A-Level Chemistry: Oxidation Reactions of Two Carboxylic Acids | A-Level 化学：两种羧酸的氧化反应解析

📚 A-Level Chemistry: Oxidation Reactions of Two Carboxylic Acids | A-Level 化学：两种羧酸的氧化反应解析

Carboxylic acids are generally considered resistant to oxidation — the carboxyl group (−COOH) is already in a highly oxidised state. However, two specific carboxylic acids, methanoic acid (HCOOH) and ethanedioic acid (HOOC−COOH), display distinctive and examinable oxidation behaviours due to their unique structural features.

一般来说，羧酸被认为难以被氧化——羧基（−COOH）已处于高度氧化态。然而，有两种特定的羧酸：甲酸（HCOOH）和乙二酸（HOOC−COOH），由于其独特的结构特征，表现出显著且常考的特殊氧化行为。

1. Structural Background: Why These Two Acids? | 结构背景：为什么是这两种酸？

Methanoic acid is the simplest carboxylic acid, but its structure contains a formyl group (−CHO) directly attached to a hydroxyl group (−OH). In effect, the molecule behaves as both a carboxylic acid and an aldehyde.

甲酸是最简单的羧酸，但它的结构中含有一个直接与羟基（−OH）相连的甲酰基（−CHO）。实际上，该分子既表现出羧酸的性质，又表现出醛的性质。

Ethanedioic acid, commonly known as oxalic acid, contains two adjacent carboxyl groups. The central carbon–carbon bond is electron-deficient and relatively weak, making the molecule prone to oxidative cleavage.

乙二酸，俗称草酸，含有两个相邻的羧基。中心的碳–碳键缺电子且相对较弱，使该分子容易被氧化断裂。

HCOOH → H–COOH (methanoic acid, contains –CHO unit)

HOOC–COOH (ethanedioic acid, two –COOH units)

2. Oxidation of Methanoic Acid by Acidified KMnO₄ | 甲酸与酸性高锰酸钾的氧化

When methanoic acid is warmed with acidified potassium manganate(VII), the purple solution decolourises. The methanoic acid is oxidised to carbon dioxide and water.

当甲酸与酸化的高锰酸钾（VII）共热时，紫色溶液褪色。甲酸被氧化为二氧化碳和水。

HCOOH + [O] → CO₂ + H₂O

The half-equation for the manganate(VII) ion in acidic solution is:

酸性溶液中高锰酸根离子的半反应为：

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Methanoic acid provides the electrons by being oxidised. This reaction is a classic test for methanoic acid, distinguishing it from other simple carboxylic acids.

甲酸通过自身被氧化而提供电子。该反应是检验甲酸的经典方法，可将其与其他简单羧酸区分开来。

Students should note that this oxidation is essentially the oxidation of the aldehyde group embedded within the formic acid structure. The product CO₂ represents a loss of two hydrogen atoms and one oxygen atom relative to the original molecule.

学生应注意，该氧化实质上是甲酸结构中嵌入的醛基被氧化。产物CO₂相当于原分子失去两个氢原子和一个氧原子。

3. Oxidation of Methanoic Acid by Tollens’ Reagent | 甲酸与托伦试剂的氧化

Tollens’ reagent is ammoniacal silver nitrate, Ag(NH₃)₂⁺. Methanoic acid reduces the silver complex to metallic silver, producing a characteristic silver mirror on the test tube wall.

托伦试剂是银氨溶液，即Ag(NH₃)₂⁺。甲酸能将银氨配离子还原为金属银，在试管壁上形成特征性的银镜。

HCOOH + 2Ag(NH₃)₂⁺ + 2OH⁻ → CO₂ + 2Ag↓ + 4NH₃ + 2H₂O

This is identical in outcome to the oxidation of an aldehyde by Tollens’ reagent. The formyl hydrogen in methanoic acid is the hydrogen that gets lost as the carbon is oxidised from the +2 oxidation state in HCOOH to the +4 state in CO₂.

