📚 Year 11 AQA Chemistry: Case Study Practice | 十一年级AQA化学案例分析实战演练
Case studies are an essential part of the AQA GCSE Chemistry course, helping you apply theoretical knowledge to real-world situations. In this article, we'll work through a series of case study drills that mirror exam-style scenarios. Each case will challenge you to think like a chemist, linking concepts from atomic structure, bonding, quantitative chemistry, and more. By practicing these, you'll sharpen your problem-solving skills and boost your confidence for the Year 11 assessments.
案例分析是AQA GCSE化学课程的重要组成部分，帮助你将理论知识应用到实际情境中。本文将通过一系列模拟考试风格的案例实战演练，带你逐一攻克难题。每个案例都会挑战你的化学思维，联系原子结构、化学键、定量化学等知识。通过练习，你将提升解题能力，增强应对十一年级评估的信心。
1. Electrolysis in Aluminium Extraction | 电解在铝提取中的应用
g point to about 950 °C, which saves energy. The electrolytic cell uses carbon electrodes. At the cathode, aluminium ions gain electrons: Al³⁺ + 3e⁻ → Al. At the anode, oxide ions are discharged: 2O²⁻ → O₂ + 4e⁻. The oxygen produced reacts with the carbon anode, forming carbon dioxide, so the anode gradually burns away.
铝是通过电解熔融氧化铝（Al₂O₃）制备的。将氧化铝溶解在冰晶石（Na₃AlF₆）中可将其熔点降至约950°C，从而节省能源。电解槽使用碳电极。在阴极，铝离子得到电子：Al³⁺ + 3e⁻ → Al。在阳极，氧离子放电：2O²⁻ → O₂ + 4e⁻。生成的氧气与碳阳极反应产生二氧化碳，因此阳极会逐渐消耗。
A local industry noticed that the anode blocks needed replacement much more frequently than expected. Suggest a reason for this observation and write the relevant chemical equation.
某地方工厂发现阳极块的更换频率远高于预期。请提出可能的原因并写出相关的化学方程式。
minium oxide contains impurities that also produce oxygen, or if the current is too high, more oxygen is generated, accelerating anode consumption. Alternatively, if the anode material is less pure, it may burn faster. The equation shows that each mole of O₂ consumes one mole of carbon.
阳极与产生的氧气反应：C(s) + O₂(g) → CO₂(g)。如果氧化铝中含有也能产生氧气的杂质，或者电流过大，则会产生更多氧气，加速阳极消耗。另外，如果阳极材料纯度较低，也可能燃烧得更快。该方程式表明，每摩尔O₂消耗一摩尔碳。
of 10 A flows for 2 hours. (Faraday constant = 96 500 C/mol, molar mass of Al = 27 g/mol)
计算当10 A电流通过2小时时产生的铝的质量。（法拉第常数=96 500 C/mol，Al的摩尔质量=27 g/mol）
rons produce 1 mole of Al. Moles of Al = 0.746 / 3 ≈ 0.249 mol. Mass = 0.249 × 27 ≈ 6.7 g.
电荷量 Q = I × t = 10 A × (2 × 3600 s) = 72 000 C。电子的物质的量 = 72 000 / 96 500 ≈ 0.746 mol。根据阴极半反应，3摩尔电子生成1摩尔Al。Al的物质的量 = 0.746 / 3 ≈ 0.249 mol。质量 = 0.249 × 27 ≈ 6.7 g。
2. Acid-Base Titration and Analysis | 酸碱滴定分析
A student titrated 25.0 cm³ of hydrochloric acid (HCl) against sodium hydroxide solution (NaOH) of concentration 0.100 mol/dm³, using phenolphthalein indicator. The results of the rough and accurate titrations are recorded below.
一位学生用0.100 mol/dm³的氢氧化钠溶液（NaOH）滴定25.0 cm³的盐酸（HCl），以酚酞作指示剂。初步滴定和精确滴定的结果记录如下。
Titre volumes: 23.50 cm³, 23.10 cm³, 23.20 cm³ (concordant). Calculate the mean titre and then the concentration of the acid in mol/dm³.
