WAEC SSCE - Chemistry - 2000

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Allotropes of an element are different forms or structures of the same element in the same physical state. They differ in their physical properties, such as color, melting point, boiling point, and density, due to the variations in the arrangement of atoms or molecules. For example, diamond and graphite are allotropes of carbon, and they have different physical properties because of their different atomic arrangements. Although allotropes have different physical properties, they have the same chemical properties, as they are composed of the same element with the same number of protons and electrons. Therefore, the correct answer is "physical properties".

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A finely divided form of a metal burns more readily in air than the rod form because of its larger surface area. When the surface area of a metal is increased, more of it is exposed to the air, which can lead to increased reactivity. The increased surface area of the finely divided form allows more oxygen molecules to collide with the metal atoms, resulting in a faster and more vigorous reaction. On the other hand, the rod form of a metal has a smaller surface area, making it less reactive as fewer metal atoms are exposed to the air. Therefore, the correct answer is "smaller surface area".

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Silver nitrate (AgNO₃) is the salt that is stable to heat. Stability to heat refers to a salt's ability to resist decomposition or breakdown when exposed to high temperatures. Silver nitrate is a relatively stable salt that does not decompose easily even when exposed to high heat. This property makes it useful in various industrial and laboratory applications that involve heating, such as in the production of photographic film. In contrast, potassium carbonate (K ₂CO ₃), ammonium sulfate ((NH ₄) ₂SO ₄), and sodium bicarbonate (NaHCO₃) are not as stable to heat and may decompose or break down when exposed to high temperatures.

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Sucrose needs to be hydrolysed before it shows reducing property. Sucrose is a non-reducing sugar, meaning it does not have a free aldehyde or ketone group that can participate in oxidation-reduction reactions. To show reducing property, it needs to be hydrolyzed into its component monosaccharides, glucose, and fructose. This is because glucose and fructose are reducing sugars as they have free aldehyde and ketone groups that can participate in oxidation-reduction reactions. Thus, hydrolysis of sucrose is necessary to break the glycosidic bond between glucose and fructose and convert them to free sugars that can show reducing properties.

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To identify the food coloring added to a pastry sample, the suitable procedure is to do a chromatogram of the coloring extract. Chromatography is a technique used to separate and analyze the components of a mixture. The various components will travel at different rates based on their interactions with the stationary phase and mobile phase. In this case, the food coloring extract can be separated into its components and the individual components can be identified based on their characteristic retention times.

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The molar volume of gases at standard temperature and pressure (STP) is the same for all gases. This means that equal volumes of gases at STP contain equal numbers of molecules. The molar volume of a gas at STP is 22.4 L/mol. From the question, we know that: mass of gas X = 9.60 g mass of hydrogen = 0.30 g We can use the molar masses of hydrogen and gas X to calculate the number of moles of each gas: moles of hydrogen = mass/molar mass = 0.30 g/1 g/mol = 0.30 mol moles of gas X = mass/molar mass = 9.60 g/molar mass Since the two gases occupy the same volume, we can equate their molar volumes: moles of hydrogen/volume = moles of gas X/volume Substituting the values we know: 0.30/22.4 = (9.60/molar mass)/22.4 Solving for the molar mass of gas X: molar mass = (9.60/0.30) x 1 = 32 g/mol Therefore, the molar mass of gas X is 32 g/mol. Answer: 32g mol ^-1

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When 20 cm³ of distilled water is added to 80 cm³ of 0.50 mol dm^-3 hydrochloric acid, the total volume of the solution becomes 100 cm³. However, the amount of HCl molecules in the solution remains the same before and after the dilution. Therefore, we can use the formula: C1V1 = C2V2 where C1 and V1 are the initial concentration and volume of the acid, and C2 and V2 are the final concentration and volume of the acid after dilution. Substituting the values, we get: C1V1 = C2V2 (0.50 mol dm^-3) x (80 cm³) = C2 x (100 cm³) C2 = (0.50 mol dm^-3) x (80 cm³) / (100 cm³) C2 = 0.40 mol dm^-3 Therefore, the concentration of the acid will change to 0.40 mol dm^-3 after the dilution. Thus, the correct option is (B) 0.40 mol dm^-3.