这一结果与醛被托伦试剂氧化的结果完全相同。甲酸中的甲酰氢正是失去的那个氢——碳的氧化态从HCOOH中的+2升高至CO₂中的+4。

Fehling’s or Benedict’s solution can also be used: methanoic acid gives a brick-red precipitate of Cu₂O. However, this reaction requires heating and is less reliable for formic acid than for simple aldehydes.

斐林试剂或本尼迪特试剂也可使用：甲酸产生砖红色的Cu₂O沉淀。不过该反应需要加热，且对甲酸而言不如对简单醛类可靠。

4. Oxidation of Ethanedioic Acid by Acidified KMnO₄ | 乙二酸与酸性高锰酸钾的氧化

Ethanedioic acid reacts readily with acidified potassium manganate(VII). This is a well-known redox titration used in A-Level practical examinations.

乙二酸可与酸性高锰酸钾（VII）迅速反应。这是A-Level实验考试中著名的氧化还原滴定反应。

5HOOC–COOH + 2MnO₄⁻ + 6H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O

The reaction requires gentle warming to initiate, but once started it is autocatalytic: the Mn²⁺ ions produced catalyse the reaction, so the decolourisation accelerates.

该反应需要微热引发，但一旦开始便呈自催化特征：生成的Mn²⁺离子会催化反应，因此褪色速度逐渐加快。

The half-equation for ethanedioic acid in acidic solution is:

乙二酸在酸性溶液中的半反应为：

HOOC–COOH → 2CO₂ + 2H⁺ + 2e⁻

注意，草酸被氧化时，碳的氧化态从+3升到+4，每个草酸分子失去2个电子。

5. Oxidation of Ethanedioic Acid by K₂Cr₂O₇ | 乙二酸与重铬酸钾的氧化

Acidified potassium dichromate(VI) can also oxidise ethanedioic acid, though more slowly than KMnO₄. The orange dichromate solution turns green as Cr₂O₇²⁻ is reduced to Cr³⁺.

酸化的重铬酸钾（VI）也能氧化乙二酸，但速率比高锰酸钾慢。橙色重铬酸根溶液逐渐变绿，因为Cr₂O₇²⁻被还原为Cr³⁺。

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Combined with the ethanedioic acid half-equation, the overall equation is:

结合乙二酸的半反应，总反应式为：

3HOOC–COOH + Cr₂O₇²⁻ + 8H⁺ → 6CO₂ + 2Cr³⁺ + 7H₂O

This reaction is less commonly tested than the KMnO₄ route, but it reinforces the key idea that oxalic acid is a strong reducing agent.

该反应虽然不如高锰酸钾路线常考，但它强化了关键概念：草酸是一种强还原剂。

6. Comparing the Two Carboxylic Acids | 两种羧酸的氧化对比

Feature \| 特征	Methanoic Acid \| 甲酸	Ethanedioic Acid \| 乙二酸
Formula \| 分子式	HCOOH	HOOC–COOH
Functional groups \| 官能团	−COOH plus −CHO character	Two −COOH groups in close proximity
Oxidation product \| 氧化产物	CO₂ + H₂O	2CO₂ + H₂O
Electrons lost per molecule \| 每分子失去电子数	2	2
Tollens’ test \| 托伦试剂	Silver mirror formed	No reaction
KMnO₄ test \| 高锰酸钾试验	Decolourises	Decolourises
Carbon oxidation state change \| 碳氧化态变化	+2 → +4	+3 → +4

7. Why Does Ethanedioic Acid Not Give a Silver Mirror? | 为什么乙二酸不能产生银镜？

Unlike methanoic acid, ethanedioic acid does not contain an aldehyde (−CHO) group. The Tollens’ reagent specifically requires a terminal formyl hydrogen atom to oxidise. Ethanedioic acid has no such hydrogen directly attached to a carbonyl carbon.

与甲酸不同，乙二酸不含醛基（−CHO）。托伦试剂需要末端的甲酰氢原子才能发生氧化。乙二酸并没有直接连在羰基碳上的氢原子。

However, ethanedioic acid can reduce warmed Fehling’s solution slightly, though this is not a reliable qualitative test and is rarely examined. The KMnO₄ decolourisation test is far more important for oxalic acid.

