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d)Influence 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 pieces/chips then crushing the chips into powder. Chips thus have a higher surface ... | {
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Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 1.8 1.4 1.0 0.8 0.5 0.5 0.5 0.5 Loss in mass 0.0 0.7 1.1 1.5 1.7 2.0 2.0 2.0 2.0 Sample questions: 1.Calculate the loss in mass made at the end of each time from the original to complete table I,II and III 2.On the same axes plot a graph o... | {
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Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion. 6.Calculate the volume of carbon(IV)oxide evolved(molar gas volume at roo... | {
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Excess marble chips were put in a beaker containing 100cm3 of 0.2M 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. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a lo... | {
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CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-incre... | {
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Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. 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.... | {
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Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture ... | {
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This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is... | {
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Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask. (ii)Put 2.0g of Manganese (IV) oxide into the clean conical/round bottomed/flat bottomed flask. Stopper the flask. Transfer the second portion of the 20cm3volume hydrogen peroxide into a conical/round bo... | {
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Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e. A + B -> C + D Reactions in which the reactants form products and the products can r... | {
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Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid. Melting boiling Freezing condensing (iv)Dissolving/ crystallization/distillation Solid crystals of soluble substances (solut... | {
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Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white. (ii)colourless liquid forms ... | {
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Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made. When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid th... | {
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A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of forma... | {
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II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH- (aq) ions . More of the CrO42- i... | {
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Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium ex... | {
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If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/le... | {
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i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When ... | {
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The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with mor... | {
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Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process. The yield of ammonia is thus favoured by high pressures Chemical equation : N2(g) + 3H2 (g) -> 2NH3 (g) Gay... | {
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(v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures. Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (g)... | {
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The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reactio... | {
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CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufactur... | {
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(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield o... | {
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Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation. 3.Optimum condition in Ostwalds process Chemical equat... | {
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C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2. Explain the effect of increase in pressure on the following: (i) N2(g) + O2(g)... | {
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(d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the
react is complete. No more hydrogen is evolved.The mass of flask remain constant. (d)At what time was the loss in mass equal to: (i)1.20g Read... | {
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v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid). The chlorine is reduced (gain hydrogen) to hydrogen chlorine gas. Cl2 (g) + H2S (g) -> S(... | {
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In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves t... | {
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Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s... | {
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b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution. Brown colour of Fe3+(aq) ... | {
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iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applie... | {
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v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides... | {
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(b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1e... | {
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Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number o... | {
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Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers o... | {
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Sulphur in; -SO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Sulphur(IV)oxide -SO3 => x + (-2 x3)= 0 thus x = 0 – (-6) = + 6 The chemical name of this compound is thus Sulphur(VI)oxide -H2SO4 = ((+1 x 2) + x + (-2 x 4)) thus x= 0-( +2 +-8) =+6 The chemical name of this compound ... | {
