📚 Case Study Practice: Mastering Applied Chemistry | 案例分析实战演练：掌握应用化学
Case study questions in CCEA Year 13 Chemistry challenge you to apply knowledge to unfamiliar situations, interpret data, and perform multi-step calculations. This article presents a series of real-worked examples covering titrations, kinetics, equilibria, energetics, organic synthesis, and spectroscopy. By working through these cases, you will sharpen your analytical thinking and become confident in tackling the Applied Chemistry section of the examination.
CCEA 13 年级化学的案例分析题要求你把知识运用到陌生情境中，解读数据并完成多步计算。本文通过一系列真实例题，涵盖滴定、动力学、平衡、能量学、有机合成与光谱。通过演练这些案例，你将锻炼分析思维，自信应对考试中的应用化学部分。
1. Acid-Base Titration and Percentage Purity | 酸碱滴定与纯度计算
A batch of sodium hydroxide pellets is suspected to have absorbed carbon dioxide from the air, forming sodium carbonate on the surface. A student dissolves 2.50 g of the pellets in water and makes up to 250.0 cm³. A 25.0 cm³ portion requires 23.85 cm³ of 0.100 mol dm⁻³ hydrochloric acid for neutralisation using phenolphthalein indicator. The reaction occurs in two stages: first all NaOH and half the carbonate react, then the remaining carbonate reacts with more acid. With phenolphthalein, the endpoint corresponds to conversion of NaOH to NaCl and Na₂CO₃ to NaHCO₃. Calculate the percentage purity of the original NaOH pellets assuming the only impurity is Na₂CO₃.
一批氢氧化钠颗粒被怀疑从空气中吸收了二氧化碳，表面生成了碳酸钠。某学生将 2.50 g 颗粒溶于水并定容至 250.0 cm³。移取 25.0 cm³ 溶液，以酚酞为指示剂，用 0.100 mol dm⁻³ 盐酸滴定，消耗 23.85 cm³。反应分两步进行：首先所有 NaOH 和一半的碳酸盐反应，然后剩余的碳酸盐再与酸反应。酚酞终点对应 NaOH 转变为 NaCl、Na₂CO₃ 转变为 NaHCO₃。假设杂质仅为 Na₂CO₃，计算原 NaOH 颗粒的纯度百分比。
In the 25.0 cm³ aliquot, let the amount of NaOH be x mol and Na₂CO₃ be y mol. The reaction with HCl to the phenolphthalein endpoint consumes HCl: for NaOH, mole ratio 1:1; for Na₂CO₃ to NaHCO₃, mole ratio 1:1. Total HCl used = x + y = (23.85/1000) × 0.100 = 0.002385 mol. In the original 250.0 cm³ solution, these amounts are ten times larger: 10x mol NaOH and 10y mol Na₂CO₃. Mass balance: 40.0 × 10x + 106.0 × 10y = 2.50 g. Solve the simultaneous equations. From x + y = 0.002385, we have 10x + 10y = 0.02385. Let A = 10x, B = 10y, then A + B = 0.02385 and 40A + 106B = 2.50. Substitute A = 0.02385 – B into the mass equation: 40(0.02385 – B) + 106B = 2.50 → 0.954 – 40B + 106B = 2.50 → 66B = 1.546 → B = 0.02342 mol; A = 0.00043 mol. Mass of NaOH = 40 × 0.00043 = 0.0172 g; mass of Na₂CO₃ = 106 × 0.02342 = 2.4828 g. That sum is almost 2.50 g, but note the NaOH amount is very small, indicating almost complete conversion to carbonate. Percentage purity of NaOH = (0.0172 / 2.50) × 100 = 0.69%. However, this result is unrealistic because the sample mainly consists of carbonate; the question likely expects recognizing that the effective NaOH content is calculated based on the acid consumed by the NaOH portion alone. Alternatively, purity may be defined as mass of NaOH equivalent present. Often, the calculation yields purity = (mass of pure NaOH / total mass) × 100. Using the solved A = 0.00043 mol gives very low purity; perhaps the question expects calculating the mass of NaOH that would be present if all alkali were NaOH, but the presence of carbonate complicates. A common approach is to determine the total moles of HCl required for complete neutralisation (to methyl orange endpoint) and then deduce composition. Since only phenolphthalein data is given, the calculated purity is indeed very low. The key point is the methodology: setting up mole balances. For a typical exam case, the numbers would be more balanced. Let’s adjust the scenario with more realistic figures: if 2.50 g sample required 23.85 cm³ of 0.100 mol dm⁻³ HCl in the 25 cm³ aliquot, the total HCl for whole solution = 0.2385 mol. In another trial, a second titration with methyl orange would be needed. But here, we can simply present the principle: purity of NaOH = (mass of NaOH determined / mass of sample) × 100. In this case, due to extreme conversion, purity is low, which could be a valid conclusion. Thus, the answer demonstrates the analytical approach.
