WAEC SSCE - Chemistry - 2004 (Essay)
Question 1 Report
(a)(i) Give the reason why copper turnings dissolve in AgNO\(_3\) solution but remain insoluble in Pb(NO\(_3\) )\(_2\) solution.
(ii) Copper turnings of mass 1.06g were placed in 250 cm\(^3\) of 0.20 mol dm\(^{-3}\) AgNO\(_3\). Calculate the amourt of silver ions present. [Cu = 63.5]
(iii) Determine whether all the copper in (a)(ii) above will discolvo in the solution. The equation for the reaction is CU\(_{(S)}\) + 2Ag\(^+_{(aq)}\) --> Ag\(_{(aq)}\) + Cu\(^{2+}_{(aq)}\)
(b)(i) List cheicil properties of acids
(ii) Give two large scale uses of HNO\(_3\)
(iii) Write an equation for the action of heat on each of the following compounds: I. Pb(NO\(_3\))\(_2\)
Il. AgNO\(_3\).
(c)(i) State what would be observed if a piece of damp blue litmus paper is dropped into a glass jar of chlorine.
(ii) Name the type of reaction which occurs in (c)(i) above.
(iii) Give the property of chlorine which is exhibited in the reaction in (c)(i) above.
(i) Name two products obtained directly from the destructive distillation of coal.
(ii) Give one use of each product named in (d)(i) above.
(a)(i) Copper is more reactive than (above) silver in the activity series, so it displaces silver from AgNO3 solution (\(Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag\)). Copper is less reactive than (below) lead, so it cannot displace lead from Pb(NO3)2 solution and stays insoluble.
(ii) Amount of Ag+ \(= 0.20 \times \dfrac{250}{1000} = 0.050\ \text{mol}\).
(iii) Amount of Cu \(= \dfrac{1.06}{63.5} = 0.0167\ \text{mol}\).
From \(Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag\), 0.0167 mol Cu needs \(2 \times 0.0167 = 0.0334\ \text{mol}\) Ag+.
Only 0.050 mol Ag+ is available, which is more than 0.0334 mol, so Ag+ is in excess and all the copper dissolves.
(b)(i) Chemical properties of acids: turn blue litmus red; react with metals above hydrogen to liberate hydrogen; react with bases (neutralization) to form salt and water; react with trioxocarbonates(IV) to give carbon(IV) oxide.
(ii) Large-scale uses of HNO3: manufacture of fertilizers (ammonium trioxonitrate(V)); manufacture of explosives (for example TNT).
(iii) Action of heat:
I. \[2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) The damp blue litmus paper first turns red and is then bleached white.
(ii) The reaction is oxidation (bleaching).
(iii) The property shown is that chlorine is a bleaching (oxidizing) agent.
(d)(i) Products from destructive distillation of coal: coke and coal tar (also coal gas, ammoniacal liquor).
(ii) Coke is used as a fuel and as a reducing agent in the blast furnace; coal tar is used as a source of drugs and dyes (and for surfacing roads).
Answer Details
(a)(i) Copper is more reactive than (above) silver in the activity series, so it displaces silver from AgNO3 solution (\(Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag\)). Copper is less reactive than (below) lead, so it cannot displace lead from Pb(NO3)2 solution and stays insoluble.
(ii) Amount of Ag+ \(= 0.20 \times \dfrac{250}{1000} = 0.050\ \text{mol}\).
(iii) Amount of Cu \(= \dfrac{1.06}{63.5} = 0.0167\ \text{mol}\).
From \(Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag\), 0.0167 mol Cu needs \(2 \times 0.0167 = 0.0334\ \text{mol}\) Ag+.
Only 0.050 mol Ag+ is available, which is more than 0.0334 mol, so Ag+ is in excess and all the copper dissolves.
(b)(i) Chemical properties of acids: turn blue litmus red; react with metals above hydrogen to liberate hydrogen; react with bases (neutralization) to form salt and water; react with trioxocarbonates(IV) to give carbon(IV) oxide.
(ii) Large-scale uses of HNO3: manufacture of fertilizers (ammonium trioxonitrate(V)); manufacture of explosives (for example TNT).
(iii) Action of heat:
I. \[2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) The damp blue litmus paper first turns red and is then bleached white.
(ii) The reaction is oxidation (bleaching).
