📚 Case Study Practice for Edexcel Year 10 Chemistry | Edexcel 十年级化学案例分析实战演练
This article presents a series of real-world case studies that link together key topics in the Year 10 Edexcel Chemistry curriculum. Each case study is designed as a mini-investigation, with background information and questions that mirror the style of exam papers. By working through these examples, you will practise applying concepts such as extraction of metals, making salts, electrolysis, equilibrium, corrosion, energy and environmental chemistry. Detailed solutions are provided in a step-by-step bilingual format, helping you build both scientific understanding and the confidence to tackle longer structured questions.
本文围绕 Edexcel 十年级化学教学大纲，精心设计了一系列贴近真实情境的案例分析。每个案例都以小探究的形式呈现，提供背景信息与考题风格的问题。通过逐一演练，你将练习如何灵活运用金属提取、盐的制备、电解、平衡、腐蚀、能量变化与环境化学等核心概念。解答部分采用中英双语的逐步解析，旨在帮助你巩固科学原理，同时提升应对长结构化题的信心。
1. Case Study: Extracting Copper from Malachite | 案例一：从孔雀石提取铜
A student heats powdered malachite, a green mineral with the formula CuCO₃·Cu(OH)₂, strongly in air. The solid turns black. She then mixes this black solid with carbon powder and heats again, obtaining a reddish-brown metal. Finally, she purifies the metal by electrolysis using copper(II) sulfate solution as the electrolyte. Explain each step, write relevant chemical equations and discuss the environmental impact of this extraction process.
一名学生将绿色的孔雀石粉末（化学式 CuCO₃·Cu(OH)₂）在空气中强热，固体变为黑色。随后她把黑色固体与碳粉混合并再次加热，得到一种红棕色金属。最后她以硫酸铜溶液为电解质，通过电解对该金属进行精炼。请解释每一步骤，写出相关化学方程式，并讨论该提取过程对环境的影响。
Solution
Step 1 – Thermal decomposition: Malachite decomposes on heating to form copper(II) oxide, carbon dioxide and water. The colour change from green to black indicates the formation of CuO.
步骤 1 – 热分解：孔雀石受热分解生成氧化铜、二氧化碳和水。颜色由绿变黑表明生成了 CuO。
CuCO₃·Cu(OH)₂ → 2CuO + CO₂ + H₂O
Step 2 – Reduction with carbon: The black copper(II) oxide is reduced by carbon to impure copper metal. Carbon is oxidised to carbon dioxide. The reddish-brown solid is crude copper.
步骤 2 – 用碳还原：黑色的氧化铜被碳还原为不纯的金属铜，碳则被氧化成二氧化碳。红棕色固体即为粗铜。
2CuO + C → 2Cu + CO₂
Step 3 – Electrolytic refining: (impurities drop as anode sludge). At the cathode: Cu²⁺ + 2e⁻ → Cu (pure copper deposits).
步骤 3 – 电解精炼：粗铜作阳极，纯铜薄片作阴极，电解质为 CuSO₄ 溶液。阳极反应：Cu → Cu²⁺ + 2e⁻（杂质沉为阳极泥）。阴极反应：Cu²⁺ + 2e⁻ → Cu（纯铜析出）。
Environmental considerations: The process requires high temperatures, consuming fossil fuels and releasing CO₂. Sulfur impurities in the ore may produce SO₂, contributing to acid rain. Electrolytic refining uses large amounts of electricity. Proper management of anode sludge, which often contains precious metals, is essential to minimise waste.
环境考量：该过程需要高温，消耗化石燃料并排放 CO₂。矿石中的硫杂质可能产生 SO₂，导致酸雨。电解精炼耗电量大。阳极泥常含有贵金属，需妥善管理以最大限度减少废弃物。
2. Case Study: Making Copper Sulfate Crystals | 案例二：制备硫酸铜晶体
A laboratory technician needs to prepare dry copper(II) sulfate crystals (CuSO₄·5H₂O) from copper(II) oxide and dilute sulfuric acid. Describe a safe method, including the key steps of heating, filtration and crystallisation. Explain why an excess of copper(II) oxide is used and how the purity of the crystals can be tested.