在 25.0 cm³ 等分溶液中，设 NaOH 的物质的量为 x mol，Na₂CO₃ 为 y mol。至酚酞终点，HCl 消耗量：NaOH 按 1:1，Na₂CO₃ 变成 NaHCO₃ 也是 1:1。总 HCl 量 = x + y = (23.85/1000) × 0.100 = 0.002385 mol。在原 250.0 cm³ 溶液中，这些量扩大十倍：10x mol NaOH 和 10y mol Na₂CO₃。质量关系：40.0 × 10x + 106.0 × 10y = 2.50 g。解联立方程：由 x + y = 0.002385 得 10x + 10y = 0.02385。令 A = 10x, B = 10y，则 A + B = 0.02385，40A + 106B = 2.50。代入 A = 0.02385 – B 得 40(0.02385 – B) + 106B = 2.50 → 0.954 – 40B + 106B = 2.50 → 66B = 1.546 → B = 0.02342 mol；A = 0.00043 mol。NaOH 质量 = 40 × 0.00043 = 0.0172 g；Na₂CO₃ 质量 = 106 × 0.02342 = 2.4828 g。此结果显示 NaOH 含量极低，几乎全部转化为碳酸盐。NaOH 纯度 = (0.0172 / 2.50) × 100 = 0.69%。这一结果看似反常，但在吸收二氧化碳严重的情况下可能成立。重点在于建立摩尔平衡的解题方法。实际考试中数据会更均衡，但原理相同。
2. Redox Titration: Iron in Iron Tablets | 氧化还原滴定：铁片中的铁含量
An iron tablet is dissolved in dilute sulfuric acid, reducing all iron to Fe²⁺ ions. The solution is titrated with 0.0200 mol dm⁻³ potassium manganate(VII) in acidic medium. The average titre is 24.50 cm³. The reaction is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Determine the mass of iron per tablet, given that the whole solution was made from one tablet weighing 0.450 g. Also calculate the percentage by mass of iron in the tablet.
将一片铁片溶于稀硫酸，使全部铁转化为 Fe²⁺ 离子。在酸性介质中用 0.0200 mol dm⁻³ 高锰酸钾溶液滴定，平均用量为 24.50 cm³。反应为：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。已知一片药片质量为 0.450 g，求每片中铁的质量，并计算铁的质量分数。
Moles of MnO₄⁻ used = (24.50/1000) × 0.0200 = 0.000490 mol. From stoichiometry, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺. Thus moles of Fe²⁺ = 5 × 0.000490 = 0.00245 mol. Mass of iron = moles × 55.8 g mol⁻¹ = 0.00245 × 55.8 = 0.1367 g. Percentage by mass = (0.1367 / 0.450) × 100 = 30.4%. This type of redox calculation is a cornerstone of applied analytical chemistry, requiring careful interpretation of mole ratios and consistent use of units.
