WAEC SSCE - Chemistry - 1989

Bayanin Amsa

Bayanin Amsa

The correct answer is allotropy. Allotropy is a phenomenon in which an element can exist in two or more different physical forms, known as allotropes, in the same physical state (i.e., solid, liquid, or gas) under different conditions of temperature and pressure. The different allotropes of an element have different physical and chemical properties, such as density, melting point, and reactivity, due to differences in their molecular or crystal structures. Some examples of elements that exhibit allotropy include carbon (which has the allotropes graphite, diamond, and fullerenes), oxygen (which has the allotropes O2 and O3), and sulfur (which has the allotropes S8, S6, and S4).

Bayanin Amsa

The molar mass of magnesium is 24.0 g/mol, which means that one mole of magnesium has a mass of 24.0 g. To find the number of atoms in 24.0 g of magnesium, we need to first determine the number of moles of magnesium in 24.0 g. We can do this using the formula: moles = mass/molar mass moles of magnesium = 24.0 g / 24.0 g/mol moles of magnesium = 1.0 mol So, there is 1.0 mole of magnesium in 24.0 g. We can use Avogadro's number (6.02 x 10^23) to find the number of atoms in 1.0 mole of magnesium: number of atoms = Avogadro's number x moles of magnesium number of atoms = 6.02 x 10^23 x 1.0 number of atoms = 6.02 x 10^23 Therefore, there are 6.02 x 10^23 atoms in 24.0 g of magnesium. The answer is 1 x 6.02 x 10^23.

Bayanin Amsa

Silver chloride is used extensively in manufacturing photographic chemicals. It is a light-sensitive compound that is used in making photographic paper and film. When exposed to light, it undergoes a chemical reaction that results in the formation of a visible image. Silver chloride is also used in making emulsions and toners for black and white photography.

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The oxidation number of an atom is the charge it would have if electrons were transferred completely to the more electronegative atom. Chlorine is more electronegative than most other elements, and thus has a high tendency to gain electrons. The oxidation number of chlorine in KCIO₃ is +5, in CI₂O₇ is +7, in ZnCI₂ is -1, and in HCI is -1. Therefore, the only option where the oxidation number of chlorine is +1 is NaCIO.

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The electron configuration of X is 1s² 2s² 2p⁶ 3s¹ which indicates that X has one valence electron in its outermost shell. The valence electron of X is in the 3s subshell, which means that X is in period 3 of the periodic table. The electron configuration of Y is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², indicating that Y has two valence electrons in its outermost shell. The valence electrons of Y are in the 4s subshell, which means that Y is in period 4 of the periodic table. Therefore, X belongs to group 1, and Y belongs to period 4, which means that the correct option is: "x belongs to group 1, Y belongs to period 4."

Bayanin Amsa

Among the given options, ethylethanoate, methanol, ethanoic acid, methanoic acid, and ethanol are all polar substances, meaning they have a partially positive and partially negative end due to the arrangement of their atoms. These partial charges allow them to interact with water molecules through hydrogen bonding, making them soluble or miscible in water. However, ethylethanoate is the only exception since it is a nonpolar substance. It lacks partial charges, and as such, cannot form hydrogen bonds with water molecules, making it insoluble or immiscible in water.

Bayanin Amsa

Octane number is a measure of the performance of petrol in internal combustion engines. It is a rating based on the ability of petrol to resist 'knocking' or 'pinging' during combustion. Knocking or pinging is an undesirable condition that occurs when the air-fuel mixture in the engine's cylinder detonates prematurely, which can cause engine damage. A high octane number indicates that the petrol can resist knocking, allowing the engine to operate at higher compression ratios, which typically results in more power and efficiency. Therefore, petrol containing a high proportion of compounds with high resistance to knocking, such as 2,2,4-trimethylpentane (also known as iso-octane), will have a higher octane number. Hence, the correct option is 2,2,4-trimethylpentane.

Bayanin Amsa

The most suitable method to use when separating an insoluble solid from a liquid is filtration. Filtration is a process of separating a solid from a liquid by passing the mixture through a porous material, such as filter paper or a sand bed. The solid is trapped on the filter paper, while the liquid passes through and is collected in a separate container. This method is effective for separating a solid that does not dissolve in the liquid, such as sand from water. Evaporation, magnetization, sublimation, and distillation are methods used for separating a solid from a liquid when the solid is either soluble in the liquid or has a different boiling point than the liquid.