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A radioactive solid should be stored in a box lined with lead. This is because lead is a dense material that can block radiation, preventing it from escaping and harming people or the environment. Storing the radioactive solid under paraffin oil or in a cool, dark cupboard does not provide sufficient protection from the harmful effects of radiation. Additionally, storing it under ultraviolet light is not recommended as it could potentially increase the radioactive decay of the substance.

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The shape of a molecule is determined by its electron pair geometry, which is based on the number of electron groups around the central atom. In the case of CCl₄, there are four bonding pairs of electrons and no lone pairs around the central carbon atom. This gives it a tetrahedral electron pair geometry, which corresponds to a tetrahedral molecular shape. Therefore, the correct answer is "Tetrahedral".

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The mass number of an element is the sum of the number of protons and neutrons in the nucleus of an atom. Given that the atom of the element has 10 protons and 12 neutrons, the mass number of the element would be: Mass number = Number of protons + Number of neutrons Mass number = 10 + 12 Mass number = 22 Therefore, the answer is: 22.

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Among the given options, only potassium oxide dissolves in water to give an alkaline solution. When potassium oxide (K2O) is dissolved in water, it reacts with water to form potassium hydroxide (KOH), which is a strong base. The chemical equation for this reaction is: K2O + H2O → 2KOH Hence, the solution formed is alkaline due to the presence of hydroxide ions (OH-) from the dissociation of the potassium hydroxide. The other compounds listed do not produce hydroxide ions when dissolved in water, and therefore do not give an alkaline solution.

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C8H18 is a hydrocarbon, which means it consists of hydrogen and carbon atoms only.
(i) Cracking: It is a process of breaking down larger hydrocarbons into smaller hydrocarbons. C8H18 can undergo cracking, which means it can break down into smaller hydrocarbons.
(ii) Combustion: It is a reaction between a substance and oxygen in the presence of a flame or spark. C8H18 can undergo combustion and produce carbon dioxide and water.
(iii) Substitution: It is a reaction in which an atom or group of atoms is replaced by another atom or group of atoms. C8H18 can undergo substitution if it reacts with a suitable reagent.
(iv) Addition: It is a reaction in which two or more reactants combine to form a larger product. C8H18 can undergo addition if it reacts with a suitable reagent.
Therefore, C8H18 can undergo all of the above reactions. So, the answer is None.

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S₈, H₂S, SO₂, and SO₃^2- are all compounds that contain sulfur, but the oxidation number of sulfur is only -2 in one of them. The oxidation number of an atom is a measure of the number of electrons that it has gained or lost in a chemical reaction. In S₈, the oxidation number of sulfur is zero because it is in its elemental state and has not gained or lost any electrons. In H₂S, the oxidation number of sulfur is -2 because hydrogen has an oxidation number of +1 and the overall charge of the compound is 0. In SO₂, the oxidation number of sulfur is +4 because oxygen has an oxidation number of -2 and the overall charge of the compound is 0. In SO₃^2-, the oxidation number of sulfur is +4 because oxygen has an oxidation number of -2 and the overall charge of the ion is -2. Therefore, the only compound in which the oxidation number of sulfur is -2 is H₂S.

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The elements in the same group of the periodic table have the same number of valence electrons. Valence electrons are the electrons located in the outermost shell of an atom that are involved in chemical bonding. The valence electrons determine the reactivity and chemical properties of an element. In this case, all four metals are in the same group and have the same number of valence electrons. However, as we move from left to right across a period, the atomic radius decreases and the effective nuclear charge (the attractive force experienced by the valence electrons from the nucleus) increases. This makes it harder for the valence electrons to be removed from the atom, so the ionization energy (the energy required to remove an electron) increases. Therefore, S, which is in period 6 and has the highest effective nuclear charge, loses electrons least readily. The correct answer is S.

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A plastic which cannot be softened by heat is described as a "thermosetting" plastic. Thermosetting plastics are a type of polymer that irreversibly hardens or "sets" when heated. They become strong and rigid and cannot be melted or reshaped by heating, unlike thermoplastics which can be melted and reshaped multiple times by heating. Examples of thermosetting plastics include Bakelite and melamine.

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The suitable procedure for identifying the food coloring added to a pastry sample is to do a chromatogram of the coloring extract. Chromatography is a technique that separates and identifies the components of a mixture. In this case, a small amount of the coloring extract can be applied to a special paper, and then a solvent is used to separate the components of the extract. The different components will travel at different speeds, allowing for the identification of the specific food coloring used in the pastry sample. This method is more specific and reliable compared to the other options provided, such as carrying out some food tests or examining the pastry sample under a microscope, which may not provide conclusive results. Determining the molar mass of the coloring is also not a suitable method for identifying the food coloring added to the pastry sample.

