2025 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

The half-cell with the more positive standard electrode potential undergoes reduction, while the other undergoes oxidation. Since \(+0.80\text{ V} > +0.77\text{ V}\), \(\text{Ag}^+(\text{aq})\) is reduced to \(\text{Ag}(\text{s})\) at the cathode, and \(\text{Fe}^{2+}(\text{aq})\) is oxidized to \(\text{Fe}^{3+}(\text{aq})\) at the anode. The overall cell reaction is: \(\text{Ag}^+(\text{aq}) + \text{Fe}^{2+}(\text{aq}) \to \text{Ag}(\text{s}) + \text{Fe}^{3+}(\text{aq})\). Since \(\text{Fe}^{2+}\) is oxidized to \(\text{Fe}^{3+}\), the concentration of \(\text{Fe}^{3+}(\text{aq})\) increases over time. The standard cell potential is \(E^\ominus_{\text{cell}} = E^\ominus_{\text{red}} - E^\ominus_{\text{ox}} = 0.80 - 0.77 = +0.03\text{ V}\). Anions in the salt bridge migrate toward the anode (the iron half-cell) to balance the positive charge build-up.

1 mark for identifying the correct statement. Correct answer is C.

According to the Born-Haber cycle: \(\Delta H^\ominus_{\text{f}} = \Delta H^\ominus_{\text{at}}(\text{M}) + IE_1(\text{M}) + IE_2(\text{M}) + \Delta H^\ominus_{\text{at}}(\text{X}_2) + 2 \times EA(\text{X}) + \Delta H^\ominus_{\text{latt}}\). Substituting the given values: \(-642 = 150 + 738 + 1450 + 244 + 2(-349) + \Delta H^\ominus_{\text{latt}}\). Simplifying the terms: \(-642 = 2582 - 698 + \Delta H^\ominus_{\text{latt}}\), which gives \(-642 = 1884 + \Delta H^\ominus_{\text{latt}}\). Therefore, \(\Delta H^\ominus_{\text{latt}} = -642 - 1884 = -2526\text{ kJ mol}^{-1}\).

1 mark for calculating the correct lattice energy using the Born-Haber cycle expression. Correct answer is A.

The presence of ligands causes the degenerate 3d orbitals of the copper(II) ion to split into two sets of non-degenerate energy levels. When visible light falls on the complex, an electron in a lower-energy d-orbital absorbs a photon of light corresponding to a specific wavelength and is promoted to a higher-energy d-orbital (d-d transition). The color of the complex is the complementary color of the light absorbed.

1 mark for the correct explanation of d-orbital splitting and d-d electron transitions. Correct answer is A.

Oxide \(\text{Y}\) must be highly acidic and react vigorously with water to produce a strongly acidic solution, which corresponds to \(\text{SO}_3\) (forming \(\text{H}_2\text{SO}_4\), pH 1-2). Oxide \(\text{Z}\) is insoluble in water but amphoteric, reacting with both acids and bases. This corresponds to aluminium oxide, \(\text{Al}_2\text{O}_3\).

1 mark for identifying both oxides based on their reactions with water, acids, and bases. Correct answer is A.

Electrophilic addition of \(\text{HBr}\) to 2-methylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{CHCH}_3\), begins with the addition of a proton (\(\text{H}^+\)). The proton can add to carbon-3 to form the more stable tertiary carbocation, \((\text{CH}_3)_2\text{C}^+\text{CH}_2\text{CH}_3\), or to carbon-2 to form the less stable secondary carbocation, \((\text{CH}_3)_2\text{CH}\text{CH}^+\text{CH}_3\). Since tertiary carbocations are more stable due to the positive inductive effect of three alkyl groups, the reaction proceeds via the tertiary carbocation, leading to the major product 2-bromo-2-methylbutane.

1 mark for identifying the correct tertiary carbocation intermediate. Correct answer is A.

The methyl group (\(-\text{CH}_3\)) on methylbenzene is an ortho/para-director (2,4-director). If we brominate first, we will get 2-bromomethylbenzene and 4-bromomethylbenzene. To get 3-bromobenzoic acid, we need a meta-director (3-director) on the ring before bromination. The carboxylic acid group (\(-\text{COOH}\)) is a meta-director. Therefore, we must first oxidize the methyl group to \(-\text{COOH}\) using alkaline \(\text{KMnO}_4\) followed by acidification, and then carry out electrophilic aromatic substitution with \(\text{Br}_2\) and \(\text{FeBr}_3\).

1 mark for choosing the correct sequence based on directing effects. Correct answer is B.

First, calculate the total electric charge passed: \(Q = I \times t = 1.50\text{ A} \times 1930\text{ s} = 2895\text{ C}\). Next, calculate the moles of electrons passed: \(n(\text{e}^-) = \frac{2895}{96500} = 0.0300\text{ mol}\). Now, calculate the moles of metal deposited: \(n(\text{X}) = \frac{1.12}{112} = 0.0100\text{ mol}\). The stoichiometry of the reduction at the cathode is \(\text{X}^{n+} + n\text{e}^- \to \text{X}\), meaning that \(n\) moles of electrons are required per mole of metal. Thus, \(n = \frac{n(\text{e}^-)}{n(\text{X})} = \frac{0.0300}{0.0100} = 3\).

1 mark for the correct calculation of moles of electrons and moles of metal, and finding the charge of the ion. Correct answer is C.

Standard entropy change (\(\Delta S^\ominus\)) is negative when the system becomes more ordered. In option B, 4 moles of gaseous reactants (\(1\text{ mol }\text{N}_2\) and \(3\text{ mol }\text{H}_2\)) react to form only 2 moles of gaseous product (\(2\text{ mol }\text{NH}_3\)). This reduction in the number of gas moles results in a significant decrease in disorder, hence \(\Delta S^\ominus < 0\). In options A, C, and D, there is an increase in disorder (gas formation or dissolution of solid), so \(\Delta S^\ominus\) is positive.

1 mark for identifying the reaction that leads to a decrease in the number of gaseous molecules. Correct answer is B.

By applying Hess's Law and setting up a Born-Haber cycle:
\(\Delta H_f^\ominus[MgO(s)] = \Delta H_{at}^\ominus[Mg(s)] + \text{1st IE}[Mg(g)] + \text{2nd IE}[Mg(g)] + \Delta H_{at}^\ominus[\frac{1}{2}O_2(g)] + \text{1st EA}[O(g)] + \text{2nd EA}[O(g)] + \Delta H_{latt}^\ominus[MgO(s)]\)

Substitute the given values into the equation:
\(-602 = 148 + 738 + 1451 + 249 + (-141) + 798 + \Delta H_{latt}^\ominus\)

Simplify the sum of the non-lattice energy terms:
\(-602 = 3243 + \Delta H_{latt}^\ominus\)

\(\Delta H_{latt}^\ominus = -602 - 3243 = -3845 \text{ kJ mol}^{-1}\)

[1] for selecting option A, which is the mathematically correct lattice energy calculation using the Born-Haber cycle.

For the silver half-cell reaction, \(Ag^+(aq) + e^- \rightleftharpoons Ag(s)\), the number of electrons transferred is \(z = 1\).
Using the Nernst equation:
\(E = E^\ominus + \frac{0.059}{1} \log[Ag^+]\)
\(E = +0.80 + 0.059 \log(0.020)\)
\(\log(0.020) = -1.699\)
\(E = +0.80 + 0.059 \times (-1.699)\)
\(E = +0.80 - 0.100 = +0.70 \text{ V}\)

[1] for selecting option A, indicating correct application of the Nernst equation for a single electrode potential calculation.

The presence of water ligands causes the d-orbitals of the copper(II) ion to split into two groups of different energy levels. When white light passes through the solution, d-electrons absorb energy in the red/orange region of the visible spectrum and are promoted to the higher energy d-orbitals. The complementary light that is not absorbed (blue light) is transmitted, giving the solution its characteristic blue colour.

[1] for selecting option A, which accurately attributes the colour to d-to-d electron transitions involving the absorption of complementary wavelengths.

Step 1 requires the partial oxidation of a primary alcohol to an aldehyde, which is achieved using acidified potassium dichromate(VI) with immediate distillation of the product to prevent further oxidation. Step 2 requires nucleophilic addition of cyanide to the carbonyl group, which is done using hydrogen cyanide in the presence of a trace amount of sodium cyanide catalyst. Step 3 is the acid-catalysed hydrolysis of a nitrile group to a carboxylic acid group, which is carried out by heating with dilute hydrochloric acid under reflux.

[1] for selecting option A, identifying the correct reagent and reaction condition combinations for each step in the organic synthetic pathway.

\(Na_2O\) is a basic metal oxide that reacts with water to form a highly alkaline solution of \(NaOH\) (solution X), giving it the highest pH. \(P_4O_{10}\) and \(SO_3\) are acidic non-metal oxides. \(SO_3\) reacts with water to form the strong acid sulfuric acid, \(H_2SO_4\) (solution Z). \(P_4O_{10}\) reacts with water to form phosphoric acid, \(H_3PO_4\) (solution Y), which is a weaker acid than sulfuric acid. Therefore, solution Z (from \(SO_3\)) has the lowest pH, followed by solution Y (from \(P_4O_{10}\)), and solution X (from \(Na_2O\)) has the highest pH. The correct order is Z < Y < X.

[1] for selecting option A, demonstrating correct application of Period 3 oxide acid-base behavior and relative acid strengths.

The methyl group in methylbenzene is electron-donating due to the positive inductive effect. This increases the electron density of the delocalised pi-system in the benzene ring, making the aromatic ring more nucleophilic and thus more attractive to electrophiles.

[1] for selecting option B, which correctly describes the electron-donating inductive effect of the alkyl group on the reactivity of the aromatic ring.

An increase in entropy (positive \(\Delta S^\ominus\)) occurs when there is an increase in disorder, typically accompanied by an increase in the number of moles of gas. Reaction C involves the thermal decomposition of a solid reactant to produce two gaseous molecules (\(NH_4Cl(s) \rightarrow NH_3(g) + HCl(g)\)), which represents a major increase in gas moles and thus a positive entropy change. Reactions A and B lead to a decrease in the number of moles of gas, and Reaction D is a condensation process, all of which correspond to negative entropy changes.