消耗的 MnO₄⁻ 物质的量 = (24.50/1000) × 0.0200 = 0.000490 mol。由化学计量关系，1 mol MnO₄⁻ 与 5 mol Fe²⁺ 反应，因此 Fe²⁺ 物质的量 = 5 × 0.000490 = 0.00245 mol。铁的质量 = 0.00245 × 55.8 = 0.1367 g。质量分数 = (0.1367 / 0.450) × 100 = 30.4%。此类氧化还原计算是应用分析化学的基础，要求准确运用摩尔比并统一单位。
3. Determining Rate Equations from Initial Rates | 由初始速率确定速率方程
The reaction between peroxodisulfate ions and iodide ions was studied: S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂. Initial rate data are given in the table.
过二硫酸根与碘离子的反应 S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ 被研究，初始速率数据如下表。
| Experiment | [S₂O₈²⁻] / mol dm⁻³ | [I⁻] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 1.5 × 10⁻⁵ |
| 2 | 0.20 | 0.10 | 3.0 × 10⁻⁵ |
| 3 | 0.10 | 0.20 | 3.0 × 10⁻⁵ |
Determine the order with respect to each reactant, write the rate equation, and calculate the rate constant with units.
确定各反应物的反应级数，写出速率方程，并计算速率常数及其单位。
Comparing experiments 1 and 2: [S₂O₈²⁻] doubles, [I⁻] constant, rate doubles → first order in S₂O₈²⁻. Experiments 1 and 3: [I⁻] doubles, [S₂O₈²⁻] constant, rate doubles → first order in I⁻. So the rate equation is: rate = k [S₂O₈²⁻] [I⁻]. Overall order = 2. Using experiment 1: k = rate / ([S₂O₈²⁻][I⁻]) = (1.5 × 10⁻⁵) / (0.10 × 0.10) = 1.5 × 10⁻³. Units: mol⁻¹ dm³ s⁻¹. This straightforward method highlights the importance of systematically comparing rate data.
比较实验 1 和 2：[S₂O₈²⁻] 加倍，[I⁻] 不变，速率加倍 → 对 S₂O₈²⁻ 为一级。实验 1 和 3：[I⁻] 加倍，[S₂O₈²⁻] 不变，速率加倍 → 对 I⁻ 为一级。因此速率方程：rate = k [S₂O₈²⁻] [I⁻]。总级数为 2。代入实验 1：k = (1.5 × 10⁻⁵) / (0.10 × 0.10) = 1.5 × 10⁻³，单位 mol⁻¹ dm³ s⁻¹。这种系统比较数据的方法至关重要。
4. Activation Energy from Arrhenius Plot | 阿仑尼乌斯图求活化能
The rate constant for a reaction was measured at different temperatures. The plot of ln k against 1/T gave a straight line with gradient –1.20 × 10⁴ K. Calculate the activation energy, Ea, in kJ mol⁻¹. (R = 8.31 J mol⁻¹ K⁻¹).
某反应的速率常数在不同温度下测定。以 ln k 对 1/T 作图得到一条直线，斜率为 –1.20 × 10⁴ K。计算活化能 Ea，单位为 kJ mol⁻¹（R = 8.31 J mol⁻¹ K⁻¹）。
The Arrhenius equation linear form is: ln k = ln A – (Ea/R)(1/T). Thus gradient = –Ea/R. Therefore, –Ea/R = –1.20 × 10⁴ K. So Ea = 1.20 × 10⁴ K × 8.31 J mol⁻¹ K⁻¹ = 99720 J mol⁻¹ = 99.7 kJ mol⁻¹. The graph provides a direct route to activation energy, demonstrating how the Arrhenius equation connects kinetic data with the energy barrier of a reaction.
阿仑尼乌斯方程的线性形式为：ln k = ln A – (Ea/R)(1/T)。因此斜率 = –Ea/R。可得 –Ea/R = –1.20 × 10⁴ K，所以 Ea = 1.20 × 10⁴ K × 8.31 J mol⁻¹ K⁻¹ = 99720 J mol⁻¹ = 99.7 kJ mol⁻¹。图形法直接给出活化能，展示了阿仑尼乌斯方程如何将动力学数据与反应的能垒联系起来。
5. Equilibrium Constant Kc for an Esterification | 酯化反应的平衡常数 Kc
Ethanoic acid (0.50 mol) and ethanol (0.50 mol) were mixed with a small amount of acid catalyst, and the mixture allowed to reach equilibrium at 298 K. The equilibrium mixture was found to contain 0.33 mol of ethyl ethanoate. The total volume of the mixture was V dm³. Derive the equilibrium constant Kc and state its units.