(iii) The property shown is that chlorine is a bleaching (oxidizing) agent.
(d)(i) Products from destructive distillation of coal: coke and coal tar (also coal gas, ammoniacal liquor).
(ii) Coke is used as a fuel and as a reducing agent in the blast furnace; coal tar is used as a source of drugs and dyes (and for surfacing roads).
Question 2 Report
(i) State Two assumptions of the kinetic theory of gases
(ii) When some solids are heated, they change directly into the gaseous state. What narne is given to this phenomenon?
(iii) List two substances which exhibit the phenomenon referred to in (a)(ii) above
(iv) Write an expression to show the mathematical relationship between the rate of diffusion of a gas and its vapour density.
(b) Consider the following equilibrium reaction:
3Fe\(_{(s)}\) + 4H\(_2\)O\(_{(g)}\) \(\rightleftharpoons\) FeO\(_3\)O\(_{4(s)}\) + 4H\(_{2(g)}\), \(\Delta\)H = - 150KJ mol\(^{-1}\)
Explain the effect of the following factors on the position of equilibrium: (i) tecrease in temperature; (ii) Increase in pressure; (iii) Removal of hydrogen.
(c) . Three beakers labelled P, Q and S each contained zinc metal of the same mass but in different forms. P contained a length of zinc rod, Q contained zinc dust while S contained zinc foil. 100cm\(^3\) of 5.0 mol dm\(^{-3}\) hydrochloric acid was added to each beaker to react with all the zinc.
(i) State the order in which the reaction came to completion in beakers P,Q and S starting with the fastest.
(ii) Give reason for your answer in (c)(i) above
(iii) Write an equation to represent the reaction between zinc rid the hydrochloric acid.
(d) (i) What is meant by pH of a solution?
(ii)(I) State with reason in each case whether the pH would increase, decrease or remain constant if the following experiments were carried out Neutralizing bench HNO\(_3\);
II. Diluting 25.0 cm\(^3\) of a given NaOH solution to 100.0cm\(^3\) Concentrating a solution of NaCI.
(a)(i) Two assumptions of the kinetic theory of gases: gas molecules are in constant, rapid, random motion; the actual volume of the molecules is negligible compared with the volume of the container (and there are negligible forces between molecules).
(ii) The phenomenon is sublimation.
(iii) Substances that sublime: iodine and ammonium chloride (also solid CO2, naphthalene, camphor).
(iv) Rate of diffusion is inversely proportional to the square root of the vapour density:
\[R \propto \frac{1}{\sqrt{d}}\]
(b) For \(3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_4_{(s)} + 4H_{2(g)}\), \(\Delta H = -150\ \text{kJ mol}^{-1}\) (exothermic):
- (i) Decrease in temperature: favours the exothermic forward reaction, so the equilibrium shifts to the right (more hydrogen).
- (ii) Increase in pressure: there are 4 moles of gas on each side, so pressure has no effect on the position of equilibrium.
- (iii) Removal of hydrogen: the equilibrium shifts to the right to replace the hydrogen removed.
(c)(i) Order (fastest first): Q (dust) > S (foil) > P (rod).
(ii) The rate depends on surface area. Zinc dust has the largest surface area exposed to the acid (fastest), the foil an intermediate area, and the rod the smallest surface area (slowest).
(iii) \[Zn + 2HCl \rightarrow ZnCl_2 + H_2\]
(d)(i) The pH of a solution is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii)
- I. Neutralizing bench HNO3 with a base: the pH increases as H+ ions are removed.
- II. Diluting 25.0 cm3 of NaOH to 100.0 cm3: the pH decreases as the alkali becomes more dilute (lower [OH-]).
- III. Concentrating a solution of NaCl: the pH remains constant, because NaCl is a neutral salt and does not change [H+].
Answer Details
(a)(i) Two assumptions of the kinetic theory of gases: gas molecules are in constant, rapid, random motion; the actual volume of the molecules is negligible compared with the volume of the container (and there are negligible forces between molecules).
(ii) The phenomenon is sublimation.
(iii) Substances that sublime: iodine and ammonium chloride (also solid CO2, naphthalene, camphor).
