What Are the Mechanisms of Chemical Change?化学变化的机理是什么？
Every reaction is, at heart, a transfer or a sharing: protons hop between acids and bases, electrons transfer wholesale in redox chemistry, or electrons are shared — one at a time in radical mechanisms, or in pairs as nucleophiles and electrophiles trade curly arrows.每一个反应，归根结底都是某种"转移"或"共享"：质子在酸碱之间跳转、电子在氧化还原（redox）反应中整体转移，或者电子被共享——自由基机理中一次一个，亲核试剂与亲电试剂用弯箭头（curly arrow）成对交换电子对。
Guiding Question导引问题What determines the direction of proton transfer in an acid-base reaction? A proton always moves from the stronger acid to the stronger base, and the position of that transfer is quantified by the acid and base dissociation constants.是什么决定了酸碱反应中质子转移的方向？质子总是从较强的酸转移到较强的碱，而这一转移的"程度"可以用酸、碱的解离常数来定量描述。
Brønsted–Lowry Acids and Bases布朗斯特-劳里酸碱
A Brønsted–Lowry acid is a proton (H⁺) donor; a Brønsted–Lowry base is a proton acceptor. Every acid-base reaction is a proton transfer, and it generates a conjugate acid-base pair on each side: the acid becomes its conjugate base (having lost H⁺), and the base becomes its conjugate acid (having gained H⁺).
HCl / Cl⁻ is one conjugate pair; H₂O / H₃O⁺ is the other. A conjugate acid always has one more H (and one higher charge) than its conjugate base.HCl / Cl⁻ 是一对共轭酸碱；H₂O / H₃O⁺ 是另一对。共轭酸总是比其共轭碱多一个 H（电荷也高一个单位）。
Amphiprotic Species两性质子物种An amphiprotic species can act as either a proton donor or a proton acceptor depending on what it reacts with. Key examples: $\mathrm{H_2O}$ (donates H⁺ to a base, accepts H⁺ from an acid) and $\mathrm{HCO_3^-}$ (can lose a proton to form $\mathrm{CO_3^{2-}}$, or gain one to form $\mathrm{H_2CO_3}$). Don't confuse "amphiprotic" (about H⁺ transfer specifically) with the broader term "amphoteric" (can react as both acid and base, including non-proton mechanisms, e.g. $\mathrm{Al_2O_3}$).两性质子（amphiprotic）物种既能作质子给体，也能作质子受体，具体取决于与之反应的对象。典型例子：$\mathrm{H_2O}$（对碱给出 H⁺，对酸接受 H⁺）以及 $\mathrm{HCO_3^-}$（可以失去质子生成 $\mathrm{CO_3^{2-}}$，也可以得到质子生成 $\mathrm{H_2CO_3}$）。注意区分"两性质子"（专指 H⁺ 转移）与更广义的"两性"（amphoteric，可通过非质子机理表现出酸碱两重性，如 $\mathrm{Al_2O_3}$）。
Strong vs Weak Acids and Bases强酸/强碱 vs 弱酸/弱碱
Strong acids and bases dissociate completely in water; weak ones dissociate only partially, setting up an equilibrium between the intact molecule and its ions. This is a statement about the extent of dissociation, not about concentration — a dilute strong acid can still have a lower pH range effect than a concentrated weak acid in terms of total acidic protons available, but its dissociation is always complete.
The pH Scale & the Ionic Product of WaterpH 标度与水的离子积
The pH scale compresses the huge range of $\mathrm{[H^+]}$ values into a manageable number. Water itself self-ionises very slightly, and the product $\mathrm{[H^+][OH^-]}$ — the ionic product of water, $K_w$ — is constant at a given temperature.
$K_w$ increases with temperature (water's self-ionisation is endothermic), so neutral pH is only exactly 7 at 298 K.$K_w$ 随温度升高而增大（水的自电离是吸热过程），因此中性 pH 恰好为 7 只在 298 K 时成立。
Worked Example — pH of Strong Acids & Bases例题 — 强酸、强碱的 pH
Find (a) the pH of $0.0100~\mathrm{mol\,dm^{-3}}$ HCl, and (b) the pH of $0.00500~\mathrm{mol\,dm^{-3}}$ NaOH, both at 298 K.求 (a) $0.0100~\mathrm{mol\,dm^{-3}}$ HCl 的 pH；(b) $0.00500~\mathrm{mol\,dm^{-3}}$ NaOH 的 pH（均在 298 K）。
Both answers land on the expected side of 7 (acid below, base above) and neither exceeds the 0–14 range.两个结果都落在 7 的预期一侧（酸小于 7，碱大于 7），且都不超出 0–14 的范围。
Reactions of Acids酸的典型反应
Acids react with several classes of substance in characteristic ways; all are, at their core, proton-transfer reactions (even the "redox-looking" reaction with metals, where H⁺ is ultimately reduced to H₂).
Neutralization & Salts中和反应与盐Neutralization is the reaction of an acid with a base to produce a salt and (usually) water: $\mathrm{acid + base \to salt + water}$. The identity of the salt depends on both the acid (which anion) and the base (which cation).中和反应（neutralization）指酸与碱反应生成盐和（通常还有）水：$\mathrm{酸 + 碱 \to 盐 + 水}$。盐的组成取决于酸提供的阴离子和碱提供的阳离子。
In the reaction $\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$, which species is the conjugate acid of $\mathrm{NH_3}$?在反应 $\mathrm{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-}$ 中，哪个物种是 $\mathrm{NH_3}$ 的共轭酸？
$\mathrm{H_2O}$
$\mathrm{NH_4^+}$
$\mathrm{OH^-}$
$\mathrm{NH_3}$
Correct! $\mathrm{NH_3}$ acts as a base (accepts H⁺) and becomes $\mathrm{NH_4^+}$ — its conjugate acid, one proton heavier.正确！$\mathrm{NH_3}$ 在此作碱（接受 H⁺），变为 $\mathrm{NH_4^+}$——即其共轭酸，比原来多一个质子。
The conjugate acid of a base is what forms when that base gains H⁺. $\mathrm{NH_3}$ gains H⁺ to form $\mathrm{NH_4^+}$. Answer: (B).碱的共轭酸是该碱获得 H⁺ 后生成的物种。$\mathrm{NH_3}$ 得到 H⁺ 生成 $\mathrm{NH_4^+}$。答案：(B)。
A solution at 298 K has $\mathrm{pOH} = 4.00$. What is $\mathrm{[H^+]}$?298 K 时某溶液 $\mathrm{pOH} = 4.00$，求 $\mathrm{[H^+]}$。
K_a, K_b and Their Logarithmic Forms$K_a$、$K_b$ 及其对数形式HL
The acid dissociation constant $K_a$ quantifies exactly how far a weak acid dissociates; the base dissociation constant $K_b$ does the same for a weak base. Because these span many orders of magnitude, $\mathrm{p}K_a$ and $\mathrm{p}K_b$ (their negative logs) are used in practice — smaller $\mathrm{p}K_a$ means a stronger acid.