将 0.50 mol 乙酸和 0.50 mol 乙醇与少量酸催化剂混合，在 298 K 下达到平衡。平衡混合物中含有 0.33 mol 乙酸乙酯。总体积为 V dm³。推导平衡常数 Kc 并说明其单位。
Reaction: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. At equilibrium: moles of ester = 0.33, thus moles of water also = 0.33. Moles of ethanoic acid remaining = 0.50 – 0.33 = 0.17, ethanol remaining = 0.17. Concentration = moles / V, so Kc = ([ester][water]) / ([acid][alcohol]) = (0.33/V × 0.33/V) / (0.17/V × 0.17/V) = (0.33²) / (0.17²) = 3.76. The V cancels, so Kc has no units. This classic organic equilibrium illustrates how Kc is unaffected by volume when the number of moles of reactants and products are equal. It also highlights the importance of stoichiometry and the presence of water as a product in esterification.
反应：CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O。平衡时，酯的物质的量 = 0.33 mol，故水的物质的量也为 0.33 mol。剩余乙酸 = 0.50 – 0.33 = 0.17 mol，乙醇剩余 = 0.17 mol。浓度 = 物质的量 / V，因此 Kc = ([酯][水]) / ([酸][醇]) = (0.33/V × 0.33/V) / (0.17/V × 0.17/V) = 0.33² / 0.17² = 3.76。V 被约去，所以 Kc 无单位。这一经典有机平衡例题表明，当反应前后分子总数不变时，Kc 不受体积影响，同时也强调了水作为产物的化学计量关系。
6. Kp Calculation for the Haber Process | 哈伯法合成氨的 Kp 计算
In the Haber process, nitrogen and hydrogen react to form ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). A 1:3 molar mixture of N₂ and H₂ at 400 °C and a total pressure of 200 atm reaches equilibrium with 15% by volume of NH₃. Calculate Kp, stating its units.
哈伯法中，氮与氢反应生成氨：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。氮与氢以 1:3 摩尔比混合，在 400 °C、总压 200 atm 下达到平衡，氨的体积分数为 15%。计算 Kp 并注明单位。
At equilibrium, mole fraction of NH₃ = 0.15. Since the initial ratio was 1:3, the remaining N₂ and H₂ also maintain that ratio. Total mole fraction of N₂ + H₂ = 1 – 0.15 = 0.85. The ratio N₂ : H₂ = 1 : 3, so mole fraction of N₂ = 0.85 × 1/4 = 0.2125, H₂ = 0.85 × 3/4 = 0.6375. Partial pressures: p(NH₃) = 0.15 × 200 = 30 atm; p(N₂) = 0.2125 × 200 = 42.5 atm; p(H₂) = 0.6375 × 200 = 127.5 atm. Kp = p(NH₃)² / [p(N₂) × p(H₂)³] = (30)² / (42.5 × (127.5)³). Calculate: 127.5³ ≈ 2.07 × 10⁶. Denominator = 42.5 × 2.07 × 10⁶ = 8.80 × 10⁷. Numerator = 900. Kp = 900 / 8.80 × 10⁷ = 1.02 × 10⁻⁵. Units: atm⁻² (since numerator atm² / (atm × atm³) = atm⁻²). Kp expressions are pivotal in industrial chemistry, linking equilibrium composition to pressure.
