Oxidation number of Cr in $\text{Cr}_2\text{O}_7^{2-}$?$\text{Cr}_2\text{O}_7^{2-}$ 中 Cr 的氧化数？
Answer:答案：(B) $+6$
(a) Set up the oxidation-number equation建立氧化数方程M1·A1·A1
Each oxygen in a polyatomic ion has oxidation number $-2$. Let the oxidation number of Cr $= x$. The ion carries a charge of $-2$:多原子离子中每个氧的氧化数为 $-2$。设 Cr 的氧化数为 $x$，离子总电荷为 $-2$：
$$ 2x + 7(-2) \;=\; -2 \;\Longrightarrow\; 2x - 14 \;=\; -2 \;\Longrightarrow\; 2x \;=\; 12 \;\Longrightarrow\; x \;=\; +6. $$
Option (B).选 (B)。
Why the distractors fail.干扰项分析。
(A) $+3$:this is the oxidation state of Cr in $\text{Cr}^{3+}$, the product after reduction, not in the dichromate ion.这是 $\text{Cr}^{3+}$ 中 Cr 的氧化态（还原后的产物），不是重铬酸根中的值。
(C) $+7$:confuses Cr with Mn; $\text{MnO}_4^-$ gives Mn $= +7$, not dichromate.将 Cr 与 Mn 混淆；$\text{MnO}_4^-$ 中 Mn 为 $+7$，重铬酸根中不是。
(D) $-2$:this is the oxidation number of oxygen, not chromium.这是氧的氧化数，不是铬的。
OX-number rule for polyatomic ions: sum of all oxidation numbers equals the ion charge.多原子离子氧化数规则：所有氧化数之和等于离子电荷。The mnemonic is: (i) oxygen is $-2$ in almost all compounds (except peroxides); (ii) hydrogen is $+1$ when bonded to nonmetals; (iii) the sum of oxidation numbers equals the ion charge. For $\text{Cr}_2\text{O}_7^{2-}$: two Cr atoms and seven O atoms, charge $= -2$. Setting $2x + 7(-2) = -2$ is the standard setup. Chromium is one of the few elements with multiple common oxidation states ($+2, +3, +6$), so it appears frequently in redox problems at every level from high school through AP and IB. Never guess an oxidation number from a formula without checking the algebra.口诀：(i) 氧在几乎所有化合物中为 $-2$（过氧化物除外）；(ii) 与非金属键合时氢为 $+1$；(iii) 所有氧化数之和等于离子电荷。对 $\text{Cr}_2\text{O}_7^{2-}$：两个 Cr 原子和七个 O 原子，总电荷 $= -2$。建立方程 $2x + 7(-2) = -2$ 是标准做法。铬是少数具有多种常见氧化态（$+2, +3, +6$）的元素之一，因此在各级别氧化还原题中频繁出现。切勿凭直觉猜测，务必验算代数。
Assign oxidation numbers: Fe goes from $0$ (elemental) to $+2$ in $\text{FeSO}_4$ (loss of electrons). Cu goes from $+2$ in $\text{CuSO}_4$ to $0$ (elemental, gain of electrons). Loss of electrons $=$ oxidation (OIL), so Fe is oxidised, changing from $0$ to $+2$.分配氧化数：Fe 从 $0$（单质）变为 $\text{FeSO}_4$ 中的 $+2$（失去电子）；Cu 从 $\text{CuSO}_4$ 中的 $+2$ 变为 $0$（单质，得到电子）。失去电子 $=$ 氧化（OIL），故 Fe 被氧化，氧化数从 $0$ 变为 $+2$。
(b) Identify oxidising and reducing agents确定氧化剂和还原剂A1·A1
The oxidising agent is the species that accepts electrons and is itself reduced: $\text{Cu}^{2+}$ (reduced from $+2$ to $0$). The reducing agent is the species that donates electrons and is itself oxidised: $\text{Fe}$ (oxidised from $0$ to $+2$).氧化剂是接受电子、自身被还原的物质：$\text{Cu}^{2+}$（从 $+2$ 还原至 $0$）。还原剂是提供电子、自身被氧化的物质：$\text{Fe}$（从 $0$ 氧化至 $+2$）。
OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain of electrons.OIL RIG：氧化是失去电子，还原是得到电子。The reducing agent is the species that loses electrons (it reduces the other); the oxidising agent is the species that gains electrons (it oxidises the other). A useful cross-check: the reducing agent is oxidised, and the oxidising agent is reduced. In this single-displacement reaction, the more active metal (Fe) displaces the less active metal ion ($\text{Cu}^{2+}$) from solution, a pattern predictable from the activity series. The $\text{SO}_4^{2-}$ ion is a spectator and does not change oxidation state.还原剂是失去电子的物质（它使对方被还原）；氧化剂是得到电子的物质（它使对方被氧化）。有用的交叉验证：还原剂本身被氧化，氧化剂本身被还原。在这个单置换反应中，活动性更强的金属（Fe）将活动性较弱的金属离子（$\text{Cu}^{2+}$）从溶液中置换出来，这一规律可从活动性顺序中预测。$\text{SO}_4^{2-}$ 是旁观离子，氧化数不变。
Q3MEDIUM中🇺🇸 US美AP-style MCQAP 风格选择题§3 Half-reaction electron count半反应中的电子数[3 marks][3 分]
Electrons transferred per $\text{MnO}_4^-$ reduced to $\text{Mn}^{2+}$ in acidic solution?酸性溶液中每个 $\text{MnO}_4^-$ 还原为 $\text{Mn}^{2+}$ 时转移的电子数？
Answer:答案：(C) 5
(a) Determine the change in oxidation number of Mn确定 Mn 氧化数的变化M1·A1·A1
In $\text{MnO}_4^-$: letting the oxidation number of Mn $= x$, $x + 4(-2) = -1$, so $x = +7$. In $\text{Mn}^{2+}$: $x = +2$. Change $= +7 \to +2$, a decrease of 5. Reduction = gain of electrons, so 5 electrons are gained per $\text{MnO}_4^-$ ion.在 $\text{MnO}_4^-$ 中：设 Mn 氧化数为 $x$，$x + 4(-2) = -1$，得 $x = +7$。在 $\text{Mn}^{2+}$ 中：$x = +2$。变化为 $+7 \to +2$，减少了 5。还原 $=$ 得到电子，故每个 $\text{MnO}_4^-$ 离子得到 5 个电子。
$$\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \;\rightarrow\; \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$$
Why the distractors fail.干扰项分析。
(A) 2:the number of electrons in the $\text{Cu}^{2+}/\text{Cu}$ half-reaction, not permanganate.这是 $\text{Cu}^{2+}/\text{Cu}$ 半反应中的电子数，不是高锰酸根的。
(D) 7:the oxidation number of Mn in $\text{MnO}_4^-$, confused with electron count.这是 $\text{MnO}_4^-$ 中 Mn 的氧化数，与电子数混淆。
Electrons transferred per ion $=$ change in oxidation number of the central atom (in magnitude).每个离子转移的电子数 $=$ 中心原子氧化数变化量的绝对值。The permanganate ion ($\text{MnO}_4^-$) is one of the most powerful and common oxidising agents in laboratory chemistry. Its reduction to $\text{Mn}^{2+}$ (colourless) in acidic solution requires 5 electrons per $\text{MnO}_4^-$ and 8 H$^+$ to consume the 4 oxygen atoms as water. This half-reaction must be memorised for AP and IB chemistry. The purple-to-colourless colour change is used as a titration endpoint (permanganate is its own indicator).高锰酸根（$\text{MnO}_4^-$）是实验室化学中最强也是最常见的氧化剂之一。在酸性溶液中还原为 $\text{Mn}^{2+}$（无色）时，每个 $\text{MnO}_4^-$ 需要 5 个电子和 8 个 H$^+$（将 4 个氧原子转化为水）。这个半反应是 AP 和 IB 化学的必背内容。紫色变无色的颜色变化用作滴定终点（高锰酸钾是自身指示剂）。
Q4MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§4 Activity series · single displacement活动性顺序 · 单置换[6 marks][6 分]
Answer:答案：(a)Reaction occurs: Zn is above Ag in the activity series发生反应：Zn 在活动性顺序中位于 Ag 之上 · (b) $\text{Zn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s)$ · (c)Silver solid deposits on zinc / blue/colourless solution forms锌表面出现银色固体沉积 / 溶液变为蓝色或无色
(a) Predict using the activity series用活动性顺序预测M1·A1·A1
Activity order given: $\text{Mg} > \text{Zn} > \text{Fe} > \text{Cu} > \text{Ag}$. Zn is higher (more active) than Ag. A more active metal can displace a less active metal ion from solution. Therefore, a reaction occurs: Zn reduces $\text{Ag}^+$ to Ag metal, and Zn is itself oxidised to $\text{Zn}^{2+}$.给定活动性顺序：$\text{Mg} > \text{Zn} > \text{Fe} > \text{Cu} > \text{Ag}$。Zn 排在 Ag 上方（活动性更强）。活动性更强的金属能将活动性更弱的金属离子从溶液中置换出来。因此，反应发生：Zn 将 $\text{Ag}^+$ 还原为 Ag 金属，Zn 自身被氧化为 $\text{Zn}^{2+}$。
