📚 IB Chemistry: Redox Processes | IB化学：氧化还原反应过程
Redox (reduction–oxidation) reactions involve the transfer of electrons between chemical species. These processes are fundamental to chemistry, biology and industry, from cellular respiration to battery technology. In IB Chemistry, students learn to identify, balance and apply redox reactions through the systematic use of oxidation numbers and half equations.
氧化还原反应指化学物种之间发生电子转移的反应。此类过程是化学、生物及工业的基础，涉及细胞呼吸到电池技术等方方面面。在 IB 化学中，学生通过系统地使用氧化数和半反应方程来识别、配平及应用氧化还原反应。
1. Oxidation and Reduction Definitions | 氧化与还原的定义
Oxidation is the loss of electrons, while reduction is the gain of electrons. The mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain) helps remember the electron-transfer definitions. Oxidation can also be described as an increase in oxidation number, and reduction as a decrease in oxidation number.
氧化是失电子，还原是得电子。记忆口诀 OIL RIG（Oxidation Is Loss, Reduction Is Gain）有助于牢记电子转移的定义。氧化也可描述为氧化数升高，还原描述为氧化数降低。
For example, when iron(III) ions react with iodide ions, iron(III) is reduced to iron(II) and iodide is oxidised to iodine.
例如，铁(III)离子与碘离子反应时，铁(III)被还原为铁(II)，碘离子被氧化为碘单质。
2. Oxidation Numbers | 氧化数的确定
Oxidation numbers are assigned by a set of rules: all free elements have oxidation number 0; a monatomic ion has an oxidation number equal to its charge; fluorine is always –1 in compounds; oxygen is usually –2 (except in peroxides where it is –1, and in OF₂ where it is +2); hydrogen is +1 when bonded to non-metals and –1 when bonded to metals; the sum of oxidation numbers equals the overall charge of the molecule or ion.
氧化数的确定遵循一系列规则：单质中元素的氧化数为 0；单原子离子的氧化数等于其电荷；氟在化合物中恒为 –1；氧通常为 –2（但在过氧化物中为 –1，在 OF₂ 中为 +2）；氢与非金属结合时为 +1，与金属结合时为 –1；分子或离子的氧化数总和等于其总电荷。
For example, in the dichromate ion Cr₂O₇²⁻, the overall charge is –2. Since each oxygen is –2, the seven oxygen atoms contribute –14. Therefore, the two chromium atoms must sum to +12, so each chromium has oxidation number +6.
例如，在重铬酸根离子 Cr₂O₇²⁻ 中，总电荷为 –2。每个氧为 –2，七个氧原子共贡献 –14。因此，两个铬原子必须合计为 +12，故每个铬的氧化数为 +6。
3. Recognising Redox Reactions | 识别氧化还原反应
A reaction is redox if the oxidation numbers of elements change between reactants and products. For example, in the reaction between copper(II) oxide and hydrogen, CuO + H₂ → Cu + H₂O, copper is reduced (+2 to 0) and hydrogen is oxidised (0 to +1). Identifying these changes allows us to determine which species is oxidised and which is reduced.
如果反应前后元素的氧化数发生变化，则该反应是氧化还原反应。例如，氧化铜与氢气的反应 CuO + H₂ → Cu + H₂O 中，铜被还原（从 +2 到 0），氢被氧化（从 0 到 +1）。通过识别氧化数变化，可判断哪种物质被氧化、哪种被还原。
Some reactions that appear to involve atom transfer are also redox. For instance, in Fe₂O₃ + 3CO → 2Fe + 3CO₂, iron is reduced and carbon is oxidised, even though no simple electron transfer is visible at first glance.
一些看似涉及原子转移的反应同样是氧化还原反应。例如，Fe₂O₃ + 3CO → 2Fe + 3CO₂ 中，铁被还原、碳被氧化，尽管乍看之下没有明显的电子转移。
4. Oxidising and Reducing Agents | 氧化剂与还原剂
An oxidising agent is a species that causes oxidation by being reduced itself; it accepts electrons. A reducing agent causes reduction by being oxidised itself; it donates electrons. Examples of strong oxidising agents include O₂, halogens, Fe³⁺, MnO₄⁻ and Cr₂O₇²⁻. Common reducing agents include H₂, Na, I⁻ and Fe²⁺.