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The chemical bond that exists between potassium and oxygen in potassium oxide is ionic. In an ionic bond, electrons are transferred from one atom to another resulting in the formation of positively charged cations and negatively charged anions. In potassium oxide, potassium (K) loses one electron to form a K+ cation, while oxygen (O) gains two electrons to form an O2- anion. The resulting electrostatic attraction between the oppositely charged ions leads to the formation of an ionic bond. Therefore, the bond between potassium and oxygen in potassium oxide is ionic because electrons are transferred from potassium to oxygen to form K+ and O2- ions.

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The correct formula for the compound of element Z would be Z₂O₃. This is because Z has a combining power of 3, meaning that it can form compounds by bonding with three other elements. In Z₂O₃, each Z atom is bonded to three O atoms, satisfying Z's combining power of 3. None of the other options satisfy this requirement, as they either have too few or too many atoms of Z.

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The correct equation that represents the reaction is: Fe(s) + Cu^2+ (aq) → Fe^2+ (aq) + Cu(s) Explanation: When an iron nail is placed in a solution of copper(II) sulfate, the iron atoms on the surface of the nail lose electrons and become Fe^2+ ions, which go into the solution. At the same time, the Cu^2+ ions from the copper(II) sulfate solution gain electrons from the iron nail and form a solid copper deposit on the surface of the nail. This is a redox reaction, in which iron is oxidized and copper is reduced. The correct equation must show the transfer of electrons from the iron to the copper ions, as well as the formation of solid copper. Only option (D) shows the correct transfer of electrons and formation of a solid copper deposit, making it the correct equation.

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0.10 mol dm^-3 NaCI conducts electricity better than 0.10 mol dm^-3 CH₃COOH because the solution of NaCI contains more ions. When an ionic compound like NaCI dissolves in water, it dissociates into its component ions, Na⁺ and Cl^-. These ions are able to carry an electric charge and are therefore able to conduct electricity. In contrast, CH₃COOH is a weak acid that only partially dissociates in water to form CH₃COO^- and H⁺ ions. Since there are fewer ions in the solution of CH₃COOH compared to NaCI, it conducts electricity less efficiently. Therefore, the correct answer to the question is that the solution of NaCI contains more ions.

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Sulphur, when burned in air, contributes to the formation of sulphur dioxide gas (SO2). This gas, when released into the atmosphere, can react with water and oxygen to form sulphuric acid (H2SO4), which is a major contributor to acid rain. Therefore, sulphur is the correct answer. The other options, magnesium, aluminium, and zinc do not contribute to acid rain when burned in air.

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The compound that contains a co-ordinate covalent bond is NH₄. A co-ordinate covalent bond, also known as a dative bond, is a type of chemical bond where both electrons in the bond originate from the same atom. In NH₄, the nitrogen atom donates a pair of electrons to a hydrogen ion, creating an NH₄⁺ ion with a co-ordinate covalent bond. HCI is a polar covalent molecule with a shared pair of electrons. CH₄ is a covalent compound where the carbon atom shares electrons with the hydrogen atoms. Na⁺CI is an ionic compound consisting of a positively charged sodium ion and a negatively charged chloride ion, which are held together by ionic bonds.

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When a solution of CuSO₄ is electrolyzed using copper electrodes, the cathode becomes coated with copper. This happens because during electrolysis, positively charged copper ions (Cu²⁺) migrate towards the negatively charged cathode, where they receive electrons from the electrode and are reduced to neutral copper atoms. These copper atoms then combine and deposit on the surface of the cathode, causing it to become coated with copper. The other options are incorrect because: - Both electrodes do not gain electrons; rather, the cathode gains electrons while the anode loses electrons. - The blue color of the solution fading gradually is not a specific outcome of the electrolysis of CuSO₄. - The anode does not increase in mass; rather, it loses mass due to the oxidation reaction that occurs at the anode during electrolysis.

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Chemical equilibrium is attained when the concentrations of the reactants and products remain constant, and the forward and reverse reactions are occurring at the same rate. In other words, the concentrations of the reactants and products do not change over time. This does not mean that the reactants have stopped forming products; it simply means that the rate of the forward reaction is equal to the rate of the reverse reaction. At equilibrium, the system is balanced, and the reaction is said to be in a steady state.