平衡时 NH₃ 的摩尔分数 = 0.15。由于初始比例为 1:3，剩余的 N₂ 和 H₂ 也保持此比例。N₂ + H₂ 的总摩尔分数 = 0.85，其中 N₂ 占 1/4，即 0.85 × 1/4 = 0.2125，H₂ 占 0.85 × 3/4 = 0.6375。分压：p(NH₃) = 0.15 × 200 = 30 atm；p(N₂) = 0.2125 × 200 = 42.5 atm；p(H₂) = 0.6375 × 200 = 127.5 atm。Kp = (30)² / [42.5 × (127.5)³]。127.5³ ≈ 2.07 × 10⁶，分母 = 42.5 × 2.07 × 10⁶ = 8.80 × 10⁷，分子 = 900，Kp = 1.02 × 10⁻⁵，单位 atm⁻²。Kp 计算在工业化学中十分关键，它将平衡组成与压力联系起来。
7. Enthalpy Change Using Hess’s Law | 盖斯定律求焓变
Given the following standard enthalpy changes of combustion: C(s) + O₂(g) → CO₂(g) ΔHc° = –394 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l) ΔHc° = –286 kJ mol⁻¹; C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔHc° = –1367 kJ mol⁻¹. Determine the standard enthalpy of formation of ethanol.
已知下列标准燃烧焓变：C(s) + O₂(g) → CO₂(g) ΔHc° = –394 kJ mol⁻¹；H₂(g) + ½O₂(g) → H₂O(l) ΔHc° = –286 kJ mol⁻¹；C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔHc° = –1367 kJ mol⁻¹。计算乙醇的标准生成焓。
Formation reaction: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). According to Hess’s law, the enthalpy of formation equals the sum of combustion enthalpies of the constituent elements minus the combustion enthalpy of the compound. ΔHf° = [2 × ΔHc°(C) + 3 × ΔHc°(H₂)] – ΔHc°(C₂H₅OH) = [2 × (–394) + 3 × (–286)] – (–1367) = (–788 – 858) + 1367 = –1646 + 1367 = –279 kJ mol⁻¹. Notice the sign conventions; care is essential when manipulating thermodynamic cycles. This example demonstrates the practical use of Hess’s law to indirectly find a value that is difficult to measure directly.
生成反应：2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)。根据盖斯定律，生成焓等于各元素燃烧焓之和减去化合物的燃烧焓。ΔHf° = [2 × (–394) + 3 × (–286)] – (–1367) = (–788 – 858) + 1367 = –1646 + 1367 = –279 kJ mol⁻¹。注意符号规则，处理热力学循环时格外小心。该例题展示了如何运用盖斯定律间接求得难以直接测量的数值。
8. Born-Haber Cycle for an Ionic Compound | 离子化合物的玻恩-哈伯循环
Sodium chloride has a lattice enthalpy that can be determined via a Born-Haber cycle. Use the following data: Enthalpy of atomisation of Na = +108 kJ mol⁻¹; First ionisation energy of Na = +496 kJ mol⁻¹; Bond dissociation enthalpy of Cl₂ = +242 kJ mol⁻¹; Electron affinity of Cl = –349 kJ mol⁻¹; Enthalpy of formation of NaCl(s) = –411 kJ mol⁻¹. Calculate the lattice enthalpy of NaCl(s).
氯化钠的晶格焓可通过玻恩-哈伯循环确定。请使用下列数据：Na 的原子化焓 = +108 kJ mol⁻¹；Na 的第一电离能 = +496 kJ mol⁻¹；Cl₂ 的键解离焓 = +242 kJ mol⁻¹；Cl 的电子亲和势 = –349 kJ mol⁻¹；NaCl(s) 的生成焓 = –411 kJ mol⁻¹。计算 NaCl(s) 的晶格焓。
The Born-Haber cycle pathway from elements to ionic solid: 1. Atomisation of Na(s) to Na(g): +108. 2. Ionisation of Na(g) to Na⁺(g): +496. 3. Dissociation of ½Cl₂(g) to Cl(g): ½ × 242 = +121. 4. Electron gain by Cl(g) to Cl⁻(g): –349. 5. Formation of NaCl(s) from gaseous ions: lattice enthalpy (ΔHL). According to Hess’s law, sum of steps 1–5 equals the enthalpy of formation: +108 + 496 + 121 – 349 + ΔHL = –411. Solving: 376 + ΔHL = –411 → ΔHL = –411 – 376 = –787 kJ mol⁻¹. The large negative value reflects the strong electrostatic attraction in the NaCl lattice. The Born-Haber cycle is a powerful tool for linking various energetic terms.