一位实验室技术员要用氧化铜和稀硫酸制备干燥的五水合硫酸铜晶体 (CuSO₄·5H₂O)。请描述安全的方法，包括加热、过滤和结晶的关键步骤。解释为何要使用过量的氧化铜，以及如何检验晶体的纯度。
Solution
Reaction principle: Copper(II) oxide, a base, reacts with sulfuric acid to form soluble copper sulfate and water. The balanced equation is:
反应原理：碱性氧化物氧化铜与硫酸反应，生成可溶的硫酸铜和水。配平的化学方程式为：
CuO + H₂SO₄ → CuSO₄ + H₂O
Procedure: (1) Warm dilute sulfuric acid in a beaker (do not boil). (2) Add black copper(II) oxide powder a little at a time, stirring constantly, until no more dissolves and some solid remains. This ensures all acid is neutralised. (3) Stop heating and allow the mixture to cool slightly, then filter to remove the unreacted CuO. (4) Transfer the filtrate (blue copper sulfate solution) to an evaporating dish and heat gently to concentrate it. Stop heating when crystals begin to form at the edge, then leave the solution to cool and crystallise slowly. (5) Filter the crystals, wash with a small amount of cold distilled water and dry between filter papers.
操作步骤：(1) 在烧杯中微热稀硫酸（勿沸腾）。(2) 分批加入黑色氧化铜粉末，持续搅拌，直至固体不再溶解并有少量剩余，确保酸被完全中和。(3) 停止加热，稍冷后过滤，除去未反应的 CuO。(4) 将滤液（蓝色硫酸铜溶液）转移至蒸发皿中，用小火加热浓缩。待边缘出现晶膜时停止加热，静置冷却结晶。(5) 过滤所得晶体，用少量冷蒸馏水洗涤，置于滤纸间吸干。
Reason for excess CuO: Adding excess copper(II) oxide guarantees that all the sulfuric acid is used up, avoiding contamination of the crystals with unreacted acid. It also simplifies the purification because any excess solid base can be removed by filtration.
过量氧化铜的原因：加入过量的氧化铜可以保证硫酸完全反应，避免晶体中残留未反应的酸。同时，多余的固体碱可通过过滤除去，简化纯化步骤。
Testing purity: The purity of the copper sulfate crystals can be checked by measuring a fixed melting point or by dissolving a sample in water and adding barium chloride solution – a white precipitate of BaSO₄ indicates the presence of sulfate ions. The intensity of the blue colour in solution can also be compared with a known standard using colorimetry.
纯度检验：可通过测定固定熔点来检验纯度，或将样品溶于水，加入氯化钡溶液，若能生成白色硫酸钡沉淀表明硫酸根存在。也可用比色法比较溶液蓝色深浅，与已知标准对照。
3. Case Study: Electrolysis of Brine | 案例三：电解食盐水
Brine is concentrated aqueous sodium chloride. When electrolysed with inert electrodes, three useful products are obtained: chlorine gas, hydrogen gas and sodium hydroxide solution. Identify the ions present, predict the products at each electrode, write half‑equations and explain why this process is important for the chemical industry.
食盐水是氯化钠的浓溶液。使用惰性电极进行电解时，可以得到三种有用产品：氯气、氢气和氢氧化钠溶液。请指出溶液中存在的离子，预测两极产物，写出对应的半反应方程式，并解释该过程在化学工业中的重要性。
Solution
Ions present: From NaCl: Na⁺ and Cl⁻. From water: a small concentration of H⁺ and OH⁻ from the self‑ionisation of water. The electrolyte also contains H₂O molecules.
存在的离子：来自 NaCl：Na⁺ 和 Cl⁻。来自水的自电离：极少量的 H⁺ 和 OH⁻。溶液中还有水分子。
At the anode (positive electrode): Chloride ions are discharged in preference to hydroxide ions because of their higher concentration. The half‑equation is: 2Cl⁻ → Cl₂ + 2e⁻. Chlorine gas bubbles off.