(iv) Rate of diffusion is inversely proportional to the square root of the vapour density:
\[R \propto \frac{1}{\sqrt{d}}\]
(b) For \(3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_4_{(s)} + 4H_{2(g)}\), \(\Delta H = -150\ \text{kJ mol}^{-1}\) (exothermic):
- (i) Decrease in temperature: favours the exothermic forward reaction, so the equilibrium shifts to the right (more hydrogen).
- (ii) Increase in pressure: there are 4 moles of gas on each side, so pressure has no effect on the position of equilibrium.
- (iii) Removal of hydrogen: the equilibrium shifts to the right to replace the hydrogen removed.
(c)(i) Order (fastest first): Q (dust) > S (foil) > P (rod).
(ii) The rate depends on surface area. Zinc dust has the largest surface area exposed to the acid (fastest), the foil an intermediate area, and the rod the smallest surface area (slowest).
(iii) \[Zn + 2HCl \rightarrow ZnCl_2 + H_2\]
(d)(i) The pH of a solution is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii)
- I. Neutralizing bench HNO3 with a base: the pH increases as H+ ions are removed.
- II. Diluting 25.0 cm3 of NaOH to 100.0 cm3: the pH decreases as the alkali becomes more dilute (lower [OH-]).
- III. Concentrating a solution of NaCl: the pH remains constant, because NaCl is a neutral salt and does not change [H+].
Question 3 Report
(a)(i) Mention two types of bond present in the ammonium ion
(ii) Give three characteristic properties of electrovalent compounds
(iii) State. two differences between chemical reactions and nuclear reactions
(b) Two elements represented by the letters and Y have atomic numbers 9 and 12 respectively.
(i) Write the electronic configuration of X using the s,p,d, notation
(ii) To what group does Y belong in the periodic table?
(iii) Write the formula of the compound formed when X copibines with Y
(iv) Explain wily X is a good oxidizing agent
(v) State with reason, whether Y would be expected to form acidic or basic oxide
(c) Balance the following nuclear equations and identify the particles represented by X and Y.
(i) \(^{14}_6C\) \(\to\) X + \(^{14}_7N\)
(ii) \(^{14}_7C\)
Y \(\to\) \(^1_1H\) + \(^{17}_8O\)
(d) Consider the following list of substances: Carbon (IV) oxide, hydrogen, zinc, sulphier, methane, potassium and mercury. From the list above, state the:
(i) elements that are metals
(ii) compounds that are gases at room temperature
(iii) non-metals that are solids at room temperature
(a)(i) The ammonium ion contains a covalent bond and a coordinate (dative) bond.
(ii) Three properties of electrovalent compounds: high melting and boiling points; conduct electricity when molten or in aqueous solution; soluble in water (polar solvents); exist as hard crystalline solids.
(iii) Differences between chemical and nuclear reactions:
- A chemical reaction involves only the outer electrons (rearrangement of atoms); a nuclear reaction involves changes in the nucleus (protons and neutrons).
- A chemical reaction forms no new element; a nuclear reaction forms new elements or isotopes and releases far more energy.
(b) X (Z = 9) is fluorine; Y (Z = 12) is magnesium.
(i) X: \(1s^2 2s^2 2p^5\).
(ii) Y belongs to Group II.
(iii) Compound of X with Y: MgF2.
(iv) X (fluorine) is a good oxidizing agent because it readily gains one electron to complete its octet, owing to its small size and high effective nuclear charge (high electronegativity); in doing so it oxidizes other species.
(v) Y is a metal (electropositive), so it forms a basic oxide (MgO), as metallic oxides are basic.
(c)(i) \(^{14}_{6}C \rightarrow X + ^{14}_{7}N\): mass \(14 = 14 + A\) so A = 0; charge \(6 = 7 + Z\) so Z = -1. X = \(^{0}_{-1}e\) (a beta particle).
(ii) \(^{14}_{7}N + Y \rightarrow ^{1}_{1}H + ^{17}_{8}O\): mass \(14 + A = 18\) so A = 4; charge \(7 + Z = 9\) so Z = 2. Y = \(^{4}_{2}He\) (an alpha particle).
(d) From the list:
- (i) Metals: zinc, potassium, mercury.
- (ii) Compounds that are gases at room temperature: carbon(IV) oxide and methane.
- (iii) Non-metals that are solids at room temperature: sulphur.
Answer Details
(a)(i) The ammonium ion contains a covalent bond and a coordinate (dative) bond.
