from copper(II) sulphate(VI) solution and H+(aq) from water (H2O) Anode- SO42-(aq) from copper(II) sulphate(VI) solution and OH- (aq) from water (H2O) III. Write the equation for the reaction during the electrolytic process at the: Cathode Cu2+ (aq) + 2e -> Cu(s) Cu2+ ions are lower than H+ ions in the electrochemical series therefore selectively discharged at the cathode.) Anode Cu (s) -> Cu2+(aq) + 2e (Both OH- ions and SO42- ions move to the anode but none is discharged. The copper anode itself ionizes/dissolves/dissociate as less energy is used to remove an electron/ionize /dissociate copper atoms than OH- ions. IV. Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1.25 g of copper metal as brown solid coat/deposits Anode-1.25 g of copper metal erodes/decrease in size V. (i)How many moles of electrons are used to deposit/erode one mole of copper metal at the cathode/anode?
Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1.25 g of copper metal as brown solid coat/deposits Anode-1.25 g of copper metal erodes/decrease in size V. (i)How many moles of electrons are used to deposit/erode one mole of copper metal at the cathode/anode? From the equation at anode/cathode= 2 moles (ii)How many Faradays are used to deposit/erode one mole of copper metal at the cathode/anode? From the equation at anode/cathode : 2 moles = 2 Faradays (iii)Calculate the quantity of electric charge used Working:
46 Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 2 x 20 x 60 = 2400C VI. (i) Calculate the quantity of electricity required to deposit/erode one mole of copper at the cathode/anode(Cu=63.5) Since 1.25 g of copper -> 2400C Then 63.5 g (1mole of copper) -> 63.5 x 2400 = 121920 C 1.25 (ii)Determine the Faraday constant from the results in V(i) above From the equation at; Cathode Cu2+ (aq) + 2e -> Cu(s) Anode Cu (s) -> Cu2+(aq) + 2e 2 moles = 2 Faradays -> 121920 C 1 moles = 1 Faradays -> 121920 = 60960 C 2 (iii) The faraday constant obtained above is far lower than theoretical.Explain -high resistance of the wires used. -temperatures at 25oC were not kept constant -plates/electrodes used were not made of pure copper -plates/electrodes used were not thoroughly clean copper Further practice 1.An element P has a relative atomic mass of 88. When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44g of P were deposited at the cathode. Determine the charge on an ion of P. (1 faraday = 96500 Coulombs).
When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44g of P were deposited at the cathode. Determine the charge on an ion of P. (1 faraday = 96500 Coulombs). 2.During 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) 3.A nitrate of a metal M was electrolysed .1.18 g of metal was deposited when a current of 4 ampheres flow for 16 minutes.Determine the formula of the sulphate(VI)salt of the metal. (Faraday constant = 96500 , RAM of X = 59.0) Working Q = It =>( 4 x 16 x 60) = 3840 C 1.18 g of X => 3840 C 59.0 g => 59.0 x 3840 = 192000 C 1.18 96500 C = 1Faraday 192000 C= 192000 C x1 = 2F thus charge of M = M2+ 96500 C Valency of M is 2 thus formula of sulphate(VI)salt MSO4
47 4. Below is the results obtained when a current of 2.0ampheres is passed through copper(II)sulphate(VI)solution for 15 minutes during electrolysis using copper electrode. Initial mass of cathode = 1.0 g Final mass of cathode = 1.6 g Change in mass of cathode = 0.60 g (i)Determine the change in mass at the anode. Explain your answer. Mass decrease = 0.6g.