For a conjugate acid-base pair at 298 K: $\mathrm{p}K_a + \mathrm{p}K_b = 14$. A stronger acid ($K_a$ large, $\mathrm{p}K_a$ small) always has a weaker conjugate base, and vice versa.对于同一共轭酸碱对，在 298 K 时：$\mathrm{p}K_a + \mathrm{p}K_b = 14$。较强的酸（$K_a$ 大、$\mathrm{p}K_a$ 小）其共轭碱一定较弱，反之亦然。
pH of a Weak Acid from K_a由 $K_a$ 求弱酸的 pHHL
For a weak acid of initial concentration $c$, the standard approximation assumes dissociation is small enough that the equilibrium concentration of undissociated acid is still $\approx c$:
Valid when $c/K_a \gtrsim 400$ (dissociation < ~5%). Otherwise, solve the full quadratic from the $K_a$ expression.当 $c/K_a \gtrsim 400$（解离度 < 约 5%）时适用。否则需要用 $K_a$ 表达式解完整的一元二次方程。
Worked Example — pH of a Weak Acid例题 — 弱酸的 pH
Find the pH of $0.100~\mathrm{mol\,dm^{-3}}$ ethanoic acid, $K_a = 1.8\times10^{-5}$.求 $0.100~\mathrm{mol\,dm^{-3}}$ 乙酸溶液的 pH，$K_a = 1.8\times10^{-5}$。
Check the approximation检验近似条件
$c/K_a = 0.100 / 1.8\times10^{-5} \approx 5\,600 \gg 400$, so the approximation is valid.$c/K_a = 0.100 / 1.8\times10^{-5} \approx 5\,600 \gg 400$，近似条件成立。
Much higher than pH 1.00 (what a strong acid of the same concentration would give) — consistent with only partial dissociation.远高于同浓度强酸的 pH 1.00——与只有部分解离相符。
Buffer Solutions缓冲溶液HL
A buffer solution resists changes in pH when small amounts of acid or base are added. An acidic buffer is made from a weak acid and a salt of its conjugate base (e.g. $\mathrm{CH_3COOH / CH_3COO^-}$); a basic buffer from a weak base and a salt of its conjugate acid (e.g. $\mathrm{NH_3 / NH_4^+}$).
How Buffer Action Works缓冲作用的原理Both the weak acid (HA) and its conjugate base (A⁻) are present in significant amounts. Added H⁺ is mopped up by A⁻ (forming HA); added OH⁻ is neutralised by HA (forming A⁻ + H₂O). Because both reservoirs are large relative to the amount added, the ratio $[\mathrm{A^-}]/[\mathrm{HA}]$ — and hence the pH — barely shifts.溶液中同时存在大量弱酸（HA）及其共轭碱（A⁻）。加入的 H⁺ 被 A⁻ 消耗（生成 HA）；加入的 OH⁻ 被 HA 中和（生成 A⁻ + H₂O）。由于两个"储备池"相对于加入量都很大，比值 $[\mathrm{A^-}]/[\mathrm{HA}]$——从而 pH——几乎不发生变化。
Close to $\mathrm{p}K_a$ (4.74) as expected, since $[\mathrm{A^-}]$ and $[\mathrm{HA}]$ are comparable in magnitude — a well-poised buffer.结果接近 $\mathrm{p}K_a$（4.74），符合预期，因为 $[\mathrm{A^-}]$ 与 $[\mathrm{HA}]$ 数量级相当——是一个配比良好的缓冲液。
Acid-Base Titration Curves酸碱滴定曲线HL
Plotting pH against volume of titrant added produces one of four characteristic shapes, depending on the strength of the acid and base involved. The equivalence point is where moles of acid = moles of base (stoichiometric neutralisation); for a weak/strong pairing this is not at pH 7. The half-equivalence point (half the volume needed to reach equivalence) is special: at that point $[\mathrm{HA}] = [\mathrm{A^-}]$, so $\mathrm{pH} = \mathrm{p}K_a$.
Buffer region before a smaller jump; half-eqv. pH = p$K_a$
Strong acid + weak base
Very low
< 7
Smaller jump; buffer region after
Weak acid + weak base
Moderate
≈ 7 (depends on p$K_a$/p$K_b$)
No sharp jump — gradual S-shape
滴定类型
初始 pH
等当点 pH
曲线形状
强酸 + 强碱
很低
= 7
陡峭且较长的垂直跃升
弱酸 + 强碱
中等
> 7
跃升前有缓冲区；半等当点 pH = $\mathrm{p}K_a$
强酸 + 弱碱
很低
< 7
跃升较小；跃升后有缓冲区
弱酸 + 弱碱
中等
≈ 7（取决于 $\mathrm{p}K_a$/$\mathrm{p}K_b$）
无明显跃升——平缓的 S 形
Choosing an Indicator指示剂的选择An indicator's colour change occurs over a pH range roughly $\mathrm{p}K_{\mathrm{In}} \pm 1$. Choose an indicator whose colour-change range falls within the vertical jump of the titration curve, ideally with $\mathrm{p}K_{\mathrm{In}}$ close to the equivalence pH. This is why weak acid/weak base titrations (no sharp jump) have no suitable single indicator — a pH meter is needed instead.指示剂的变色范围大约是 $\mathrm{p}K_{\mathrm{In}} \pm 1$。应选择变色范围落在滴定曲线垂直跃升段内的指示剂，理想情况下 $\mathrm{p}K_{\mathrm{In}}$ 接近等当点 pH。这也是为什么弱酸/弱碱滴定（无明显跃升）找不到合适的单一指示剂——只能改用 pH 计。
Salt Hydrolysis盐的水解HL
Salts formed from a strong acid and a weak base give an acidic solution (the cation, e.g. $\mathrm{NH_4^+}$, is itself a weak acid). Salts formed from a weak acid and a strong base give a basic solution (the anion, e.g. $\mathrm{CH_3COO^-}$, is itself a weak base). Salts of a strong acid and strong base (e.g. NaCl) are neutral — neither ion hydrolyses.