滴定管读数：23.50 cm³、23.10 cm³、23.20 cm³（数据吻合）。计算平均滴定体积并求酸的浓度（mol/dm³）。
002315 mol. Moles of HCl in 25.0 cm³ = 0.002315 mol. Concentration of HCl = 0.002315 / (25.0 / 1000) = 0.0926 mol/dm³.
平均体积 = (23.10 + 23.20) ÷ 2 = 23.15 cm³。中和反应方程式为：HCl + NaOH → NaCl + H₂O。所用NaOH的物质的量 = (23.15 / 1000) × 0.100 = 0.002315 mol。25.0 cm³ HCl的物质的量 = 0.002315 mol。盐酸浓度 = 0.002315 / (25.0 / 1000) = 0.0926 mol/dm³。
The student then repeated the titration but forgot to add the indicator before starting. They added it after adding 20 cm³ of NaOH. The solution turned pink immediately. Explain how this procedural error could affect the calculated acid concentration.
该学生随后重复滴定实验，但在开始前忘记加入指示剂。在加入20 cm³ NaOH后才加入指示剂，溶液立即变为粉红色。解释此操作错误将如何影响计算出的酸的浓度。
At 20 cm³ of NaOH, the solution was already neutral or slightly alkaline because the endpoint occurs around 23.15 cm³. Adding indicator at this stage and seeing pink means the endpoint had already been passed. The student would wrongly think the endpoint was at 20 cm³, leading to a smaller titre volume. A smaller titre gives a lower calculated number of moles of NaOH, so the calculated acid concentration would be lower than the true value.
在加入20 cm³ NaOH时，溶液已经呈中性或微碱性，因为终点约在23.15 cm³。此时加入指示剂并看到粉红色意味着已经超过了终点。学生会错误地认为终点是20 cm³，导致滴定体积偏小。较小的滴定体积得出较少的NaOH物质的量，因此计算出的酸浓度将低于真实值。
3. Reaction Rates and Collision Theory | 反应速率与碰撞理论
Marble chips (calcium carbonate) were added to excess hydrochloric acid, and the volume of carbon dioxide gas evolved was measured every 30 seconds. The same experiment was repeated at 40 °C. The graph shows a steeper initial gradient for the 40 °C experiment. Using collision theory, explain the shape of both curves.
将大理石碎片（碳酸钙）加入过量盐酸中，每隔30秒测量二氧化碳气体的体积。同一实验在40°C下重复进行。图表显示40°C实验的初始斜率更陡。用碰撞理论解释两条曲线的形状。
(g) + H₂O(l). At higher temperature, the particles have greater average kinetic energy. A greater proportion of collisions have energy equal to or exceeding the activation energy, and the collision frequency is higher. Therefore, the rate of reaction increases, giving a steeper initial gradient. Both curves level off when the limiting reactant (CaCO₃) is used up; the final volume of CO₂ is the same because the same amount of marble was used.
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)。温度较高时，粒子的平均动能更大。有更大比例的碰撞具有等于或超过活化能的能量，且碰撞频率更高。因此反应速率增加，初始斜率更陡。两种情况下曲线都趋于水平，因为限制反应物（CaCO₃）被消耗完；由于大理石用量相同，最终CO₂体积相同。
The student then broke the marble chips into powder and ran the experiment at the original temperature. Predict and explain the effect on the rate, using collision theory.
然后学生将大理石碎片磨成粉末，并在原温度下进行实验。预测并解释对速率的影响，使用碰撞理论。
Powdering the solid increases its surface area. More solid particles are exposed to the acid, so the frequency of collisions between reactant particles increases. This raises the rate of reaction, so the initial gradient would be steeper than with chips, but the final volume of CO₂ would be unchanged.
