Oxidation of Two Carboxylic Acids | 两种羧酸的氧化

📚 Oxidation of Two Carboxylic Acids | 两种羧酸的氧化

In A-level Chemistry, carboxylic acids are usually the final oxidation products of primary alcohols and aldehydes. However, two common carboxylic acids – methanoic acid and ethanedioic acid – can themselves be oxidised further. This article explains their oxidation, gives balanced redox equations, and highlights the key exam points.

在 A-level 化学中，羧酸通常是一级醇和醛氧化的最终产物。但有两种常见羧酸——甲酸和乙二酸——自身还能被进一步氧化。本文解释它们的氧化反应、给出配平的氧化还原方程式，并强调关键考点。

1. Why Most Carboxylic Acids Resist Oxidation | 为什么大多数羧酸难以被氧化

Carboxylic acids are generally resistant to oxidation because the carboxyl carbon is already in a fairly high oxidation state. For a typical acid such as ethanoic acid, CH₃COOH, the carbon chain lacks a hydrogen atom directly attached to the carbonyl carbon, so there is no aldehyde-like C–H bond to oxidise easily.

羧酸通常难以被氧化，因为羧基碳已经处于较高的氧化态。以乙酸 CH₃COOH 为代表的普通羧酸，其碳链上没有直接连在羰基碳上的氢原子，因此缺少类似醛的易氧化 C–H 键。

---

2. The Two Important Exceptions | 两个重要例外

The two carboxylic acids that are commonly oxidised are methanoic acid, HCOOH, and ethanedioic acid, HOOC–COOH or (COOH)₂. Methanoic acid has a hydrogen atom bonded to the carbonyl carbon, so it behaves like an aldehyde. Ethanedioic acid has two adjacent carboxyl groups, making its C–C bond susceptible to oxidative cleavage.

两种通常可被氧化的羧酸是甲酸 HCOOH 和乙二酸 HOOC–COOH 或 (COOH)₂。甲酸的羰基碳上连有氢原子，因此具有类似醛的性质。乙二酸的两个羧基相邻，使其 C–C 键容易被氧化断裂。

---

3. Oxidation of Methanoic Acid | 甲酸的氧化

Methanoic acid is oxidised to carbon dioxide and water. Using [O] to represent oxygen supplied by the oxidising agent, the overall equation is:

甲酸被氧化为二氧化碳和水。用 [O] 表示氧化剂提供的氧原子，总方程式为：

HCOOH + [O] → CO₂ + H₂O

The carbon oxidation state increases from +2 in HCOOH to +4 in CO₂, so each molecule loses two electrons.

碳的氧化态从 HCOOH 中的 +2 升高到 CO₂ 中的 +4，因此每个分子失去两个电子。

---

4. Oxidation of Ethanedioic Acid | 乙二酸的氧化

Ethanedioic acid is oxidised to carbon dioxide and water. Its overall oxidation using [O] is:

乙二酸被氧化为二氧化碳和水。使用 [O] 表示的总氧化反应为：

(COOH)₂ + [O] → 2CO₂ + H₂O

Both carbon atoms are oxidised from +3 to +4. Although the molecule has two carbon atoms, the total electron loss is still two electrons per molecule.

两个碳原子都从 +3 氧化到 +4。尽管该分子有两个碳原子，但每个分子的总电子损失仍为两个电子。

---

5. Oxidising Agents and Conditions | 氧化剂与反应条件

The usual oxidising agents are acidified potassium manganate(VII), KMnO₄, and acidified potassium dichromate(VI), K₂Cr₂O₇. The reaction is normally carried out by heating under reflux with dilute sulfuric acid as the acid source.

常用氧化剂是酸化高锰酸钾 KMnO₄ 和酸化重铬酸钾 K₂Cr₂O₇。反应通常在稀硫酸提供的酸性条件下加热回流进行。

• Acidified KMnO₄: MnO₄⁻ is reduced to Mn²⁺; the purple solution turns colourless. | 酸化 KMnO₄：MnO₄⁻ 被还原为 Mn²⁺；紫色溶液褪为无色。
• Acidified K₂Cr₂O₇: Cr₂O₇²⁻ is reduced to Cr³⁺; the orange solution turns green. | 酸化 K₂Cr₂O₇：Cr₂O₇²⁻ 被还原为 Cr³⁺；橙色溶液变为绿色。

---

6. Redox Half-Equations and Full Equations | 氧化还原半反应与总方程式

For acidified manganate(VII), the reduction half-equation is:

对于酸化高锰酸根，还原半反应为：

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

The oxidation half-equations for the two carboxylic acids are:

两种羧酸的氧化半反应分别为：

HCOOH → CO₂ + 2H⁺ + 2e⁻

(COOH)₂ → 2CO₂ + 2H⁺ + 2e⁻

Multiplying the oxidation half-equation by 5 and the reduction half-equation by 2 gives the balanced full ionic equations:

将氧化半反应乘以 5、还原半反应乘以 2，可得到配平的总离子方程式：

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

2

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）