[1] for selecting option C, correctly identifying the reaction that results in a net increase in the number of gas phase species.

The stability constant for \([Cu(en)_2(H_2O)_2]^{2+}\) is significantly higher than that for \([Cu(NH_3)_4(H_2O)_2]^{2+}\) (\(1.0 \times 10^{20}\) vs \(1.2 \times 10^{13}\)), meaning the bidentate 1,2-diaminoethane complex is much more thermodynamically stable (due to the chelate effect). Therefore, if excess 1,2-diaminoethane is added to a solution containing the tetraamminecopper(II) complex, a ligand exchange reaction will occur to form the more stable bis(1,2-diaminoethane)copper(II) complex.

[1] for selecting option B, correctly deducing the outcome of ligand substitution based on the relative magnitudes of the stability constants.

Using the Born-Haber cycle equation:
\(\Delta H_f^\ominus = \Delta H_{at}^\ominus(\text{Ca}) + \text{IE}_1(\text{Ca}) + \text{IE}_2(\text{Ca}) + 2 \times \Delta H_{at}^\ominus(\text{Cl}) + 2 \times \text{EA}_1(\text{Cl}) + \Delta H_{latt}^\ominus(\text{CaCl}_2)\)

Substitute the given values into the equation:
\(-796 = 178 + 590 + 1145 + 2(121) + 2(-349) + \Delta H_{latt}^\ominus(\text{CaCl}_2)\)

\(-796 = 178 + 590 + 1145 + 242 - 698 + \Delta H_{latt}^\ominus(\text{CaCl}_2)\)
\(-796 = 1457 + \Delta H_{latt}^\ominus(\text{CaCl}_2)\)
\(\Delta H_{latt}^\ominus(\text{CaCl}_2) = -796 - 1457 = -2253\text{ kJ mol}^{-1}\)

1 mark for the correct calculation showing the correct multiplication of chlorine's atomisation energy and electron affinity by 2, yielding -2253 kJ/mol.

The Nernst equation at \(298\text{ K}\) is:
\(E = E^\ominus + \frac{0.059}{z} \log\left(\frac{[\text{oxidised}]}{[\text{reduced}]}\right)\)

Here, \(z = 1\), \([\text{oxidised}] = [\text{Fe}^{3+}] = 0.10\text{ mol dm}^{-3}\), and \([\text{reduced}] = [\text{Fe}^{2+}] = 0.0010\text{ mol dm}^{-3}\).

\(E = +0.77 + \frac{0.059}{1} \log\left(\frac{0.10}{0.0010}\right)\)
\(E = +0.77 + 0.059 \log(100)\)
\(E = +0.77 + 0.059 \times 2\)
\(E = +0.77 + 0.118 = +0.888 \approx +0.89\text{ V}\)

1 mark for substituting the correct concentrations and number of electrons into the Nernst equation to obtain +0.89 V.

The target reaction is obtained by reversing the first equilibrium and adding it to the second equilibrium:

Reversed 1: \([\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O} \rightleftharpoons [\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \quad K_c' = \frac{1}{K_1}\)

Second: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} \quad K_2\)

Adding these two equations gives the target reaction, so:
\(K_c = K_2 \times \frac{1}{K_1} = \frac{K_2}{K_1} = \frac{1.2 \times 10^{13}}{4.0 \times 10^5} = 3.0 \times 10^7\text{ dm}^{12}\text{ mol}^{-4}\)

1 mark for identifying that the equilibrium constant is the ratio of K2 to K1 and calculating the correct value of 3.0 x 10^7.

To obtain 3-bromobenzoic acid, the meta-directing group (\(-\text{COOH}\)) must be present on the ring before bromination, because bromine is an ortho/para-director, whereas the carboxylic acid group is a meta-director.

1. Benzene is first alkylated with \(\text{CH}_3\text{Cl}\) in the presence of catalyst \(\text{AlCl}_3\) to form methylbenzene.
2. Methylbenzene is then oxidised using alkaline \(\text{KMnO}_4\) followed by dilute acid to yield benzoic acid.
3. Benzoic acid is then brominated with \(\text{Br}_2\) in the presence of \(\text{FeBr}_3\) to give 3-bromobenzoic acid as the major product because the \(-\text{COOH}\) group directs the incoming bromine substituent to the meta (3-) position.

1 mark for choosing the sequence that oxidises the methyl side chain to a meta-directing carboxylic acid group before brominating.

\(\text{NaCl}\) dissolves to form a neutral solution (pH ~7).
\(\text{MgCl}_2\) dissolves to form a slightly acidic solution due to weak hydration (pH ~6.5).
\(\text{AlCl}_3\) undergoes significant hydrolysis because of the high charge density of the hydrated aluminium ion, giving an acidic solution (pH ~3).
\(\text{SiCl}_4\) undergoes rapid and complete hydrolysis with water, releasing a high concentration of strong acid: \(\text{SiCl}_4(l) + 2\text{H}_2\text{O}(l) \rightarrow \text{SiO}_2(s) + 4\text{HCl}(aq)\). This produces the maximum amount of \(\text{H}^+\) ions per mole of chloride, yielding a highly acidic solution (pH ~1-2).

1 mark for identifying Silicon(IV) chloride as the chloride that undergoes complete hydrolysis to produce the lowest pH.

3-methylpent-2-ene is \(\text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_3\).

When reacting with \(\text{HCl}\), electrophilic addition occurs via the most stable carbocation intermediate. The tertiary carbocation formed at C3 is more stable than the secondary carbocation at C2. Therefore, the chloride ion attacks C3, giving the major product: 3-chloro-3-methylpentane.

The structural formula of 3-chloro-3-methylpentane is \(\text{CH}_3-\text{CH}_2-\text{C}(\text{Cl})(\text{CH}_3)-\text{CH}_2-\text{CH}_3\).

The central carbon (C3) is bonded to: a chlorine atom (\(-\text{Cl}\)), a methyl group (\(-\text{CH}_3\)), and two identical ethyl groups (\(-\text{CH}_2\text{CH}_3\)). Because it is not bonded to four different groups, C3 is achiral. No other carbon in this molecule has four different groups either, so the total number of chiral carbon atoms is 0.

1 mark for identifying the structure of the major product as 3-chloro-3-methylpentane and correctly deducing that it has 0 chiral carbon atoms.

* Butan-1-ol (primary alcohol) is oxidised by acidified potassium dichromate(VI), changing the solution from orange to green. It does not react with alkaline aqueous iodine.
* Butan-2-ol (secondary alcohol with the \(\text{CH}_3\text{CH(OH)}-\) group) is oxidised by acidified potassium dichromate(VI), changing the solution from orange to green. It also reacts with alkaline aqueous iodine to produce a yellow precipitate of triiodomethane.
* 2-Methylpropan-2-ol (tertiary alcohol) is resistant to oxidation by acidified potassium dichromate(VI) (solution remains orange). It does not react with alkaline aqueous iodine.

Thus, both tests used together allow all three alcohols to be uniquely distinguished from one another.

1 mark for selecting the correct pair of reagents that differentiate primary, secondary, and tertiary structures via oxidation and the iodoform test.

1. Calculate total charge passed, \(Q\):
\(Q = I \times t = 1.50\text{ A} \times (40.0 \times 60\text{ s}) = 3600\text{ C}\)

2. Calculate the moles of electrons, \(n(e^-)\):
\(n(e^-) = \frac{Q}{F} = \frac{3600\text{ C}}{96500\text{ C mol}^{-1}} = 0.0373\text{ mol}\)

3. Calculate the moles of copper metal deposited, \(n(\text{Cu})\):
\(n(\text{Cu}) = \frac{\text{mass}}{A_r} = \frac{1.18\text{ g}}{63.5\text{ g mol}^{-1}} = 0.0186\text{ mol}\)

4. Determine the charge, \(n\):
\(n = \frac{n(e^-)}{n(\text{Cu})} = \frac{0.0373}{0.0186} = 2.01 \approx 2\)

Therefore, the charge on the copper ion is +2.

1 mark for calculating the total charge, translating it to moles of electrons, and dividing by the moles of copper to obtain +2.

Lattice energy is directly proportional to the product of the ionic charges and inversely proportional to the sum of the ionic radii: \(L.E. \propto \frac{q_+ q_-}{r_+ + r_-}\). Magnesium oxide (\(\text{MgO}\)) and calcium oxide (\(\text{CaO}\)) contain \(+2\) and \(-2\) ions, so they have significantly more exothermic lattice energies than lithium fluoride (\(\text{LiF}\)) and sodium fluoride (\(\text{NaF}\)), which contain \(+1\) and \(-1\) ions. Between \(\text{MgO}\) and \(\text{CaO}\), the \(\text{Mg}^{2+}\) ion has a smaller ionic radius than the \(\text{Ca}^{2+}\) ion. This smaller distance results in stronger electrostatic attractions and thus a more exothermic lattice energy for \(\text{MgO}\).

[1 mark] for identifying C as the correct answer by correctly evaluating the impact of higher ionic charges and smaller ionic radius on lattice energy.

A reaction is spontaneous under standard conditions if the standard cell potential is positive: \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} > 0\). For option B, the reduction is \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\) with \(E^\ominus = +0.77\text{ V}\), and the oxidation is \(2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2\text{e}^-\rightleftharpoons\) with \(E^\ominus = +0.54\text{ V}\). This gives \(E^\ominus_{\text{cell}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\). Since the cell potential is positive, the reaction is spontaneous.

[1 mark] for choosing B, supported by standard electrode potential calculations showing a positive overall cell potential.

The structural formula of 2,3-dihydroxy-3-methylpentanoic acid is \(\text{CH}_3\text{-CH}_2\text{-C}(\text{CH}_3)(\text{OH})\text{-CH}(\text{OH})\text{-COOH}\). Carbon 2 (\(\text{-CH(OH)-}\)) is bonded to four different groups: \(-\text{COOH}\), \(-\text{OH}\), \(-\text{H}\), and the large group on carbon 3. Carbon 3 is bonded to four different groups: \(-\text{CH}_3\), \(-\text{OH}\), \(-\text{CH}_2\text{CH}_3\), and the group containing carbon 1 and 2. Therefore, there are exactly 2 chiral carbon atoms in this molecule.