Initial mass of cathode = 1.0 g Final mass of cathode = 1.6 g Change in mass of cathode = 0.60 g (i)Determine the change in mass at the anode. Explain your answer. Mass decrease = 0.6g. Electrode ionization take place where the cathode increase in mass form the erosion of the anode (ii)Calculate the quantity of electricity required to deposit one mole of copper.(Cu =63.5) Q =It => 2 x 15 x 60 = 1800 coulombs Method 1 0.60 g of copper ->1800 coulombs 63.5 g -> 63.5 x 1800 = 190500 Coulombs 0.60 Method 2 Moles of Copper = Mass => 0.60 = 9.4488 x10 -3 moles Molar mass 63.5 9.4488 x10 -3 moles -> 1800 coulombs 1 Mole -> 1 x 1800 coulombs = 190500.381 coulombs 9.4488 x10 -3 moles (iii)Determine the oxidation number of copper produced at the cathode and hence the formula of its nitrate (V)salt (1 Faraday = 96500 Coulombs) 96500 Coulombs -> 1 Faraday 190500.381 coulombs -> 190500.381 coulombs x 1 96500 Coulombs = 1.9741 Faradays => 2F(whole number) Charge of copper = 2+ = Oxidation number => Valency of copper = 2 hence chemical formula of nitrate (V)salt = Cu (NO3)2
1 22.0.0 METALS (20 LESSONS) a)Introduction to metals The rationale of studying metals cannot be emphasized.Since ages, the world over, metals like gold and silver have been used for commercial purposes. The periodicity of alkali and alkaline earth metals was discussed in year 2 of secondary school education. This topic generally deals with: (a)Natural occurrence of the chief ores of the most useful metals for industrial /commercial purposes. (b)Extraction of these metals from their ores for industrial/ commercial purposes. (c)industrial/ commercial uses of these metals. (d)main physical and chemical properties /characteristic of the metals.
(b)Extraction of these metals from their ores for industrial/ commercial purposes. (c)industrial/ commercial uses of these metals. (d)main physical and chemical properties /characteristic of the metals. The metals given detailed emphasis here are; Sodium, Aluminium, Iron, Zinc, Lead and Copper. 2 The main criteria used in extraction of metals is based on its position in the electrochemical/reactivity series and its occurrence on the earth’s crust. 1.SODIUM a) Natural occurrence Sodium naturally occurs as: (i)Brine-a concentrated solution of sodium chloride(NaCl(aq)) in salty seas and oceans. (ii)Rock salt-solid sodium chloride(NaCl(s) (iii)Trona-sodium sesquicarbonate(NaHCO3.Na2CO3.2H2O) especially in lake Magadi in Kenya. (iv)Chile saltpeter-sodium nitrate(NaNO3) Position on the earth’s crust If near the surface ,open cast mining / quarrying is used If deep on the earth’s crust deep mining is used If the ore is low grade oil, water, and air is blown forming a froth(froth flotation) to concentrate The ore first roasted if it is a carbonate or sulphide of Zinc, Iron, Tin, Lead, and Copper to form the oxide Electrolysis of the ore is used for reactive metals; Potassium, Sodium, Magnesium, Calcium, Aluminium The oxide is reduced using carbon/ carbon(II) oxide in a furnace if it is made of Zinc ,Tin, Lead ,Copper and Iron
3 b)(i) Extraction of Sodium from brine/Manufacture of Sodium hydroxide/The flowing mercury cathode cell/ TheCaster-Keller process I.Raw materials (i) Brine-concentrated solution of sodium chloride (NaCl (aq)) from salty seas and oceans. (ii)Mercury (iii)Water from river/lakes II. Chemical processes Salty lakes, seas and oceans contain large amount of dissolved sodium chloride (NaCl (aq)) solution. This solution is concentrated to form brine which is fed into an electrolytic chamber made of suspended Carbon graphite/titanium as the anode and a continuous flow of Mercury as the cathode.Note Mercury is the only naturally occurring known liquid metal at room temperature and pressure Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process.