将固体磨成粉末增加了其表面积。更多的固体颗粒与酸接触，因此反应物粒子之间的碰撞频率增加。这提高了反应速率，所以初始斜率会比使用碎片时更陡，但最终CO₂体积不变。
4. Organic Chemistry: Alcohols and Their Reactions | 有机化学：醇及其反应
Ethanol (C₂H₅OH) can be produced by fermentation of glucose using yeast at 30–40 °C under anaerobic conditions. It can also be made by direct hydration of ethene with steam in the presence of a phosphoric acid catalyst. A company is considering which method to use for a new sustainable fuels plant. Compare the two processes.
乙醇（C₂H₅OH）可以通过使用酵母在30–40°C厌氧条件下发酵葡萄糖制得，也可以在磷酸催化剂存在下，通过乙烯与水蒸气的直接水合反应制得。一家公司正在考虑为新的可持续燃料工厂选择哪种工艺。比较这两种工艺。
emperature and normal pressure, but produces a relatively dilute ethanol solution (about 15% de oil (non-renewable), requires high temperature and pressure, and a catalyst, but the reaction is fast and gives high yield of pure ethanol. For sustainability, fermentation is preferable because it uses renewable biomass and is carbon-neutral over its life cycle.
发酵：C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂。使用可再生原料（植物中的葡萄糖），在低温和常压下操作，但产生相对稀的乙醇溶液（约15%），需要分馏。反应速率慢。水合：C₂H₄ + H₂O → C₂H₅OH。使用来自原油的乙烯（不可再生），需要高温高压和催化剂，但反应快且纯乙醇产率高。从可持续性角度看，发酵更可取，因为它使用可再生生物质，并且在其生命周期内是碳中性的。
When ethanol is left open to air, it can turn sour due to oxidation. Write the equation for the formation of ethanoic acid and suggest how the alcohol functional group can be confirmed by a chemical test.
当乙醇敞开放置时，会因氧化而变酸。写出生成乙酸的方程式，并说明如何通过化学试验确认醇的官能团。
(using oxidising agent like acidified potassium dichromate(VI)). The alcohol functional group (– idified potassium dichromate solution; the colour changes from orange to green as the alcohol is oxidised to a carboxylic acid.
氧化反应：C₂H₅OH + O₂ → CH₃COOH + H₂O（使用酸性重铬酸钾(VI)等氧化剂）。醇官能团（–OH）可通过与酸性重铬酸钾溶液加热来检测；颜色由橙色变为绿色，表明醇被氧化为羧酸。
5. Equilibrium and the Haber Process | 化学平衡与哈柏法
The Haber process is used to manufacture ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = – these conditions are chosen, referring to reaction rate, equilibrium yield, and economics.
哈柏法用于制造氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = –92 kJ/mol。典型条件是450°C、200 atm和铁催化剂。解释为何选择这些条件，涉及反应速率、平衡产率和经济因素。
t low temperatures the rate is too slow to be economical. 450 °C is a compromise that gives a reasonable rate while still achieving an acceptable equilibrium yield. High pressure favours the forward reaction because there are fewer gas molecules on the product side (4 mol → 2 mol). 200 atm is high enough to shift equilibrium right and increase rate, but very high pressures are costly and require stronger equipment. The iron catalyst speeds up the reaction without affecting the equilibrium position, allowing the equilibrium to be reached faster at the chosen temperature.
正向反应是放热的，因此降低温度会使平衡向右移动，增加氨的产率。然而在低温下反应速率太慢，不经济。450°C是一个折中选择，既能提供合理的速率，又能获得可接受的平衡产率。高压有利于正向反应，因为产物气体分子数更少（4 mol → 2 mol）。200 atm足以使平衡右移并提高速率，但极高压力成本高且需要更坚固的设备。铁催化剂加速反应但不影响平衡位置，使得在选定温度下能更快达到平衡。
If a factory switches from a conventional catalyst to a newly developed, more efficient catalyst, state and explain the effect on the equilibrium yield of ammonia at the same temperature and pressure.