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Carbon in; -CO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus carbon(IV)oxide -CO => x + -2 = 0 thus x = 0 – -2 = + 2 The chemical name of this compound is thus carbon(II)oxide -H2CO3 = ((+1 x 2) + x + (-2 x 3)) thus x= 0-( +2 +-6) =+4 The chemical name of this compound is thus Ca... | {
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Using this rule ,the oxidation number of unknown atom of an element in a charged radical/complex ion can be determined as in the examples below; a) SO42- has- -one atom of S with oxidation number +6( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 1) Sum of oxidation numbers of atoms in S... | {
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Determine the oxidation number of: I.Nitrogen in; -NO2- => x + (-2 x2)= -1 thus x = -1 – (-4) = + 3 The chemical name of this compound/ion/radical is thus Nitrate(III)ion II. Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon... | {
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Sulphur in; -SO32- => x + (-2 x3)= -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Sulphate(IV)ion III. Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 th... | {
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Carbon in; -CO32- = x + (-2 x 3) = -2 thus x = -2 – (-6) = + 4 The chemical name of this compound/ion/radical is thus Carbonate(IV)ion IV.Manganese in; -MnO4 - = x + (-2 x 4)= -1 thus x= -1-(-2 +-8) =+7 The chemical name of this compound/ion/radical is thus manganate(VII) ion V.Chromium in -Cr2O72- => 2x + (-2 x7)= -2 ... | {
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When a metal rod/plate is put in a solution of its own salt, some of the metal ionizes and dissolve into the solution i.e. M(s) -> M+(aq) + e ( monovalent metal) M(s) -> M2+(aq) + 2e ( divalent metal) M(s) -> M3+(aq) + 3e ( Trivalent metal) The ions move into the solution leaving electrons on the surface of the metal r... | {
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Put a clean copper rod/plate of equal area (length x width) with Zinc into the solution. Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the t... | {
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Connect/join the two metals(to a voltmeter) using connecting wires. Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above proc... | {
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Dip a folded filter paper into a solution of Potassium nitrate(V) or sodium(I) chloride(I) until it soaks. Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbo... | {
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Use the folded soaked filter paper to connect/join the two solutions in the two beakers.The whole set up should be as below Repeat the above procedure by replacing: (i)Zinc half cell with Magnesium rod/plate/ribbon dipped in 50cm3 of IM magnesium (II) sulphate(VI) solution V
11 (ii)Zinc half cell with Silver rod/plate/... | {
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Brown solid/residue/ deposit 0.8 (Theoretical value=1.10V) Using Mg/Cu half cell -The rod decrease in size /mass /dissolves/ erodes -copper rod /plate increase in size /mass/ deposited Magnesium(II) sulphate(VI) colour remain colourless Blue Copper (II)sulphate (VI)colour fades Brown solid/residue/ deposit 1.5 (Theoret... | {
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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 accumulation/increase 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... | {
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(i)Zn/Cu cell
13 1. Zinc rod ionizes /dissolves to form Zn2+ ions at the negative terminal Zn(s) -> Zn2+(aq) + 2e 2. Copper ions in solution gain the donated electrons to form copper atoms/metal Cu2+(aq) + 2e -> Cu(s) 3.Overall redox equation Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) 4.cell representation. Zn(s) / 1M, Zn2+(... | {
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Fe(s) / 1M, Fe2+(aq) // 1M,Cu2+(aq) / Cu(s) E0 = +0.78 V 5.cell diagram. Fe Fe++ V
15 (iv)Ag/Cu cell 1. Copper rod ionizes /dissolves to form Cu2+ ions at the negative terminal Cu(s) -> Cu2+(aq) + 2e 2. Silver ions in solution gain the donated electrons to form silver atoms/metal 2Ag+(aq) + 2e -> 2Ag(s) 3.Overall redox... | {
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Hydrogen gas is bubbled on the platinum electrodes at: (i)a temperature of 25oC (ii)atmospheric pressure of 101300Pa/101300Nm-2/1atm/760mmHg/76cmHg (iii)a concentration of 1M(1moledm-3) of sulphuric(VI) acid/ H+ ions and 1M(1moledm-3) of the other half cell. Hydrogen is adsorbed onto the surface of the platinum. An equ... | {
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½ H2 (g) ==== H+ (aq) + e The half cell representation is: Pt,½ H2 (g) / H+ (aq), 1M Cu Ag V Cu++ Ag+ Cu(s) + 2Ag+ (aq) Cu2+(aq) + 2Ag(s)
16 The standard electrode potential (Eᶿ) 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 ... | {
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(ii)The element/species in the half cell with the highest negative Eᶿ value easily gain / acquire electrons. It is thus the strongest oxidizing agent and its reduction process is highly possible/feasible. The element/species in the half cell with the lowest positive Eᶿ value easily donate / lose electrons. 17 It is thu... | {
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17 It is thus the strongest reducing agent and its reduction process is the least possible/feasible. (iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ... | {
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(iii)The overall redox reaction is possible/feasible is it has a positive (+) Eᶿ. If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(highe... | {
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If the overall redox reaction is not possible/ not feasible/ forced, it has a negative (-) Eᶿ Calculation examples on Eᶿ Calculate the Eᶿ value of a cell made of: a)Zn and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Zn2+ (aq) + 2e ->Zn(s) Eᶿ = -0.77V(lower Eᶿ/ Left H... | {
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b)Mg and Cu From the table above: Cu2+ (aq) + 2e -> Cu(s) Eᶿ = +0.34V(higher Eᶿ /Right Hand Side diagram) Mg2+ (aq) + 2e ->Mg(s) Eᶿ = -2.37V(lower Eᶿ/ Left Hand Side diagram) Mg(s) ->Mg2+ (aq) + 2e Eᶿ = +2.37(reverse lower Eᶿ to derive cell reaction / representation) Overall Eᶿ = Eᶿ higher- Eᶿ lower / Eᶿ RHS - Eᶿ LHS/ ... | {
