2023 Cambridge IGCSE Chemistry (0620) 模擬試題連答案詳解

(a) Excess zinc oxide is added to ensure that all of the dilute sulfuric acid has completely reacted, leaving no unreacted acid in the solution.

(b) The mixture is filtered. The unreacted zinc oxide is the residue on the filter paper, while the zinc sulfate solution passes through as the filtrate.

(c) To obtain crystals from the filtrate, the solution is heated to evaporate some water until it is saturated (the crystallization point, which can be tested by seeing if crystals form on a cold glass rod). The solution is then left to cool slowly, during which zinc sulfate crystals form. The crystals are separated from the remaining liquid by filtration, washed with a small amount of cold distilled water, and finally dried by gently pressing them between sheets of filter paper or placing them in a warm oven.

(a) To ensure all the (sulfuric) acid is reacted / used up / neutralized [1] (Reject: 'to speed up the reaction')

(b) Filtration / filtering [1]

(c) Any four from:
- Heat / evaporate the filtrate / solution [1]
- ...until saturated / to the point of crystallization / until crystals start to form on a cold glass rod / to reduce volume [1] (Reject: 'heat to dryness')
- Leave / allow to cool (to crystallize) [1]
- Filter (to separate the crystals from the remaining solution) [1]
- Dry crystals between filter paper / in a warm oven / in a desiccator [1] (Reject: 'heat strongly to dry')

(a) Excess zinc oxide is added to ensure that all of the dilute sulfuric acid has completely reacted, leaving no unreacted acid in the solution.

(b) The mixture is filtered. The unreacted zinc oxide is the residue on the filter paper, while the zinc sulfate solution passes through as the filtrate.

(c) To obtain crystals from the filtrate, the solution is heated to evaporate some water until it is saturated (the crystallization point, which can be tested by seeing if crystals form on a cold glass rod). The solution is then left to cool slowly, during which zinc sulfate crystals form. The crystals are separated from the remaining liquid by filtration, washed with a small amount of cold distilled water, and finally dried by gently pressing them between sheets of filter paper or placing them in a warm oven.

(a) To ensure all the (sulfuric) acid is reacted / used up / neutralized [1] (Reject: 'to speed up the reaction')

(b) Filtration / filtering [1]

(c) Any four from:
- Heat / evaporate the filtrate / solution [1]
- ...until saturated / to the point of crystallization / until crystals start to form on a cold glass rod / to reduce volume [1] (Reject: 'heat to dryness')
- Leave / allow to cool (to crystallize) [1]
- Filter (to separate the crystals from the remaining solution) [1]
- Dry crystals between filter paper / in a warm oven / in a desiccator [1] (Reject: 'heat strongly to dry')

(a) Aqueous nickel(II) sulfate contains ions from the dissolved salt itself (\(\text{Ni}^{2+}\) and \(\text{SO}_4^{2-}\)) and from the water solvent (\(\text{H}^+\) and \(\text{OH}^-\)).

(b)(i) At the cathode, the metal ion \(\text{Ni}^{2+}\) is reduced because nickel is less reactive than hydrogen under these electrolysis conditions. Nickel metal is deposited as a silver-grey or pinkish solid: \(\text{Ni}^{2+} + 2\text{e}^- \rightarrow \text{Ni}\).

(b)(ii) At the anode, hydroxide ions (\(\text{OH}^-\)) from water are preferentially discharged over sulfate ions (\(\text{SO}_4^{2-}\)) because hydroxide ions are easier to oxidize. This forms oxygen gas and water: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\).

(b)(iii) As \(\text{OH}^-\)-ions are discharged to form oxygen gas, the concentration of \(\text{H}^+\) ions increases relative to hydroxide ions. This makes the solution more acidic, meaning the pH decreases.

(c) When active nickel electrodes are used, the nickel anode undergoes oxidation in preference to hydroxide discharge: \(\text{Ni} \rightarrow \text{Ni}^{2+} + 2\text{e}^-\), causing its mass to decrease. At the cathode, \(\text{Ni}^{2+}\) ions from the electrolyte are reduced to nickel metal: \(\text{Ni}^{2+} + 2\text{e}^- \rightarrow \text{Ni}\), depositing on the cathode and increasing its mass.

(a) [2 marks total]
- Award 1 mark for naming/formula of nickel and sulfate ions: \(\text{Ni}^{2+}\) (nickel) and \(\text{SO}_4^{2-}\) (sulfate).
- Award 1 mark for naming/formula of hydrogen and hydroxide ions: \(\text{H}^+\) (hydrogen) and \(\text{OH}^-\) (hydroxide).