然而，乙二酸可以微弱地还原温热的斐林溶液，但这不是可靠的定性检验，考查较少。对于草酸而言，高锰酸钾褪色试验要重要得多。

This distinction is a favourite exam point: students must know that only methanoic acid among the simple carboxylic acids gives a positive Tollens’ test.

这一区别是考试热点：学生必须知道，在简单羧酸中，只有甲酸能给出托伦试剂阳性反应（银镜）。

8. Stoichiometry in Redox Titrations | 氧化还原滴定中的化学计量

In A-Level practical papers, the oxidation of ethanedioic acid by KMnO₄ is used to determine concentration. The key mole ratio is:

在A-Level实验卷中，常用乙二酸与高锰酸钾的氧化反应来测定浓度。关键摩尔比为：

2 mol KMnO₄ ⇌ 5 mol HOOC–COOH

Worked example: 25.0 cm³ of ethanedioic acid solution required 21.50 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate the concentration of the acid.

例题：25.0 cm³的乙二酸溶液恰好与21.50 cm³、0.0200 mol dm⁻³的KMnO₄完全反应。计算乙二酸的浓度。

n(KMnO₄) = 0.02150 × 0.0200 = 4.30 × 10⁻⁴ mol

n(HOOC–COOH) = 4.30 × 10⁻⁴ × 5/2 = 1.075 × 10⁻³ mol

c = 1.075 × 10⁻³ / 0.0250 = 0.0430 mol dm⁻³

For methanoic acid titrations, the ratio with KMnO₄ is different. Each HCOOH molecule loses 2 electrons, and each MnO₄⁻ gains 5 electrons, so the LCM is 10:

对于甲酸滴定，与高锰酸钾的配比不同。每个HCOOH分子失2个电子，每个MnO₄⁻得5个电子，最小公倍数为10：

5HCOOH + 2MnO₄⁻ + 6H⁺ → 5CO₂ + 2Mn²⁺ + 8H₂O

Note the similarity: both acids share the same 5:2 stoichiometric ratio with manganate(VII), because both lose two electrons per molecule.

注意两者的相似之处：两种酸与高锰酸根（VII）的化学计量比均为5:2，因为每分子都失去2个电子。

9. Common Exam Questions and Pitfalls | 常考题型与常见错误

Question 1: A student observes that warm acidified KMnO₄ decolourises when added separately to methanoic acid and ethanedioic acid. Explain this observation.

题目1：某同学观察到温热的酸性高锰酸钾分别加入甲酸和乙二酸后都褪色。请解释这一现象。

Answer: Both acids act as reducing agents. Methanoic acid contains an aldehyde unit that is oxidised to CO₂; ethanedioic acid undergoes oxidative cleavage of the central C–C bond, also forming CO₂. In both cases, MnO₄⁻ is reduced to Mn²⁺, losing its purple colour.

答案：两种酸都充当还原剂。甲酸含有醛基单元，被氧化为CO₂；乙二酸的中心C–C键发生氧化断裂，同样生成CO₂。两种情况下，MnO₄⁻均被还原为Mn²⁺，紫色消失。

Question 2: Which carboxylic acid gives a silver mirror with Tollens’ reagent? Why?

题目2：哪种羧酸能与托伦试剂产生银镜？为什么？

Answer: Methanoic acid only, because it is the only carboxylic acid containing an aldehyde functional group (−CHO) capable of reducing Ag⁺ to Ag.