Bayanin Amsa

Locally produced soap in which wood ash is used as the base is usually soft because the wood ash contains a lot of potassium ions. Potassium hydroxide, also known as caustic potash, is formed when wood ash is mixed with water. It is a strong alkali that reacts with fats and oils in the soap-making process to form soap molecules. Potassium ions are responsible for making the soap soft because they have a weaker attraction to water molecules than sodium ions, which are used in the production of hard soap. Therefore, soap made from potassium ions tends to be softer and more soluble in water.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The voltage of the cell can be calculated using the formula: E_cell = E_{reduction, cathode} - E_{oxidation, anode} The half-reactions for the cell are: Cathode: Cu²⁺ (aq) + 2e^- → Cu(s) (E_{reduction, cathode} = 0.337V) Anode: Zn(s) → Zn²⁺ (aq) + 2e^- (E_{oxidation, anode} = -0.763V) Substituting the values into the formula, we get: E_cell = 0.337V - (-0.763V) = 1.1V Therefore, the voltage of the cell represented at Zn_(s) Zn ²⁺_(aq) /Cu²⁺_(aq) /Cu_(s) is +1.10V. Answer: +1.10V

Bayanin Amsa

Bayanin Amsa

Cellulose does not usually occur in the crystalline form. Cellulose is a complex carbohydrate that forms long, linear chains of glucose molecules, which are linked by beta-glycosidic bonds. The chains form bundles, which are stabilized by hydrogen bonds. These bundles further form microfibrils, which make up the structural material of plants. Although cellulose has a regular and repeating structure, it does not form crystals because of its complex and branched structure. In contrast, glucose and sucrose can form crystals, and maltose and fructose can occur in both crystalline and amorphous forms.

Bayanin Amsa

Copper can best be purified by electrolyzing a copper (ll) salt solution using the impure copper as the anode. Electrolysis is a process in which an electric current is passed through a solution containing ions, causing them to move towards the electrodes. In this case, the impure copper is used as the anode, and the cathode is made of pure copper. As the current passes through the solution, the impure copper anode dissolves, and the copper ions move towards the cathode and are deposited as pure copper. This process allows for the removal of impurities from the copper, resulting in a higher level of purity than other methods such as roasting or heating with coke.

Bayanin Amsa

Bayanin Amsa

The molar mass of a gas can be calculated using the following formula: molar mass = mass ÷ number of moles At standard temperature and pressure (STP), the molar volume of a gas is 22.4 dm³, which is equivalent to 22,400 cm³. Therefore, the number of moles of the gas can be calculated as follows: number of moles = volume of gas at STP ÷ molar volume of gas at STP number of moles = 200 cm³ ÷ 22,400 cm³/mol number of moles = 0.00893 mol Now, we can use the formula for molar mass to find the answer: molar mass = mass ÷ number of moles molar mass = 0.268 g ÷ 0.00893 mol molar mass = 30.03 g/mol Therefore, the molar mass of the gas is approximately 30 g/mol. The correct answer is.

Bayanin Amsa

Bayanin Amsa

The solubility of a substance is the maximum amount of solute that can be dissolved in a given amount of solvent at a specific temperature. In this question, we are given the volume and concentration of a saturated solution of potassium trioxonitrate (V) at 40°C. We can use this information to calculate the solubility of the salt at 40°C. First, we need to convert the given volume from cm³ to dm³ (litres) since the concentration is given in mol dm^-3. 50 cm³ = 50 ÷ 1000 dm³ = 0.05 dm³ The number of moles of potassium trioxonitrate (V) in the solution can be calculated using the formula: moles = concentration × volume The concentration is given as saturated, which means that the solution contains the maximum amount of potassium trioxonitrate (V) that can dissolve at 40°C. Therefore, the concentration is equal to the solubility of the salt at that temperature. So, we have: moles of KNO₃ = solubility × volume = solubility × 0.05 dm³ We are also given the mass of KNO₃ in the solution, which is 5.05 g. We can use the molar mass of KNO₃ (101 g/mol) to calculate the number of moles: moles of KNO₃ = mass ÷ molar mass = 5.05 ÷ 101 = 0.05 mol Now, we can equate the two expressions for the number of moles of KNO₃: solubility × 0.05 = 0.05 solubility = 0.05 ÷ 0.05 = 1 mol dm^-3 Therefore, the solubility of potassium trioxonitrate (V) at 40°C is 1.0 mol dm^-3. Answer: 1.0 mol dm^-3.