(b)(i) [2 marks total]
- Award 1 mark for observation: grey / silver / pink / shiny solid / deposit (Reject: bubbles / effervescence at cathode).
- Award 1 mark for correct ionic half-equation: \(\text{Ni}^{2+} + 2\text{e}^- \rightarrow \text{Ni}\) (Accept state symbols, but not required).

(b)(ii) [2 marks total]
- Award 1 mark for gas name: oxygen (Accept \(\text{O}_2\)).
- Award 1 mark for correct ionic half-equation: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\) (or \(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-\)).

(b)(iii) [2 marks total]
- Award 1 mark for stating that the pH decreases / becomes more acidic.
- Award 1 mark for explanation: hydroxide (\(\text{OH}^-\)) ions are discharged / removed, leaving an excess of hydrogen (\(\text{H}^+\)) ions in solution (or sulfuric acid is formed).

(c) [2 marks total]
- Award 1 mark for stating both electrode mass changes: anode mass decreases AND cathode mass increases.
- Award 1 mark for explanation: nickel atoms from the anode dissolve/oxidize to form nickel ions (\(\text{Ni} \rightarrow \text{Ni}^{2+} + 2\text{e}^-\)) AND these nickel ions are reduced/deposited as metal back onto the cathode.

(a) In a system at dynamic equilibrium: 1. The rate of the forward reaction is equal to the rate of the reverse reaction. 2. The concentrations of the reactants and products remain constant in a closed system. (b)(i) Increasing pressure: The position of equilibrium shifts to the right (forward direction). This is because there are fewer moles of gas on the product side (1 mole of gas) compared to the reactant side (3 moles of gas). Shifting to the side with fewer gas molecules opposes the increase in pressure. (b)(ii) Increasing temperature: The position of equilibrium shifts to the left (reverse direction). This is because the forward reaction is exothermic (\(\Delta H\) is negative). Shifting to the endothermic side (left) absorbs the added thermal energy and opposes the temperature increase. (c)(i) The catalyst has no effect on the position of equilibrium. (c)(ii) The catalyst increases the rate of the reverse reaction (it increases the rate of both forward and reverse reactions equally). (d) 1. Calculate the relative formula mass (\(M_r\)) of \(\text{CO}\): \(12 + 16 = 28\). 2. Find the moles of \(\text{CO}\): \(\text{moles} = \frac{14.0}{28} = 0.50\text{ mol}\). 3. Use the stoichiometric ratio from the balanced equation (\(1\text{ CO} : 2\text{ H}_2\)): \(\text{moles of } \text{H}_2 = 0.50 \times 2 = 1.00\text{ mol}\). 4. Calculate the volume of \(\text{H}_2\) at r.t.p.: \(\text{volume} = 1.00\text{ mol} \times 24\text{ dm}^3/\text{mol} = 24.0\text{ dm}^3\).

(a) [2 marks total] - Rate of forward reaction equals rate of reverse reaction [1] - Concentration of reactants and products remains constant (accept: closed system required) [1] (b)(i) [2 marks total] - Position of equilibrium shifts to the right / forward direction [1] - Fewer gas moles/molecules on the right-hand side (1 mole vs 3 moles on the left) [1] (b)(ii) [2 marks total] - Position of equilibrium shifts to the left / backward direction [1] - Forward reaction is exothermic / reverse reaction is endothermic [1] (c) [2 marks total] - (i) No effect on position of equilibrium [1] - (ii) Increases the rate [1] (d) [2 marks total] - Moles of CO calculated correctly: \(14.0 / 28 = 0.50\text{ mol}\) [1] - Volume of hydrogen gas calculated correctly: \(0.50 \times 2 \times 24 = 24.0\text{ dm}^3\) (accept 24) [1]

### Detailed Solution

**Part (a)(i)**
* **Anode:** In molten \( \text{ZnCl}_2 \), the only anion present is chloride, \( \text{Cl}^- \). It is oxidized at the anode to form **chlorine gas**.
* **Cathode:** The only cation present is zinc, \( \text{Zn}^{2+} \). It is reduced at the cathode to form **zinc metal**.