Chemical processes Salty lakes, seas and oceans contain large amount of dissolved sodium chloride (NaCl (aq)) solution. This solution is concentrated to form brine which is fed into an electrolytic chamber made of suspended Carbon graphite/titanium as the anode and a continuous flow of Mercury as the cathode.Note Mercury is the only naturally occurring known liquid metal at room temperature and pressure Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process. H2O(l) H+(aq) + OH-(aq) NaCl(aq) Na+(aq) + Cl-(aq) II. Name the ions present in brine that moves to the: (i)Mercury cathode; H+(aq) , Na+(aq) (ii)Titanium/graphite; OH-(aq), Cl-(aq) III. Write the equation for the reaction that take place during the electrolytic process at the; Cathode; 2Na+(aq) + 2e 2Na(s) Anode; 2Cl-(aq) Cl2(g) + 2e Note (i)Concentration of 2Cl-(aq) ions is higher than OH- ions causing overvoltage thus blocking OH- ions from being discharged at the anode. (ii)Concentration of Na+(aq) ions is higher than H+ ions causing overvoltage thus blocking H+ ions from being discharged at the cathode. 4 IV. Name the products of electrolysis in the flowing mercury-cathode cell. (i)Mercury cathode; Sodium metal as grey soft metal/solid (ii)Titanium/graphite; Chlorine gas as a pale green gas that turns moist blue/red litmus papers red then bleaches both. Chlorine gas is a very useful by-product in; (i)making (PVC)polyvinylchloride(polychloroethene) pipes. (ii)chlorination/sterilization of water to kill germs. (iii)bleaching agent (iv)manufacture of hydrochloric acid. Sodium produced at the cathode immediately reacts with the mercury at the cathode forming sodium amalgam(NaHg) liquid that flow out of the chamber. Na(s) + Hg(l) Na Hg (l) Sodium amalgam is added distilled water and reacts to form sodium hydroxide solution, free mercury and Hydrogen gas.
(iii)bleaching agent (iv)manufacture of hydrochloric acid. Sodium produced at the cathode immediately reacts with the mercury at the cathode forming sodium amalgam(NaHg) liquid that flow out of the chamber. Na(s) + Hg(l) Na Hg (l) Sodium amalgam is added distilled water and reacts to form sodium hydroxide solution, free mercury and Hydrogen gas. 2Na Hg (l) + 2H2O(l) 2NaOH (aq) + 2Hg(l) + H2(g) Hydrogen gas is a very useful by-product in; (i)making ammonia gas in the Haber process (ii)manufacture of hydrochloric acid (iii)in weather balloons to forecast weather (iv)as rocket fuel As the electrolysis of brine continues, the concentration of Cl-ions decreases and oxygen gas start being liberated. Continuous feeding of the electrolyte is therefore very necessary. III.Uses of sodium hydroxide The sodium hydroxide produced is very pure and is used mainly in: (i)Making soapy and soapless detergents. (ii)making cellulose acetate/rayon IV. Diagram showing the Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell. 5 3 V. Environmental effects of Manufacture of Sodium hydroxide from the flowing Mercury-cathode cell. 1.Most of the Mercury used at the cathode is recycled ; (i)to reduce the cost because mercury is expensive (ii)to reduce pollution because mercury kills marine life. (iii)because it causes chromosomal/genetic mutation to human beings. 2.Chlorine produced at the anode; (i)has a pungent irritating smell that causes headache to human beings. (ii)bleaches any wet substance. (iii)dissolves water to form both hydrochloric acid and chloric(I)acid Both cause marine pollution and stomach upsets. b)(ii) Extraction of sodium from rock salt/The Downs cell/process
6 I. Raw materials (i)Rock salt/solid sodium chloride (ii)calcium(II)chloride II. Chemical processes. Rock salt/ solid sodium chloride is heated to molten state in a chamber lined with fire bricks on the outside. Sodium chloride has a melting point of about 800oC.