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Chlorine atom has 7 electrons in its outermost shell, and it needs 1 more electron to complete its octet and become stable. Therefore, it tends to gain 1 electron from another atom to form a stable chloride ion, Cl-. This process is called gaining one electron, and it is represented by the following equation: Cl + e- → Cl- So the correct answer is "Gaining one electron".

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From the given information, it can be concluded that K is a carbohydrate and L is a protein. This is because the blue-black colour with iodine solution is a characteristic test for the presence of starch, which is a type of carbohydrate. On the other hand, the deep red precipitate with Million's reagent is a characteristic test for the presence of proteins. Therefore, based on the tests carried out, K is a carbohydrate and L is a protein.

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Biotechnology is applied in the production of various useful products using biological organisms or their components. Baking industry and wine production both use biotechnology in their processes. Production of antibiotics involves the use of biotechnology to produce these drugs using microorganisms like bacteria and fungi. However, the manufacture of soap does not involve the use of biotechnology. Soap is typically made from natural or synthetic ingredients using chemical reactions rather than biological processes. Therefore, the answer is "manufacture of soap."

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Electrovalent compounds, also known as ionic compounds, are characterized by a high melting point. This is because they have a lattice structure made up of positive and negative ions that are held together by strong electrostatic forces of attraction. These forces require a lot of energy to break, hence the high melting point of electrovalent compounds. Additionally, electrovalent compounds tend to be brittle, have high boiling points, and are good conductors of electricity in their molten or aqueous states due to the presence of charged ions. Solubility in ethanol or strong oxidizing ability are not defining characteristics of electrovalent compounds.

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The reaction between dilute tetraoxosulphate (VI) acid (H₂SO₄) and sodium hydrogen trioxocarbonate (IV) (NaHCO₃) produces carbon (IV) oxide (CO₂) gas. This can be explained by the reaction equation: H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2CO₂ + 2H₂O As can be seen from the equation, carbon (IV) oxide gas is produced as one of the products. Therefore, the gas evolved when dilute tetraoxosulphate (VI) acid reacts with sodium hydrogen trioxocarbonate (IV) is carbon (IV) oxide gas.

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The electronic configuration of an element X is given as 1s²2s²2p⁶3s ²3p⁴. This indicates that the element X has a total of 16 electrons. In the electronic configuration, the last electron enters the 3p subshell. Since there are 4 electrons present in the 3p subshell, this means that the element X belongs to the 6th group of the periodic table (also known as group VI), which contains elements with 6 valence electrons. Therefore, the correct option is "belongs to group VI in the Periodic Table". The other options are incorrect as there is no information in the given electronic configuration to support those statements.

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Out of the given options, chlorine reacts most readily with metals. Chlorine is a highly reactive non-metal that readily reacts with most of the metals to form ionic compounds called metal chlorides. When chlorine gas comes into contact with a metal, it gains one electron to form chloride ions, which have a stable electronic configuration of a noble gas. The reaction is exothermic and produces a white solid or colorless gas, depending on the metal. Chlorine gas is used in many industries, such as water treatment, chemical manufacturing, and paper production, due to its strong oxidizing properties and ability to react with metals. Therefore, the correct answer is "Chlorine".

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None

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(a)

(i) Esterification is the reaction between an alkanol and an alkanoic acid, in the presence of a mineral acid catalyst such as concentrated sulphuric acid, to form an ester and water.

(ii) Saponification is the alkaline hydrolysis of a fat, oil or ester with an alkali such as sodium hydroxide to produce soap, which is the sodium salt of a fatty acid, and glycerol in the case of fats and oils.

(b)

(i) The general molecular formula of alkynes is: \[\mathrm{C_nH_{2n-2}}\]

(ii) Ethyl ethanoate has molecular formula \(\mathrm{C_4H_8O_2}\). Its empirical formula is \(\mathrm{C_2H_4O}\).

(iii) The displayed structure of 1,1,2,2-tetrabromoethane is:

(iv)

\[2\mathrm{C_2H_5OH}+2\mathrm{Na}\rightarrow2\mathrm{C_2H_5ONa}+\mathrm{H_2}\]

(c)

(i)

1. Fractional distillation.
2. Cracking, usually catalytic cracking.
3. Addition reaction, specifically chlorination of ethene.
4. Addition polymerisation.