**Part (a)(ii)**
* **Anode half-equation:** \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \)
* **Cathode half-equation:** \( \text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn} \)

**Part (b)(i)**
* **Anode observation:** Bubbles of a colourless gas (oxygen gas is produced from the discharge of hydroxide ions).
* **Cathode observation:** A pink-brown / reddish-brown solid is deposited on the electrode surface (copper metal).

**Part (b)(ii)**
* **Anode half-equation:** \( 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \)
* **Cathode half-equation:** \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \)

**Part (b)(iii)**
* The pH **decreases** (becomes acidic).
* Explanation: Both \( \text{Cu}^{2+} \) and \( \text{OH}^- \) ions are discharged from the solution. This leaves \( \text{H}^+ \) and \( \text{SO}_4^{2-} \) ions behind in the electrolyte, which effectively forms sulfuric acid, increasing the concentration of hydrogen ions.

**Part (c)(i)**
* **Anode:** Mass **decreases** (copper atoms lose electrons and dissolve into solution as copper ions).
* **Cathode:** Mass **increases** (copper ions in solution gain electrons and deposit as copper metal).

**Part (c)(ii)**
* With copper electrodes, copper atoms from the anode dissolve to form \( \text{Cu}^{2+} \) ions at the same rate that \( \text{Cu}^{2+} \) ions are discharged and deposited at the cathode. Therefore, the overall concentration of \( \text{Cu}^{2+} \) ions (which cause the blue colour) remains constant.
* In contrast, with inert electrodes, \( \text{Cu}^{2+} \) ions are discharged but are not replenished by the anode, so their concentration decreases, causing the blue colour to fade.

**Part (c)(iii)**
* Refining/purification of copper OR electroplating.

**(a)(i)** [Total: 2 marks]
* Anode: chlorine / \( \text{Cl}_2 \) [1]
* Cathode: zinc / \( \text{Zn} \) [1]

**(a)(ii)** [Total: 2 marks]
* Anode: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \) [1] (accept: \( 2\text{Cl}^- - 2\text{e}^- \rightarrow \text{Cl}_2 \))
* Cathode: \( \text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn} \) [1]

**(b)(i)** [Total: 2 marks]
* Anode observation: bubbles / effervescence / gas evolved [1] (reject: 'chlorine/oxygen gas formed' without physical observation; accept 'colourless gas')
* Cathode observation: pink / brown / pink-brown / red-brown solid [1] (reject: 'copper forms' on its own; accept 'pink-brown deposit')

**(b)(ii)** [Total: 2 marks]
* Anode: \( 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \) [1]
* Cathode: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \) [1]

**(b)(iii)** [Total: 3 marks]
* pH decreases / becomes acidic [1]
* \( \text{Cu}^{2+} \) and \( \text{OH}^- \) ions are discharged / removed [1]
* leaving \( \text{H}^+ \) (and \( \text{SO}_4^{2-} \)) ions in solution [1]

**(c)(i)** [Total: 2 marks]
* Anode: mass decreases / dissolves [1]
* Cathode: mass increases [1]

**(c)(ii)** [Total: 2 marks]
* Rate of copper dissolution at the anode is equal to rate of copper deposition at the cathode / concentration of copper ions remains constant [1]
* With inert electrodes, copper ions are removed from solution and not replaced [1]

**(c)(iii)** [Total: 1 mark]
* Refining of copper / purification of copper OR electroplating [1]

(a) A reversible reaction is a reaction that can occur in both directions, where reactants react to form products, and the products can react to reform the reactants. (b) At dynamic equilibrium: (1) The rate of the forward reaction equals the rate of the reverse reaction. (2) The concentrations of all reactants and products remain constant (in a closed system). (c)(i) The forward reaction is exothermic (\(\Delta H = -41\text{ kJ/mol}\)). Increasing the temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the heat. (c)(ii) There are 2 moles of gas on the reactant side (1 mol CO + 1 mol \(\text{H}_2\text{O}\)) and 2 moles of gas on the product side (1 mol \(\text{CO}_2\) + 1 mol \(\text{H}_2\)). Since the number of moles of gas is equal on both sides, a change in pressure has no effect on the position of equilibrium. (c)(iii) A catalyst increases the rates of both the forward and reverse reactions by the same factor. Therefore, it does not change the position of equilibrium, it only helps the system reach equilibrium faster. (d)(i) Initially, \(2.0\text{ mol}\) of CO is present. Since \(1.4\text{ mol}\) of \(\text{CO}_2\) is formed, \(1.4\text{ mol}\) of CO must have reacted. Moles of CO at equilibrium = \(2.0 - 1.4 = 0.6\text{ mol}\). The mole ratio of \(\text{CO}_2\) to \(\text{H}_2\) is 1:1, so \(1.4\text{ mol}\) of \(\text{H}_2\) is also formed. Moles of \(\text{H}_2\) at equilibrium = \(1.4\text{ mol}\). (d)(ii) Theoretical yield of \(\text{CO}_2\) is \(2.0\text{ mol}\) (since 1 mol of CO produces 1 mol of \(\text{CO}_2\)). Percentage yield = \((\text{actual yield} / \text{theoretical yield}) \times 100 = (1.4 / 2.0) \times 100 = 70.0\%\). (d)(iii) The percentage yield will decrease. Since the forward reaction is exothermic, increasing the temperature shifts the equilibrium to the left, which decreases the amount of products (and thus decreases the yield of \(\text{CO}_2\)).