如果一家工厂由常规催化剂改用新开发的更高效催化剂，在相同温度和压力下，陈述并解释对氨的平衡产率有何影响。
A catalyst does not alter the position of equilibrium; it only lowers the activation energy for both the forward and backward reactions equally. Therefore, the equilibrium yield of ammonia remains exactly the same. The advantage is that the equilibrium is reached in a shorter time, increasing production rate.
催化剂不会改变平衡位置；它仅同等降低正反应和逆反应的活化能。因此氨的平衡产率完全相同。优点是能在更短时间内达到平衡，从而提高生产速率。
6. Mole Calculations and Percentage Yield | 摩尔计算与产率
A student heated 10.0 g of calcium carbonate and collected 2.20 dm³ of carbon dioxide at room temperature and pressure (RTP, where 1 mol of gas occupies 24 dm³). Calculate the percentage yield of the reaction.
石灰石的热分解用于生产氧化钙（生石灰）：CaCO₃(s) → CaO(s) + CO₂(g)。某学生加热10.0 g碳酸钙，在室温和常压（RTP，1摩尔气体占24 dm³）下收集到2.20 dm³的二氧化碳。计算反应的百分产率。
O₂ = 0.0999 mol. Theoretical volume of CO₂ at RTP = 0.0999 × 24 = 2.40 dm³. Actual volume collected = 2.20 dm³. Percentage yield = (actual / theoretical) × 100 = (2.20 / 2.40) × 100 ≈ 91.7%.
CaCO₃的摩尔质量 = 40.1 + 12.0 + (3 × 16.0) = 100.1 g/mol。所用CaCO₃的物质的量 = 10.0 / 100.1 ≈ 0.0999 mol。1:1摩尔比对应的理论CO₂物质的量为0.0999 mol。RTP下CO₂的理论体积 = 0.0999 × 24 = 2.40 dm³。实际收集到的体积 = 2.20 dm³。百分产率 = (实际 / 理论) × 100 = (2.20 / 2.40) × 100 ≈ 91.7%。
Suggest two reasons why the yield is less than 100%. Explain each reason.
提出产率低于100%的两个原因，并分别解释。
1. Incomplete decomposition: Some calcium carbonate may not have decomposed fully if the heating time was insufficient or temperature too low, so not all the CO₂ was released. 2. Gas leakage or collection inefficiency: Some CO₂ may have escaped before being collected, or some dissolved in the water if collected over water, leading to a lower measured volume. Impurities in the limestone could also mean the actual mass of pure CaCO₃ was less than 10.0 g.
1. 不完全分解：如果加热时间不足或温度过低，部分碳酸钙可能未能完全分解，因此并未完全释放出CO₂。2. 气体泄漏或收集效率不佳：部分CO₂可能在收集前逸出，或者如果采用排水法收集，有些气体溶于水，导致测量体积偏低。石灰石中的杂质也可能意味着纯CaCO₃的实际质量不足10.0 g。
7. Chemical Analysis: Testing for Ions | 化学分析：离子鉴定
An unknown solution contains one or more of the following ions: Cu²⁺, Fe²⁺, Fe³⁺, Cl⁻, SO₄²⁻, CO₃²⁻. Describe a series of tests that could be carried out to identify the ions present, giving expected observations and ionic equations where appropriate.
某未知溶液含有以下一种或多种离子：Cu²⁺、Fe²⁺、Fe³⁺、Cl⁻、SO₄²⁻、CO₃²⁻。描述一系列可以用来鉴别这些离子的试验，给出预期的观察结果和适当的离子方程式。
f effervescence occurs, CO₃²⁻ is present: CO₃²⁻ + 2H⁺ → CO₂ + H₂O. Then add barium chloride solution to a fresh sample acidified with dilute HCl: a white precipitate indicates SO₄²⁻: Ba²⁺ + SO₄²⁻ → BaSO₄(s). For halides, add nitric acid then silver nitrate solution: a white precipitate soluble in dilute ammonia indicates Cl⁻: Ag⁺ + Cl⁻ → AgCl(s). To distinguish between Fe²⁺ and Fe³⁺, add sodium hydroxide solution: a green precipitate that turns brown on standing indicates Fe²⁺; a brown precipitate indicates Fe³⁺. Cu²⁺ gives a blue precipitate with NaOH.