(HL) A weak acid has $\mathrm{p}K_a = 3.75$. What is $\mathrm{p}K_b$ of its conjugate base at 298 K?（HL）某弱酸 $\mathrm{p}K_a = 3.75$。求其共轭碱在 298 K 时的 $\mathrm{p}K_b$。
Use $\mathrm{p}K_a + \mathrm{p}K_b = 14$ for a conjugate pair at 298 K: $\mathrm{p}K_b = 14 - 3.75 = 10.25$. Answer: (A).对同一共轭对，298 K 时 $\mathrm{p}K_a + \mathrm{p}K_b = 14$：$\mathrm{p}K_b = 14 - 3.75 = 10.25$。答案：(A)。
(HL) In a titration of a weak acid with a strong base, the pH at the half-equivalence point equals（HL）用强碱滴定弱酸时，半等当点处的 pH 等于
7.00
the equivalence-point pH等当点的 pH
$\mathrm{p}K_a$ of the weak acid（弱酸的）
0.00
Correct! At the half-equivalence point, exactly half the weak acid has been converted to its conjugate base, so $[\mathrm{HA}] = [\mathrm{A^-}]$ and the Henderson–Hasselbalch equation collapses to $\mathrm{pH} = \mathrm{p}K_a$.正确！在半等当点处，恰好一半的弱酸转化为其共轭碱，故 $[\mathrm{HA}] = [\mathrm{A^-}]$，此时亨德森-哈塞尔巴赫方程简化为 $\mathrm{pH} = \mathrm{p}K_a$。
At half-equivalence, $[\mathrm{A^-}]/[\mathrm{HA}] = 1$, so $\log_{10}(1) = 0$ and $\mathrm{pH} = \mathrm{p}K_a + 0 = \mathrm{p}K_a$. Answer: (C).在半等当点，$[\mathrm{A^-}]/[\mathrm{HA}] = 1$，故 $\log_{10}(1) = 0$，$\mathrm{pH} = \mathrm{p}K_a + 0 = \mathrm{p}K_a$。答案：(C)。
Reactivity 3.2
Electron Transfer Reactions电子转移反应
Guiding Question导引问题What causes electron transfer in a redox reaction, and how can it be harnessed to do useful work? Electrons flow spontaneously from a species that holds them less tightly to one that holds them more tightly — and if that flow is forced through an external circuit, it becomes electricity.是什么驱使氧化还原（redox）反应中的电子转移，又如何利用它做有用功？电子会自发地从束缚较弱的物种流向束缚较强的物种——如果强制这股电子流经外部电路，就能转化为电能。
Oxidation States & the Language of Redox氧化态与氧化还原的语言
An oxidation state (oxidation number) tracks how many electrons an atom has "gained" or "lost" relative to the free element, assuming fully ionic bonding. Oxidation is loss of electrons (oxidation state increases); reduction is gain of electrons (oxidation state decreases) — remembered by the mnemonic OIL RIG.
氧化态（oxidation state，又称氧化数）追踪某原子相对于单质状态"得到"或"失去"了多少电子（假设全部为离子键）。氧化是失去电子（氧化态升高）；还原是得到电子（氧化态降低）——可用口诀 OIL RIG（Oxidation Is Loss, Reduction Is Gain）记忆。
Oxidizing & Reducing Agents氧化剂与还原剂The oxidizing agent is the species that gets reduced (it causes oxidation in the other species by accepting its electrons). The reducing agent is the species that gets oxidized (it causes reduction by donating electrons). Every redox reaction has both, working simultaneously.氧化剂是被还原的物种（它接受电子，从而使另一物种被氧化）。还原剂是被氧化的物种（它给出电子，从而使另一物种被还原）。每一个氧化还原反应中，两者总是同时存在、同时发生作用。
Half-Equations & Balancing Redox in Acidic Solution半反应方程式与酸性介质中的配平
A redox equation can be split into two half-equations — one for oxidation, one for reduction — each balanced separately for atoms, then charge, before being combined so electrons cancel exactly.
Balancing Recipe (Acidic Solution)配平步骤（酸性介质）1. Balance the atom that changes oxidation state. 2. Balance O by adding $\mathrm{H_2O}$. 3. Balance H by adding $\mathrm{H^+}$. 4. Balance charge by adding electrons ($e^-$). 5. Scale the two half-equations so electrons cancel, then add.1. 先配平氧化态改变的原子。2. 用 $\mathrm{H_2O}$ 配平 O。3. 用 $\mathrm{H^+}$ 配平 H。4. 加电子（$e^-$）配平电荷。5. 将两个半反应按比例放大使电子数相等，再相加。
The reactivity (activity) series ranks metals by how readily they lose electrons. A more reactive metal will displace a less reactive one from solution (e.g. $\mathrm{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}$). Redox titrations (like $\mathrm{MnO_4^-}$ against $\mathrm{Fe^{2+}}$, self-indicating by the disappearance of purple colour) let you quantify the amount of a reducing or oxidizing agent in solution using the stoichiometry of the balanced half-equations.
A voltaic cell converts spontaneous chemical energy into electrical energy by physically separating the oxidation and reduction half-reactions, forcing electrons to travel through an external wire.
A salt bridge (often a KNO₃ or KCl agar gel) completes the internal circuit and maintains electrical neutrality in both half-cells by letting ions migrate without the two solutions mixing directly. Cell diagram notation lists the anode half-cell on the left, cathode on the right, with a single vertical line ( | ) for a phase boundary and a double line ( ‖ ) for the salt bridge:
Zn (anode, oxidised) on the left; Cu (cathode, reduced) on the right. Electrons flow through the external wire from Zn to Cu.左边为 Zn（阳极，被氧化）；右边为 Cu（阴极，被还原）。电子经外部导线由 Zn 流向 Cu。
Worked Example — Balancing a Redox Equation例题 — 配平氧化还原方程式
Balance the reaction of acidified dichromate with iodide: $\mathrm{Cr_2O_7^{2-} + I^- \to Cr^{3+} + I_2}$ (acidic solution).配平酸性条件下重铬酸根与碘离子的反应：$\mathrm{Cr_2O_7^{2-} + I^- \to Cr^{3+} + I_2}$（酸性溶液）。
In the reaction $\mathrm{Mg(s) + 2HCl(aq) \to MgCl_2(aq) + H_2(g)}$, what is the oxidizing agent?反应 $\mathrm{Mg(s) + 2HCl(aq) \to MgCl_2(aq) + H_2(g)}$ 中，氧化剂是什么？
$\mathrm{Mg}$
$\mathrm{H^+}$
$\mathrm{Cl^-}$
$\mathrm{H_2}$
Correct! H⁺ is reduced (from +1 in HCl to 0 in H₂), so it accepts electrons from Mg — making it the oxidizing agent (it oxidizes Mg while itself being reduced).正确！H⁺ 被还原（从 HCl 中的 +1 变为 H₂ 中的 0），它从 Mg 接受电子——因此是氧化剂（它使 Mg 被氧化，自身被还原）。
The oxidizing agent is the species that is itself reduced. Mg goes from 0 to +2 (oxidized, so Mg is the reducing agent); H goes from +1 to 0 (reduced). Answer: (B).氧化剂是自身被还原的物种。Mg 从 0 变为 +2（被氧化，故 Mg 是还原剂）；H 从 +1 变为 0（被还原）。答案：(B)。
In a voltaic cell, which statement about the anode is correct?在原电池中，关于阳极下列说法正确的是？
Oxidation occurs there, and it is the negative electrode.氧化在此发生，且它是负极。
Reduction occurs there, and it is the positive electrode.还原在此发生，且它是正极。
Oxidation occurs there, and it is the positive electrode.氧化在此发生，且它是正极。
No electron transfer occurs there.此处不发生电子转移。
Correct! In a voltaic (galvanic) cell, the anode is where oxidation occurs and electrons are released into the external circuit — making it the negative terminal.正确！在原电池中，阳极是氧化发生、电子释放进入外电路之处——因此它是负极。
Remember "An Ox, Red Cat": Anode = Oxidation, Cathode = Reduction. In a voltaic cell (spontaneous), the anode is negative. Answer: (A).记住 "An Ox, Red Cat"：阳极（Anode）= 氧化（Oxidation），阴极（Cathode）= 还原（Reduction）。在原电池（自发）中，阳极为负极。答案：(A)。
Standard Electrode Potentials标准电极电势HL
The standard hydrogen electrode (SHE) — $\mathrm{H^+(aq, 1\,mol\,dm^{-3}) \mid H_2(g, 100\,kPa) \mid Pt(s)}$ — is defined as $E^{\ominus} = 0.00~\mathrm{V}$ by convention. Every other standard electrode potential $E^{\ominus}$ is measured relative to it, as the reduction potential of a half-cell under standard conditions (298 K, 100 kPa, 1 mol dm⁻³).