(b) Balanced ionic equation配平离子方程式A1·A1
Each Zn loses 2 electrons; each $\text{Ag}^+$ gains 1 electron. Two $\text{Ag}^+$ ions are needed per Zn atom:每个 Zn 失去 2 个电子；每个 $\text{Ag}^+$ 得到 1 个电子。每个 Zn 原子需要两个 $\text{Ag}^+$：
$$ \text{Zn}(s) + 2\text{Ag}^+(aq) \;\rightarrow\; \text{Zn}^{2+}(aq) + 2\text{Ag}(s) $$
(c) Observable sign可观察证据A1
A grey/silvery solid (Ag metal) deposits on the surface of the zinc, and the solution may become slightly milky or colourless as $\text{Ag}^+$ is consumed. Accept any one: solid depositing on Zn / solution lightening / Zn appearing to corrode.锌表面出现灰色/银白色固体（Ag 金属）沉积；随着 $\text{Ag}^+$ 被消耗，溶液可能变得略微浑浊或变为无色。接受以下任一答案：Zn 表面有固体析出 / 溶液颜色变浅 / Zn 表面被腐蚀。
The activity series predicts single-displacement: a higher metal displaces the ion of a lower metal from solution.活动性顺序预测单置换反应：位置更高的金属能将位置更低金属的离子从溶液中置换出来。The rule is: reaction occurs if and only if the metal (solid) is higher in the activity series than the metal ion in solution. Here Zn $>$ Ag, so reaction proceeds. Note that the equation requires 2:1 stoichiometry ($\text{Zn}:\text{Ag}^+$) because the oxidation state changes are +2 vs. +1. The $\text{NO}_3^-$ ion is a spectator and is omitted from the net ionic equation. In the reverse direction (Ag placed in $\text{ZnSO}_4$), no reaction would occur because Ag $<$ Zn.规则：当且仅当固体金属在活动性顺序中高于溶液中的金属离子时，反应才发生。此处 Zn $>$ Ag，故反应进行。注意方程式需要 $2:1$ 的化学计量比（$\text{Zn}:\text{Ag}^+$），因为氧化态变化分别为 $+2$ 和 $+1$。$\text{NO}_3^-$ 是旁观离子，在净离子方程式中省略。反向情形（将 Ag 置于 $\text{ZnSO}_4$ 中）则不发生反应，因为 Ag $<$ Zn。
Zn is more active than Cu and is oxidised: the anode is the Zn electrode (oxidation occurs at the anode). Cu$^{2+}$ ions are reduced at the cathode (Cu electrode). Memory aid: An Ox / Red Cat (Anode = Oxidation; Cathode = Reduction).Zn 比 Cu 活泼，被氧化：阳极为 Zn 电极（氧化在阳极发生）。Cu$^{2+}$ 离子在阴极（Cu 电极）被还原。记忆口诀：阳氧阴还（阳极 $=$ 氧化；阴极 $=$ 还原）。
Electrons are released at the Zn anode and flow through the external circuit (wire) to the Cu cathode where they are consumed. Electrons always flow from anode to cathode in the external circuit.电子在 Zn 阳极产生，经外部电路（导线）流向 Cu 阴极，在那里被消耗。外部电路中电子始终从阳极流向阴极。
(d) Salt bridge role and anion direction盐桥作用与阴离子方向A1·A1
The salt bridge maintains electrical neutrality in both half-cells. As Zn dissolves (adding $\text{Zn}^{2+}$ cations to its compartment), negative ions (anions) from the salt bridge migrate toward the Zn/anode compartment to balance the positive charge. Simultaneously, cations migrate toward the cathode compartment where $\text{Cu}^{2+}$ is being removed.盐桥在两个半电池中维持电中性。随着 Zn 溶解（向其半电池中加入 $\text{Zn}^{2+}$ 阳离子），盐桥中的阴离子向 Zn 阳极一侧迁移以中和多余正电荷。同时，阳离子向阴极一侧迁移，补充被消耗的 $\text{Cu}^{2+}$。