[1 mark] for selecting B, demonstrating correct identification of both chiral carbon centres.

When ligands bond to a transition metal ion, the degenerate d-orbitals split into two sets of different energy levels. When white light passes through the solution, an electron absorbs energy equal to the energy gap (\(\Delta E\)) to promote itself from a lower d-orbital to a higher d-orbital (a d-d transition). The absorbed light is in the visible region, and the remaining non-absorbed light is transmitted and observed as the complementary colour.

[1 mark] for choosing A, which accurately describes d-orbital splitting, d-d transition absorption, and complementary colour transmission.

\(\text{SiCl}_4\) is a simple molecular covalent substance that exists as a volatile liquid at room temperature. It undergoes rapid, vigorous hydrolysis in water to produce silicon dioxide precipitate and white fumes of hydrogen chloride (\(\text{HCl}\)) gas. \(\text{NaCl}\) and \(\text{MgCl}_2\) are solid ionic chlorides, and \(\text{PCl}_5\) is a solid covalent chloride at room temperature.

[1 mark] for selecting C by identifying the physical state at room temperature and the nature of the hydrolysis reaction.

Oxidation of an alkene with hot, concentrated acidified \(\text{KMnO}_4\) cleaves the double bond. A terminal \(\text{=CH}_2\) group is oxidised fully to carbon dioxide (\(\text{CO}_2\)), and a \(\text{=CR}_1\text{R}_2\) group is oxidised to a ketone. Since the products are propanone (\(\text{CH}_3\text{COCH}_3\)) and carbon dioxide (\(\text{CO}_2\)), the starting alkene must be 2-methylpropene, \(\text{(CH}_3)_2\text{C}=\text{CH}_2\).

[1 mark] for B, identifying 2-methylpropene from the oxidative cleavage products.

Spontaneity occurs when \(\Delta G^\ominus < 0\). Using the relation \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), the transition point where \(\Delta G^\ominus = 0\) is calculated as: \(T = \frac{\Delta H^\ominus}{\Delta S^\ominus}\). Converting \(\Delta H^\ominus\) to Joules gives \(142000\text{ J mol}^{-1}\). Thus, \(T = \frac{142000}{245} \approx 579.6\text{ K}\). Therefore, the reaction is spontaneous above approximately \(580\text{ K}\).

[1 mark] for selecting A, showing correct unit conversion of enthalpy change and calculation using the Gibbs free energy formula.

Calculate charge: \(Q = I \times t = 2.50\text{ A} \times (45.0 \times 60)\text{ s} = 6750\text{ C}\). Calculate moles of electrons: \(n(\text{e}^-) = \frac{6750}{96500} = 0.06995\text{ mol}\). The sulfate formula \(X\text{SO}_4\) shows the cation is \(X^{2+}\). The reduction equation is \(X^{2+}(\text{aq}) + 2\text{e}^- \rightarrow X(\text{s})\). Moles of metal \(X = \frac{n(\text{e}^-)}{2} = 0.03497\text{ mol}\). Relative atomic mass: \(A_r(X) = \frac{\text{mass}}{\text{moles}} = \frac{2.22\text{ g}}{0.03497\text{ mol}} \approx 63.5\text{ g mol}^{-1}\). This value corresponds to copper.

[1 mark] for C, correctly calculating the relative atomic mass using stoichiometry and Faraday's constant.

The standard electrode potential for the \(\text{S}_2\text{O}_8^{2-}/2\text{SO}_4^{2-}\) half-cell (\(E^\ominus = +2.01\text{ V}\)) is more positive than that for the \(\text{MnO}_4^-/\text{Mn}^{2+}\) half-cell (\(E^\ominus = +1.51\text{ V}\)). This means \(\text{S}_2\text{O}_8^{2-}\) is a stronger oxidising agent and will undergo reduction (\(\text{S}_2\text{O}_8^{2-} + 2\text{e}^- \rightarrow 2\text{SO}_4^{2-}\)), forcing the other half-cell reaction to go in the reverse direction as oxidation (\(\text{Mn}^{2+} + 4\text{H}_2\text{O} \rightarrow \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^-\)). Thus, \(\text{Mn}^{2+}\) is oxidised to \(\text{MnO}_4^-\). The cell potential is \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 2.01 - 1.51 = +0.50\text{ V}\). Electrons flow from the negative electrode (where oxidation occurs, Half-cell 1) to the positive electrode (where reduction occurs, Half-cell 2).

1 mark for identifying the correct relationship between standard electrode potentials and feasibility of redox reactions.

According to the Born–Haber cycle: \(\Delta H^\ominus_{\text{f}}[\text{CaO}(\text{s})] = \Delta H^\ominus_{\text{at}}[\text{Ca}(\text{s})] + \text{IE}_1[\text{Ca}(\text{g})] + \text{IE}_2[\text{Ca}(\text{g})] + \Delta H^\ominus_{\text{at}}[\text{O}(\text{g})] + \text{EA}_1[\text{O}(\text{g})] + \text{EA}_2[\text{O}(\text{g})] + \Delta H^\ominus_{\text{latt}}\). Rearranging the formula to find the lattice energy: \(\Delta H^\ominus_{\text{latt}} = \Delta H^\ominus_{\text{f}}[\text{CaO}(\text{s})] - \left(\Delta H^\ominus_{\text{at}}[\text{Ca}(\text{s})] + \text{IE}_1[\text{Ca}(\text{g})] + \text{IE}_2[\text{Ca}(\text{g})] + \Delta H^\ominus_{\text{at}}[\text{O}(\text{g})] + \text{EA}_1[\text{O}(\text{g})] + \text{EA}_2[\text{O}(\text{g})]\right)\). Substituting the given values: \(\Delta H^\ominus_{\text{latt}} = -635 - \left(178 + 590 + 1145 + 249 - 141 + 798\right) = -635 - 2819 = -3454\text{ kJ mol}^{-1}\).

1 mark for the correct application of the Born-Haber cycle equation and mathematical evaluation.

Transition metal ions are coloured because of d-orbital splitting in the presence of ligands. When visible light is absorbed, an electron is promoted from a lower energy d-orbital to a higher energy d-orbital (d–d transition). This is only possible if there is a partially filled d-subshell, as in \(\text{Cu}^{2+}\) (\(3\text{d}^9\)), which absorbs red/orange light and transmits its complementary colour, blue. Because \(\text{Zn}^{2+}\) has a \(3\text{d}^{10}\) configuration, its d-subshell is completely full, meaning no electrons can be promoted to a higher d-orbital. Thus, it cannot undergo d–d transitions and its solutions are colourless.

1 mark for explaining colour in transition metals using d–d transition and electronic configurations.

\(\text{NaCl}\) (the chloride of Na) dissolves in water without hydrolysis, yielding a neutral solution of pH 7. \(\text{PCl}_5\) or \(\text{PCl}_3\) (chlorides of P) react violently with water to produce phosphoric/phosphorous acid and hydrogen chloride gas (white fumes), resulting in a very acidic solution (pH 2). \(\text{MgCl}_2\) (the chloride of Mg) undergoes slight hydrolysis due to the charge density of the magnesium ion, forming a weakly acidic solution of approximately pH 6.5.

1 mark for identifying the elements based on the pH of their chloride aqueous solutions.

During the electrophilic addition of \(\text{HCl}\) to \((\text{CH}_3)_2\text{C}=\text{CHCH}_3\), protonation can occur at either C2 or C3. Protonation at C3 gives the tertiary carbocation \((\text{CH}_3)_2\text{C}^+\text{CH}_2\text{CH}_3\), while protonation at C2 gives the secondary carbocation \((\text{CH}_3)_2\text{CH}\text{CH}^+\text{CH}_3\). The tertiary carbocation is more stable due to the greater positive inductive effect of three electron-donating alkyl groups attached to the positively charged carbon. The nucleophilic attack of \(\text{Cl}^-\) on this more stable tertiary carbocation yields the major product, 2-chloro-2-methylbutane.

1 mark for identifying the tertiary carbocation as the more stable intermediate and 2-chloro-2-methylbutane as the major product.

To synthesize 2-hydroxypropanenitrile (a hydroxynitrile) from ethanol (a primary alcohol), ethanol must first be oxidised to ethanal (an aldehyde). This is achieved by heating ethanol with acidified potassium dichromate(VI) and distilling off the volatile ethanal immediately to prevent further oxidation to ethanoic acid. Ethanal then undergoes nucleophilic addition with hydrogen cyanide (\(\text{HCN}\)) in the presence of a trace amount of sodium cyanide (\(\text{NaCN}\)) catalyst at room temperature to form 2-hydroxypropanenitrile.

1 mark for selecting the correct reagents and reaction conditions for oxidation of primary alcohol to aldehyde and subsequent nucleophilic addition.

The basicity of nitrogen-containing compounds depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton. In ethanamide (4), the lone pair is strongly delocalised over the carbonyl group, making it extremely non-basic (weakest). In phenylamine (1), the lone pair is delocalised into the benzene ring's pi-electron system, reducing its availability. Ammonia (3) has no delocalisation, so it is stronger than phenylamine. In ethylamine (2), the electron-donating ethyl group increases the electron density on the nitrogen atom through the positive inductive effect, making its lone pair highly available (strongest). Hence, the order is 4 < 1 < 3 < 2.

1 mark for correctly ordering the four compounds in terms of their basic strength based on electron delocalisation and inductive effects.

From Exp 1 and Exp 2: Doubling \([\text{CH}_3\text{COCH}_3]\) while keeping other concentrations constant doubles the rate, meaning the reaction is first-order with respect to \(\text{CH}_3\text{COCH}_3\). From Exp 1 and Exp 3: Doubling \([\text{I}_2]\) while keeping other concentrations constant has no effect on the rate, meaning the reaction is zero-order with respect to \(\text{I}_2\). From Exp 1 and Exp 4: Doubling \([\text{H}^+]\) while keeping other concentrations constant doubles the rate, meaning the reaction is first-order with respect to \(\text{H}^+\). Combining these gives the rate equation: \(\text{Rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]\).