Reading E° Tables解读 $E^{\ominus}$ 表All values in the data booklet are written as reduction half-equations. A more positive $E^{\ominus}$ means a greater tendency to be reduced (a stronger oxidizing agent); a more negative $E^{\ominus}$ means a greater tendency for the reverse (oxidation) to occur (a stronger reducing agent).数据手册中所有数值均按还原半反应书写。$E^{\ominus}$ 越正，表示越容易被还原（氧化能力越强）；$E^{\ominus}$ 越负，表示逆反应（氧化）越容易发生（还原能力越强）。
Both electrode potentials are taken as written (reduction values). $F = 96\,500~\mathrm{C\,mol^{-1}}$ (Faraday constant); $n$ = moles of electrons transferred. A spontaneous cell reaction requires $E^{\ominus}_{\text{cell}} > 0$, equivalently $\Delta G^{\ominus} < 0$.两个电极电势均按原始（还原）数值代入。$F = 96\,500~\mathrm{C\,mol^{-1}}$（法拉第常数）；$n$ 为转移的电子摩尔数。自发的电池反应要求 $E^{\ominus}_{\text{cell}} > 0$，等价于 $\Delta G^{\ominus} < 0$。
Worked Example — Cell Potential & Spontaneity例题 — 电池电势与自发性
Given $E^{\ominus}(\mathrm{Cu^{2+}/Cu}) = +0.34~\mathrm{V}$ and $E^{\ominus}(\mathrm{Zn^{2+}/Zn}) = -0.76~\mathrm{V}$, find $E^{\ominus}_{\text{cell}}$ for the Daniell cell and confirm it is spontaneous.已知 $E^{\ominus}(\mathrm{Cu^{2+}/Cu}) = +0.34~\mathrm{V}$，$E^{\ominus}(\mathrm{Zn^{2+}/Zn}) = -0.76~\mathrm{V}$，求丹尼尔电池的 $E^{\ominus}_{\text{cell}}$，并确认其自发性。
Identify cathode/anode判断阴、阳极
The more positive $E^{\ominus}$ is reduced (cathode): Cu²⁺/Cu. The more negative is oxidized (anode): Zn²⁺/Zn.$E^{\ominus}$ 较正者被还原（阴极）：Cu²⁺/Cu。较负者被氧化（阳极）：Zn²⁺/Zn。
$E^{\ominus}_{\text{cell}} > 0$, so $\Delta G^{\ominus} = -nFE^{\ominus}_{\text{cell}} < 0$ — the reaction is spontaneous as written, matching everyday experience with a Daniell cell.$E^{\ominus}_{\text{cell}} > 0$，故 $\Delta G^{\ominus} = -nFE^{\ominus}_{\text{cell}} < 0$——按此书写方向的反应是自发的，与丹尼尔电池的实际表现一致。
Electrolysis电解HL
Electrolysis uses an external power source to force a non-spontaneous redox reaction to occur ($E^{\ominus}_{\text{cell}} < 0$ becomes possible by supplying electrical energy). In an electrolytic cell the electrode polarities flip relative to a voltaic cell: the anode is positive (still oxidation) and the cathode is negative (still reduction) — because the external supply is now driving the electron flow rather than the cell itself.
Predicting Electrolysis Products in Aqueous Solution预测水溶液电解产物At the cathode, whichever species is easiest to reduce wins — compare $E^{\ominus}$ values for the metal ion vs. water ($\mathrm{2H_2O + 2e^- \to H_2 + 2OH^-}$, $E^{\ominus} = -0.83~\mathrm{V}$). At the anode, whichever is easiest to oxidize wins, though concentrated halide solutions often give the halogen for kinetic reasons even when the thermodynamic prediction favours O₂.在阴极，最容易被还原的物种优先反应——比较金属离子与水（$\mathrm{2H_2O + 2e^- \to H_2 + 2OH^-}$，$E^{\ominus} = -0.83~\mathrm{V}$）的 $E^{\ominus}$。在阳极，最容易被氧化的物种优先反应，但高浓度卤化物溶液常因动力学原因优先生成卤素单质，即使热力学预测应生成 O₂。
Faraday's Law — Quantitative Electrolysis法拉第定律 — 电解的定量计算HL
The amount of charge passed through an electrolytic cell determines exactly how much product forms, via the number of moles of electrons transferred.