(ii) \(X\) is ethene, \(\mathrm{C_2H_4}\), and \(Y\) is chlorine, \(\mathrm{Cl_2}\).

(iii) The IUPAC name of the product in stage III is 1,2-dichloroethane.

\[\mathrm{CH_2{=}CH_2}+\mathrm{Cl_2}\rightarrow\mathrm{ClCH_2CH_2Cl}\]

(iv) Ethene is polymerised at a high temperature of about \(150\text{ to }300^\circ\mathrm{C}\) and high pressure of about \(500\text{ to }1500\) atmospheres, using a trace of oxygen or an organic peroxide as initiator.

(d)

(i) Palm wine froths because yeast ferments the sugars present to ethanol and carbon dioxide. The bubbles of carbon dioxide cause the foaming.

\[\mathrm{C_6H_{12}O_6}\xrightarrow{\text{yeast}}2\mathrm{C_2H_5OH}+2\mathrm{CO_2}\]

(ii) After some days, bacteria oxidise the ethanol in palm wine, in the presence of atmospheric oxygen, to ethanoic acid. Ethanoic acid gives the wine its sour taste.

\[\mathrm{C_2H_5OH}+\mathrm{O_2}\xrightarrow{\text{bacteria}}\mathrm{CH_3COOH}+\mathrm{H_2O}\]

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(a) Laboratory procedures:

• (i) Remove sediment from water: by filtration (or decantation/sedimentation followed by filtration).
• (ii) Soften temporary hard water without heating: add a calculated amount of calcium hydroxide (slaked lime) - Clark's method - or add sodium trioxocarbonate(IV) (washing soda) to precipitate the Ca²⁺/Mg²⁺ ions.
• (iii) Obtain pure water from muddy water: by (simple) distillation (filter first, then distil).
• (iv) Remove oxygen and moisture from air: pass the air over heated copper to remove oxygen, then through a drying agent such as concentrated H₂SO₄ (or fused CaCl₂) to remove moisture.

(b) (i) Alums are double salts. (ii) In water treatment, alum acts as a coagulant/flocculant: it causes the fine suspended clay/colloidal particles to clump together and settle out, clarifying the water. (iii) Rain water that has passed through limestone deposits dissolves calcium hydrogentrioxocarbonate(IV), becoming temporary hard water; with soap it gives little lather and forms an insoluble scum (because Ca²⁺ reacts with the soap) before it can lather.

(c) Chlorine

(i) Laboratory preparation: \(\text{MnO}_2 + 4\text{HCl} \to \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}\) (conc. HCl, warmed).

(ii) Products of chlorine with hot concentrated NaOH: sodium chloride, sodium trioxochlorate(V) (chlorate, NaClO₃) and water: \(3\text{Cl}_2 + 6\text{NaOH} \to 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\).

(iii) Moist blue litmus paper turns red and is then bleached (decolourised) by the chlorine.

(iv) Volume of Cl₂ at s.t.p. to react with 3.70 g of slaked lime, Ca(OH)₂ (molar mass = 74):

The ratio Ca(OH)₂ : Cl₂ is 1 : 1, so moles Cl₂ = 0.05 mol.

(d) (i) Warming ammonium trioxonitrate(V) with NaOH: a pungent gas (ammonia) is evolved that turns moist red litmus blue: \(\text{NH}_4\text{NO}_3 + \text{NaOH} \to \text{NaNO}_3 + \text{H}_2\text{O} + \text{NH}_3\). (ii) Ammonium trioxocarbonate(IV) leaves no residue on heating because it decomposes completely into gaseous products only (ammonia, carbon(IV) oxide and steam): \((\text{NH}_4)_2\text{CO}_3 \to 2\text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O}\), all of which escape.

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(a)(i) Graham's law of diffusion states that, at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or relative molecular mass.

\[r \propto \frac{1}{\sqrt{d}}\qquad \text{or}\qquad r \propto \frac{1}{\sqrt{M}}\]

(ii) Since X is triatomic, its formula is \(X_3\).

\[M_r(X_3)=3\times16=48\]

\[\text{Vapour density}=\frac{M_r}{2}=\frac{48}{2}=24\]

Vapour density of X = 24.