Chemical processes. Rock salt/ solid sodium chloride is heated to molten state in a chamber lined with fire bricks on the outside. Sodium chloride has a melting point of about 800oC. A little calcium (II) chloride is added to lower the melting point of the electrolyte to about 600oC. The molten electrolyte is the electrolyzed in a carbon graphite anode suspended at the centre and surrounded by steel cathode. Questions I. Write the equation for the decomposition of the electrolyte during the electrolytic process. NaCl(l) Na+(l) + Cl-(l) Note: In absence of water, the ions are in liquid state. II. Name the ions present in molten rock salt that move to the; (i)Steel cathode -Na+(l) (ii)Carbon graphite anode- Cl-(l) III. Write the equation for the reaction that take place during the electrolytic process at the; (i)Steel cathode 2Na+(l) + 2e 2Na(l) (ii)Carbon graphite anode 2Cl-(l) Cl2(g) + 2e IV. Name the products of electrolysis in the Downs cell at; (i)Cathode: Grey solid Sodium metal is less dense than the molten electrolyte and therefore float on top of the cathode to be periodically tapped off. (ii)Anode: Pale green chlorine gas that turns moist/damp/wet blue/red litmus papers red then bleaches/decolorizes both. Chlorine gas is again a very useful by-product in; (i)making (PVC)polyvinylchloride(polychloroethene) pipes. (ii)chlorination/sterilization of water to kill germs. (iii)bleaching agent (iv)manufacture of hydrochloric acid. 7 A steel diaphragm/gauze is suspended between the electrodes to prevent recombination of sodium at the cathode and chlorine gas at the anode back to sodium chloride. III. Diagram showing the Downs cell/process for extraction of sodium IV. Uses of sodium. 1.Sodium vapour is used as sodium lamps to give a yellow light in street lighting.
Diagram showing the Downs cell/process for extraction of sodium IV. Uses of sodium. 1.Sodium vapour is used as sodium lamps to give a yellow light in street lighting. 2.Sodium is used in making very useful sodium compounds like; (i)Sodium hydroxide(NaOH) (ii)Sodium cyanide(NaCN) (iii)Sodium peroxide(Na2O2) (iv)Sodamide(NaNH2) 3.An alloy of Potassium and Sodium is used as coolant in nuclear reactors. V. Environmental effects of Downs cell. 1.Chlorine produced at the anode;
8 (i)has a pungent irritating smell that causes headache to human beings. (ii)bleaches any wet substance. (iii)dissolves water to form both hydrochloric acid and chloric(I)acid Both cause marine pollution and stomach upsets. 2.Sodium metal rapidly react with traces of water to form alkaline Sodium hydroxide(NaOH(aq))solution. This raises the pH of rivers/lakes killing aquatic lifein case of leakages. VI. Test for presence of Na. If a compound has Na+ ions in solid/molten/aqueous state then it changes a non-luminous clear/colourless flame to a yellow coloration but does not burn Experiment Scoop a portion of sodium chloride crystals/solution in a clean metallic spatula. Introduce it to a clear /colourless Bunsen flame. Observation Inference Yellow coloration Na+ Practice (i)Calculate the time taken in hours for 230kg of sodium to be produced in the Downs cell when a current of 120kA is used. (ii)Determine the volume of chlorine released to the atmosphere.
10 (a)Calculate the concentration of the resulting solution in moles per litre. (b)The volume of gaseous products formed at s.t.p(1 mole of gas =22.4 dm3 at s.t.p) Chemical equation at Caster-Keller tank 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g) Mole ratio Na:NaOH = 2 : 2 => 1:1 Moles Na =10000moles=10000moles of NaOH 25000dm3 ->10000moles of NaOH 1dm3 -> 10000 x 1 = 0.4M / 0.4 moles/dm3 25000 Mole ratio Na: H2 (g) = 2 : 1 Moles Na = 10000moles = 5000moles of H2 (g) Volume of H2 (g) = moles x molar gas volume at s.t.p => 5000moles x 22.4 dm3 =120,000dm3 (iv)The solution formed was further diluted with water for a titration experiment. 25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration.