Bayanin Amsa

Bayanin Amsa

The reaction that represents the laboratory preparation of hydrogen is Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). In this reaction, zinc metal reacts with hydrochloric acid to produce hydrogen gas and zinc chloride. The other reactions do not produce hydrogen gas as one of their products or involve hydrogen gas as a reactant.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Cracking is a process in which large hydrocarbon molecules are broken down into smaller molecules. This is usually achieved by heating the hydrocarbon molecules to high temperatures and then passing them over a catalyst. Cracking is an important process in the petroleum industry because it allows for the production of smaller, more useful hydrocarbons, such as gasoline and diesel fuel, from larger, less useful ones, such as heavy oils and waxes. The smaller hydrocarbons produced by cracking have higher commercial value and are in greater demand for use as fuels, solvents, and raw materials for the production of other chemicals. Therefore, the answer is in which large hydrocarbon molecules are broken into smaller units.

Bayanin Amsa

The formation of allotropes of sulphur is dependent on the concentration of the sulphur solution. If the concentration of the sulphur solution is high, then the allotrope of sulphur formed would be rhombic sulphur. However, if the concentration of the sulphur solution is low, then the allotrope of sulphur formed would be amorphous sulphur. Therefore, the concentration of the solution determines the allotrope of sulphur formed during evaporation.

Bayanin Amsa

The major raw material in a plastic industry is ethene. Ethene, also known as ethylene, is a colorless gas that is derived from natural gas or petroleum. It is the most commonly used monomer, or building block, in the production of plastic. Ethene is polymerized to form polyethylene, which is used in a wide variety of plastic products, including packaging materials, toys, and household items. The production of plastic from ethene is a highly energy-intensive process and contributes to the emission of greenhouse gases, which can have negative impacts on the environment and climate change.

Bayanin Amsa

Bayanin Amsa

The condition for a spontaneous reaction is that the Gibbs free energy change (∆G) of the system should be negative. ∆G is related to enthalpy change (∆H), entropy change (∆S), and temperature (T) through the equation: ∆G = ∆H - T∆S For a spontaneous reaction, ∆G must be negative, which can only occur when one of the following conditions is met: - ∆H is negative (exothermic reaction) and ∆S is positive (increase in disorder) - ∆H is positive (endothermic reaction) and ∆S is negative (decrease in disorder), but T∆S is larger than ∆H, so the overall ∆G is negative. Therefore, the option that represents a condition for a spontaneous reaction is: ∆H-T∆S is negative.

Bayanin Amsa

The s-block elements are the elements in Group 1 and Group 2 of the periodic table, plus hydrogen and helium. They are called "s-block" elements because their valence (outermost) electrons are in the s subshell of their atoms. Aluminium is not an s-block element. It is located in Group 13, which is part of the p-block elements. Its valence electrons are in the p subshell of its atoms. Therefore, the answer is aluminium.

Bayanin Amsa

The statement that is not true of tin is that it combines with copper to form the alloy brass. Brass is actually an alloy of copper and zinc, not copper and tin. All the other statements are true of tin. Tin is a p-block element, it is extracted from cassiterite SnO₂, it is used for protecting iron containers from corrosion, and it shows two oxidation states, +2 and +4 in most of its compounds.

Bayanin Amsa

There are 6 carbon atoms in a benzene ring. Benzene is a colorless, highly flammable liquid that is composed of carbon and hydrogen atoms. Its molecular formula is C6H6, which means it contains six carbon atoms and six hydrogen atoms. The carbon atoms are arranged in a unique hexagonal ring structure, which is sometimes referred to as an aromatic ring. This ring structure gives benzene its distinctive chemical and physical properties and makes it an important building block for the production of many chemicals and materials, including plastics, rubber, and dyes.

Bayanin Amsa

Petrol is a complex mixture of hydrocarbons obtained from petroleum, and the composition of petrol can vary depending on its source and refining process. However, hydrocarbons with a higher number of carbon atoms tend to have higher boiling points and are more likely to be present in diesel or other heavier fuels rather than petrol. Therefore, the hydrocarbon that is not likely to be present in petrol is C₁₄H₃₀.

Bayanin Amsa

(a) A sample of water is said to be hard when it does not readily form a lather (foam) with soap, but instead forms an insoluble scum, because it contains dissolved calcium and magnesium salts.

