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AS Chemistry: Redox Reactions | 氧化还原 考点精讲

2026-07-13 04:51 · 本文部分内容由AI辅助创作

📚 AS Chemistry: Redox Reactions | 氧化还原 考点精讲

Redox chemistry lies at the heart of AS-level chemistry, linking electron transfer, oxidation states, and energy changes. Mastering redox concepts is essential for understanding everything from electrolysis to titrations.

氧化还原是AS化学的核心,连接电子转移、氧化态和能量变化。掌握氧化还原概念是从电解到滴定等理解的关键。

1. Defining Oxidation and Reduction | 氧化与还原的定义

Oxidation is the loss of electrons; reduction is the gain of electrons. This is the fundamental electron-transfer definition.

氧化是失去电子;还原是获得电子。这是电子转移的基本定义。

Oxidation can also be defined as an increase in oxidation number; reduction as a decrease in oxidation number.

氧化也可以定义为氧化数增加;还原为氧化数减少。

A helpful mnemonic: OIL RIG – Oxidation Is Loss, Reduction Is Gain.

记忆口诀:OIL RIG – 氧化是失电子,还原是得电子。


2. Oxidation Number Rules | 氧化数规则

Rule规则
Free elements have oxidation number 0 (e.g., O₂, Na).游离态单质氧化数为0(如O₂, Na)。
For monatomic ions, oxidation number equals the charge (e.g., Na⁺ = +1, Cl⁻ = -1).单原子离子氧化数等于电荷(如Na⁺=+1, Cl⁻=-1)。
Oxygen usually has oxidation number -2, except in peroxides (-1) and with fluorine.氧的氧化数通常为-2,过氧化物中为-1,与氟结合时例外。
Hydrogen is +1 when bonded to non-metals, -1 when bonded to metals (hydrides).氢与非金属结合时为+1,与金属结合时为-1(氢化物)。
The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion equals the ion charge.中性化合物中氧化数之和为0;多原子离子中等离子电荷。

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3. Calculating Oxidation Numbers | 氧化数的计算

number of Mn in KMnO₄. K is +1, O is -2 each, so let Mn be x: +1 + x + 4(-2) = 0 → x = +7.

计算KMnO₄中Mn的氧化数。K为+1,O为-2,设Mn为x:+1 + x + 4(-2) = 0 → x = +7。

In the ion Cr₂O₇²⁻, O is -2, let Cr be y: 2y + 7(-2) = -2 → 2y = +12 → y = +6.

在Cr₂O₇²⁻离子中,O为-2,设Cr为y:2y + 7(-2) = -2 → 2y = +12 → y = +6。

Practice determining oxidation numbers in unfamiliar compounds – this is a frequent exam skill.

练习确定陌生化合物中的氧化数——这是常见的考试技能。


4. Oxidising and Reducing Agents | 氧化剂与还原剂

An oxidising agent (oxidant) accepts electrons and is itself reduced; its oxidation number decreases.

氧化剂接受电子,自身被还原;其氧化数降低。

A reducing agent (reductant) donates electrons and is itself oxidised; its oxidation number increases.

还原剂提供电子,自身被氧化;其氧化数升高。

ample, in Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is the reducing agent and Cu²⁺ is the oxidising agent.

例如,Zn + Cu²⁺ → Zn²⁺ + Cu,Zn是还原剂,Cu²⁺是氧化剂。


5. Half-Equations | 半反应方程式

A half-equation shows either the oxidation or reduction process separately, with electrons explicitly shown.

半反应方程式分别显示氧化或还原过程,并明确写出电子。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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