(iii) Molecules of Z have the greater average velocity. At the same temperature, molecules of Y and Z have the same average kinetic energy. Since a molecule of Y has twice the mass of a molecule of Z, the lighter Z molecule moves faster.

\[\frac{1}{2}m_Yv_Y^2=\frac{1}{2}m_Zv_Z^2\]

Since \(m_Y=2m_Z\),

\[v_Z=\sqrt{2}\,v_Y\]

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(b)(i) The reaction occurred for 12 minutes, after which the curve became horizontal.

(ii) Carbon(IV) oxide, \(\mathrm{CO_2}\), was evolved and escaped from the open vessel. This caused the mass of the flask and its contents to decrease.

(iii) The reaction rate was fastest at the beginning of the reaction because the curve had its greatest gradient at the beginning. The acid concentration was then highest, so effective collisions occurred most frequently.

(iv) Conditions that affect the slope of the curve include:

1. concentration of hydrochloric acid;
2. temperature of the reaction mixture;
3. surface area or particle size of magnesium carbonate.

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(c) \[\mathrm{PCl_{5(g)}\rightleftharpoons PCl_{3(g)}+Cl_{2(g)}}\qquad \Delta H=+95\ \mathrm{kJ\,mol^{-1}}\]

(i)

\[K_c=\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}\]

(ii)

1. Increased pressure: equilibrium shifts to the left, favouring \(\mathrm{PCl_5}\), because there is one mole of gas on the left but two moles of gas on the right.
2. Increased temperature: equilibrium shifts to the right, because the forward reaction is endothermic.
3. Removal of chlorine: equilibrium shifts to the right to replace the removed \(\mathrm{Cl_2}\).

Energy profile diagram for the forward reaction

The products are at a higher energy level than the reactant; hence \(\Delta H\) is positive and the forward reaction is endothermic.

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(a)(i) Three properties that increase generally across a period: ionisation energy, electronegativity, and electron affinity (non-metallic character also increases).

(a)(ii) Two differences between a chemical reaction and a nuclear reaction:

(b) The atoms

• (i) Isotopes of the same element: P and S (both have atomic number 8).
• (ii) Contains 18 neutrons: R (35 - 17 = 18).
• (iii) Chemically unreactive: T (atomic number 10, a noble gas with a stable octet).
• (iv) Readily forms an ion with two positive charges: Q (atomic number 20, group II, forms Q\(^{2+}\)).
• (v) Attains an octet by accepting one electron: R (atomic number 17, seven outer electrons, gains one).
• (vi) Forms an ionic bond with R: Q (a metal that donates its two outer electrons to the non-metal R).
• (vii) Belongs to the s-block: Q (its outer electrons occupy the 4s sub-shell).

(c)(i) CuCO\(_3\) to Cu

Heat the copper(II) trioxocarbonate(IV) strongly to give the oxide, then reduce the oxide by passing hydrogen (or heating with carbon) over it:

(c)(ii) MgO to MgSO\(_4\)

Add the magnesium oxide (a base) to warm dilute tetraoxosulphate(VI) acid until no more dissolves; filter off excess oxide, then evaporate and crystallise the filtrate:

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(a)(i) Two transition-metal properties of iron: it exhibits variable oxidation states (Fe²⁺ and Fe³⁺) and it forms coloured compounds (it also acts as a catalyst, e.g. in the Haber process).

(a)(ii) Extraction of iron in the blast furnace (equations):

Removal of impurities (limestone decomposes and the lime removes the silica gangue as slag):

(a)(iii) Prevention of rusting (three methods): painting the surface; coating with oil/grease; galvanising (coating with zinc) or electroplating; sacrificial protection with a more reactive metal.

(a)(iv) The product is Fe₂O₃; the oxidation number of iron in it is +3.

(b)(i) Increasing reactivity: Au < Sn < Zn < Mg < Ca < Na. Those extracted by electrolysis are the most reactive: Na, Ca and Mg.

(b)(ii) Zinc is amphoteric because it (and its oxide/hydroxide) reacts with both acids and alkalis to form a salt and water.

(c)(i) Oxidation: the loss of electrons by a species.

(c)(ii) Moles of electrons \(= \dfrac{Q}{F} = \dfrac{4825}{96500} = 0.05\ \text{mol}\).

(c)(iii) For \(\text{Cu}^{2+} + 2e^- \to \text{Cu}\), 0.250 F provides 0.250 mol of electrons, discharging \(\dfrac{0.250}{2} = 0.125\ \text{mol}\) of Cu²⁺.