通过电解池的电荷量，经由转移的电子摩尔数，精确决定了生成产物的量。
Faraday's Law (data booklet)法拉第定律（数据手册）
$$Q = It \qquad n(e^-) = \dfrac{Q}{F}$$
$Q$ = charge (C); $I$ = current (A); $t$ = time (s); $F = 96\,500~\mathrm{C\,mol^{-1}}$. Combine with the half-equation's electron stoichiometry to find moles (then mass or volume) of product.$Q$ = 电荷量（C）；$I$ = 电流（A）；$t$ = 时间（s）；$F = 96\,500~\mathrm{C\,mol^{-1}}$。结合半反应中电子的化学计量数，即可求出产物的摩尔数（进而求质量或体积）。
Worked Example — Mass Deposited by Electrolysis例题 — 电解析出的质量
A current of 2.50 A is passed through molten $\mathrm{CuCl_2}$ for 30.0 minutes. Find the mass of copper deposited at the cathode ($M_r(\mathrm{Cu}) = 63.55$).用 2.50 A 电流电解熔融 $\mathrm{CuCl_2}$，持续 30.0 分钟。求阴极析出铜的质量（$M_r(\mathrm{Cu}) = 63.55$）。
(HL) Given $E^{\ominus}(\mathrm{Ag^+/Ag}) = +0.80~\mathrm{V}$ and $E^{\ominus}(\mathrm{Fe^{2+}/Fe}) = -0.44~\mathrm{V}$, what is $E^{\ominus}_{\text{cell}}$ for a cell built from these two half-cells?（HL）已知 $E^{\ominus}(\mathrm{Ag^+/Ag}) = +0.80~\mathrm{V}$，$E^{\ominus}(\mathrm{Fe^{2+}/Fe}) = -0.44~\mathrm{V}$，由这两个半电池组成的电池 $E^{\ominus}_{\text{cell}}$ 是多少？
$-1.24~\mathrm{V}$
$+0.36~\mathrm{V}$
$-0.36~\mathrm{V}$
$+1.24~\mathrm{V}$
Correct! Ag⁺/Ag (more positive) is the cathode; Fe²⁺/Fe is the anode. $E^{\ominus}_{\text{cell}} = (+0.80) - (-0.44) = +1.24~\mathrm{V}$.正确！Ag⁺/Ag（更正）为阴极；Fe²⁺/Fe 为阳极。$E^{\ominus}_{\text{cell}} = (+0.80) - (-0.44) = +1.24~\mathrm{V}$。
(HL) In the electrolysis of concentrated aqueous NaCl (brine), what forms at the anode?（HL）电解浓 NaCl 水溶液（盐水）时，阳极生成什么？
$\mathrm{O_2}$
$\mathrm{Cl_2}$
$\mathrm{Na}$
$\mathrm{H_2}$
Correct! Although water is thermodynamically easier to oxidise than Cl⁻, the high concentration of Cl⁻ in brine means Cl₂ is produced in practice — a classic case where concentration (kinetics) overrides the standard-potential prediction.正确！虽然从热力学角度水比 Cl⁻ 更容易被氧化，但盐水中 Cl⁻ 浓度很高，实际上会生成 Cl₂——这是浓度（动力学）因素压过标准电势预测的经典例子。
Na⁺ is never discharged in aqueous solution (water is always reduced first at the cathode). At the anode of concentrated brine, Cl₂ forms despite O₂ being the thermodynamically favoured product. Answer: (B).Na⁺ 在水溶液中绝不会被放电（阴极总是水优先被还原）。在浓盐水的阳极，尽管 O₂ 是热力学更有利的产物，实际生成的是 Cl₂。答案：(B)。
Reactivity 3.3
Electron Sharing Reactions电子共享反应
Guiding Question导引问题What happens when a covalent bond breaks symmetrically? Instead of one atom taking both bonding electrons (as in ionic-style heterolytic fission), each atom keeps one — producing highly reactive radicals that drive chain reactions.当共价键对称断裂时会发生什么？与异裂（一个原子拿走全部两个成键电子）不同，此时每个原子各保留一个电子——生成高活性的自由基，从而引发链式反应。
Radicals & Homolytic Fission自由基与均裂
A radical is a species with an unpaired electron, usually written with a single dot ($\mathrm{X}\!\cdot$). Radicals form by homolytic fission — a covalent bond breaks so that each fragment keeps one electron from the shared pair. This is different from heterolytic fission, where both electrons go to one fragment, producing ions. Homolytic fission is typically triggered by energy input such as UV light or heat, and is shown with a "fish-hook" single-barbed curly arrow (half-headed) representing the movement of a single electron.
Homolytic vs Heterolytic Fission均裂 vs 异裂Homolytic: $\mathrm{X{-}Y \to X^{\bullet} + Y^{\bullet}}$ (each atom keeps one electron; produces two radicals). Heterolytic: $\mathrm{X{-}Y \to X^+ + Y^-}$ (one atom keeps both electrons; produces two ions). Homolytic fission is favoured in non-polar environments (e.g. the gas phase, UV-initiated reactions); heterolytic fission dominates in polar, ionic mechanisms (see Reactivity 3.4).均裂：$\mathrm{X{-}Y \to X^{\bullet} + Y^{\bullet}}$（每个原子各保留一个电子，生成两个自由基）。异裂：$\mathrm{X{-}Y \to X^+ + Y^-}$（一个原子保留两个电子，生成两个离子）。均裂多发生在非极性环境中（如气相、紫外引发反应）；异裂则主导极性、离子型机理（见 Reactivity 3.4）。
Free-Radical Substitution of Alkanes烷烃的自由基取代反应
Alkanes are otherwise unreactive, but under UV light they undergo free-radical substitution with halogens (e.g. methane + chlorine). The mechanism proceeds through three distinct phases.
UV photons homolytically cleave the weak Cl–Cl bond, generating two chlorine radicals. This is the only step that consumes energy from outside the chain.紫外光子使较弱的 Cl–Cl 键均裂，生成两个氯自由基。这是唯一从链外部消耗能量的一步。
Each propagation step consumes one radical and produces another — the radical count is conserved, so the cycle repeats many times before termination (a "chain reaction").每个链增长步骤都消耗一个自由基、生成另一个——自由基数目守恒，因此该循环会重复很多次才发生终止（即"链式反应"）。
Two radicals combine to form a stable, electron-paired molecule, removing radicals from the system and ending that chain.两个自由基结合生成稳定的电子配对分子，从体系中移除自由基，终止该链式反应。
Why a Mixture of Products Forms为什么会生成多种产物的混合物Because termination is a random radical-radical collision, and propagation can abstract any hydrogen on the chain (or substitute further once some H atoms are already replaced), free-radical substitution never gives one clean product. Excess $\mathrm{Cl_2}$ drives further substitution ($\mathrm{CH_3Cl \to CH_2Cl_2 \to CHCl_3 \to CCl_4}$), and termination products like $\mathrm{C_2H_6}$ or $\mathrm{Cl_2}$ appear as minor side products. This lack of selectivity is a key limitation of the mechanism (contrast with the much more controllable ionic mechanisms of Reactivity 3.4).由于链终止是随机的自由基-自由基碰撞，而链增长可以夺取链上任意一个氢原子（一旦部分 H 已被取代，还可以继续取代），自由基取代反应从不生成单一纯净产物。过量的 $\mathrm{Cl_2}$ 会驱动进一步取代（$\mathrm{CH_3Cl \to CH_2Cl_2 \to CHCl_3 \to CCl_4}$），而终止产物如 $\mathrm{C_2H_6}$ 或 $\mathrm{Cl_2}$ 会作为少量副产物出现。这种选择性的缺失是该机理的一个关键局限（与 Reactivity 3.4 中可控性高得多的离子型机理形成对比）。
Worked Example — Full Mechanism for Methane + Chlorine例题 — 甲烷与氯气反应的完整机理
