📚 Redox II & Transition Metals in Edexcel A-Level Chemistry | Edexcel A-Level 化学：氧化还原 II 与过渡金属（综合主题 14 & 15）
This comprehensive revision guide covers the essential concepts from Edexcel A-Level Chemistry Topics 14 (Redox II) and 15 (Transition Metals). Both topics are deeply interconnected: Topic 14 extends your understanding of oxidation, reduction and electrochemical cells, while Topic 15 applies these principles to the unique properties of d-block elements, complexes and their reactions. Mastering these areas is vital for the final examination and builds a firm foundation for further study in chemistry.
本综合复习指南涵盖 Edexcel A-Level 化学主题 14（氧化还原 II）和 15（过渡金属）的核心知识。这两个主题紧密相连：主题 14 拓展了氧化、还原和电化学电池的理解，主题 15 则将这些原理应用于 d 区元素的独特性质、配合物及其反应。掌握这些内容对应对最终考试至关重要，并为后续化学学习奠定坚实基础。
1. Oxidation Numbers & Redox Reactions | 氧化数与氧化还原反应
In redox chemistry, oxidation is defined as the loss of electrons, while reduction is the gain of electrons. An oxidising agent accepts electrons and is itself reduced; a reducing agent donates electrons and is itself oxidised. Oxidation numbers (oxidation states) are the bookkeeping system that lets us track electron transfer, and they are indispensable for compounds and polyatomic ions in which simple ionic charges cannot be assigned.
在氧化还原化学中，氧化被定义为失去电子，还原被定义为得到电子。氧化剂接受电子、自身被还原；还原剂给出电子、自身被氧化。氧化数（氧化态）是追踪电子转移的记账系统，对于无法直接分配离子电荷的化合物和多原子离子而言不可或缺。
The examination requires you to apply a fixed set of rules confidently and quickly.
考试要求你自信、迅速地运用一套固定规则。
- Uncombined element = 0 (e.g. Fe, O₂, P₄). | 单质中元素氧化数为 0（如 Fe、O₂、P₄）。
- A monatomic ion carries an oxidation number equal to its charge (Na⁺ = +1, Cl⁻ = −1). | 单原子离子的氧化数等于其所带电荷（Na⁺ = +1，Cl⁻ = −1）。
- Group 1 = +1, Group 2 = +2, Al = +3. | 第 1 族 = +1，第 2 族 = +2，Al = +3。
- Hydrogen = +1, except in metal hydrides such as NaH where it is −1. | 氢通常为 +1，但在 NaH 等金属氢化物中为 −1。
- Oxygen = −2, except in peroxides such as H₂O₂ where it is −1, and in OF₂ where it is +2. | 氧通常为 −2，但在 H₂O₂ 等过氧化物中为 −1，在 OF₂ 中为 +2。
- Fluorine is always −1 in its compounds. | 氟在其化合物中始终为 −1。
- The sum of oxidation numbers equals 0 in a neutral compound and equals the overall charge in a polyatomic ion. | 中性化合物中各氧化数之和为 0，多原子离子中则等于离子所带电荷。
−2, so x = +6; in S₂O₃²⁻, 2x + 3(−2) = −2, so x = +2; in H₂O₂, 2(+1) + 2x = 0, so x = −1.
必须能够熟练复现的例题：MnO₄⁻ 中 x + 4(−2) = −1，故 x = +7；Cr₂O₇²⁻ 中 2x + 7(−2) = −2，故 x = +6；S₂O₃²⁻ 中 2x + 3(−2) = −2，故 x = +2；H₂O₂ 中 2(+1) + 2x = 0，故 x = −1。
on, as in 2Cu⁺ → Cu + Cu²⁺ or in the decomposition of hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂. Displacement reactions, such as Zn + Cu²⁺ → Zn²⁺ + Cu, are redox reactions in which a more reactive metal reduces the ion of a less reactive metal.
两种特殊情况常被考查。歧化反应指同一元素在同一反应中同时被氧化和被还原，例如 2Cu⁺ → Cu + Cu²⁺，或过氧化氢的分解：2H₂O₂ → 2H₂O + O₂。置换反应如 Zn + Cu²⁺ → Zn²⁺ + Cu，是较活泼金属还原较不活泼金属离子的氧化还原反应。
2. Balancing Redox Equations & Half-Equations | 配平氧化还原方程式与半反应式
Redox equations are balanced using half-equations, and Edexcel expects you to construct these systematically both in acidic and in alkaline solution. The order of operations matters, so follow the same sequence every time.