首先，观察溶液颜色。蓝色表示Cu²⁺；浅绿色表示Fe²⁺；黄/棕色表示Fe³⁺。加入稀盐酸：若产生气泡，则含CO₃²⁻：CO₃²⁻ + 2H⁺ → CO₂ + H₂O。然后向用稀盐酸酸化的新样品中加入氯化钡溶液：白色沉淀表示SO₄²⁻：Ba²⁺ + SO₄²⁻ → BaSO₄(s)。对于卤离子，先加硝酸再加硝酸银溶液：可溶于稀氨水的白色沉淀表示Cl⁻：Ag⁺ + Cl⁻ → AgCl(s)。为区分Fe²⁺和Fe³⁺，加入氢氧化钠溶液：在空气中变为棕色的绿色沉淀表示Fe²⁺；棕色沉淀表示Fe³⁺。Cu²⁺与NaOH生成蓝色沉淀。
A student performed a flame test on the same solution and observed a blue-green colour. Which cation is indicated? How does the flame test confirm the identity of the ion?
一名学生对同一样品进行了焰色试验，观察到蓝绿色火焰。指出是哪种阳离子，并说明焰色试验如何确认离子身份。
A blue-green flame indicates Cu²⁺ ions. In the flame test, electrons in the metal ion absorb energy and are excited to higher energy levels. When they fall back, they emit light of characteristic wavelengths. Copper(II) emits primarily in the blue-green region. This confirms the presence of copper.
蓝绿色火焰表明存在Cu²⁺离子。焰色试验中，金属离子的电子吸收能量并跃迁到高能级。当它们回迁时，发射出特征波长的光。铜(II)主要在蓝绿区域发射。这证实铜的存在。
8. Energy Changes and Chemical Cells | 能量变化与化学电池
A simple chemical cell is constructed using a strip of zinc and a strip of copper dipped into a solution of their own ions (ZnSO₄ and CuSO₄) connected by a salt bridge. The cell produces a voltage of about 1.1 V. Identify the anode and cathode, write the half-equations, and explain why the cell generates electricity.
构建一个简单化学电池，使用锌条和铜条分别浸入其自身离子的溶液（ZnSO₄和CuSO₄）中，并用盐桥连接。该电池产生约1.1 V的电压。指出阳极和阴极，写出半反应方程式，并解释电池为何能产生电流。
aq) + 2e⁻ → Cu(s) (cathode, positive electrode). The overall reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Electrons flow from the zinc electrode to the copper electrode through the external circuit, generating a current. The salt bridge completes the circuit by allowing ions to move, maintaining electrical neutrality.
锌比铜活泼，因此锌发生氧化反应：Zn(s) → Zn²⁺(aq) + 2e⁻（阳极，负极）。铜离子被还原：Cu²⁺(aq) + 2e⁻ → Cu(s)（阴极，正极）。总反应为Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。电子通过外电路从锌电极流向铜电极，产生电流。盐桥允许离子移动以保持电中性，从而构成完整回路。
If the copper half-cell is replaced by a silver half-cell (Ag⁺/Ag), predict how the cell voltage would change. Justify your answer using the reactivity series.
若将铜半电池替换为银半电池（Ag⁺/Ag），预测电池电压将如何变化，并用金属活动性顺序说明理由。
er the difference in reactivity of the two metals, the greater the cell potential. Therefore, the voltage would be higher than 1.1 V. The half-equations would be: Zn(s) → Zn²⁺ + 2e⁻; 2Ag⁺ + 2e⁻ → 2Ag(s).
银不如铜活泼，因此锌与银的活泼性差异大于锌与铜的差异。两种金属的活泼性差异越大，电池电势就越高。因此电压将高于1.1 V。半反应式为：Zn(s) → Zn²⁺ + 2e⁻；2Ag⁺ + 2e⁻ → 2Ag(s)。
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