Write the initiation, propagation, and termination steps for the free-radical substitution of methane by chlorine to form chloromethane, and classify each step.写出甲烷经自由基取代生成氯甲烷的链引发、链增长、链终止步骤，并说明每一步的类型。
Any two radicals combining removes them from the chain — this is one of three possible termination pairings.任意两个自由基结合都会将它们从链中移除——这是三种可能的终止配对之一。
Which step in free-radical substitution both consumes and regenerates a radical, keeping the chain going?自由基取代反应中，哪一步既消耗又再生自由基，从而维持链式反应？
Initiation链引发
Termination链终止
Propagation链增长
None of these steps regenerate a radical以上均不再生自由基
Correct! Each propagation step consumes one radical as a reactant and produces exactly one radical as a product, so the total radical count is conserved and the chain can repeat.正确！每个链增长步骤都以一个自由基作反应物，同时生成恰好一个自由基作产物，因此自由基总数守恒，链式反应得以重复。
Initiation creates the first radicals (0 → 2); termination destroys radicals (2 → 0 stable molecules). Only propagation conserves the radical count while advancing the chain. Answer: (C).链引发生成最初的自由基（0 → 2）；链终止消灭自由基（2 → 0，生成稳定分子）。只有链增长在推进反应的同时守恒自由基数目。答案：(C)。
Why does free-radical substitution of an alkane typically give a mixture of mono-, di-, and further-substituted products rather than a single clean product?为什么烷烃的自由基取代反应通常生成一混合物（单取代、双取代及更多取代产物），而非单一纯净产物？
The reaction is under thermodynamic control, favouring the most stable product.反应受热力学控制，偏向生成最稳定的产物。
Propagation can abstract any available H atom (including those on already-substituted carbons), and termination is a random radical-radical collision.链增长可以夺取任何可及的 H 原子（包括已被取代碳上的 H），且链终止是随机的自由基碰撞。
Only initiation determines the product distribution.只有链引发决定产物分布。
Chlorine radicals only ever react once.氯自由基只反应一次。
Correct! Propagation has no way to "choose" a particular hydrogen selectively, and once some product forms it can itself be attacked again by Cl₂/Cl• — plus termination products form at random. All of this produces a statistical mixture.正确！链增长没有办法"选择性"地夺取某个特定氢原子，而且一旦生成部分产物，它自身也可能再次被 Cl₂/Cl• 攻击——再加上终止产物是随机生成的，这一切共同导致产物呈统计混合。
The lack of selectivity comes from the radical mechanism itself: propagation attacks any accessible C-H bond, substituted products can react further, and termination is random. Answer: (B).选择性的缺失源于自由基机理本身：链增长会攻击任何可及的 C-H 键，取代产物还能继续反应，且链终止是随机的。答案：(B)。
Reactivity 3.4
Electron-Pair Sharing Reactions电子对共享反应HL
Guiding Question导引问题How do electron-rich and electron-poor species find each other? A nucleophile's lone pair attacks an electrophile's empty or partially-positive orbital, and the resulting mechanism's exact pathway (one step or two) depends on the structure of the substrate.富电子物种与缺电子物种是如何"找到彼此"的？亲核试剂用孤对电子进攻亲电试剂的空轨道或带部分正电荷的位点，而反应具体经由哪条路径（一步还是两步）取决于底物的结构。
This entire sub-topic is Higher Level only.本子主题全部内容仅限 HL。
Nucleophiles & Electrophiles亲核试剂与亲电试剂HL
A nucleophile ("nucleus-loving") is an electron-pair donor — a species with a lone pair or a $\pi$ bond that attacks a region of positive or partial-positive charge. An electrophile ("electron-loving") is an electron-pair acceptor — a species with an empty orbital or a positively polarised atom. This maps directly onto the Lewis acid-base framework: nucleophiles are Lewis bases, electrophiles are Lewis acids.
Curly Arrow Notation弯箭头记法A full-headed curly arrow represents the movement of an electron pair (as opposed to the single-barbed "fish-hook" arrow used for single electrons in radical mechanisms, Reactivity 3.3). The arrow always starts at the electron source (a lone pair or a bond) and points to where the electrons end up.完整箭头（双尾）表示一对电子的移动（区别于 Reactivity 3.3 自由基机理中表示单个电子移动的单尾"鱼钩"箭头）。箭头总是从电子来源（孤对电子或化学键）出发，指向电子的去处。
Nucleophilic Substitution: S_N2 vs S_N1亲核取代反应：S_N2 与 S_N1HL
Halogenoalkanes undergo nucleophilic substitution because the C–X bond is polarised ($\mathrm{C}^{\delta+}{-}\mathrm{X}^{\delta-}$), making the carbon electrophilic. Two limiting mechanisms operate depending on the class of the substrate.
The nucleophile attacks from the side opposite the leaving group (backside attack) in a single transition state, simultaneously forming the new bond and breaking the old one. This inverts the configuration at carbon (like an umbrella flipping inside out).亲核试剂从离去基团的反面（背面）进攻，经过单一过渡态，新键形成的同时旧键断裂。这会使碳原子的构型发生翻转（如同雨伞被风吹得内外翻转）。
The C–Br bond breaks first (heterolytic fission, rate-determining), giving a planar $\mathrm{sp^2}$ carbocation. The nucleophile can then attack from either face with equal probability, giving a racemic (50:50) mixture of both configurations.C–Br 键先断裂（异裂，决速步骤），生成平面 $\mathrm{sp^2}$ 碳正离子。此后亲核试剂可从两侧以相等概率进攻，生成两种构型各占 50% 的外消旋混合物。
Why Substrate Class Decides the Mechanism为什么底物类型决定反应机理Primary carbons are too sterically hindered for a stable carbocation to persist, but open enough for backside attack (favours S_N2). Tertiary carbons form a highly stabilised carbocation (three alkyl groups donate electron density by hyperconjugation/induction) but are too hindered for backside attack (favours S_N1). Secondary substrates are borderline and can go by either pathway depending on the nucleophile, solvent, and leaving group.伯碳的位阻使碳正离子难以稳定存在，但空间足够开阔，便于背面进攻（有利于 S_N2）。叔碳能形成高度稳定的碳正离子（三个烷基通过超共轭/诱导效应提供电子云密度），但位阻过大不利于背面进攻（有利于 S_N1）。仲碳底物处于两者之间，具体走哪条路径取决于亲核试剂、溶剂和离去基团。
Electrophilic Addition to Alkenes烯烃的亲电加成反应HL
The $\pi$ bond of an alkene is electron-rich and acts as a nucleophile toward electrophiles such as $\mathrm{HBr}$. Addition of an asymmetric reagent to an asymmetric alkene follows Markovnikov's rule: the more stable (more substituted) carbocation intermediate forms preferentially, so H adds to the carbon that already has more hydrogens.