(b) Temporary hardness is caused by dissolved hydrogen carbonates of calcium and magnesium and can be removed simply by boiling, whereas permanent hardness is caused by the sulphates and chlorides of calcium and magnesium and cannot be removed by boiling.

(c) Hardness can be completely removed by distillation (or by the ion-exchange/permutit method, which removes both temporary and permanent hardness).

(d) Two local materials used for making soap: a vegetable oil or animal fat (for example palm oil or palm-kernel oil) and wood ash (which supplies the alkali/caustic potash).

Bayanin Amsa

(a) Zinc with dilute HNO₃ vs dilute HCl

Dilute hydrochloric acid is a non-oxidizing acid. Because zinc lies above hydrogen in the activity series, it displaces hydrogen ions as hydrogen gas: \[Zn_{(s)} + 2HCl_{(aq)} \rightarrow ZnCl_{2(aq)} + H_{2(g)}\]

Trioxonitrate(V) acid (nitric acid) is a strong oxidizing acid. The nitrate ion, rather than the hydrogen ion, is reduced, so hydrogen gas is not evolved; instead oxides of nitrogen (or ammonium salt with very dilute acid) are formed, for example: \[3Zn + 8HNO_3 \rightarrow 3Zn(NO_3)_2 + 2NO + 4H_2O\]

(b) Ammonia to trioxonitrate(V) acid (Ostwald process)

\[4NH_3 + 5O_2 \xrightarrow{Pt} 4NO + 6H_2O\]

\[2NO + O_2 \rightarrow 2NO_2\]

\[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]

(c) Mass of sodium trioxonitrate(V)

\[NaOH + HNO_3 \rightarrow NaNO_3 + H_2O\]

Moles of HNO₃ = \(1.00 \times \frac{100}{1000} = 0.100\,mol\). Moles of NaOH = \(\frac{30}{40} = 0.75\,mol\).

The acid is the limiting reagent, so moles of NaNO₃ = 0.100 mol. Molar mass of NaNO₃ = 23 + 14 + 48 = 85 g mol^-1.

\[\text{mass} = 0.100 \times 85 = 8.5\,g\]

(d) Decomposition by heat

• (i) \[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
• (ii) \[2Cu(NO_3)_2 \rightarrow 2CuO + 4NO_2 + O_2\]
• (iii) \[2Hg(NO_3)_2 \rightarrow 2Hg + 4NO_2 + O_2\]

Bayanin Amsa

(a) Le Chatelier's principle

When a system at equilibrium is subjected to a change in conditions (concentration, temperature or pressure), the equilibrium shifts in the direction that tends to oppose or reduce the effect of that change.

(b) Conditions for a high yield of ammonia in the Haber process

The reaction is: \[N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \quad \Delta H = -ve\]

• High pressure (about 200 to 250 atm): the forward reaction produces fewer gas molecules (4 volumes to 2 volumes), so raising the pressure shifts the equilibrium to the right, increasing the yield of ammonia.
• Low temperature: the forward reaction is exothermic, so a lower temperature favours a higher yield. However, too low a temperature makes the reaction very slow, so a moderate (compromise) temperature of about 450 to 500 °C is used together with an iron catalyst to reach equilibrium quickly.
• Continuous removal of ammonia (by liquefaction) drives the equilibrium forward.

(c) Chemical test for ammonia

Ammonia turns moist red litmus paper blue, and it gives dense white fumes of ammonium chloride when a glass rod dipped in concentrated hydrochloric acid is held in it: \[NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}\]

(d) Observations with aqueous ammonia

• (i) Zinc chloride solution: a white gelatinous precipitate of zinc hydroxide forms, which dissolves in excess ammonia to give a colourless solution.
• (ii) Copper(II) tetraoxosulphate(VI) solution: a pale blue precipitate of copper(II) hydroxide forms, which dissolves in excess ammonia to give a deep royal-blue solution.

(e) H-N-H bond angle less than H-C-H

In methane the carbon atom has four bonding pairs of electrons and no lone pair, so the equal bond-pair repulsions give the ideal tetrahedral angle of 109.5°. In ammonia the nitrogen atom has three bonding pairs and one lone pair. A lone pair repels more strongly than a bonding pair, so it compresses the bonding pairs closer together, reducing the H-N-H angle to about 107°.

(f) Two uses of ammonia

• Manufacture of nitrogenous fertilizers such as ammonium salts and urea.
• Manufacture of trioxonitrate(V) acid (nitric acid) in the Ostwald process. (It is also used as a refrigerant.)