(a) a reaction that can occur in both the forward and reverse directions [1] (b) rate of forward reaction equals rate of reverse reaction [1]; concentrations / amounts of reactants and products remain constant [1] (c)(i) shifts to the left / reactants side [1]; forward reaction is exothermic / reverse reaction is endothermic [1] (c)(ii) no effect [1]; equal moles / volumes of gas on both sides of the equation [1] (c)(iii) no effect [1]; catalyst increases the rate of both forward and reverse reactions by the same amount [1] (d)(i) moles of CO = 0.6 mol [1]; moles of \(\text{H}_2\) = 1.4 mol [1] (d)(ii) \((1.4 / 2.0) \times 100\) [1]; 70% / 70.0% [1] (d)(iii) (percentage yield) decreases [1]; forward reaction is exothermic / equilibrium shifts to left [1]

For (a)(i), the discharge of copper(II) ions at the cathode forms copper metal, which is visible as a pink/red-brown coating. Hydroxide ions are discharged at the anode to release oxygen gas, visible as bubbling. For (a)(ii), the oxidation of hydroxide ions is represented by \(4\text{OH}^{-}(\text{aq}) \rightarrow \text{O}_{2}(\text{g}) + 2\text{H}_{2}\text{O}(\text{l}) + 4\text{e}^{-}\). For (a)(iii), since \(\text{Cu}^{2+}\) and \(\text{OH}^{-}\) ions are preferentially discharged, the remaining ions in solution are \(\text{H}^{+}\) and \(\text{SO}_{4}^{2-}\), which form sulfuric acid. For (b)(i), copper atoms from the anode dissolve as copper ions (oxidation), while copper ions from the solution deposit onto the cathode as copper metal (reduction). Thus, the anode loses mass and the cathode gains mass. For (b)(ii), the rate at which copper ions are produced at the anode is equal to the rate at which they are discharged at the cathode, keeping the concentration of copper(II) ions constant, so the blue color does not fade. For (c)(i), moles of Cu = \(0.48\text{ g} / 64\text{ g/mol} = 0.0075\text{ mol}\). For (c)(ii), number of atoms = \(0.0075\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 4.515 \times 10^{21}\text{ atoms}\).

Part (a)(i): 1 mark for pink/red-brown solid (on cathode) or blue solution fading. 1 mark for bubbles/effervescence (at anode). (Max 2 marks). Part (a)(ii): 1 mark for correct reactants and products: \(4\text{OH}^{-} \rightarrow \text{O}_{2} + 2\text{H}_{2}\text{O} + 4\text{e}^{-}\). 1 mark for correct state symbols: \(\text{aq}\), \(\text{g}\), \(\text{l}\). Part (a)(iii): 1 mark for naming sulfuric acid. 1 mark for explaining that \(\text{H}^{+}\) and \(\text{SO}_{4}^{2-}\) ions are left in solution (as \(\text{Cu}^{2+}\) and \(\text{OH}^{-}\) are discharged). Part (b)(i): 1 mark for stating that the anode mass decreases. 1 mark for stating that the cathode mass increases. Part (b)(ii): 1 mark for stating that the color remains unchanged/blue. 1 mark for explaining that the rate of copper dissolving at the anode is equal to the rate of copper discharging at the cathode (concentration of \(\text{Cu}^{2+}\) remains constant). Part (c)(i): 1 mark for \(0.0075\text{ mol}\). Part (c)(ii): 1 mark for \(4.515 \times 10^{21}\) or \(4.52 \times 10^{21}\) (allow ecf from c(i)).