(ii) Three properties of electrovalent compounds: high melting and boiling points; conduct electricity when molten or in aqueous solution; soluble in water (polar solvents); exist as hard crystalline solids.
(iii) Differences between chemical and nuclear reactions:
- A chemical reaction involves only the outer electrons (rearrangement of atoms); a nuclear reaction involves changes in the nucleus (protons and neutrons).
- A chemical reaction forms no new element; a nuclear reaction forms new elements or isotopes and releases far more energy.
(b) X (Z = 9) is fluorine; Y (Z = 12) is magnesium.
(i) X: \(1s^2 2s^2 2p^5\).
(ii) Y belongs to Group II.
(iii) Compound of X with Y: MgF2.
(iv) X (fluorine) is a good oxidizing agent because it readily gains one electron to complete its octet, owing to its small size and high effective nuclear charge (high electronegativity); in doing so it oxidizes other species.
(v) Y is a metal (electropositive), so it forms a basic oxide (MgO), as metallic oxides are basic.
(c)(i) \(^{14}_{6}C \rightarrow X + ^{14}_{7}N\): mass \(14 = 14 + A\) so A = 0; charge \(6 = 7 + Z\) so Z = -1. X = \(^{0}_{-1}e\) (a beta particle).
(ii) \(^{14}_{7}N + Y \rightarrow ^{1}_{1}H + ^{17}_{8}O\): mass \(14 + A = 18\) so A = 4; charge \(7 + Z = 9\) so Z = 2. Y = \(^{4}_{2}He\) (an alpha particle).
(d) From the list:
- (i) Metals: zinc, potassium, mercury.
- (ii) Compounds that are gases at room temperature: carbon(IV) oxide and methane.
- (iii) Non-metals that are solids at room temperature: sulphur.
Question 4 Report
(a)(i) State one physical method and one chemical method in each case by which the following can be removed:
I. Permanent hardness in water
II. A suspension of CaCO\(_3\) in water.
(ii) Give one disadvantage of hard water.
(b) Explain the following observations: (i) Crystals of washing soda become powdery on exposure to air for a long time
(ii) The concentration of chloride ions in 0.02 mol dm\(^{-3}\) calcium chloride solution is not the same s in 0.02 mol dm\(^{-3}\) sodium chloride solution.
(iii) Iron filings corrode faster than iron nails of the same mass.
(c)(i) Classify each of the following oxides as acidic, basic, neutral or amphoteric. I. ZnO II. CO III. NO\(_2\)
(ii) Give the formula of the acid anhydride of each of the following: I. H\(_2\)CO\(_3\) II. H\(_2\)SO\(_4\)
(iii) Give the IUPAC name of the following salts: I. COCl\(_2\) II. Mg(NO\(_3\))\(_2\).
(d)(i) Mention one pollutant associated with depletion of ozone layer in the atmosphere
(ii) Calculate the volume occupied by 0.125 mole of oxygen at 27°C and a pressure of 2.02 x 105 Nm\(^{-2}\) [I mole of gas occupies 22.4 dm\(^3\) at s.t.p; standard pressure = 1.01 x 105 Nm\(^{-2}\)]
(iii) State one process used for the industrial preparation of oxygen.
Question 5 Report
(a) Draw an energy profile diagram to illustrate a catalysed exothermic reaction and label parts of the curves representing the following:
(i) activated complex (without catalyst);
(ii) activated energy (with catalyst)
(iii) enthalpy change
(b) Give the reasons for the following observations:
(i) A balloon filled with liyilrogen becomes deflated faster than a balloon filled with air under the same conditions.
(ii) Hydrogen peroxide decomposes slowly at room temperature but when a pinch of MnO, is added, bubbles form rapidly.
(iii) A solution of hydrogen chloride as in methylbenzene has no effect on `litmus but a solution of the gas in water turns blue litmus paper red.
(c) Consider the reaction represented by the following equation: 2MnO\(^-_{4(aq)}\) + 5C\(_2\)O\(^{2-}_4\) + 16H\(^+\) \(\to\) 2Mn\(^{2+}_{(aq)}\) + 8H\(_2\)O\(_{(l)}\) + 10C\(_{2(g)}\) .
Write down: (i) the species undergoing reduction giving reasons;
(ii) the reducing agent giving reasons;
(iii) the reduction half equation;
(iv) one observation made during the reaction.