在阳极（正极）：由于 Cl⁻ 浓度远高于 OH⁻，氯离子优先放电。半反应式：2Cl⁻ → Cl₂ + 2e⁻。氯气逸出。
At the cathode (negative electrode): Hydrogen ions are discharged in preference to sodium ions because H⁺ is more easily reduced. The half‑equation is: 2H⁺ + 2e⁻ → H₂, which can also be written as 2H₂O + 2e⁻ → H₂ + 2OH⁻. Hydrogen gas is released, leaving behind OH⁻ ions, so the solution around the cathode becomes rich in sodium hydroxide.
在阴极（负极）：H⁺ 比 Na⁺ 更容易被还原，因此氢离子放电。半反应式：2H⁺ + 2e⁻ → H₂，也可写成 2H₂O + 2e⁻ → H₂ + 2OH⁻。氢气逸出，同时 OH⁻ 留在溶液中，使阴极附近形成氢氧化钠溶液。
Industrial importance: Chlorine is used for PVC production, water treatment and disinfectants. Hydrogen is used in the Haber process and as a clean fuel. Sodium hydroxide is a vital base for making soap, paper and textiles. The simultaneous production of three bulk chemicals from a cheap raw material makes this membrane‑cell process highly economical.
工业重要性：氯气用于生产 PVC、水处理和消毒剂；氢气用于哈伯法制氨和清洁燃料；氢氧化钠是制造肥皂、纸张和纺织品的重要碱。利用廉价原料同时制得三种大宗化学品，使得膜电解法极具经济效益。
4. Case Study: The Haber Process – Maximising Ammonia Yield | 案例四：哈伯法 — 最大化氨的产率
Ammonia is produced by the reaction: N₂ + 3H₂ ⇌ he reactor operates at 450 °C and 200 atm with an iron catalyst. Analyse why these conditions are chosen, how a yield of about 30% he percentage yield if 850 kg of ammonia is obtained from a theoretical maximum of 1000 kg under the same feed of nitrogen.
工业上用反应 N₂ + 3H₂ ⇌ 2NH₃（ΔH = −92 kJ/mol）合成氨。工厂的合成塔在 450 °C、200 atm 以及铁催化剂条件下运行。请分析为何选择这些条件，解释约 30% 的单程产率为何仍然具有利润，并计算：若一定量氮气理论上最多可产 1000 kg 氨，实际只得到 850 kg，求百分产率。
Solution
Equilibrium considerations: sure favours the side with fewer gas molecules (4 moles → 2 moles), also increasing yield.
平衡因素：正反应放热。降低温度会使平衡向放热方向移动，提高氨的产率。高压有利于气体分子数较少的一方（4 mol → 2 mol），也能提高产率。
Rate considerations: A very low temperature would make the reaction far too slow. The iron catalyst speeds up the reaction but does not affect the equilibrium position. 450 °C is a compromise temperature that gives a reasonable rate while still allowing a useful equilibrium yield.
速率因素：温度过低会导致反应速率极慢。铁催化剂能加快反应，但不改变平衡位置。450 °C 是一个折中温度，既能保证较快的速率，又兼顾了可接受的平衡产率。
Pressure and cost: 200 atm is high enough to push the equilibrium towards ammonia but not so high that the cost of building and running compressors becomes unbearable. The unreacted nitrogen and hydrogen are recycled, raising the overall yield to above 95%.
压强与成本：200 atm 足够将平衡推向氨，但又不会因压强过高而使压缩机成本和运营费用过于昂贵。未反应的氮气和氢气循环使用，使总产率超过 95%。
Yield calculation: Percentage yield = (actual mass / theoretical mass) × 100%. Plug in the numbers: (850 kg / 1000 kg) × 100% = 85%. Although the single‑pass yield is around 30%, recycling of reactants enables the overall plant yield to be much higher.
产率计算：百分产率 = (实际质量 / 理论质量) × 100%。代入数据：(850 kg / 1000 kg) × 100% = 85%。虽然单程产率约 30%，但通过循环利用反应物，工厂的总产率可以大大提高。
5. Case Study: Preventing Rust on Ships | 案例五：船舶的防锈保护
The hull of a steel ship is constantly exposed to seawater, a strong electrolyte that accelerates rusting. To protect the hull, blocks of zinc are attached at regular intervals. Explain the chemistry of rusting, why zinc prevents corrosion even when the hull surface is scratched, and compare this method with painting and galvanising.