氧化还原方程式通过半反应式配平，Edexcel 要求你在酸性溶液和碱性溶液中都能系统地构建它们。操作顺序非常重要，因此每次都遵循同一流程。
- Step 1: write the unbalanced half-equations for oxidation and reduction. | 第一步：写出未配平的氧化与还原半反应式。
- Step 2: balance all atoms except oxygen and hydrogen. | 第二步：配平除氧和氢以外的所有原子。
- Step 3: balance oxygen by adding H₂O. | 第三步：通过加入 H₂O 配平氧原子。
- Step 4: balance hydrogen by adding H⁺. | 第四步：通过加入 H⁺ 配平氢原子。
- Step 5: balance charge by adding electrons, e⁻. | 第五步：通过加入电子 e⁻ 配平电荷。
- Step 6: multiply the half-equations so the electrons cancel, then add them and cancel species appearing on both sides. | 第六步：将两个半反应式乘以适当系数使电子数相等，相加后消去两边都出现的物种。
For alkaline conditions, first balance as though the solution were acidic, then add OH⁻ ions to both sides in sufficient number to neutralise every H⁺, combine H⁺ and OH⁻ into H₂O, and cancel any surplus water molecules.
在碱性条件下，先按酸性条件配平，然后向两边加入足量的 OH⁻ 以中和所有 H⁺，将 H⁺ 与 OH⁻ 结合为 H₂O，并消去多余的水分子。
The two most heavily examined half-equations in acidic solution are:
酸性溶液中最常考的两个半反应式为：
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O E° = +1.51 V
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O E° = +1.33 V
Combining the manganate(VII) half-equation with the oxidation of iron(II) gives the classic titration equation:
将高锰酸根半反应式与铁(II)的氧化半反应式合并，得到经典滴定方程式：
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
The iodine–thiosulfate system is equally important, because it underpins the indirect titration of copper(II) and of many oxidising agents:
碘–硫代硫酸盐体系同样重要，因为它是铜(II)以及许多氧化剂间接滴定的基础：
I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
Note the change in oxidation number of sulfur from +2 in thiosulfate to +2.5 in tetrathionate, which shows that this is genuinely a redox process even though the numerical change is small.
注意硫的氧化数从硫代硫酸盐中的 +2 变为连四硫酸盐中的 +2.5，这说明尽管数值变化很小，这确实是一个氧化还原过程。
3. Electrochemical Cells & Standard Electrode Potentials | 电化学电池与标准电极电势
An electrochemical cell converts chemical energy into electrical energy by separating the two half-reactions of a redox process. The standard electrode potential, E°, of a half-cell is measured relative to the standard hydrogen electrode (SHE), which is arbitrarily assigned a value of exactly 0.00 V.
电化学电池通过把氧化还原过程的两半反应分开，将化学能转化为电能。半电池的标准电极电势 E° 是相对于标准氢电极（SHE）测得的，后者被人为规定为恰好 0.00 V。
The standard conditions are 298 K, a pressure of 100 kPa for any gas, and a concentration of 1.00 mol dm⁻³ for every aqueous ion. In the SHE, hydrogen gas at 100 kPa is bubbled over a platinised platinum electrode immersed in 1.00 mol dm⁻³ H⁺(aq). The equation is:
标准条件为 298 K、任何气体压强 100 kPa、每种水溶液离子浓度 1.00 mol dm⁻³。在标准氢电极中，100 kPa 的氢气通过镀铂铂电极，电极浸入 1.00 mol dm⁻³ H⁺(aq) 中。其方程式为：
2H⁺(aq) + 2e⁻ ⇌ H₂(g) E° = 0.00 V
Cell notation uses a single vertical line for a phase boundary and a double vertical line for the salt bridge. A platinum electrode is shown where a half-cell has no solid conductor.
电池符号中，单竖线表示相界面，双竖线表示盐桥。若某半电池中没有固体导体，则以铂电极表示。
Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)
The cell potential is always calculated in the same way, and the sign convention must be handled carefully:
电池电动势的计算方式始终一致，但正负号约定必须谨慎处理：
E°cell = E°(right-hand electrode) − E°(left-hand electrode)
E°cell = E°(reduction) − E°(oxidation) = +0.34 − (−0.76) = +1.10 V
The electrode with the more negative E° is the negative terminal and is the site of
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