(b)The volume of gaseous products formed at s.t.p(1 mole of gas =22.4 dm3 at s.t.p) Chemical equation at Caster-Keller tank 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2 (g) Mole ratio Na:NaOH = 2 : 2 => 1:1 Moles Na =10000moles=10000moles of NaOH 25000dm3 ->10000moles of NaOH 1dm3 -> 10000 x 1 = 0.4M / 0.4 moles/dm3 25000 Mole ratio Na: H2 (g) = 2 : 1 Moles Na = 10000moles = 5000moles of H2 (g) Volume of H2 (g) = moles x molar gas volume at s.t.p => 5000moles x 22.4 dm3 =120,000dm3 (iv)The solution formed was further diluted with water for a titration experiment. 25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration. Chemical equation 2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + H2O(l) Moles ratio NaOH : H2SO4 = 2 : 1 Moles ratio H2SO4 = molarity x volume => 0.2M x 20 1000 1000 =4.0 x 10-3 moles Moles NaOH = 2 x 4.0 x 10-3 moles= 8.0 x 10-3 moles Molarity of NaOH= Moles x 1000=> 8.0 x 10-3 moles x 1000 volume 25 =0.16 molesdm-3 /M Volume used during dilution C1V1 = C2V2 => 0.4M x V1 = 0.16 M x 25
11 = 0.16 M x 25 = 10cm3 0.4 (a) Below is a simplified diagram of the Downs Cell used for the manufacture of sodium.
25.0 cm3 of the diluted solution required 20.0cm3 of 0.2M sulphuric(VI)acid for complete neutralization. Calculate the volume of water added to the diluted solution before titration. Chemical equation 2NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + H2O(l) Moles ratio NaOH : H2SO4 = 2 : 1 Moles ratio H2SO4 = molarity x volume => 0.2M x 20 1000 1000 =4.0 x 10-3 moles Moles NaOH = 2 x 4.0 x 10-3 moles= 8.0 x 10-3 moles Molarity of NaOH= Moles x 1000=> 8.0 x 10-3 moles x 1000 volume 25 =0.16 molesdm-3 /M Volume used during dilution C1V1 = C2V2 => 0.4M x V1 = 0.16 M x 25
11 = 0.16 M x 25 = 10cm3 0.4 (a) Below is a simplified diagram of the Downs Cell used for the manufacture of sodium. Study it and answer the questions that follow (i)What material is the anode made of? Give a reason (2 mks) Carbon graphite/Titanium This because they are cheap and inert/do not influence/affect the products of electrolysis (ii) What precaution is taken to prevent chlorine and sodium from re- combination? ( 1 mks) Using a steel gauze/diaphragm separating the cathode from anode (iii) Write an ionic equation for the reaction in which chlorine gas is formed ( 1mk) 2Cl-(l) -> Cl2(g) + 2e (b) In the Downs process, (used for manufacture of sodium), a certain salt is added to lower the melting point of sodium chloride from about 8000C to about 6000C.
Study it and answer the questions that follow (i)What material is the anode made of? Give a reason (2 mks) Carbon graphite/Titanium This because they are cheap and inert/do not influence/affect the products of electrolysis (ii) What precaution is taken to prevent chlorine and sodium from re- combination? ( 1 mks) Using a steel gauze/diaphragm separating the cathode from anode (iii) Write an ionic equation for the reaction in which chlorine gas is formed ( 1mk) 2Cl-(l) -> Cl2(g) + 2e (b) In the Downs process, (used for manufacture of sodium), a certain salt is added to lower the melting point of sodium chloride from about 8000C to about 6000C. (i) Name the salt that is added (1mk) Calcium chloride (ii) State why it is necessary to lower the temperature(1mk) To reduce the cost of production
12 (c) Explain why aqueous sodium chloride is not suitable as an electrolyte for the manufacture of sodium in the Downs process( 2mk) The sodium produced react explosively/vigorously with water in the aqueous sodium chloride (d) Sodium metal reacts with air to form two oxide. Give the formulae of two oxides ( 1mk) Na2O Sodium oxide(in limited air) Na2O2 Sodium peroxide(in excess air) 2.ALUMINIUM a)Natural occurrence Aluminium is the most common naturally occurring metal. It makes 7% of the earths crust as: (i)Bauxite ore- Hydrated aluminium oxide(Al2O3.2H2O) (ii)Mica ore-Potassium aluminium silicate(K2Al2Si6O16) (iii)China clay ore- aluminium silicate (Al2Si6O16) (iv)Corrundum-Anhydrous aluminium oxide(Al2O3) b)Extraction of aluminium from Bauxite/Halls cell/process) The main ore from which aluminium is extracted is Bauxite ore- hydrated aluminium oxide(Al2O3.2H2O). The ore is mined by open-caste mining method/quarrying where it is scooped together with silica/sand/silicon(IV)oxide (SiO2) and soil/ iron(III)oxide (Fe2O3) as impurities.