从单质到离子固体的玻恩-哈伯路径：1. Na(s) 原子化为 Na(g)：+108。2. Na(g) 电离为 Na⁺(g)：+496。3. ½Cl₂(g) 解离为 Cl(g)：½ × 242 = +121。4. Cl(g) 获得电子生成 Cl⁻(g)：–349。5. 气态离子形成 NaCl(s) 的晶格焓 (ΔHL)。由盖斯定律，步骤 1–5 之和等于生成焓：+108 + 496 + 121 – 349 + ΔHL = –411。计算得 376 + ΔHL = –411，故 ΔHL = –787 kJ mol⁻¹。大负值反映了 NaCl 晶格中强大的静电引力。玻恩-哈伯循环是联系各项能量术语的有力工具。
9. Organic Synthesis Pathways: From Alkene to Ester | 有机合成路径：从烯烃到酯
Starting from ethene, propose a two-step synthesis to ethyl ethanoate, giving reagents and conditions for each step. Also write balanced equations for the reactions.
以乙烯为起始原料，提出两步合成乙酸乙酯的路线，给出各步骤的试剂和条件，并写出配平的反应方程式。
Step 1: hydration of ethene to ethanol. Reagents: steam, H₃PO₄ catalyst on silica, 300 °C, 60–70 atm. C₂H₄ + H₂O → C₂H₅OH. Step 2: esterification of ethanol with ethanoic acid. Reagents: ethanoic acid, concentrated H₂SO₄ catalyst, heat under reflux. C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O. The synthesis requires careful control of conditions to maximise yield and minimise side reactions. The use of an acid catalyst in both steps is typical of industrial routes to esters.
第一步：乙烯水合生成乙醇。试剂：水蒸气，磷酸/硅藻土催化剂，300 °C，60–70 atm。C₂H₄ + H₂O → C₂H₅OH。第二步：乙醇与乙酸的酯化。试剂：乙酸，浓硫酸催化剂，加热回流。C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O。该合成需要小心控制条件以使产率最大化并减少副反应。两步均使用酸催化剂是工业生产酯的典型路线。
10. Electrochemical Cells and Standard Electrode Potentials | 电化学电池与标准电极电势
A galvanic cell is constructed using Zn²⁺/Zn and Cu²⁺/Cu half-cells. The standard electrode potentials are: Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E° = –0.76 V; Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E° = +0.34 V. Identify the anode and cathode, write the cell reaction, calculate the standard cell potential, and state the direction of electron flow in the external circuit.
使用 Zn²⁺/Zn 和 Cu²⁺/Cu 半电池构成原电池。标准电极电势：Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E° = –0.76 V；Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E° = +0.34 V。指出阳极和阴极，写出电池反应，计算标准电池电势，并说明外电路电子流动方向。
The more negative electrode (Zn) undergoes oxidation; it is the anode. The copper electrode is the cathode. Cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). E°cell = E°cathode – E°anode = 0.34 – (–0.76) = 1.10 V. Electrons flow from the zinc electrode (anode) to the copper electrode (cathode) through the external wire. This type of cell is the basis of a Daniell cell and illustrates the relationship between electrode potentials and spontaneous redox reactions.
电极电势更负的 Zn 发生氧化，为阳极；铜电极为阴极。电池反应：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。E°cell = 0.34 – (–0.76) = 1.10 V。电子经外电路由锌电极（阳极）流向铜电极（阴极）。这种电池是丹尼尔电池的基础，体现了电极电势与自发氧化还原反应之间的关系。
11. Interpretation of NMR and IR Spectra | 核磁共振与红外光谱解析
An unknown organic compound X has the molecular formula C₃H₆O. Its IR spectrum shows a strong absorption at 1720 cm⁻¹. The ¹H NMR spectrum shows a singlet at δ 2.1 (3H) and a singlet at δ 9.8 (1H), with no other signals. Identify compound X and explain
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