(d)(i) What is an electrochemical cell?
(ii) State three differences between an electrochemical cell and an electrolytic cell.
(a) Energy profile for a catalysed exothermic reaction
In an exothermic reaction the products lie at a lower energy than the reactants, so the enthalpy change \(\Delta H\) is negative. A catalyst provides an alternative reaction pathway of lower activation energy, so its energy hump (and the activated complex on it) is lower than that of the uncatalysed path. The catalyst changes neither the energy of the reactants nor that of the products, so \(\Delta H\) is exactly the same for both paths.
Reading of the labelled parts:
- (i) Activated complex (without catalyst) - the highest point (top of the upper, uncatalysed curve). It is the unstable transition state through which the reactants pass on the uncatalysed path.
- (ii) Activation energy (with catalyst), \(E_{a(cat)}\) - the smaller energy gap from the reactant level up to the top of the lower (catalysed) curve.
- (iii) Enthalpy change, \(\Delta H\) - the vertical drop from the reactant level down to the product level; it is negative because the reaction is exothermic.
(b) Reasons for the observations
(i) Hydrogen gas is much lighter (less dense) than air. By Graham's law the rate of diffusion of a gas is inversely proportional to the square root of its density, so the light hydrogen molecules diffuse (escape) through the tiny pores of the balloon wall much faster than the heavier oxygen and nitrogen molecules of air. The hydrogen balloon therefore loses its gas and deflates faster.
(ii) Manganese(IV) oxide, \(MnO_2\), acts as a catalyst. It provides an alternative pathway of lower activation energy for the decomposition of hydrogen peroxide, so the reaction that is very slow on its own now proceeds rapidly, giving off oxygen gas quickly as bubbles:
\[2H_2O_2\;\xrightarrow{\;MnO_2\;}\;2H_2O\;+\;O_2\]
The \(MnO_2\) is recovered chemically unchanged at the end.
(iii) Hydrogen chloride is a covalent molecule. In a non-polar solvent such as methylbenzene it stays as neutral \(HCl\) molecules and produces no ions, so there are no free \(H^{+}\) ions and litmus is unaffected. In water it ionises completely, giving hydrogen ions (hydroxonium ions):
\[HCl(g)\;+\;H_2O(l)\;\to\;H_3O^{+}(aq)\;+\;Cl^{-}(aq)\]
These \(H^{+}/H_3O^{+}\) ions make the solution acidic, so it turns blue litmus paper red.
(c) The reaction \(2MnO_4^{-}(aq) + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+}(aq) + 8H_2O(l) + 10CO_2(g)\)
(i) Species undergoing reduction: the permanganate ion, \(MnO_4^{-}\). Reason: the oxidation number of manganese decreases from \(+7\) in \(MnO_4^{-}\) to \(+2\) in \(Mn^{2+}\); a decrease in oxidation number is reduction (gain of electrons).
(ii) Reducing agent: the oxalate ion, \(C_2O_4^{2-}\). Reason: it is itself oxidised (the oxidation number of carbon increases from \(+3\) in \(C_2O_4^{2-}\) to \(+4\) in \(CO_2\)), and in being oxidised it supplies the electrons that reduce the \(MnO_4^{-}\).
(iii) Reduction half equation:
\[MnO_4^{-}\;+\;8H^{+}\;+\;5e^{-}\;\to\;Mn^{2+}\;+\;4H_2O\]
(iv) One observation: the purple (pink) colour of the permanganate is discharged, leaving an almost colourless (very pale pink) solution, and effervescence of a colourless gas (\(CO_2\)) is seen.
(d) Electrochemical cell
(i) An electrochemical cell is a device in which a spontaneous chemical (redox) reaction is used to produce electrical energy; that is, it converts chemical energy into electrical energy.
(ii) Three differences between an electrochemical cell and an electrolytic cell:
| Electrochemical (galvanic) cell | Electrolytic cell |
|---|---|
| A spontaneous chemical reaction produces electricity. | An external electric current is supplied to drive a non-spontaneous reaction. |
| Converts chemical energy into electrical energy. | Converts electrical energy into chemical energy. |
| The anode is the negative electrode and the cathode is the positive electrode. | The anode is the positive electrode and the cathode is the negative electrode. |
Answer Details
(a) Energy profile for a catalysed exothermic reaction
In an exothermic reaction the products lie at a lower energy than the reactants, so the enthalpy change \(\Delta H\) is negative. A catalyst provides an alternative reaction pathway of lower activation energy, so its energy hump (and the activated complex on it) is lower than that of the uncatalysed path. The catalyst changes neither the energy of the reactants nor that of the products, so \(\Delta H\) is exactly the same for both paths.