1 mark for determining the reaction orders for each species from the initial rates data and correctly writing the rate equation.

(a) (i) \( 2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)} \)
(a) (ii) \( \text{Mg(s)} + \text{H}_2\text{O(g)} \rightarrow \text{MgO(s)} + \text{H}_2\text{(g)} \)

(b) The thermal stability of Group 2 nitrates increases down the group. As you go down the group, the ionic radius of the Group 2 cation increases while keeping the same \( 2+ \) charge, so its charge density decreases. As a result, the larger cation has a lower polarizing power and polarizes the electron cloud of the nitrate anion less. This makes the N-O bonds in the nitrate ion more stable, requiring a higher temperature to decompose.

(c) The solubility of Group 2 hydroxides increases down the group. Both the lattice energy and the hydration energy of the hydroxides decrease down the group. However, because the hydroxide ion is relatively small, the lattice energy decreases more rapidly than the hydration energy of the cations. Consequently, the standard enthalpy change of solution, \( \Delta H^\ominus_{\text{sol}} \), becomes more exothermic (or less endothermic) down the group, leading to an increase in solubility.

(a) (i)
- M1: Balanced chemical equation: \( 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \) [1]
- M2: Correct state symbols: (s), (l), (aq), (g) [1]

(a) (ii)
- M1: Balanced chemical equation: \( \text{Mg} + \text{H}_2\text{O} \rightarrow \text{MgO} + \text{H}_2 \) [1]
- M2: Correct state symbols: (s), (g), (s), (g) [1]

(b)
- M1: Identifies that thermal stability increases down the group [1]
- M2: Explains that cationic radius increases / charge density of \( \text{M}^{2+} \) decreases down the group [1]
- M3: Explains that there is less polarization/distortion of the nitrate anion/electron cloud [1]

(c)
- M1: Identifies that solubility of hydroxides increases down the group [1]
- M2: Mentions that both lattice energy and hydration energy decrease down the group [1]
- M3: Explains that lattice energy decreases more rapidly than hydration energy (making \( \Delta H^\ominus_{\text{sol}} \) more exothermic) [1]

(a) (i) Stereoisomerism (specifically cis-trans / geometrical isomerism) arises in but-2-ene because of the restricted rotation about the \( \text{C}=\text{C} \) double bond due to the presence of a \( \pi \) bond, and because each carbon atom of the double bond is attached to two different groups (a methyl group, \( -\text{CH}_3 \), and a hydrogen atom, \( -\text{H} \)).

(ii) Cis-but-2-ene has both methyl groups on the same side of the double bond:
H3C-C(H)=C(H)-CH3
Trans-but-2-ene has the methyl groups on opposite sides:
H3C-C(H)=C(CH3)-H

(b) (i) Electrophilic addition mechanism:
1. The electron density of the \( \pi \) bond in propene attacks the hydrogen atom of the polar \( \text{H}^{\delta+} - \text{Br}^{\delta-} \) molecule. The H-Br bond breaks heterolytically, producing a bromide ion (\( \text{Br}^- \)) with a lone pair.
2. This forms a secondary carbocation intermediate: \( \text{CH}_3\text{CH}^+\text{CH}_3 \).
3. The bromide ion uses its lone pair to attack the positively charged carbon atom of the secondary carbocation, forming 2-bromopropane.

(ii) The reaction proceeds via a secondary carbocation intermediate rather than a primary carbocation (\( \text{CH}_3\text{CH}_2\text{CH}_2^+ \)). The secondary carbocation has two electron-releasing alkyl (methyl) groups that stabilize the positive charge through the positive inductive effect, making it more stable and faster to form than the primary carbocation which only has one alkyl group.

(a) (i)
- M1: Restricted rotation around the \( \text{C}=\text{C} \) bond / due to \( \pi \) bond [1]
- M2: Each carbon of the double bond is bonded to two different groups (\( -\text{H} \) and \( -\text{CH}_3 \)) [1]

(a) (ii)
- M1: Correct displayed formula and label of cis-but-2-ene [1]
- M2: Correct displayed formula and label of trans-but-2-ene [1]

(b) (i)
- M1: Curly arrow from \( \text{C}=\text{C} \) double bond to H of H-Br, with correct dipoles on \( \text{H}^{\delta+}-\text{Br}^{\delta-} \) [1]
- M2: Curly arrow from H-Br bond to Br [1]
- M3: Correct structure of the secondary carbocation intermediate, \( \text{CH}_3\text{CH}^+\text{CH}_3 \) [1]
- M4: Curly arrow from the lone pair of the bromide ion, \( \text{Br}^- \), to the positive carbon of the carbocation [1]

(b) (ii)
- M1: States that the major product goes via a secondary carbocation, which is more stable than a primary carbocation [1]
- M2: Explains that the two alkyl groups in the secondary carbocation exert a positive inductive effect / release electron density to stabilize the positive charge [1]

(a) (i) The enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (100 kPa and 298 K).

(ii) The enthalpy change when 1 mole of a substance is burned completely in excess oxygen under standard conditions (100 kPa and 298 K).

(b) Using Hess's law:
\( \Delta H^\ominus_{\text{c}} = \sum \Delta H^\ominus_{\text{f}} \text{(products)} - \sum \Delta H^\ominus_{\text{f}} \text{(reactants)} \)
\( \Delta H^\ominus_{\text{c}} = [2 \times \Delta H^\ominus_{\text{f}}(\text{CO}_2) + 3 \times \Delta H^\ominus_{\text{f}}(\text{H}_2\text{O})] - [\Delta H^\ominus_{\text{f}}(\text{C}_2\text{H}_5\text{OH}) + 3 \times \Delta H^\ominus_{\text{f}}(\text{O}_2)] \)
\( \Delta H^\ominus_{\text{c}} = [2 \times (-393.5) + 3 \times (-285.8)] - [-277.6 + 0] \)
\( \Delta H^\ominus_{\text{c}} = [-787.0 - 857.4] - [-277.6] \)
\( \Delta H^\ominus_{\text{c}} = -1644.4 + 277.6 = -1366.8\text{ kJ mol}^{-1} \)

(c) Average bond energies are determined in the gaseous state. Under standard conditions, ethanol and water are liquids. Therefore, the standard enthalpy values account for intermolecular forces of attraction (such as hydrogen bonding) and the energy changes associated with phase changes (vaporization/condensation), whereas average bond energy calculations ignore these intermolecular interactions.

(a) (i)
- M1: Enthalpy change when 1 mole of a compound is formed from its elements in their standard states [1]
- M2: Under standard conditions (specified as 298 K and 1 bar/100 kPa/1 atm) [1]

(a) (ii)
- M1: Enthalpy change when 1 mole of a substance is completely burned in oxygen [1]
- M2: Under standard conditions (298 K and 100 kPa) [1]

(b)
- M1: Correctly sets up Hess's cycle or expression: \( [2 \times (-393.5) + 3 \times (-285.8)] - [-277.6] \) [1]
- M2: Calculates correct numerical value: \( -1366.8 \) [1]
- M3: Correct units (\( \text{kJ mol}^{-1} \)) and negative sign [1]

(c)
- M1: State that bond energies are average values / calculated for gaseous state only [1]
- M2: State that ethanol and/or water are liquids in their standard states [1]
- M3: Explain that energy is required to overcome intermolecular forces / hydrogen bonding when transitioning from liquid to gas [1]

(a) (i) The mechanism is \( \text{S}_\text{N}1 \):
1. Slow step: The highly polar \( \text{C}-\text{Br} \) bond breaks heterolytically. A curly arrow is drawn from the \( \text{C}-\text{Br} \) bond to the bromine atom. This forms a stable tertiary carbocation intermediate, \( (\text{CH}_3)_3\text{C}^+ \), and a bromide ion, \( \text{Br}^- \).
2. Fast step: The hydroxide ion, \( \text{OH}^- \), attacks the positively charged carbon of the carbocation. A curly arrow is drawn from the lone pair on the oxygen of the \( \text{OH}^- \) ion to the positive carbon atom.
3. The product, 2-methylpropan-2-ol, is formed.

(ii) Mechanism type: \( \text{S}_\text{N}1 \) (Nucleophilic substitution unimolecular).

(b) Hydrolysis involves breaking the carbon-halogen bond. The bond enthalpy of the \( \text{C}-\text{Cl} \) bond (\( 338\text{ kJ mol}^{-1} \)) is much higher than that of the \( \text{C}-\text{I} \) bond (\( 238\text{ kJ mol}^{-1} \)). Because the \( \text{C}-\text{I} \) bond is weaker, it requires less activation energy and breaks much more easily during the reaction, leading to a faster rate of reaction for 1-iodobutane.

(c) The two structural isomers formed by the elimination of HBr from 2-bromobutane are:
1. But-1-ene (double bond between C1 and C2): \( \text{CH}_2=\text{CHCH}_2\text{CH}_3 \)
2. But-2-ene (double bond between C2 and C3): \( \text{CH}_3\text{CH}=\text{CHCH}_3 \)

(a) (i)
- M1: Curly arrow from \( \text{C}-\text{Br} \) bond to Br [1]
- M2: Correct structure of tertiary carbocation intermediate, \( (\text{CH}_3)_3\text{C}^+ \), and \( \text{Br}^- \) [1]
- M3: Curly arrow from the lone pair of \( \text{OH}^- \) to the positive carbon of the carbocation [1]
- M4: Correct structures of reactants and final product [1]

(a) (ii)
- M1: \( \text{S}_\text{N}1 \) / unimolecular nucleophilic substitution [1]

(b)
- M1: Identifies that the rate of reaction depends on bond strength/bond enthalpy (not bond polarity) [1]
- M2: States that \( \text{C}-\text{I} \) bond is weaker than \( \text{C}-\text{Cl} \) bond [1]
- M3: Concludes that the \( \text{C}-\text{I} \) bond breaks more easily / has lower activation energy [1]

(c)
- M1: Correct skeletal formula of but-1-ene [1]
- M2: Correct skeletal formula of but-2-ene [1]

(a) Two features of dynamic equilibrium are:
1. The rate of the forward reaction is equal to the rate of the reverse reaction.
2. The concentrations of reactants and products remain constant over time (in a closed system).