答案：只有甲酸。因为它是唯一含有醛基（−CHO）的羧酸，醛基能将Ag⁺还原为Ag。

Common pitfalls:

常见易错点：

Writing the formula of methanoic acid incorrectly as COOH instead of HCOOH — the formyl hydrogen matters!
将甲酸分子式误写为COOH而不是HCOOH——甲酰氢非常重要！
Forgetting to balance H⁺ and H₂O in half-equations.
忘记在半反应中配平H⁺和H₂O。
Assuming ethanedioic acid reacts with Tollens’ reagent — it does not, as no −CHO group is present.
误认为乙二酸能与托伦试剂反应——实际不能，因为它不含−CHO基团。
Using incorrect oxidation numbers: in HCOOH the carbon of the formyl group is +2; in HOOC–COOH each carbon is +3.
氧化态计算错误：HCOOH中醛基碳为+2；HOOC–COOH中每个碳为+3。

10. Applications and Significance | 应用与意义

The reducing power of ethanedioic acid has practical uses: it is used to remove rust (iron(III) oxide) and in bleaching. Because it decolourises KMnO₄ cleanly without forming colored intermediates, it is the preferred primary standard for standardising KMnO₄ solutions.

乙二酸的还原性有实际用途：用于除锈（氧化铁）和漂白。由于它能将高锰酸钾干净地褪色且不产生有色中间体，因此是标定高锰酸钾溶液的首选基准物质。

Methanoic acid’s reducing properties are relevant in its industrial use as a preservative and antibacterial agent. Its ability to reduce Tollens’ reagent also provides a quick qualitative check in organic synthesis.

甲酸的还原性与其作为防腐剂和抗菌剂的工业用途相关。它能还原托伦试剂，也为有机合成中的快速定性检验提供了手段。

In the laboratory, distinguishing these two acids is straightforward: use Tollens’ reagent (positive only for methanoic acid) or measure the volume of CO₂ produced per mole of acid oxidised.

在实验室中，区分这两种酸非常简单：使用托伦试剂（仅甲酸呈阳性），或测量每摩尔酸氧化产生的CO₂体积。

11. Summary of Reaction Equations | 反应方程式总结

All key equations in one place for revision:

以下汇总所有关键方程式，方便复习：

Methanoic acid + KMnO₄ | 甲酸 + 高锰酸钾：

5HCOOH + 2KMnO₄ + 3H₂SO₄ → 5CO₂ + K₂SO₄ + 2MnSO₄ + 8H₂O

Methanoic acid + Tollens’ reagent | 甲酸 + 托伦试剂：

HCOOH + 2Ag(NH₃)₂⁺ + 2OH⁻ → CO₂ + 2Ag↓ + 4NH₃ + 2H₂O

Ethanedioic acid + KMnO₄ | 乙二酸 + 高锰酸钾：

5HOOC–COOH + 2KMnO₄ + 3H₂SO₄ → 10CO₂ + K₂SO₄ + 2MnSO₄ + 8H₂O

Ethanedioic acid + K₂Cr₂O₇ | 乙二酸 + 重铬酸钾：

3HOOC–COOH + K₂Cr₂O₇ + 4H₂SO₄ → 6CO₂ + K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O

Memorising these equations, together with their colour changes, gives you full command of this topic in the exam.

记住这些方程式及其颜色变化，你就能在考试中完整掌握这一专题。

12. Conclusion | 总结

The oxidation of methanoic acid and ethanedioic acid is a classic A-Level topic that connects organic structure to redox chemistry. Methanoic acid oxidises because of its hidden aldehyde group; ethanedioic acid oxidises because of its fragile C–C bond between two electron-withdrawing carboxyl groups.

甲酸与乙二酸的氧化是一个经典的A-Level专题，它将有机结构与氧化还原化学联系起来。甲酸因隐藏的醛基可被氧化；乙二酸因两个吸电子羧基之间的脆弱C–C键而可被氧化。

Remember the key tests: Tollens’ reagent identifies methanoic acid, while KMnO₄ decolourisation identifies both. Master the half-equations, the 5:2 stoichiometric ratio, and the colour changes, and you will handle any exam question with confidence.

请记住关键检验：托伦试剂鉴定甲酸，高锰酸钾褪色可鉴定两者。掌握半反应、5:2化学计量比以及颜色变化，你就能自信应对任何考题。

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A-Level Chemistry: Oxidation Reactions of Two Carboxylic Acids | A-Level 化学：两种羧酸的氧化反应解析