Give the formulae of two oxides ( 1mk) Na2O Sodium oxide(in limited air) Na2O2 Sodium peroxide(in excess air) 2.ALUMINIUM a)Natural occurrence Aluminium is the most common naturally occurring metal. It makes 7% of the earths crust as: (i)Bauxite ore- Hydrated aluminium oxide(Al2O3.2H2O) (ii)Mica ore-Potassium aluminium silicate(K2Al2Si6O16) (iii)China clay ore- aluminium silicate (Al2Si6O16) (iv)Corrundum-Anhydrous aluminium oxide(Al2O3) b)Extraction of aluminium from Bauxite/Halls cell/process) The main ore from which aluminium is extracted is Bauxite ore- hydrated aluminium oxide(Al2O3.2H2O). The ore is mined by open-caste mining method/quarrying where it is scooped together with silica/sand/silicon(IV)oxide (SiO2) and soil/ iron(III)oxide (Fe2O3) as impurities. The mixture is first dissolved in hot concentrated sodium/potassium hydroxide solution. The alkalis dissolve both bauxite and silicon(IV)oxide. This is because bauxite is amphotellic while silicon(IV)oxide is acidic. Iron(III)oxide (Fe2O3) is filtered of /removed as a residue. Carbon(IV)oxide is bubbled into the filtrate to precipitate aluminium (III) hydroxide (Al(OH)3) as residue. The aluminium (III) hydroxide (Al(OH)3) residue is filtered off. Silicon (IV)oxide remain in the solution as filtrate. Aluminium (III) hydroxide (Al(OH)3) residue is then heated to form pure aluminium (III)oxide(Al2O3) 2Al(OH)3 (s) Al2O3 (s) + 3H2O(l)
13 Pure aluminium (III)oxide (Al2O3) has a very high melting point of 2015oC. Alot of energy is required to melt the oxide.
Silicon (IV)oxide remain in the solution as filtrate. Aluminium (III) hydroxide (Al(OH)3) residue is then heated to form pure aluminium (III)oxide(Al2O3) 2Al(OH)3 (s) Al2O3 (s) + 3H2O(l)
13 Pure aluminium (III)oxide (Al2O3) has a very high melting point of 2015oC. Alot of energy is required to melt the oxide. It is therefore dissolved first in molten cryolite /sodium hexafluoroaluminate (III)/Na3AlF6 to lower the melting point to about 800oC. The molten electrolyte is put in the Hall cell made up of a steel tank lined with carbon graphite and an anode suspended into the electrolyte. During the electrolysis: (i)At the cathode; 4Al3+(l) + 12e 4Al(l) (ii) At the anode; 6O2-(l) 3O2(g) + 12e Aluminium is denser than the electrolyte therefore sink to the bottom of the Hall cell. At this temperature ,the Oxygen evolved/produced at the anode reacts with carbon anode to form carbon(IV)oxide gas that escape to the atmosphere. C(s) + O2(g) CO2(g) The anode thus should be continuously replaced from time to time. Flow chart summary of extraction of aluminium from Bauxite Bauxite(Al2O3.2H2O) ore with impurities Fe2O3 and SiO2 Powdered mixture Crush (increase surface area) Hot concentrated sodium hydroxide
14 c) Diagram showing the Hall cell / process for extraction of Bauxite Iron(III)oxide- Fe2O3 as residue Sodium aluminate (NaAl(OH)4) and sodium silicate (Na2SiO3) as filtrate Carbon(IV)oxide Aluminium hydroxide (Al(OH)3) as residue Sodium silicate (Na2SiO3) Aluminium (III) Oxide Roast at 1000oC Cryolite Electrolysis Oxygen gas at anode Pure aluminium sinks in Hall cell
15 d)Uses of aluminium (i) In making aeroplane parts, buses, tankers, furniture because aluminium is very light.