(b) The expression for the equilibrium constant is:
\( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \)
Units:
\( \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^3} = \frac{1}{(\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6 \)

(c) (i) Increasing the total pressure shifts the position of equilibrium to the right (towards the products). According to Le Chatelier's principle, the system will counteract the increase in pressure by favoring the side with fewer gas molecules (2 moles of gas on the product side compared to 4 moles on the reactant side). This increases the yield of ammonia.

(ii) Increasing the temperature shifts the position of equilibrium to the left (towards the reactants). According to Le Chatelier's principle, the system will favor the endothermic direction to absorb the added thermal energy. Since the forward reaction is exothermic (\( \Delta H = -92\text{ kJ mol}^{-1} \)), the reverse reaction is endothermic, decreasing the yield of ammonia.

(d) The catalyst increases the rate of the forward reaction by providing an alternative reaction pathway with a lower activation energy. However, it has no effect on the position of equilibrium because it increases the rates of both the forward and reverse reactions by the exact same factor.

(a)
- M1: Rate of forward reaction = rate of reverse reaction [1]
- M2: Concentrations of reactants and products remain constant / must be in a closed system [1]

(b)
- M1: Correct \( K_c \) expression: \( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \) [1]
- M2: Correct units: \( \text{mol}^{-2} \text{dm}^6 \) [1]

(c) (i)
- M1: Equilibrium shifts to the right / yield of ammonia increases [1]
- M2: Reason: System shifts to side with fewer moles of gas (2 moles on right vs 4 on left) to decrease pressure [1]

(c) (ii)
- M1: Equilibrium shifts to the left / yield of ammonia decreases [1]
- M2: Reason: System shifts in the endothermic direction to absorb heat [1]

(d)
- M1: Explains that the catalyst increases the rate of the reaction (by lowering activation energy) [1]
- M2: Explains that the catalyst has no effect on the position of equilibrium (because it increases forward and reverse rates equally) [1]

(a) (i) Observation: Steamy/misty fumes (of \( \text{HCl} \) gas).
Equation: \( \text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)} \) (or equivalent with \( \text{Na}_2\text{SO}_4 \)).

(ii) Observations: Purple fumes (vapor of \( \text{I}_2 \)) / dark grey solid / choking gas (\( \text{SO}_2 \)).
Equation: \( 2\text{NaI(s)} + 2\text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Na}_2\text{SO}_4\text{(s)} + \text{I}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \) (or ionic equivalent: \( 2\text{I}^- + 4\text{H}^+ + \text{SO}_4^{2-} \rightarrow \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \)).

(b) The reducing ability of halide ions increases down Group 17. As the group is descended, the ionic radius of the halide ion increases and there is more shielding of the outer electrons. This decreases the electrostatic attraction between the nucleus and the outermost electrons, making it easier for the halide ion to lose its outermost electron to be oxidized.

(c) (i) To confirm the presence of chloride ions: Add dilute nitric acid (\( \text{HNO}_3 \)) to acidify the solution, then add silver nitrate (\( \text{AgNO}_3 \)) solution. A white precipitate (of silver chloride, \( \text{AgCl} \)) will be observed, which dissolves in dilute aqueous ammonia.

(ii) The broad, strong absorption at \( 3350\text{ cm}^{-1} \) corresponds to the \( \text{O}-\text{H} \) alcohol group.

(a) (i)
- M1: Misty / steamy fumes [1]
- M2: Balanced equation: \( \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \) [1]

(a) (ii)
- M1: Purple fumes / dark grey solid [1]
- M2: Balanced equation: \( 2\text{NaI} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{I}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \) [2] (Deduct 1 mark if not balanced, but products are correct)

(b)
- M1: Reducing power increases down the group [1]
- M2: Explanation: ionic radius / shielding increases, meaning less attraction to outermost electron, making electron loss easier [1]

(c) (i)
- M1: Add dilute nitric acid followed by silver nitrate solution [1]
- M2: Formation of a white precipitate [1]

(c) (ii)
- M1: Alcohol / hydroxyl / \( \text{O}-\text{H} \) group [1]

(a) Table of titration results:
- Trial 1 (Rough): Initial = \(0.00\text{ cm}^3\), Final = \(25.80\text{ cm}^3\), Titre = \(25.80\text{ cm}^3\)
- Trial 2: Initial = \(0.55\text{ cm}^3\), Final = \(25.60\text{ cm}^3\), Titre = \(25.05\text{ cm}^3\)
- Trial 3: Initial = \(25.60\text{ cm}^3\), Final = \(50.60\text{ cm}^3\), Titre = \(25.00\text{ cm}^3\)

Concordant trials are Trial 2 and Trial 3 as their titres differ by no more than \(0.10\text{ cm}^3\).
\(\text{Mean titre} = \frac{25.05 + 25.00}{2} = 25.025\text{ cm}^3 \approx 25.03\text{ cm}^3\).

(b) Ionic equation:
\(\text{MnO}_4^-(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)

(c) Moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{25.025}{1000}\text{ dm}^3 = 5.005 \times 10^{-4}\text{ mol}\) (or \(5.01 \times 10^{-4}\text{ mol}\))

(d) Moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\):
According to the 1:5 ratio,
\(n(\text{Fe}^{2+}) = 5 \times 5.005 \times 10^{-4} = 2.5025 \times 10^{-3}\text{ mol}\) (or \(2.50 \times 10^{-3}\text{ mol}\))

(e) Percentage by mass of \(\text{Fe}^{2+}\):
Total moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\) of FB 2:
\(n_{\text{total}} = 10 \times 2.5025 \times 10^{-3} = 2.5025 \times 10^{-2}\text{ mol}\)

Mass of \(\text{Fe}^{2+}\) in sample:
\(m = 2.5025 \times 10^{-2}\text{ mol} \times 55.8\text{ g mol}^{-1} = 1.396\text{ g}\)

Percentage by mass:
\(\%\text{ Fe}^{2+} = \frac{1.396\text{ g}}{9.80\text{ g}} \times 100\% = 14.249\% \approx 14.3\%\)

(f) Pipette uncertainty:
\(\%\text{ uncertainty} = \frac{0.06}{25.0} \times 100\% = 0.24\%\)

(g) Beaker uncertainty:
\(\%\text{ uncertainty} = \frac{12.5}{250} \times 100\% = 5.0\%\)

Explanation:
The wide neck of the beaker results in high volumetric uncertainty (5.0%), which is over 20 times greater than that of a volumetric flask. This leads to extremely poor precision in the concentration of FB 2, making the calculated percentage of iron highly inaccurate and unreliable.

Award marks as follows (total 13 marks):
1. Tabulation: Correctly formatted table with headings and units (e.g., Initial reading / cm3) [1]
2. Precision: All burette readings recorded to 2 decimal places (to the nearest 0.05 cm3) [1]
3. Concordance: Correctly identifies Trial 2 and Trial 3 as concordant titres [1]
4. Mean Titre: Correctly calculates mean titre to 2 decimal places (25.03 cm3) [1]
5. Ionic Equation: Balanced equation with correct species and stoichiometry [1]
6. Moles of Permanganate: Correct calculation of n(MnO4-) showing working [1]
7. Moles of Iron(II): Uses 1:5 stoichiometry to find n(Fe2+) in 25 cm3 [1]
8. Scaling: Multiplies by 10 to find moles in 250 cm3 [1]
9. Mass of Iron: Correctly calculates mass of iron using Ar = 55.8 [1]
10. Percentage by Mass: Calculates 14.3% correctly to 3 significant figures [1]
11. Pipette Error: Calculates 0.24% for the pipette uncertainty [1]
12. Beaker Error: Calculates 5.0% for the beaker uncertainty [1]
13. Explanation: Explains that a beaker has a wide neck, making the meniscus position highly inaccurate and the resulting concentration unreliable [1]

(a) Extrapolation method:
- Plot a horizontal line through the initial points from \(t = 0\) to \(t = 3\text{ min}\) at \(T = 21.0^\circ\text{C}\).
- Plot the cooling line from \(t = 5\) to \(t = 10\text{ min}\). The temperature decreases linearly by \(1.5^\circ\text{C min}^{-1}\).
- Extrapolate the cooling line back to \(t = 4\text{ min}\):
\(T_{\text{extrapolated}} = 56.5 + 1.5 = 58.0^\circ\text{C}\).
- Corrected temperature rise:
\(\Delta T = 58.0^\circ\text{C} - 21.0^\circ\text{C} = 37.0^\circ\text{C}\) (or \(37.0\text{ K}\)).

(b) Heat energy produced:
\(q = m c \Delta T\)
\(m = 50.0\text{ g}\) (density = \(1.00\text{ g cm}^{-3}\))
\(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 37.0\text{ K} = 7733\text{ J} = 7.733\text{ kJ}\) (or \(7.73\text{ kJ}\)).

(c) Moles of \(\text{CuSO}_4\):
\(n(\text{CuSO}_4) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\).

(d) Enthalpy change, \(\Delta H\):
Since the reaction is exothermic, \(\Delta H\) has a negative sign:
\(\Delta H = -\frac{7.733\text{ kJ}}{0.0500\text{ mol}} = -154.7\text{ kJ mol}^{-1} \approx -155\text{ kJ mol}^{-1}\).

(e) Moles of Zn:
\(n(\text{Zn}) = \frac{5.00\text{ g}}{65.4\text{ g mol}^{-1}} = 0.0765\text{ mol}\).
Since \(0.0765\text{ mol} > 0.0500\text{ mol}\), zinc is in excess.

(f) Effect of using a copper beaker:
Copper is a good thermal conductor, which would allow heat to escape rapidly to the surroundings during the reaction. The recorded maximum temperature would be lower, resulting in a smaller measured \(\Delta T\). Thus, the calculated value of \(\Delta H\) would be less exothermic (less negative).

(g) Improvement:
Use a plastic lid on the cup to reduce heat loss by evaporation and convection, or use a digital thermometer/temperature probe with a higher resolution (e.g., \(\pm 0.1^\circ\text{C}\) or \(\pm 0.01^\circ\text{C}\)) to improve precision.