烯烃的 $\pi$ 键富含电子，对亲电试剂（如 $\mathrm{HBr}$）表现出亲核性。不对称试剂加成到不对称烯烃上遵循马尔科夫尼科夫规则（Markovnikov's rule）：优先生成较稳定（取代基更多）的碳正离子中间体，因此 H 加到原本氢原子较多的碳上。
The secondary carbocation (more substituted, more stabilised by adjacent alkyl groups) forms preferentially over the primary one, so the major product is 2-bromopropane, not 1-bromopropane.仲碳正离子（取代基更多，受相邻烷基稳定作用更强）优先于伯碳正离子生成，因此主要产物是 2-溴丙烷而非 1-溴丙烷。
Electrophilic Substitution of Benzene苯的亲电取代反应HL
Benzene's delocalised ring of $\pi$ electrons is a nucleophile, but instead of addition (which would destroy the aromatic stabilisation), benzene undergoes electrophilic substitution — a hydrogen is replaced, and the stable aromatic ring is regenerated.
Concentrated $\mathrm{H_2SO_4}$ generates the electrophile $\mathrm{NO_2^+}$ (the nitronium ion) from $\mathrm{HNO_3}$. This attacks the ring; a proton is then lost to regenerate aromaticity — the electrophile substitutes for H rather than adding across a double bond.浓 $\mathrm{H_2SO_4}$ 使 $\mathrm{HNO_3}$ 生成亲电试剂 $\mathrm{NO_2^+}$（硝鎓离子）。它进攻苯环，随后失去一个质子以恢复芳香性——亲电试剂取代了 H，而不是加成到双键上。
Reduction of Carbonyl Compounds羰基化合物的还原HL
The carbon of a carbonyl group ($\mathrm{C{=}O}$) is electrophilic (polarised $\mathrm{C}^{\delta+}$), so it can be attacked by a nucleophilic hydride ion ($\mathrm{H^-}$) delivered from a reducing agent such as $\mathrm{NaBH_4}$. Aldehydes are reduced to primary alcohols; ketones are reduced to secondary alcohols.
$\mathrm{H^-}$ attacks the electrophilic carbonyl carbon, pushing the $\pi$ electrons onto oxygen to form an alkoxide; protonation on aqueous work-up gives the alcohol.$\mathrm{H^-}$ 进攻亲电的羰基碳，将 $\pi$ 电子推向氧原子生成烷氧负离子；后处理时质子化即得到醇。
Worked Example — Predicting S_N1 vs S_N2例题 — 预测 S_N1 还是 S_N2
Predict the dominant mechanism and stereochemical outcome for the hydrolysis of (a) 1-bromobutane and (b) 2-bromo-2-methylpropane (tert-butyl bromide) with aqueous $\mathrm{OH^-}$.预测 (a) 1-溴丁烷和 (b) 2-溴-2-甲基丙烷（叔丁基溴）在水中与 $\mathrm{OH^-}$ 水解反应的主导机理及立体化学结果。
(a) 1-bromobutane(a) 1-溴丁烷
Primary substrate: unhindered for backside attack, and no stable carbocation could form. Mechanism: S_N2. Rate depends on both $[\text{substrate}]$ and $[\mathrm{OH^-}]$; product shows inversion of configuration (irrelevant here as the carbon isn't a stereocentre).伯底物：背面进攻无空间位阻，且难以形成稳定碳正离子。机理：S_N2。速率同时依赖 $[\text{底物}]$ 和 $[\mathrm{OH^-}]$；产物构型翻转（此处碳原子非手性中心，故翻转不显现）。
(b) tert-butyl bromide(b) 叔丁基溴
Tertiary substrate: too hindered for backside attack, but forms a highly stabilised tertiary carbocation. Mechanism: S_N1. Rate depends only on $[\text{substrate}]$; if the carbon were a stereocentre, the product would be racemic.叔底物：位阻过大不利于背面进攻，但能形成高度稳定的叔碳正离子。机理：S_N1。速率只依赖 $[\text{底物}]$；若该碳为手性中心，产物将呈外消旋。
Evaluate检验
This matches the general trend: primary → S_N2 dominates; tertiary → S_N1 dominates; secondary substrates are the ambiguous middle case.这与一般规律一致：伯底物 → 以 S_N2 为主；叔底物 → 以 S_N1 为主；仲底物则处于模糊的中间情形。
(HL) Which best describes an S_N2 reaction?（HL）下列哪项最准确地描述了 S_N2 反应？
One concerted step; rate depends on both substrate and nucleophile; inversion of configuration.单一协同步骤；速率同时依赖底物和亲核试剂；构型翻转。
Two steps via a carbocation; rate depends only on the substrate; racemization.经碳正离子的两步反应；速率只依赖底物；外消旋化。
One step; rate depends only on the nucleophile.单步反应；速率只依赖亲核试剂。
Two steps; retention of configuration.两步反应；构型保持。
Correct! S_N2 is bimolecular and concerted — the rate equation is second order overall, rate $= k[\text{substrate}][\text{Nu}^-]$, and backside attack inverts the stereocentre.正确！S_N2 是双分子协同反应——速率方程为总二级，rate $= k[\text{底物}][\text{Nu}^-]$，背面进攻会使手性中心构型翻转。
Option (B) describes S_N1. S_N2 is a single concerted step, second order overall, with inversion of configuration at the carbon under attack. Answer: (A).选项 (B) 描述的是 S_N1。S_N2 是单一协同步骤，总反应为二级，被进攻的碳原子构型发生翻转。答案：(A)。
(HL) When propene reacts with HBr, the major product is 2-bromopropane rather than 1-bromopropane because（HL）丙烯与 HBr 反应，主要产物是 2-溴丙烷而非 1-溴丙烷，原因是
Br⁻ always attacks the less hindered carbon.Br⁻ 总是进攻位阻较小的碳。
The reaction proceeds via S_N2.该反应经由 S_N2 进行。
HBr is a radical initiator under these conditions.在此条件下 HBr 是自由基引发剂。
The secondary carbocation intermediate is more stable than the primary one, so it forms preferentially (Markovnikov's rule).仲碳正离子中间体比伯碳正离子更稳定，因此优先生成（马尔科夫尼科夫规则）。
Correct! Electrophilic addition to an unsymmetrical alkene proceeds via whichever carbocation is more stable. The secondary carbocation (more alkyl substitution, more hyperconjugative stabilisation) forms faster, so Br⁻ ends up on the middle carbon.正确！不对称烯烃的亲电加成经由更稳定的碳正离子进行。仲碳正离子（烷基取代更多、超共轭稳定作用更强）生成更快，因此 Br⁻ 最终连接在中间的碳上。
This is electrophilic addition, not S_N2 or a radical mechanism. The regiochemistry is controlled by carbocation stability: the more substituted (secondary) carbocation is favoured. Answer: (D).这是亲电加成，不是 S_N2 或自由基机理。区域选择性由碳正离子的稳定性决定：取代更多（仲）的碳正离子更有利。答案：(D)。
Preparation备考
Exam Strategy考试策略
Paper 1 (MC, no calc, no data booklet)Paper 1（选择题，无计算器、无数据手册）
Reactivity 3 questions here lean on mechanism recognition: classify a step as initiation/propagation/termination, spot conjugate acid-base pairs, identify oxidizing vs reducing agents from oxidation-state changes, or pick out the more stable carbocation. Diagrams and curly arrows are tested conceptually, not numerically.