氧化剂是能使其他物质氧化的物种，其自身被还原（得到电子）。还原剂是能使其他物质还原的物种，其自身被氧化（失去电子）。强氧化剂的例子包括 O₂、卤素、Fe³⁺、MnO₄⁻ 和 Cr₂O₇²⁻。常见还原剂包括 H₂、Na、I⁻ 和 Fe²⁺。
In the reaction between acidified permanganate and iron(II) sulfate, MnO₄⁻ is reduced to Mn²⁺, so permanganate is the oxidising agent; Fe²⁺ is oxidised to Fe³⁺, so iron(II) is the reducing agent.
在酸化高锰酸钾与硫酸亚铁的反应中，MnO₄⁻ 被还原为 Mn²⁺，因此高锰酸根是氧化剂；Fe²⁺ 被氧化为 Fe³⁺，因此亚铁离子是还原剂。
5. Writing Half Equations | 书写半反应方程
Half equations show oxidation or reduction separately. To balance half equations in acidic medium: balance atoms other than H and O, then balance O by adding H₂O, balance H by adding H⁺, and finally balance charge by adding electrons. In basic medium, use OH⁻ and H₂O, or add H⁺ first and then neutralise with OH⁻.
半反应方程单独表示氧化或还原。在酸性介质中配平半反应：先平衡除 H 和 O 外的原子，然后用 H₂O 平衡 O，用 H⁺ 平衡 H，最后用电子 e⁻ 平衡电荷。在碱性介质中则使用 OH⁻ 和 H₂O，也可先加 H⁺ 再中和。
Example, reduction of permanganate to manganese(II) in acid:
例如，酸性条件下高锰酸根被还原为锰(II)离子：
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Example, oxidation of iodide to iodine:
例如，碘离子被氧化为碘单质：
2I⁻ → I₂ + 2e⁻
6. Combining Half Equations | 合并半反应方程
To build a full redox equation, combine the oxidation and reduction half equations after multiplying by appropriate factors so that the number of electrons lost equals the number gained. For example, the oxidation of Fe²⁺ to Fe³⁺ is Fe²⁺ → Fe³⁺ + e⁻. Multiplying this by 5 and adding it to the acidic permanganate half equation gives:
要构造完整的氧化还原方程式，需将氧化半反应和还原半反应乘以适当系数后相加，使失去电子数等于得到电子数。例如，Fe²⁺ 氧化为 Fe³⁺ 的半反应为 Fe²⁺ → Fe³⁺ + e⁻。将其乘以 5，再与酸性高锰酸根半反应相加，得：
5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
Check that atoms and charges are balanced: left side has 6 positive charges from 5Fe²⁺ and 8H⁺, plus 1 negative from MnO₄⁻, giving a net charge of +13; right side has 5Fe³⁺ (+15) and Mn²⁺ (+2), giving +17? Wait, careful: the net charge calculation should be +13 on both sides. Actually, the left net charge is 5(+2) + 8(+1) + 1(-1) = +17. Wait, 10+8-1=17. The right side: 5(+3) + 1(+2) = +17. Yes balanced. Need to correct this in text. Let’s write correctly.
检查原子和电荷是否守恒：左侧净电荷为 5×( +2) + 8×( +1) + 1×( –1) = +17；右侧为 5×( +3) + 1×( +2) = +17，电荷守恒。
In balanced equations, water molecules and H⁺ ions often appear as products or reactants depending on the acidity of the medium. In basic conditions, OH⁻ may be added to neutralise excess H⁺.
在配平方程中，水分子和 H⁺ 离子通常作为产物或反应物出现，具体取决于介质的酸碱性。在碱性条件下，可加入 OH⁻ 来中和多余的 H⁺。
7. Redox Titration and Stoichiometry | 氧化还原滴定与化学计量
Redox titrations use a known concentration of an oxidising or reducing agent to determine the amount of an analyte. Potassium permanganate is a common titrant because its purple colour disappears in the presence of reducing agents, so it acts as its own indicator. Starch is often used as an indicator for iodine titrations.