Award marks as follows (total 13 marks):
1. Graph - Axes: Scales chosen correctly, axes labelled with quantities and units [1]
2. Graph - Plotting: Accurate plotting of all points [1]
3. Graph - Lines: Two distinct best-fit straight lines drawn [1]
4. Graph - Extrapolation: Correct extrapolation of the cooling curve back to t = 4 min [1]
5. Corrected Delta T: Determination of Delta T as 37.0 degrees C (+/- 0.2 degrees C) [1]
6. Heat Calculation: Correct calculation of q = 7.73 kJ (or 7733 J) [1]
7. Moles of Reactant: Correctly calculates moles of CuSO4 as 0.0500 mol [1]
8. Enthalpy Value: Calculates magnitude of Delta H as 155 kJ/mol [1]
9. Enthalpy Sign: Assigns a negative (-) sign to Delta H [1]
10. Excess Calculation: Correctly calculates moles of Zn (0.0765 mol) and compares with CuSO4 [1]
11. Copper Beaker Explanation: Identifies copper as a conductor of heat leading to faster heat loss [1]
12. Effect on Delta H: Correctly states Delta H would be less negative / less exothermic [1]
13. Modification: Suggests using a lid, or a digital temperature probe with smaller uncertainty [1]

(a) Relative rate calculations (\(1000/t\)):
- Experiment 1: \(1000 / 40 = 25.0\text{ s}^{-1}\)
- Experiment 2: \(1000 / 53 = 18.9\text{ s}^{-1}\)
- Experiment 3: \(1000 / 80 = 12.5\text{ s}^{-1}\)
- Experiment 4: \(1000 / 160 = 6.25\text{ s}^{-1}\)

(b) Relationship:
As the volume of KI (and hence concentration of iodide ions) decreases, the relative rate of reaction decreases proportionally. For example, halving the volume of KI from \(20.0\text{ cm}^3\) to \(10.0\text{ cm}^3\) halves the relative rate from \(25.0\text{ s}^{-1}\) to \(12.5\text{ s}^{-1}\).

(c) Reaction order:
Since the rate is directly proportional to the volume (concentration) of iodide ions (i.e. doubling concentration doubles the rate), the reaction is first-order with respect to iodide ions.

(d) Role of sodium thiosulfate:
Sodium thiosulfate is used to react with the iodine produced. The time \(t\) represents the time required to produce a fixed, small amount of iodine that completely reacts with the thiosulfate. If the volume (and therefore moles) of thiosulfate varied, the amount of iodine needed to trigger the blue-black color would change, and the time \(t\) would no longer be a direct measure of the initial rate.

(e) Distilled water:
Distilled water is added to ensure that the total volume of the reaction mixture is kept constant (at \(52.0\text{ cm}^3\) ) in all runs. This ensures that the volume of KI added is directly proportional to its final concentration in the reaction mixture, and the concentration of other reactants is not changed.

(f) Source of error and modification:
- Source of error: Human reaction time/subjective determination of the exact point of the blue-black color change.
- Modification: Use a colorimeter connected to a data logger to measure light absorbance over time, providing a highly objective and precise detection of the onset of the color change.

Award marks as follows (total 13 marks):
1. Rate Calculation (Exp 1): Calculates 25.0 s-1 [1]
2. Rate Calculation (Exp 2): Calculates 18.9 s-1 [1]
3. Rate Calculation (Exp 3): Calculates 12.5 s-1 [1]
4. Rate Calculation (Exp 4): Calculates 6.25 s-1 [1]
5. Precision: All rate values recorded consistently to 3 significant figures [1]
6. Relationship Description: Explains that rate is directly proportional to KI volume [1]
7. Order Deduction: States that the reaction is first-order with respect to I- [1]
8. Order Explanation: Uses data (e.g., comparing Exp 1 and 3) to show that halving concentration halves the rate [1]
9. Thiosulfate Role: Explains that a fixed amount of thiosulfate sets a constant threshold of iodine to be produced [1]
10. Thiosulfate Impact: Explains that varying thiosulfate would make the time comparison invalid [1]
11. Water Purpose (Constant Volume): Explains that water keeps the total volume of the mixture constant [1]
12. Water Purpose (Proportionality): Explains that constant total volume ensures KI volume is directly proportional to its concentration [1]
13. Error & Modification: Correctly identifies human reaction time error in visual clock end-point and suggests using a colorimeter [1]

(a) The Ag+/Ag half-cell consists of a silver wire/electrode in a solution of 1.00 mol dm-3 Ag+(aq) ions (e.g., AgNO3). The Fe3+/Fe2+ half-cell consists of a platinum electrode in a solution containing both 1.00 mol dm-3 Fe2+(aq) and 1.00 mol dm-3 Fe3+(aq) ions. The two half-cells are connected by a salt bridge (filled with a solution of an inert electrolyte like KNO3 or NH4NO3) and an external circuit with a high-resistance voltmeter. All solutions must be at 298 K.
(b) Overall equation: Fe2+(aq) + Ag+(aq) -> Fe3+(aq) + Ag(s). E⦵cell = E⦵(reduction) - E⦵(oxidation) = E⦵(Ag+/Ag) - E⦵(Fe3+/Fe2+) = +0.80 - 0.77 = +0.03 V.
(c) E = +0.77 + (0.059/1) * log(0.050 / 0.80) = +0.77 + 0.059 * log(0.0625) = +0.77 + 0.059 * (-1.204) = +0.77 - 0.071 = +0.699 V (or +0.70 V).
(d) Adding NaCl(aq) introduces Cl- ions, which react with Ag+(aq) to form a white precipitate of AgCl(s). This causes a significant decrease in [Ag+]. Since [Ag+] decreases, the electrode potential of the Ag+/Ag half-cell decreases (becomes less positive), which in turn decreases the overall cell potential, Ecell.

(a) 1 mark for Ag electrode in 1.00 mol dm-3 Ag+(aq); 1 mark for Pt electrode in mixture of 1.00 mol dm-3 Fe2+(aq) and 1.00 mol dm-3 Fe3+(aq); 1 mark for salt bridge and voltmeter connection; 0.5 marks for specifying standard conditions of 298 K.
(b) 1 mark for correct balanced ionic equation; 2 marks for calculating +0.03 V with correct sign and units.
(c) 1 mark for substituting z = 1; 1 mark for correctly setting up the ratio log(0.050/0.80); 1 mark for the final calculated value of +0.699 V (or +0.70 V).
(d) 1 mark for identifying that Cl- precipitates Ag+ as AgCl(s); 1 mark for stating that [Ag+] decreases; 1 mark for explaining that this decreases the half-cell potential of Ag+/Ag and thus lowers the overall cell potential.

(a) Lattice energy is the enthalpy change when one mole of an ionic crystalline solid is formed from its constituent gaseous ions under standard conditions.
(b) Using the Born-Haber cycle equation: ΔH⦵f = ΔH⦵at(Ca) + I.E.1(Ca) + I.E.2(Ca) + 2 * ΔH⦵at(Cl) + 2 * E.A.1(Cl) + ΔH⦵latt. Substituting the values: -796 = +178 + 590 + 1145 + 2(121) + 2(-349) + ΔH⦵latt. -796 = 178 + 590 + 1145 + 242 - 698 + ΔH⦵latt. -796 = 1457 + ΔH⦵latt. ΔH⦵latt = -796 - 1457 = -2253 kJ mol-1.
(c) Lattice energy depends on the charges of the ions and their ionic radii (proportional to (q+ * q-) / (r+ + r-)). The charge on the oxide ion (2-) is double the charge on the chloride ion (1-). The calcium ion (2+) is common to both. Since the product of the ionic charges is much greater in CaO than in CaCl2, the electrostatic attraction between the gaseous ions is much stronger in CaO. Thus, the lattice energy of CaO is much more exothermic than that of CaCl2.

(a) 1 mark for stating 'one mole of ionic solid formed from gaseous ions'; 1 mark for specifying 'under standard conditions'.
(b) 1.5 marks for drawing or correctly setting up the algebraic Born-Haber cycle equation; 1 mark for doubling the atomisation enthalpy of chlorine; 1 mark for doubling the electron affinity of chlorine; 2 marks for correct calculation steps; 1 mark for the final answer of -2253 kJ mol-1 with negative sign and units.
(c) 1 mark for stating that lattice energy is proportional to the product of charges / inversely proportional to sum of ionic radii; 1 mark for identifying the charge of O2- is 2- and Cl- is 1-; 1 mark for stating that the product of ionic charges is greater in CaO; 1 mark for concluding that there is much stronger electrostatic attraction in CaO leading to a more exothermic lattice energy.

(a) Co is [Ar] 3d7 4s2, so Co2+ is [Ar] 3d7 (or 1s2 2s2 2p6 3s2 3p6 3d7).
(b)(i) [Co(H2O)6]2+ + 4Cl- -> [CoCl4]2- + 6H2O. The color changes from pink to blue.
(b)(ii) Chloride ligands (Cl-) are larger in size and carry a negative charge compared to the smaller, neutral water ligands. This causes significant steric hindrance and electrostatic repulsion between the chloride ligands, so only 4 chloride ligands can fit around the central Co2+ ion.
(c)(i) The trans isomer has the two Cl- ligands at opposite corners of the octahedron (180 degrees apart), with the two 'en' rings in a planar/perpendicular arrangement. The cis isomer has the two Cl- ligands adjacent to each other (90 degrees apart).
(c)(ii) The cis-isomer is optically active because it lacks a plane of symmetry (it is chiral, with non-superimposable mirror images). The trans-isomer has a plane of symmetry, so it is achiral and optically inactive.

(a) 1 mark for [Ar] 3d7 or full equivalent.
(b)(i) 2 marks for correct equation; 1 mark for color change from pink to blue.
(b)(ii) 1 mark for stating that chloride ligands are larger/have larger ionic radii; 1 mark for explaining steric hindrance or electrostatic repulsion limits the coordination number to 4.
(c)(i) 2.5 marks for cis isomer drawn clearly with bidentate ligands connected to adjacent sites; 2.5 marks for trans isomer drawn with Cl ligands opposite.
(c)(ii) 0.5 marks for identifying the cis-isomer; 1 mark for explaining that it lacks a plane of symmetry / has non-superimposable mirror images.