Paper 2 (short + long, calculator + data booklet)Paper 2（短题 + 长题，可用计算器和数据手册）
This is where the numerical Reactivity 3 work lives. Expect (a) a pH/pOH or weak-acid $K_a$ calculation, often extended into a buffer or titration-curve question at HL, (b) a redox half-equation balancing question, sometimes with an $E^{\ominus}_{\text{cell}}$ or electrolysis (Faraday's law) calculation at HL, and (c) a full free-radical or S_N1/S_N2 mechanism to draw out with curly arrows.
Reactivity 3 commonly drives Paper-3 data analysis: an acid-base titration curve to interpret (locate equivalence and half-equivalence points, choose an indicator), or a kinetics/electrochemistry dataset that overlaps with Reactivity 2. Practice reading pH-curve shapes and matching them to the four titration types.
Data Booklet — Useful Tables数据手册中的实用表格Section 1 (equilibrium/acid-base formulas: $K_w$, $K_a$/$K_b$, Henderson–Hasselbalch), Section 2 (constants — $R$, $F$), Section 24/25 (standard electrode potentials $E^{\ominus}$), and the formula sheet ($E^{\ominus}_{\text{cell}}$, $\Delta G^{\ominus} = -nFE^{\ominus}_{\text{cell}}$, Faraday's law). Bookmark these before the exam.第 1 部分（平衡/酸碱公式：$K_w$、$K_a$/$K_b$、亨德森-哈塞尔巴赫方程）、第 2 部分（常数 $R$、$F$）、第 24/25 部分（标准电极电势 $E^{\ominus}$ 表）以及公式页（$E^{\ominus}_{\text{cell}}$、$\Delta G^{\ominus} = -nFE^{\ominus}_{\text{cell}}$、法拉第定律）。考前先把这些页标好。
Watch Out小心陷阱
Common Mistakes常见错误
Confusing a conjugate pair with a simple pair of ions把共轭酸碱对与普通的离子对混淆A conjugate acid-base pair differs by exactly one H⁺ (and one unit of charge). $\mathrm{HCl}$ and $\mathrm{Cl^-}$ are a pair; $\mathrm{HCl}$ and $\mathrm{OH^-}$ are not.共轭酸碱对之间恰好相差一个 H⁺（以及一个电荷单位）。$\mathrm{HCl}$ 与 $\mathrm{Cl^-}$ 是一对；$\mathrm{HCl}$ 与 $\mathrm{OH^-}$ 不是。
Using the weak-acid approximation when it isn't valid在不满足条件时仍使用弱酸近似$[\mathrm{H^+}] \approx \sqrt{K_a c}$ only holds when dissociation is small ($c/K_a \gtrsim 400$). For more concentrated weak acids or larger $K_a$, you must solve the full quadratic.$[\mathrm{H^+}] \approx \sqrt{K_a c}$ 只在解离程度很小（$c/K_a \gtrsim 400$）时成立。对于浓度更高的弱酸或 $K_a$ 较大的情况，必须解完整的一元二次方程。
Mixing up oxidation and reduction把氧化和还原弄反OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain. The oxidizing agent is the species that is itself reduced — a very common point of confusion.OIL RIG：氧化是失去（电子），还原是得到。氧化剂是自身被还原的那个物种——这是极易混淆的一点。
Flipping anode/cathode polarity between cell types在不同电池类型间弄反阳极/阴极的极性The anode is always where oxidation occurs and the cathode is always where reduction occurs — but the anode is negative in a voltaic cell and positive in an electrolytic cell (since the external supply reverses the driving force).阳极始终是氧化发生之处，阴极始终是还原发生之处——但阳极在原电池中为负极，在电解池中为正极（因为外部电源反转了驱动力）。
Predicting electrolysis products by ignoring concentration预测电解产物时忽略浓度因素$E^{\ominus}$ tables predict what should form under standard (dilute) conditions, but concentrated halide solutions often give the halogen at the anode for kinetic reasons even though water is thermodynamically easier to oxidize.$E^{\ominus}$ 表预测的是标准（稀）条件下应该生成的产物，但高浓度卤化物溶液常因动力学原因在阳极生成卤素单质，即使从热力学角度水更容易被氧化。
Assuming free-radical substitution gives one clean product误以为自由基取代只生成一种纯净产物Propagation can abstract any hydrogen and further substitution can continue on the product, so a statistical mixture always forms — never write just one product for this mechanism.链增长可以夺取任何氢原子，且产物还能继续被取代，因此总会生成统计混合物——切勿只写出一种产物。
Mixing up S_N1 and S_N2 stereochemistry混淆 S_N1 与 S_N2 的立体化学结果S_N2 gives inversion (backside attack, one step). S_N1 gives racemization (planar carbocation intermediate, attacked from either face). These are opposite outcomes — don't swap them.S_N2 给出构型翻转（背面进攻，单步）。S_N1 给出外消旋化（平面碳正离子中间体，可从两侧被进攻）。这是两种相反的结果——切勿混淆。
Review复习
Flashcards闪卡
Identify the limiting reactant?如何判定限量试剂？
Divide each reactant's moles by its coefficient in the balanced equation; the smallest ratio = limiting.各反应物摩尔数除以方程式中的系数；比值最小者 = 限量。
Two conditions for a successful collision?有效碰撞的两个条件？
Sufficient energy ($\geq E_a$) and correct orientation.能量足够（$\geq E_a$）且取向正确。
How does a catalyst increase rate?催化剂如何加快反应速率？
Provides an alternative pathway with a lower activation energy. Doesn't affect $\Delta H$ or the equilibrium position.提供活化能更低的替代路径。不改变 $\Delta H$，也不改变平衡位置。
(HL) Linear Arrhenius equation?（HL）线性形式的阿伦尼乌斯方程？
$\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$ — plot $\ln k$ vs $1/T$ to get slope $= -E_a/R$.作 $\ln k$ 对 $1/T$ 的图，斜率 $= -E_a/R$。
Le Châtelier — what changes $K$?勒夏特列——什么会改变 $K$？
Only temperature. Pressure and concentration changes shift the position but $K$ stays the same.只有温度。压强和浓度的变化会移动平衡位置，但 $K$ 不变。
$K$ expression for $aA + bB \rightleftharpoons cC + dD$?$aA + bB \rightleftharpoons cC + dD$ 的 $K$ 表达式？
$K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ — equilibrium concentrations only, coefficients become exponents.只用平衡浓度，系数变指数。