Reading of the labelled parts:
- (i) Activated complex (without catalyst) - the highest point (top of the upper, uncatalysed curve). It is the unstable transition state through which the reactants pass on the uncatalysed path.
- (ii) Activation energy (with catalyst), \(E_{a(cat)}\) - the smaller energy gap from the reactant level up to the top of the lower (catalysed) curve.
- (iii) Enthalpy change, \(\Delta H\) - the vertical drop from the reactant level down to the product level; it is negative because the reaction is exothermic.
(b) Reasons for the observations
(i) Hydrogen gas is much lighter (less dense) than air. By Graham's law the rate of diffusion of a gas is inversely proportional to the square root of its density, so the light hydrogen molecules diffuse (escape) through the tiny pores of the balloon wall much faster than the heavier oxygen and nitrogen molecules of air. The hydrogen balloon therefore loses its gas and deflates faster.
(ii) Manganese(IV) oxide, \(MnO_2\), acts as a catalyst. It provides an alternative pathway of lower activation energy for the decomposition of hydrogen peroxide, so the reaction that is very slow on its own now proceeds rapidly, giving off oxygen gas quickly as bubbles:
\[2H_2O_2\;\xrightarrow{\;MnO_2\;}\;2H_2O\;+\;O_2\]
The \(MnO_2\) is recovered chemically unchanged at the end.
(iii) Hydrogen chloride is a covalent molecule. In a non-polar solvent such as methylbenzene it stays as neutral \(HCl\) molecules and produces no ions, so there are no free \(H^{+}\) ions and litmus is unaffected. In water it ionises completely, giving hydrogen ions (hydroxonium ions):
\[HCl(g)\;+\;H_2O(l)\;\to\;H_3O^{+}(aq)\;+\;Cl^{-}(aq)\]
These \(H^{+}/H_3O^{+}\) ions make the solution acidic, so it turns blue litmus paper red.
(c) The reaction \(2MnO_4^{-}(aq) + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+}(aq) + 8H_2O(l) + 10CO_2(g)\)
(i) Species undergoing reduction: the permanganate ion, \(MnO_4^{-}\). Reason: the oxidation number of manganese decreases from \(+7\) in \(MnO_4^{-}\) to \(+2\) in \(Mn^{2+}\); a decrease in oxidation number is reduction (gain of electrons).
(ii) Reducing agent: the oxalate ion, \(C_2O_4^{2-}\). Reason: it is itself oxidised (the oxidation number of carbon increases from \(+3\) in \(C_2O_4^{2-}\) to \(+4\) in \(CO_2\)), and in being oxidised it supplies the electrons that reduce the \(MnO_4^{-}\).
(iii) Reduction half equation:
\[MnO_4^{-}\;+\;8H^{+}\;+\;5e^{-}\;\to\;Mn^{2+}\;+\;4H_2O\]
(iv) One observation: the purple (pink) colour of the permanganate is discharged, leaving an almost colourless (very pale pink) solution, and effervescence of a colourless gas (\(CO_2\)) is seen.
(d) Electrochemical cell
(i) An electrochemical cell is a device in which a spontaneous chemical (redox) reaction is used to produce electrical energy; that is, it converts chemical energy into electrical energy.
(ii) Three differences between an electrochemical cell and an electrolytic cell:
| Electrochemical (galvanic) cell | Electrolytic cell |
|---|---|
| A spontaneous chemical reaction produces electricity. | An external electric current is supplied to drive a non-spontaneous reaction. |
| Converts chemical energy into electrical energy. | Converts electrical energy into chemical energy. |
| The anode is the negative electrode and the cathode is the positive electrode. | The anode is the positive electrode and the cathode is the negative electrode. |
Question 6 Report
(a) Methane reacts with chlorine under certain condition to produce tetrachloromethane.