The salt bridge completes the circuit by allowing ion flow, preventing charge build-up that would stop the cell.盐桥通过允许离子迁移来接通回路，防止电荷积累而使电池停止工作。Without the salt bridge, the Zn compartment would accumulate positive charge (from $\text{Zn}^{2+}$) and the Cu compartment would deplete of positive charge (as $\text{Cu}^{2+}$ is reduced). This charge imbalance would quickly oppose further electron flow, killing the cell voltage. The salt bridge (typically KCl or $\text{KNO}_3$ in agar) neutralises this by allowing anions to flow toward the anode and cations toward the cathode. This is also why removing the salt bridge stops the cell immediately.没有盐桥，Zn 半电池会因 $\text{Zn}^{2+}$ 积累而带正电，Cu 半电池则因 $\text{Cu}^{2+}$ 被还原而正电荷不足。这种电荷失衡会很快阻止电子继续流动，使电池电压降为零。盐桥（通常是琼脂中的 KCl 或 $\text{KNO}_3$）通过允许阴离子流向阳极、阳离子流向阴极来中和这种失衡。这也是为什么取走盐桥会立即使电池停止工作的原因。
(a) Unbalanced reduction half-reaction and Mn oxidation number change未配平还原半反应与 Mn 氧化数变化A1·A1
Mn goes from $+7$ (in $\text{MnO}_4^-$) to $+2$ (in $\text{Mn}^{2+}$), a decrease of 5 (reduction, gain of electrons).Mn 从 $+7$（$\text{MnO}_4^-$ 中）变为 $+2$（$\text{Mn}^{2+}$ 中），减少了 5（还原，得到电子）。
$$ \text{MnO}_4^-(aq) \;\rightarrow\; \text{Mn}^{2+}(aq) \quad \text{(unbalanced / 未配平)} $$
(b) Balance the reduction half-reaction step by step逐步配平还原半反应A1·A1·A1
Step 1 (balance Mn): 1 Mn on each side already balanced. Step 2 (balance O with $\text{H}_2\text{O}$): 4 oxygen atoms on left, add 4 $\text{H}_2\text{O}$ on right:步骤 1（配平 Mn）：两边各一个 Mn，已平衡。步骤 2（用 $\text{H}_2\text{O}$ 配平 O）：左边 4 个氧原子，右边加 4 个 $\text{H}_2\text{O}$：
$$ \text{MnO}_4^- \;\rightarrow\; \text{Mn}^{2+} + 4\text{H}_2\text{O} $$
Step 3 (balance H with $\text{H}^+$): 8 H on right (from 4 $\text{H}_2\text{O}$), add 8 $\text{H}^+$ on left:步骤 3（用 $\text{H}^+$ 配平 H）：右边 8 个 H（来自 4 个 $\text{H}_2\text{O}$），左边加 8 个 $\text{H}^+$：
$$ \text{MnO}_4^- + 8\text{H}^+ \;\rightarrow\; \text{Mn}^{2+} + 4\text{H}_2\text{O} $$
Step 4 (balance charge with $e^-$): Left charge $= -1 + 8(+1) = +7$; Right charge $= +2$. Add 5 electrons to left:步骤 4（用 $e^-$ 配平电荷）：左边电荷 $= -1 + 8(+1) = +7$；右边电荷 $= +2$。左边加 5 个电子：
$$ \text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \;\rightarrow\; \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) $$
(c) Oxidation half-reaction for $\text{Fe}^{2+} \to \text{Fe}^{3+}$$\text{Fe}^{2+} \to \text{Fe}^{3+}$ 的氧化半反应A1
Multiply the Fe half-reaction by 5 so electrons cancel ($5e^-$ each side), then add:将 Fe 半反应乘以 5 使电子相消（每边 $5e^-$），然后相加：
$$ \text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \;\rightarrow\; \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l) $$
Mole ratio $\text{MnO}_4^- : \text{Fe}^{2+} = 1 : 5$.摩尔比 $\text{MnO}_4^- : \text{Fe}^{2+} = 1 : 5$。