Bayanin Amsa

(a) Physical property

• (i) Diamond: it is extremely hard (the hardest natural substance) and transparent; it does not conduct electricity.
• (ii) Graphite: it is soft, slippery and greyish-black (opaque); it conducts electricity.

(b) Two uses of diamond: for making cutting and drilling tools (e.g. glass cutters and rock drills) and as an abrasive for grinding and polishing; it is also used in jewellery.

(c) Evidence that graphite and diamond are allotropes of carbon: both substances are composed of carbon only, and when each is completely burnt in excess oxygen the only product formed is carbon dioxide (in the same proportion). Producing the same single product on combustion shows that both are simply different physical forms of the same element, carbon, which is what allotropy means.

Bayanin Amsa

(a) The statement that the half-life of \(^{56}_{25}Mn\) is \(9.3\times10^{3}\) s means that the time required for one-half of the atoms (nuclei) present in any given sample of the radioactive manganese-56 to decay (disintegrate) is \(9.3\times10^{3}\) seconds. After each interval of \(9.3\times10^{3}\) s, the amount of the undecayed Mn-56 remaining is reduced to half of what it was at the start of that interval.

(b) The three types of radiation usually emitted by radioactive substances:

• Alpha (\(\alpha\)) radiation, positively charged helium nuclei.
• Beta (\(\beta\)) radiation, negatively charged fast-moving electrons.
• Gamma (\(\gamma\)) radiation, high-energy electromagnetic waves carrying no charge.

Bayanin Amsa

(a) Two characteristic features of boiling: it takes place at a definite, fixed temperature (the boiling point) for a given external pressure, and the temperature stays constant while boiling continues; bubbles of vapour form and rise throughout the whole (bulk) of the liquid.

(b) Effect on the boiling point of water:

• (i) Addition of sodium chloride crystals: the boiling point is raised (the dissolved salt causes an elevation of boiling point).
• (ii) Reduction of the atmospheric pressure: the boiling point is lowered (water boils at a lower temperature when the pressure above it is reduced).

(c) Two ways boiling differs from evaporation: boiling occurs only at a fixed temperature (the boiling point), whereas evaporation occurs at all temperatures below the boiling point; boiling takes place throughout the whole body of the liquid (with bubble formation), whereas evaporation takes place only at the surface of the liquid.

Bayanin Amsa

(a) Three characteristics of a homologous series

• All members can be represented by the same general formula.
• Consecutive members differ by a -CH₂- group (a difference of 14 in relative molar mass).
• They have similar chemical properties because they contain the same functional group, and show a gradual change in physical properties as molar mass increases.

(b) Examples

• (i) Alkane: methane, CH₄.
• (ii) Alkyne: ethyne, C₂H₂.

(c) Formula from composition

Empirical formula = CH (empirical mass = 12 + 1 = 13).

\[n = \frac{78}{13} = 6\]

Molecular formula = (CH)₆ = C₆H₆.

(ii) The hydrocarbon is benzene. Its structural formula is a hexagonal ring of six carbon atoms with alternating single and double bonds, each carbon carrying one hydrogen atom (often drawn as a hexagon with an inscribed circle).

(d) X decolorized acidified KMnO₄, so X is unsaturated (an alkene/alkyne); Y did not, so Y is saturated (an alkane).

• (i) Substitution only: Y.
• (ii) Both addition and substitution: X.
• (iii) Polymerization: X.

(e) Ethanol to ethanoic acid

• (i) Conditions: heat (reflux) the ethanol with an acidified oxidizing agent such as acidified potassium tetraoxomanganate(VII) or acidified potassium dichromate(VI).
• (ii) Type of reaction: oxidation. \[CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O\]

Bayanin Amsa

None

Bayanin Amsa

(a) Formation of an alkanoate (ester)

An alkanoate is formed when an alkanoic acid reacts with an alkanol in the presence of a little concentrated tetraoxosulphate(VI) acid as catalyst (esterification):

\[CH_3COOH + C_2H_5OH \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O\]

(ethanoic acid + ethanol give ethyl ethanoate and water).

(b) Proteins

• (i) The monomers of protein are called amino acids (specifically 2-amino acids / alpha-amino acids).
• (ii) Two functional groups present in the monomers: the amino group (\(-NH_2\)) and the carboxyl group (\(-COOH\)).
• (iii) The reaction that joins the amino acids into proteins is condensation polymerization (forming peptide/amide linkages with the elimination of water molecules).