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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 Eᶿ value. e)Strongest oxidizing agent and the strongest re... | {
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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. Zi... | {
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Ammonium chloride is used as paste because the solid does not conduct electricity because the ions are fused/not mobile. Since the reactants are used up, the dry /primary /Laclanche cell cannot provide continous supply of electricity.The process of restoring the reactants is called recharging. b)Wet/Secondary/Accumulat... | {
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Pb2+ (aq) + SO42- (aq) -> PbSO4(s) 5.The overall cell reaction is called discharging PbO2(s) +Pb(s) + 4H+(aq) + 2SO42- (aq)-> 2PbSO4(s) + 2H2O(l) Eᶿ = +2.0V 6.The insoluble Lead(II) sulphate (VI) formed should not be left for long since fine Lead(II) sulphate (VI) will change to a course non-reversible and inactive for... | {
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A compound that is not decomposed by an electric current is called nonelectrolyte. Non-electrolytes are those compounds /substances that exist as molecules and thus cannot ionize/dissociate into(any) ions . Common non-electrolytes include: (i) most organic solvents (e.g. petrol/paraffin/benzene/methylbenzene/ethanol) (... | {
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M(l) -> M+(l) + e (for cations from molten electrolytes) M(s) -> M+(aq) + 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
23 10. During electrolysis, free cations are attract... | {
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BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode
24 12. For a compound /salt containing only two ion/binary salt the products of electrolysis in an electrolytic cell can be determined as in the below examples: a)To determine the products of electro... | {
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II.At the anode pale green chlorine gas. b)To determine the products of electrolysis of molten Zinc bromide (i)Decomposition of electrolyte into free ions; ZnBr2 (l) -> Zn 2+(l) + 2Br-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); Zn 2+(l) + 2e -> Zn(l) (C... | {
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II.At the anode red bromine liquid / red/brown bromine gas. c)To determine the products of electrolysis of molten sodium chloride (i)Decomposition of electrolyte into free ions; NaCl (l) -> Na +(l) + Cl-(l) (Compound decomposed into free cation and anion in liquid state) (ii)At the cathode/negative electrode(-); 2Na+(l... | {
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For a compound /salt mixture containing many ions in an electrolytic cell, the discharge of ions in the cell depend on the following factors: a) 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. Th... | {
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All the other anionic radicals(SO42- ,SO32- ,CO32- ,HSO4- ,HCO3- ,NO3- ,PO43-)are not/never 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
27 4OH- (aq) -> 2H2O(l) + O2 (g) + 4e (most readily/easily discharged) ... | {
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Electrolytic cell set up during electrolysis of acidified water/dilute sulphuric(VI) acid BatteryAnode(+)Cathode(-)ElectrolyteSimple set up of electrolytic cellGaseous product at anodeGaseous product at cathode Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the el... | {
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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 sulphuric(VI) acid called “electrolysis of (acidified) water”? The ratio of H2 (g):... | {
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Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) Mg SO4(aq) -> SO42-(aq) + Mg2+(aq) II. Name the ions in Magnesium sulphate(VI) solution that are attracted/move to: Cathode- Mg2+(aq) from Magnesium sulphate(VI) solution (Mg SO4) and H+(aq) from wa... | {
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VI. Explain the changes in concentration of the electrolyte during electrolysis of Magnesium sulphate(VI) solution The concentration of dilute Magnesium sulphate(VI) solution increases. The ratio of H2 (g): O2 (g) is 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gase... | {
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Answer the following questions: I. Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) NaCl(aq) -> Cl-(aq) + Na+(aq) II. Name the ions in sodium chloride solution that are attracted/move to: Cathode- Na+(aq) from Sodium chloride solution (NaCl) and H... | {
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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 increasing/rising. 32 II. Dissolve about 20 g of pure sodium chloride crystals in 100cm3 of water. Place the solution in an elect... | {
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Two (2) electrons donated/lost by Cl- ions to form 1 molecule/1volume/1mole of Chlorine (Cl2)gas at the anode are gained/acquired/accepted by two H+(aq) ions to form 1 molecule/1volume/1mole of Hydrogen (H2)gas at the cathode. The volume of Chlorine gas at the anode is equal to the volume of Hydrogen produced at the ca... | {
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Write the equation for the decomposition of the electrolytes during the electrolytic process. H2O(l) -> OH- (aq) + H+(aq) HCl(aq) -> Cl-(aq) + H+(aq) II. Name the ions in dilute Hydrochloric acid solution that are attracted/move to: Cathode- H+(aq) from dilute Hydrochloric acid (HCl) and H+(aq) from water (H2O) Anode- ... | {
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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 2:1 as they are combined in water. Water in the electrolyte is decomposed into Hydrogen and Oxygen gases that escape as products... | {
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Write the equation for the reaction during the electrolytic process at the: Cathode 4H+(aq) + 4e -> 2H2(g) H+ ions selectively discharged instead of Na+ ions at the cathode) Anode 2Cl- (aq) -> Cl2+ 2e (OH- ions concentration is low.Cl- ions concentration is higher at the anode thus cause over voltage/block discharge of... | {