(a) Step 1: Reagents and conditions: Chlorine gas, Cl2, in the presence of UV light (or heat/sunlight) to undergo free radical substitution. Intermediate X is chloromethylbenzene, C6H5CH2Cl.
Step 2: Reagents and conditions: Potassium cyanide, KCN (or NaCN), in ethanol, heated under reflux. Intermediate Y is 2-phenylethanenitrile, C6H5CH2CN.
Step 3: Reagents and conditions: Dilute acid (e.g., HCl(aq) or H2SO4(aq)), heated under reflux. This is acid hydrolysis.
(b)(i) Reagent: Lithium tetrahydridoaluminate, LiAlH4, in dry ether at room temperature.
(b)(ii) Product: 2-phenylethan-1-ol, C6H5CH2CH2OH.
(c) Reagents: Thionyl chloride (SOCl2) or phosphorus pentachloride (PCl5) or phosphorus trichloride (PCl3) with heat. Reaction type: Nucleophilic substitution / acyl substitution.

(a) Step 1: 1 mark for Cl2 and UV light / heat; 1.5 marks for drawing C6H5CH2Cl.
Step 2: 1 mark for KCN/NaCN in ethanol and reflux/heat; 1.5 marks for drawing C6H5CH2CN.
Step 3: 1 mark for dilute acid (HCl/H2SO4) and reflux/heat; 1.5 marks for identifying the hydrolysis conversion.
(b)(i) 1 mark for LiAlH4 in dry ether (reject NaBH4).
(b)(ii) 1 mark for drawing C6H5CH2CH2OH.
(c) 2 marks for SOCl2 / PCl5 / PCl3; 1 mark for stating nucleophilic substitution or acyl substitution.

(a)(i) C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-.
(a)(ii) Order: phenylamine < ammonia < ethylamine. Explanation: Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton. In ethylamine, the ethyl group is electron-donating (+I inductive effect), which increases electron density on the nitrogen, making the lone pair more available. In phenylamine, the nitrogen lone pair overlaps with the π-delocalised ring system of the benzene ring. This delocalisation significantly reduces the electron density on the nitrogen, making the lone pair much less available to accept a proton. Ammonia is intermediate as it has no electron-donating or withdrawing groups.
(b)(i) Reagents: Nitrous acid (HNO2), typically prepared in situ from sodium nitrite (NaNO2) and excess hydrochloric acid (HCl). Conditions: Temperature maintained between 0 °C and 10 °C.
(b)(ii) Azo dye structure: C6H5-N=N-C6H4-OH (specifically 4-hydroxyphenylazobenzene). Reagents and conditions: Phenol dissolved in aqueous sodium hydroxide (forming sodium phenoxide), kept cold (below 10 °C).

(a)(i) 1 mark for correct reversible equation.
(a)(ii) 1.5 marks for correct order (phenylamine < ammonia < ethylamine); 1.5 marks for explaining the +I effect of the ethyl group in ethylamine; 1.5 marks for explaining the delocalisation of the nitrogen lone pair into the benzene ring in phenylamine.
(b)(i) 2 marks for NaNO2 + HCl (or HNO2); 1 mark for specifying the temperature must be below 10 °C (or between 0 °C and 10 °C).
(b)(ii) 2 marks for the correct structure of 4-hydroxyphenylazobenzene; 2 marks for stating phenol in aqueous NaOH (alkaline) and cold conditions.

(a) Relationship: F = L * e. Definition: The Faraday constant is the electric charge carried by one mole of electrons.
(b)(i) Q = I * t = 1.50 A * (45.0 * 60) s = 1.50 * 2700 = 4050 C.
(b)(ii) Moles of Cu = mass / Ar = 1.31 g / 63.5 g mol-1 = 0.02063 mol.
(b)(iii) The cathode reaction is: Cu2+(aq) + 2e- -> Cu(s). Therefore, 1 mole of Cu requires 2 moles of electrons. Moles of electrons passed = 2 * 0.02063 mol = 0.04126 mol. F = Q / n(e-) = 4050 C / 0.04126 mol = 98158 C mol-1 (or 9.82 x 10^4 C mol-1).
(c) Physical changes: The copper anode dissolves/decreases in mass (or becomes smaller/thinner). No gas bubbles are observed. Ionic half-equation: Cu(s) -> Cu2+(aq) + 2e-.

(a) 1 mark for F = L * e; 2 marks for defining the Faraday constant as the total charge carried by 1 mole of electrons.
(b)(i) 1 mark for formula Q = I * t; 1 mark for correct calculation of 4050 C.
(b)(ii) 1 mark for dividing mass by 63.5; 1 mark for 0.02063 mol.
(b)(iii) 1 mark for identifying that 2 moles of electrons are needed per mole of Cu; 1 mark for calculating 0.04126 mol of electrons; 1 mark for dividing Q by moles of electrons to get 98158 C mol-1 (or 9.82 x 10^4 C mol-1).
(c) 1 mark for noting the anode decreases in size/dissolves; 1.5 marks for the correct equation Cu(s) -> Cu2+(aq) + 2e-.

(a) The entropy change, ΔS⦵, will be positive because a solid reactant decomposes to produce a solid and a gas. Since gas molecules are much more highly disordered than solids, the overall disorder/entropy of the system increases.
(b) ΔH⦵reaction = ΣΔH⦵f(products) - ΣΔH⦵f(reactants) = [ΔH⦵f(BaO) + ΔH⦵f(CO2)] - [ΔH⦵f(BaCO3)] = [(-554) + (-394)] - [-1216] = -948 + 1216 = +268 kJ mol-1.
ΔS⦵reaction = ΣS⦵(products) - ΣS⦵(reactants) = [S⦵(BaO) + S⦵(CO2)] - [S⦵(BaCO3)] = [70 + 214] - 112 = 284 - 112 = +172 J K-1 mol-1.
(c) A reaction becomes thermodynamically feasible when ΔG⦵ <= 0. Since ΔG⦵ = ΔH⦵ - TΔS⦵, at the threshold: T = ΔH⦵ / ΔS⦵. Convert ΔS⦵ to kJ K-1 mol-1: +0.172 kJ K-1 mol-1. T = 268 / 0.172 = 1558 K.
(d) Calcium carbonate has a lower decomposition temperature than barium carbonate. The Ca2+ ion has a smaller ionic radius and thus a higher charge density than the Ba2+ ion, allowing it to polarise and weaken the carbon-oxygen bonds in the carbonate ion more effectively.

(a) 1 mark for stating positive sign; 1 mark for explaining increase in disorder due to gas formation.
(b) 1.5 marks for correct enthalpy calculation (+268 kJ mol-1); 1.5 marks for correct entropy calculation (+172 J K-1 mol-1) (deduct marks for sign/unit errors).
(c) 1 mark for setting ΔG⦵ <= 0; 1 mark for converting units of ΔS⦵ to kJ K-1 mol-1; 1 mark for final temperature calculation (1558 K).
(d) 0.5 marks for lower decomposition temperature for CaCO3; 1 mark for explaining higher charge density and polarising power of Ca2+ compared to Ba2+.

(a)(i) For NO: Comparing Exp 1 and Exp 2, [H2] is constant. [NO] is doubled, and the rate increases by 4 times ((4.8 x 10^-3)/(1.2 x 10^-3) = 4). Since 2^2 = 4, the reaction is second order with respect to NO. For H2: Comparing Exp 1 and Exp 3, [NO] is constant. [H2] is doubled, and the rate doubles ((2.4 x 10^-3)/(1.2 x 10^-3) = 2). Since 2^1 = 2, the reaction is first order with respect to H2.
(a)(ii) Rate = k [NO]2 [H2].
(a)(iii) Using Exp 1: 1.2 x 10^-3 = k (0.10)2 * (0.10). k = 1.2 x 10^-3 / 1.0 x 10^-3 = 1.2. Units: rate / ([NO]2[H2]) = (mol dm-3 s-1) / (mol dm-3)3 = mol-2 dm6 s-1.
(b) The rate-determining step is the slow step (Step 2), so Rate = k2 [N2O2][H2]. Since N2O2 is an intermediate, we express its concentration using the fast equilibrium in Step 1: Kc = [N2O2] / [NO]2, which gives [N2O2] = Kc [NO]2. Substituting this back into the rate expression gives: Rate = k2 * Kc [NO]2 [H2] = k [NO]2 [H2]. This matches the experimental rate equation, so the mechanism is consistent.
(c) An increase in temperature increases the value of the rate constant, k. This is because higher temperature shifts the Maxwell-Boltzmann distribution curve to the right, meaning a significantly larger fraction of molecules possess energy greater than or equal to the activation energy (Ea), leading to a higher frequency of successful collisions.

(a)(i) 1.5 marks for showing second order for NO with math comparison; 1.5 marks for showing first order for H2; 0.5 marks for clarity of reasoning.
(a)(ii) 1 mark for Rate = k [NO]2 [H2].
(a)(iii) 1.5 marks for calculating k = 1.2; 1.5 marks for correct units mol-2 dm6 s-1.
(b) 1 mark for identifying Step 2 as the rate-determining step; 1 mark for relating [N2O2] to [NO]2 via the Step 1 equilibrium; 1 mark for substituting and showing the final rate expression matches the experimental one.
(c) 1 mark for stating k increases with temperature; 1 mark for explaining that a larger fraction of molecules have energy >= Ea.

(a) The standard enthalpy change of solution represents the dissolution of 1 mole of solute in excess water to form an infinitely dilute solution under standard conditions:
\(\text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq)\) (or \(\text{CaCl}_2(s) \xrightarrow{\text{aq}} \text{CaCl}_2(aq)\))

(b) The essential quantitative measurements are:
- The mass of the anhydrous calcium chloride solid, \(m_{\text{solid}}\), using a balance.
- The volume of distilled water used, \(V_{\text{water}}\), using a measuring cylinder or pipette.
- The initial temperature of the water and the temperature of the mixture at regular time intervals before and after addition.