氧化还原滴定使用已知浓度的氧化剂或还原剂来测定分析物的含量。高锰酸钾是常用滴定剂，因为其紫色在还原剂存在时褪色，可作为自身指示剂。碘量法滴定中常使用淀粉作指示剂。
Stoichiometric calculations use the relationship n = c × V, then apply the mole ratio from the balanced redox equation. For example, 20.0 cm³ of 0.100 mol dm⁻³ iron(II) solution reacts with acidified permanganate according to the 5:1 ratio. The amount of Fe²⁺ is 0.00200 mol, so the amount of MnO₄⁻ is 0.000400 mol. If the titre volume is 10.0 cm³, the concentration of MnO₄⁻ is:
化学计量计算使用关系式 n = c × V，再应用配平氧化还原方程中物质的量的比例。例如，20.0 cm³ 的 0.100 mol dm⁻³ 亚铁溶液与酸化高锰酸钾按 5:1 比例反应。Fe²⁺ 的物质的量为 0.00200 mol，因此 MnO₄⁻ 的物质的量为 0.000400 mol。若滴定体积为 10.0 cm³，则 MnO₄⁻ 的浓度为：
c(MnO₄⁻) = 0.000400 mol / 0.0100 dm³ = 0.0400 mol dm⁻³
8. Electrochemical Cells | 电化学电池
Electrochemical cells convert chemical energy into electrical energy. In a galvanic cell, oxidation occurs at the anode (the negative electrode) and reduction occurs at the cathode (the positive electrode). The salt bridge completes the circuit by allowing ions to flow and maintains electrical neutrality in the half-cells.
电化学电池将化学能转化为电能。在原电池中，氧化发生在阳极（负极），还原发生在阴极（正极）。盐桥通过使离子流动来闭合电路，并维持各半电池中的电荷平衡。
The shorthand cell diagram follows the convention: anode half-cell on the left, cathode half-cell on the right, with a double vertical line for the salt bridge. For a zinc–copper cell, the diagram is:
电池简式的约定：阳极半电池写在左侧，阴极半电池写在右侧，双竖线代表盐桥。对于锌铜电池，简式为：
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
The electrode reaction at the zinc electrode is Zn(s) → Zn²⁺(aq) + 2e⁻; at the copper electrode, Cu²⁺(aq) + 2e⁻ → Cu(s).
锌电极上的电极反应为 Zn(s) → Zn²⁺(aq) + 2e⁻；铜电极上为 Cu²⁺(aq) + 2e⁻ → Cu(s)。
9. Standard Electrode Potentials | 标准电极电势
Standard electrode potential (E°) measures a half-cell’s tendency to be reduced under standard conditions (1 mol dm⁻³ solutions, 298 K, 1 atm gases). The more positive the E°, the stronger the oxidising agent. Standard hydrogen electrode (SHE) is assigned E° = 0.00 V and is used as the reference half-cell.
标准电极电势（E°）衡量半电池在标准条件（溶液 1 mol dm⁻³，298 K，气体 1 atm）下被还原的倾向。E° 越正，氧化能力越强。标准氢电极（SHE）的 E° 被规定为 0.00 V，作为参比半电池。
| Half reaction | E° / V |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | –0.76 |
| Cu²⁺(aq) + 2e⁻ → Cu(s) | +0.34 |
| MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) | +1.51 |
The overall cell potential is calculated as E°cell = E°cathode – E°anode. For the zinc–copper cell:
电池总电动势的计算式为 E°cell = E°cathode – E°anode。对于锌铜电池：
E°cell = (+0.34 V) – (–0.76 V) = +1.10 V
A positive E°cell indicates a spontaneous redox reaction under standard conditions.
若 E°cell 为正，则表明氧化还原反应在标准条件下可以自发进行。
10. Predicting Feasibility of Reactions | 预测反应能否进行
Using standard electrode potentials, we can predict whether a redox reaction will occur. Compare the two half reactions: the half reaction with the more positive E° will proceed as reduction, while the one with the less positive E° will proceed as oxidation. The resulting E°cell must be positive for the reaction to be feasible.
利用标准电极电势可以预测氧化还原反应能否发生。比较两个半反应：E° 较正的半反应发生还原，E° 较负的半反应发生氧化。最终计算得到的 E°cell 必须为正，反应才可行。
For example, acidified permanganate (E° = +1.51 V) can oxidise Fe²⁺ (Fe³⁺/Fe²⁺ E° = +0.77 V) because the overall cell potential is 1.51 – 0.77 = +0.74 V. Therefore, the reaction is spontaneous.
例如，酸化高锰酸根（E° = +1.51 V）能氧化 Fe²⁺（Fe³⁺/Fe²⁺ E° = +0.77 V），因为电池总电动势为 1.51 – 0.77 = +0.74 V，反应自发进行。
Keep in mind that a positive E° predicts thermodynamic feasibility, not the rate of reaction. Many thermodynamically feasible reactions are slow without a catalyst.
需要注意，E° 为正仅预测热力学可行性，并不代表反应速率。许多热力学可行的反应在无催化剂时速率很慢。
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