钢制船体长期浸泡在海水（强电解质）中，会加速生锈。为保护船体，人们定期安装锌块。请解释铁生锈的化学原理，说明为何锌块即便在船体表面刮伤后仍能防止腐蚀，并将此法与涂漆和镀锌进行对比。
Solution
Rusting of iron: tal to a cathode area where oxygen is reduced (O₂ + 2H₂O + 4e⁻ → 4OH⁻). The Fe²⁺ ions then react further with oxygen and water to form hydrated iron(III) oxide, rust.
铁的生锈：铁在氧气和水存在下生锈。这是一个电化学过程：铁作阳极（Fe → Fe²⁺ + 2e⁻），电子通过金属流向阴极区，氧气在那里被还原（O₂ + 2H₂O + 4e⁻ → 4OH⁻）。亚铁离子进一步与氧和水反应，生成水合氧化铁，即铁锈。
Sacrificial protection with zinc: : Zn → Zn²⁺ + 2e⁻. The electrons flow to the iron hull, which acts as the cathode and is therefore protected from oxidation. Even if the iron surface is scratched, the zinc continues to corrode sacrificially.
锌的牺牲保护：锌比铁活泼（在金属活动性顺序中位于铁之上）。当两者相连时，锌优先作为阳极被氧化：Zn → Zn²⁺ + 2e⁻。电子流向铁制船体，使其变成阴极，从而得到保护。即使船体表面被刮伤，锌仍会继续牺牲自己，保护铁不被腐蚀。
Comparison with other methods: Painting provides a barrier that blocks oxygen and water. However, once the paint is scratched, bare iron is exposed and rusting starts. Galvanising (coating with zinc) offers both a barrier and sacrificial protection because the underlying zinc will corrode first if the coating is damaged. The zinc blocks on ships work purely via sacrificial protection and are easier to replace than re-galvanising a large hull.
与其他方法对比：涂漆可形成阻隔氧气和水的屏障，但一旦漆膜划伤，裸露的铁就会开始生锈。镀锌（表面覆盖锌层）兼具屏障和牺牲保护双重作用，即使镀层破损，底层的锌仍会优先腐蚀。船体安装的锌块则完全依靠牺牲保护原理，且更换比大规模重新镀锌更为方便。
6. Case Study: Hydrogen Fuel Cells for Vehicles | 案例六：车用氢燃料电池
A car manufacturer is comparing a hydrogen‑ ate electricity, which powers an electric motor. Evaluate the advantages and disadvantages of the fuel‑ell car in terms of energy efficiency, environmental impact and infrastructure challenges.
某汽车制造商正在比较氢氧燃料电池系统与传统汽油发动机。燃料电池通过反应 2H₂ + O₂ → 2H₂O 产生电能，驱动电动机。请从能量效率、环境影响和基础设施挑战三方面，评价燃料电池汽车的优缺点。
Solution
Energy efficiency: Fuel cells convert chemical energy directly into electrical energy, typically achieving 40–60% efficiency, compared with about 20–30% for internal combustion engines. The motor then converts electrical energy into motion more efficiently. Overall, fuel‑cell vehicles can deliver around twice the energy‑to‑wheels efficiency of petrol cars.
能量效率：燃料电池直接将化学能转化为电能，效率通常在 40–60% 之间，而内燃机的效率仅约 20–30%。电动机再将电能高效转化为机械运动。整体而言，燃料电池汽车从储能到车轮的效率约为汽油车的两倍。
Environmental impact: When running, the only exhaust product is water, so there are no CO₂, NOₓ or SO₂ emissions at the point of use. However, the environmental friendliness depends on how the hydrogen is produced. If hydrogen is made by steam reforming of methane, CO₂ is still released. Electrolysis of water using renewable energy makes the entire cycle cleaner.