(c) The apparatus should consist of:
- A polystyrene cup (expanded polystyrene is an excellent thermal insulator with low heat capacity).
- A glass beaker (to hold the polystyrene cup securely and provide an insulating pocket of air).
- A tight-fitting lid with two holes (one for the thermometer/temperature probe and one for the stirrer) to prevent evaporation and heat loss via convection.
- A precise thermometer (graduated to at least \(0.1\,^{\circ}\text{C}\)) or digital temperature probe.
- A glass or plastic stirrer to ensure even temperature distribution.

(d) Step-by-Step Procedure:
1. Accurately measure \(50.0\text{ cm}^3\) of distilled water into the polystyrene cup using a volumetric pipette.
2. Weigh approximately \(3.00\text{ g}\) of anhydrous \(\text{CaCl}_2\) in a weighing bottle and record the initial mass.
3. Insert the thermometer and start a timer. Record the temperature of the water every minute for 3 minutes.
4. At exactly 4.0 minutes, add the anhydrous \(\text{CaCl}_2\) to the water. Do not record the temperature at this point. Stir the mixture rapidly and continuously to ensure rapid and complete dissolution.
5. Reweigh the empty weighing bottle to determine the exact mass of solid transferred by difference.
6. Record the temperature of the solution every minute from 5.0 minutes to 10.0 minutes.

(e) Temperature-time Graph analysis:
- Plot a graph of Temperature (y-axis) against Time (x-axis).
- Draw a straight line of best fit through the points from 0 to 3 minutes (pre-mixing trend) and extrapolate this line forward to the 4th minute.
- Draw a line of best fit through the cooling portion of the curve (from 5 to 10 minutes) and extrapolate this line backward to the 4th minute.
- The corrected temperature rise, \(\Delta T\), is the vertical distance between these two extrapolated lines at exactly the 4th minute.

(f) Calculations and Assumptions:
- Heat energy released: \(q = m_{\text{water}} \times c \times \Delta T\) where \(m_{\text{water}} = 50.0\text{ g}\) (assuming density is \(1.00\text{ g cm}^{-3}\)) and \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\).
- Moles of solute: \(n = \frac{m_{\text{solid}}}{111.0}\).
- Enthalpy of solution: \(\Delta H^\ominus_{\text{soln}} = -\frac{q}{1000 \times n}\text{ kJ mol}^{-1}\) (the negative sign represents an exothermic process).
- Key Assumptions: (1) The specific heat capacity of the resulting solution is equal to that of pure water (\(4.18\text{ J g}^{-1}\text{ K}^{-1}\)); (2) The heat capacity of the polystyrene cup is negligible; (3) The density of the solution is \(1.00\text{ g cm}^{-3}\).

(a) [1 Mark] Correct chemical equation with correct state symbols: \(\text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq)\) (or equivalent).

(b) [2 Marks]
- 1 Mark for measuring mass of anhydrous \(\text{CaCl}_2\) and volume/mass of water.
- 1 Mark for measuring initial temperature of water and subsequent temperature readings at regular time intervals.

(c) [3 Marks]
- 1 Mark for describing a polystyrene cup placed inside a glass beaker (to insulate and stabilise).
- 1 Mark for describing a lid with holes and a stirrer.
- 1 Mark for explaining how the materials minimise heat loss (polystyrene and air gap serve as insulators; lid minimises convective/evaporative heat loss).

(d) [4 Marks]
- 1 Mark for measuring a specific, suitable volume of water (e.g. \(50.0\text{ cm}^3\)) into the cup using a pipette or burette.
- 1 Mark for measuring initial temperatures of water at regular intervals (e.g. every minute) for at least 3 minutes before addition.
- 1 Mark for adding a known mass of solid at a defined time (e.g. 4th minute) without taking a temperature reading at that exact instant, followed by continuous, rapid stirring.
- 1 Mark for continuing temperature measurements at regular intervals (e.g. every minute) for several minutes after addition (e.g. up to 10 minutes) to establish the cooling trend.

(e) [2 Marks]
- 1 Mark for describing a plot of temperature against time.
- 1 Mark for explaining the extrapolation of both the pre-addition and post-addition lines to the time of mixing (e.g. 4 minutes) to find the difference, \(\Delta T\).

(f) [3 Marks]
- 1 Mark for calculating \(q = m c \Delta T\) and \(n = \text{mass}/111.0\).
- 1 Mark for stating \(\Delta H^\ominus_{\text{soln}} = -\frac{q}{1000 \times n}\) (with negative sign or stating 'exothermic' explicitly).
- 1 Mark for stating two valid assumptions (e.g., solution density is \(1.00\text{ g cm}^{-3}\); solution specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\); negligible heat capacity of the calorimeter).

(a) Calculating \(\log([\text{Cu}^{2+}])\) to 3 decimal places:
- Exp 1: \(\log(1.00) = 0.000\)
- Exp 2: \(\log(0.250) = -0.602\)
- Exp 3: \(\log(0.0500) = -1.301\)
- Exp 4: \(\log(0.0100) = -2.000\)
- Exp 5: \(\log(0.00200) = -2.699\)
- Exp 6: \(\log(0.000500) = -3.301\)
- Exp 7: \(\log(0.000100) = -4.000\)

(b) Graph plotting description:
- The y-axis represents \(E_{\text{cell}} / \text{V}\), ranging from \(+0.20\text{ V}\) to \(+0.36\text{ V}\) with sensible grid divisions (e.g., \(2\text{ cm} = 0.02\text{ V}\)).
- The x-axis represents \(\log([\text{Cu}^{2+}(aq)] / \text{mol dm}^{-3})\), ranging from \(-4.5\) to \(+0.5\) with sensible grid divisions (e.g., \(2\text{ cm} = 0.5\text{ units}\)).
- All seven data points must be plotted as small crosses or circled dots.
- Draw a single straight line of best fit. The line should pass directly through the points for Experiments 1, 2, 3, 4, 6, and 7, while completely ignoring the anomalous point of Experiment 5.

(c) Experiment 5 is anomalous because the measured potential of \(+0.295\text{ V}\) is significantly higher than the expected value of \(\approx +0.261\text{ V}\) predicted by the trend line. A higher potential indicates that the actual concentration of \(\text{Cu}^{2+}\) in the half-cell was higher than \(0.00200\text{ mol dm}^{-3}\). This could be caused by:
- Incomplete rinsing of the copper metal electrode (or the half-cell container) with distilled water after it was used in a more concentrated solution (e.g., \(0.0100\text{ mol dm}^{-3}\)), leading to contamination.
- An error during the serial dilution of the copper(II) sulfate stock solution, resulting in a more concentrated solution than intended.

(d) Calculation of the gradient:
\[ \text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} \]
Using the coordinates \(A(-4.000, 0.222)\) and \(B(0.000, 0.340)\):
\[ \text{Gradient} = \frac{0.340 - 0.222}{0.000 - (-4.000)} = \frac{0.118}{4.000} = +0.0295\text{ V} \]

(e) Calculating the number of electrons transferred, \(z\):
According to the Nernst equation:
\[ \text{Gradient} = \frac{0.059}{z} \]
\[ 0.0295 = \frac{0.059}{z} \implies z = \frac{0.059}{0.0295} = 2.0 \]
Since \(z = 2\), this perfectly supports the theoretical reduction half-reaction: \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\).

(f) Standard Electrode Potential, \(E^\ominus\):
According to the Nernst equation, when \(\log([\text{Cu}^{2+}(aq)]) = 0.000\) (i.e., at a standard concentration of \(1.00\text{ mol dm}^{-3}\)), the term \(\frac{0.059}{z} \log[\text{Cu}^{2+}(aq)]\) becomes zero. Thus, \(E_{\text{cell}} = E^\ominus\).
This corresponds to the y-intercept of the graph. From the table/graph, at \(\log([\text{Cu}^{2+}]) = 0.000\), \(E_{\text{cell}} = +0.340\text{ V}\). Therefore, \(E^\ominus_{\text{Cu}^{2+}/\text{Cu}} = +0.340\text{ V}\).

(a) [2 Marks]
- 1 Mark for calculating all seven log values correctly with negative signs where appropriate.
- 1 Mark for expressing all values to exactly 3 decimal places (0.000, -0.602, -1.301, -2.000, -2.699, -3.301, -4.000).

(b) [4 Marks]
- 1 Mark for correct axes labeling including units (y-axis: \(E_{\text{cell}} / \text{V}\); x-axis: \(\log([\text{Cu}^{2+}(aq)] / \text{mol dm}^{-3})\)).
- 1 Mark for choosing a scale where the plotted points cover more than half of the graph paper area in both dimensions.
- 1 Mark for plotting all 7 points accurately.
- 1 Mark for drawing a straight line of best fit that ignores the anomalous point (Experiment 5).

(c) [2 Marks]
- 1 Mark for identifying Experiment 5 as the anomalous point.
- 1 Mark for suggesting a valid experimental error (e.g. copper electrode contaminated with a more concentrated solution due to lack of rinsing, or dilution error resulting in a higher \(\text{Cu}^{2+}\) concentration than recorded).

(d) [3 Marks]
- 1 Mark for showing the correct calculation of \(\Delta y\) (\(0.340 - 0.222 = 0.118\)).
- 1 Mark for showing the correct calculation of \(\Delta x\) (\(0.000 - (-4.000) = 4.000\)).
- 1 Mark for calculating the gradient as \(+0.0295\) (accept range \(+0.0290\) to \(+0.0300\)).

(e) [2 Marks]
- 1 Mark for equating their gradient to \(0.059 / z\) and solving for \(z = 2\).
- 1 Mark for stating that this supports the theoretical equation because \(z = 2\) represents the gain of two electrons by the \(\text{Cu}^{2+}\) ion.

(f) [2 Marks]
- 1 Mark for stating that at the y-intercept, \(\log([\text{Cu}^{2+}]) = 0\), which corresponds to standard conditions, making \(E_{\text{cell}} = E^\ominus\).
- 1 Mark for identifying \(E^\ominus = +0.340\text{ V}\) (accept range \(+0.338\text{ V}\) to \(+0.342\text{ V}\)).