(i) State the condition for the reaction
(ii) Name the type of reaction
(iii) Give two uses of methane
(iv) Name one major natural source of methane
(b)(i) Mention one similarity between the reaction of ethanol with sodium and that of sodium with water
(ii) Write the structure of two isomers of C\(_3\)H\(_8\)O
(iii) Differentiate between a fine chemical and a heavy chemical
(iv) Give one example of each in (b)(iii) above
(c) Two compounds X and Y have the same percentage composition by mass of 92.3% carbon and 7.7% hydrogen. Calculate the:
(i) empirical formula of X and Y;
(ii) molecular formula of each compound if molar mass of X is 26 g and Y is 78g.
(d) A protein is boiled for a long time with dilute HCI and a reaction occurred.
(i) State the type of reaction that occurred
(ii) Name the major product formed
(iii) Give the functional groups present in (d)(ii) above.
(a)(i) Condition: presence of (ultraviolet) sunlight / diffused light.
(ii) Type of reaction: substitution (photochemical, free-radical substitution).
(iii) Uses of methane: as a domestic and industrial fuel (natural gas); manufacture of hydrogen; manufacture of carbon black / methanol.
(iv) Major natural source: natural gas (also biogas/marsh gas from decaying organic matter).
(b)(i) Similarity: both reactions liberate hydrogen gas. (\(2Na + 2C_2H_5OH \rightarrow 2C_2H_5ONa + H_2\); \(2Na + 2H_2O \rightarrow 2NaOH + H_2\)).
(ii) Two isomers of C3H8O
Propan-1-ol: CH3-CH2-CH2-OH
Propan-2-ol: CH3-CH(OH)-CH3(iii) A fine chemical is made in small quantities to a high degree of purity and is costly; a heavy chemical is made in large (bulk) quantities and is of lower purity.
(iv) Fine chemical: a drug (for example aspirin) or a dye. Heavy chemical: tetraoxosulphate(VI) acid (or sodium hydroxide, ammonia).
(c) C : H by mass = 92.3% : 7.7%.
C: \(\dfrac{92.3}{12} = 7.69\); H: \(\dfrac{7.7}{1} = 7.7\). Ratio \(\approx 1:1\).
(i) Empirical formula of X and Y = CH (empirical mass = 13).
(ii) X: \(\dfrac{26}{13} = 2\), so molecular formula = C2H2 (ethyne). Y: \(\dfrac{78}{13} = 6\), so molecular formula = C6H6 (benzene).
(d)(i) Type of reaction: hydrolysis.
(ii) Major product: amino acids.
(iii) Functional groups present: the amino group (-NH2) and the carboxyl group (-COOH).
Answer Details
(a)(i) Condition: presence of (ultraviolet) sunlight / diffused light.
(ii) Type of reaction: substitution (photochemical, free-radical substitution).
(iii) Uses of methane: as a domestic and industrial fuel (natural gas); manufacture of hydrogen; manufacture of carbon black / methanol.
(iv) Major natural source: natural gas (also biogas/marsh gas from decaying organic matter).
(b)(i) Similarity: both reactions liberate hydrogen gas. (\(2Na + 2C_2H_5OH \rightarrow 2C_2H_5ONa + H_2\); \(2Na + 2H_2O \rightarrow 2NaOH + H_2\)).
(ii) Two isomers of C3H8O
Propan-1-ol: CH3-CH2-CH2-OH
Propan-2-ol: CH3-CH(OH)-CH3(iii) A fine chemical is made in small quantities to a high degree of purity and is costly; a heavy chemical is made in large (bulk) quantities and is of lower purity.
(iv) Fine chemical: a drug (for example aspirin) or a dye. Heavy chemical: tetraoxosulphate(VI) acid (or sodium hydroxide, ammonia).
(c) C : H by mass = 92.3% : 7.7%.
C: \(\dfrac{92.3}{12} = 7.69\); H: \(\dfrac{7.7}{1} = 7.7\). Ratio \(\approx 1:1\).
(i) Empirical formula of X and Y = CH (empirical mass = 13).
(ii) X: \(\dfrac{26}{13} = 2\), so molecular formula = C2H2 (ethyne). Y: \(\dfrac{78}{13} = 6\), so molecular formula = C6H6 (benzene).
(d)(i) Type of reaction: hydrolysis.
(ii) Major product: amino acids.
(iii) Functional groups present: the amino group (-NH2) and the carboxyl group (-COOH).