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c) Nature of electrodes used in the electrolytic cell Inert electrodes (carbon-graphite and platinum) do not alter the expected products of electrolysis in an electrolytic cell. If another/different electrode is used in the electrolytic cell it alters/influences/changes the expected products of electrolysis. The exampl... | {
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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)
36 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 wat... | {
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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 co... | {
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Sample results Mass of cathode before electrolysis 23.4 g Mass of anode before electrolysis 22.4 g Mass of cathode after electrolysis 25.4 g Mass of anode after electrolysis 20.4 g Brown solid deposit at the cathode after electrolysis - Anode decrease insize/erodes/wear off - Blue colour of electrolyte remain blue - Bl... | {
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Name the products of electrolysis of 1M copper(II) sulphate(VI) solution using copper electrodes. Cathode-1 moles of copper metal as brown solid coat (Cathode increase/deposits) Anode-Anode erodes/decrease in size V. Explain the changes that take place during the electrolytic process (i)Cathode -Cu2+ ions are lower tha... | {
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(b)Purifying copper after exraction from copper pyrites ores. Copper obtained from copper pyrites ores is not pure. After extraction, the copper is refined by electrolysing copper(II)sulphate(VI) solution using the impure copper as anode and a thin strip of pure copper as cathode. Electrode ionization take place there:... | {
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During electroplating,the cathode is made of the metal to be coated/impure. 39 Example: During the electroplating of a spoon with silver (i)the spoon/impure is placed as the cathode(negative terminal of battery) (ii)the pure silver is placed as the anode(positive terminal of battery) (iii)the pure silver erodes/ionizes... | {
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Working: Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 96500 2 = 48250 seconds 3. 96500 coulombs of charge were produced after 10 minutes in an electrolytic cell . Calculate the amount of current used. Working: Current(I) in amperes = Quantity of electricity(in Coulombs)... | {
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Al3+ require to gain 3 moles of electrons=3 Faradays =3 x 96500 coulombs of electricity at the cathode Na+ require to gain 1 moles of electrons=1 Faradays =1 x 96500 coulombs of electricity at the cathode 2H+ require to gain 2 moles of electrons=2 Faradays =2 x 96500 coulombs of electricity at the cathode to form 1mole... | {
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4OH- require to lose/donate 4 moles of electrons=4 Faradays =4 x 96500 coulombs of electricity at the anode to form 1molecule of Oxygen gas and 2 molecules of water. 41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperatu... | {
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41 (c)The mass/amount of products at the cathode/anode is related to the molar mass of the substance and/or the volume of gases at standard/room temperature and pressure as in the below examples: Practice examples 1.Calculate the mass of copper deposited at the cathode when a steady current of 4.0 amperes is passed thr... | {
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(Cu=63.5, 1F = 96500C) Working: Quantity of electricity(in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 4 x (30 x 60) = 7200 C Equation at the cathode: Cu2+ (aq) + 2e -> Cu(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus; 2 x 96500C -> 63.5 g 720... | {
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Calculate the current passed if a mass of 0.234 g of copper is deposited in 4 minutes during electrolysis of a solution of copper (II)sulphate(VI). (Cu= 63.5 ,1F = 96500C) Working: Equation at the cathode: Cu(s) -> Cu2+ (aq) + 2e 2 mole of electrons = 2 Faradays = 2 x 96500 C produce a mass =molar mass of copper thus;
... | {
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(a)What quantity of electricity will deposit a mass of 2.43 g of Zinc during electrolysis of a solution of Zinc (II)sulphate(VI). (Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 965... | {
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(Zn= 65 ,1F = 96500C) Working: Equation at the cathode: Zn2+ (aq) + 2e -> Zn(s) 2 mole of electrons = 2 Faradays = 2 x 96500 C erode/dissolve a mass =molar mass of Zinc thus; 65 g -> 2 x 96500 2.43 g -> 2.43 x 2 x 96500 = 7215.2308 C 65 (b)Calculate the time (in minutes) it would take during electrolysis of the solutio... | {
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Time(t) in seconds = Quantity of electricity(in Coulombs) Current(I) in amperes Substituting = 7215.2308 = 1803.8077 seconds = 30.0635 minutes 4 60 6.When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass at cathode increased by 0.26 g.(Faraday constant = 96500C ... | {
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Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 1.5 x (15 x 60) = 1350 C b) Determine the relative atomic mass of metal M Equation at the cathode: M3+ (aq) + 3e -> M(s) 1350 C of electricity -> 0.26 g of metal M 3 mole of electrons = 3 Faradays = 3 x 96500 C produc... | {
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Determine the charge on an ion of “P”(Faraday constant = 96500C) Working: Quantity of electricity (in Coulombs) = Current(I) x time(t) Substituting /converting time to second = 0.5 x ((32 x 60) + 10) = 965C 0.44 g of metal “P” are deposited by 965C 88g of of metal “P” are deposited by: 88 x 965= 193000 C 0.44 96500 C =... | {
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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 ... | {
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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? Fr... | {
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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 ... | {
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