环境影响：运行时唯一排放物是水，因此使用阶段不产生 CO₂、NOₓ 或 SO₂。然而其环保程度取决于氢气的生产方式。若氢气通过甲烷蒸汽重整制取，仍会排放 CO₂；若使用可再生能源电解水，则整个循环更加清洁。
Infrastructure and storage: Hydrogen must be stored under high pressure (350–700 bar) in strong, heavy tanks, which adds weight and reduces vehicle range. A refuelling network comparable to petrol stations does not yet exist in most countries, making uptake slow. Safety concerns about hydrogen's flammability also require rigorous standards.
基础设施与储存：氢气需要在高压（350–700 bar）下储存于坚固沉重的罐体中，增加车重并限制续航。目前多数国家尚未建立起与加油站相当规模的加氢网络，推广受阻。此外，氢气的可燃性引发安全顾虑，必须执行严格标准。
Overall evaluation: Fuel‑cell vehicles offer zero‑emission driving and high efficiency but are currently limited by hydrogen production methods, storage difficulties and infrastructure. They are most promising for heavy transport and long‑distance journeys where battery electric vehicles face range and charging‑time issues.
总体评价：燃料电池汽车实现零排放驾驶且效率高，但目前在氢气制取方式、储存技术及充能基础设施方面受到限制。它在重型运输和长距离行驶中最具前景，可弥补纯电动汽车在续航和充电时间上的不足。
7. Case Study: Acid Rain and Limestone Buildings | 案例七：酸雨与石灰岩建筑
A historic cathedral built from limestone (calcium carbonate) shows severe surface weathering. Scientists identify the main pollutant as sulfur dioxide from coal‑fired power stations. Describe the sequence of reactions that convert SO₂ into acid rain, explain how the acid attacks limestone, write relevant equations and suggest measures to reduce the damage.
一座用石灰岩（碳酸钙）建造的古老大教堂表面风化严重。科学家确定主要污染物来自燃煤电厂的二氧化硫。请描述 SO₂ 转化为酸雨的反应过程，解释酸如何侵蚀石灰岩，写出相关方程式，并提出减轻损害的措施。
Solution
Formation of acid rain: Sulfur dioxide is released when fossil fuels containing sulfur impurities are burned. In the atmosphere, SO₂ is oxidised to sulfur trioxide: 2SO₂ + O₂ → 2SO₃. SO₃ then reacts with rainwater to form sulfuric acid: SO₃ + H₂O → H₂SO₄. Some nitrogen oxides from vehicle exhausts also contribute, forming nitric acid.
酸雨的形成：含硫杂质的化石燃料燃烧时释放出 SO₂。在大气中，SO₂ 被氧化为三氧化硫：2SO₂ + O₂ → 2SO₃。SO₃ 再与雨水反应生成硫酸：SO₃ + H₂O → H₂SO₄。汽车尾气中的氮氧化物也可形成硝酸，加剧酸雨。
Attack on limestone: Limestone is mainly CaCO₃. Acid rain (containing H₂SO₄) reacts with calcium carbonate in a neutralisation reaction, producing soluble calcium sulfate, carbon dioxide and water: CaCO₃ + H₂SO₄ → CaSO₄ + CO₂ + H₂O. The soluble sulfate is washed away, gradually wearing away the stone and blurring carved details.
对石灰岩的侵蚀：石灰岩的主要成分是 CaCO₃。酸雨中的硫酸与之发生中和反应，生成可溶的硫酸钙、二氧化碳和水：CaCO₃ + H₂SO₄ → CaSO₄ + CO₂ + H₂O。可溶性硫酸盐被雨水冲走，使石料逐渐流失，雕刻细节模糊不清。
Mitigation measures: (1) Install flue‑gas desulfurisation units (scrubbers) in power stations to remove SO₂ before release. (2) Switch to low‑sulfur fuels or renewable energy to reduce SO₂ emissions. (3) Apply a protective consolidant to historic stonework, though this must be done carefully to avoid trapping moisture. (4) Reduce nitrogen oxide emissions from vehicles using catalytic converters.
减轻措施：(1) 在电厂安装烟气脱硫装置（洗涤器），排放前去除 SO₂。(2) 改用低硫燃料或可再生能源，减少
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