Bayanin Amsa

(a)

(i) Carbon(IV) oxide, CO₂, is evolved.

(ii) Sulphur(IV) oxide, SO₂, is evolved.

(iii) Hydrogen sulphide, H₂S, is evolved.

(b) Sulphur(IV) oxide decolourises acidified potassium tetraoxomanganate(VII) solution without forming a precipitate. Alternatively, it changes acidified potassium heptaoxodichromate(VI) solution from orange to green.

Bayanin Amsa

Both A and B have the same molecular formula, C2H2Cl2, with one H and one Cl on each doubly bonded carbon. They differ only in how the two chlorine atoms are arranged about the rigid C=C double bond.

• A: the two Cl atoms are on opposite sides of the double bond (one up, one down).
• B: the two Cl atoms are on the same side of the double bond (both down).

(a) Relationship between A and B

They are geometric (cis-trans) isomers, a form of stereoisomerism. Restricted rotation about the C=C double bond fixes the positions of the substituents, so the two arrangements are distinct compounds with the same structural formula.

(b) Names

• A = trans-1,2-dichloroethene (Cl atoms on opposite sides).
• B = cis-1,2-dichloroethene (Cl atoms on the same side).

(c) Are the chemical properties the same?

Yes, A and B have essentially the same chemical properties. Reason: they possess the same functional group (a C=C double bond) and the same number and type of atoms and bonds, so they undergo the same chemical reactions. Their differences are mainly physical (for example melting point, boiling point and polarity: the cis form is polar while the trans form is non-polar), not chemical.

Bayanin Amsa

(a) Components

• (i) Producer gas: carbon(II) oxide (CO) and nitrogen (N₂), together with a small amount of carbon(IV) oxide (CO₂). It is made by passing air over red-hot coke.
• (ii) Water gas: carbon(II) oxide (CO) and hydrogen (H₂). It is made by passing steam over red-hot coke: \[C_{(s)} + H_2O_{(g)} \rightarrow CO_{(g)} + H_{2(g)}\]

(b) Why water gas is the better fuel

Water gas is a better fuel than producer gas because both of its components, carbon(II) oxide and hydrogen, are combustible and release heat when burnt. Producer gas, on the other hand, contains a large proportion (about half) of nitrogen, which is chemically inert and does not burn. This non-combustible nitrogen dilutes the fuel and carries away heat, so producer gas has a much lower calorific (heat) value than water gas.

Bayanin Amsa

Type of chemical process involved:

• (a) Polythene from ethene: addition polymerization.
• (b) Ethene from the kerosene fraction of petroleum: cracking (thermal or catalytic cracking).
• (c) Soap from vegetable oils: saponification (alkaline hydrolysis of the oil).
• (d) Margarine from vegetable oils: hydrogenation (addition of hydrogen to the unsaturated oil).

Bayanin Amsa

(a) Conductor vs electrolyte

A conductor is a substance that allows electricity to pass through it without undergoing any chemical change; it conducts by the movement of free electrons (for example, a metal). An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it; it conducts by the movement of ions.

(b)(i) Faraday's first law

The mass of a substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.

(ii) Investigation

• Weigh a clean, dry copper cathode and record its mass.
• Set up an electrolytic cell containing copper(II) tetraoxosulphate(VI) solution with two copper electrodes, connected in series with an ammeter, a rheostat and a switch.
• Pass a steady current I (read on the ammeter) for a measured time t (with a stopwatch); the charge is Q = It.
• Remove, wash, dry and re-weigh the cathode to find the mass of copper deposited.
• Repeat for different values of Q, then plot a graph of mass deposited against Q. A straight line through the origin confirms that mass is proportional to charge.

(c) Relative atomic mass

\[Q = It = 0.45 \times (25 \times 60) = 675\,C\]

For a divalent metal, M²⁺ + 2e^- → M, so 2 × 96500 C deposit 1 mole of metal.

\[\text{moles of metal} = \frac{675}{2 \times 96500} = 3.497 \times 10^{-3}\,mol\]

\[\text{RAM} = \frac{0.222}{3.497 \times 10^{-3}} \approx 63.5\]

The relative atomic mass of the metal is about 63.5 (the metal is copper).

(d) Two applications of electrolysis

• Electroplating of metals (for example, coating with silver, chromium or nickel).
• Purification (refining) of metals such as copper; also the extraction of reactive metals such as aluminium and sodium.