The four-step half-reaction method (balance atom by atom, then charge) is the systematic approach for all acidic redox equations.四步半反应法（逐一配平原子，再配平电荷）是所有酸性氧化还原方程式的系统方法。The order must be Mn (heavy atom) first, then O (using $\text{H}_2\text{O}$), then H (using $\text{H}^+$), then charge (using electrons). Reversing steps leads to incorrect equations. The electron-multiplication step ensures the electrons cancel exactly: the LCM of 5 (Mn half-reaction) and 1 (Fe half-reaction) is 5, so the Fe half-reaction is multiplied by 5. Check atoms and charges on both sides after combining: Mn (1 each), O (4 right as water), H (8 left as $\text{H}^+$, 8 right as water), Fe (5 each), charge: left $= -1 + 5(+2) + 8(+1) = +17$; right $= +2 + 5(+3) + 0 = +17$. Balanced.顺序必须是：先配平重原子（Mn），然后用 $\text{H}_2\text{O}$ 配平 O，用 $\text{H}^+$ 配平 H，最后用电子配平电荷。顺序颠倒会导致错误方程。电子倍乘步骤确保电子恰好相消：5（Mn 半反应）和 1（Fe 半反应）的最小公倍数为 5，故 Fe 半反应乘以 5。合并后核对两边的原子数和电荷：Mn（各 1）、O（右边 4 个水中）、H（左边 8 个 $\text{H}^+$，右边 4 个水中 8 个 H）、Fe（各 5）；电荷：左 $= -1 + 5(+2) + 8(+1) = +17$，右 $= +2 + 5(+3) + 0 = +17$，平衡。
The cathode is where reduction occurs. The half-cell with the higher (more positive) standard reduction potential is preferentially reduced. $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V} > E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$, so the Cu half-cell is the cathode. Accordingly, the Zn half-cell is the anode (oxidation).阴极是还原发生的地方。标准还原电势较高（更正）的半电池优先被还原。$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V} > E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$，故 Cu 半电池为阴极。相应地，Zn 半电池为阳极（氧化）。
(c) Spontaneity and sign of $E^\circ_\text{cell}$自发性与 $E^\circ_\text{cell}$ 符号的关系A1·A1
The cell reaction is spontaneous under standard conditions. A positive $E^\circ_\text{cell}$ ($> 0$) indicates a spontaneous reaction; a negative $E^\circ_\text{cell}$ ($< 0$) indicates a non-spontaneous reaction. This connects to thermodynamics: $\Delta G^\circ = -nFE^\circ_\text{cell}$, so positive $E^\circ_\text{cell}$ means negative $\Delta G^\circ$ (spontaneous).在标准条件下，电池反应是自发的。$E^\circ_\text{cell} > 0$ 表示反应自发；$E^\circ_\text{cell} < 0$ 表示反应非自发。这与热力学相联系：$\Delta G^\circ = -nFE^\circ_\text{cell}$，故 $E^\circ_\text{cell} > 0$ 对应 $\Delta G^\circ < 0$（自发）。
(d) Overall balanced cell equation总配平电池反应方程式A1·A1
Both half-reactions involve 2 electrons, so combine directly:两个半反应各涉及 2 个电子，可直接合并：
$$ \text{Zn}(s) + \text{Cu}^{2+}(aq) \;\rightarrow\; \text{Zn}^{2+}(aq) + \text{Cu}(s) $$
The sign of $E^\circ_\text{cell}$ is the quickest test for spontaneity: positive $=$ spontaneous galvanic cell.$E^\circ_\text{cell}$ 的符号是判断自发性的最快方法：正值 $=$ 自发的原电池。The Daniel cell (Zn-Cu) is the classic galvanic cell example, producing $+1.10$ V under standard conditions. This is one of the most-tested cells in high school and AP chemistry. Key pitfall: when applying $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$, the anode potential is subtracted (not flipped in sign). Never flip the sign of a tabulated reduction potential; only use the formula as written. A second pitfall: standard reduction potentials do not change with stoichiometric coefficients ($E^\circ$ is intensive, not extensive), so you never multiply $E^\circ$ values when balancing electron counts.丹尼尔电池（Zn-Cu）是经典原电池示例，标准条件下产生 $+1.10$ V。这是高中和 AP 化学中最常考的电池之一。关键陷阱：使用 $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ 时，是减去阳极电势（而非翻转符号后加）。切勿翻转表格中的还原电势符号；只需按公式书写即可。第二个陷阱：标准还原电势不随化学计量系数变化（$E^\circ$ 是强度量，不是广度量），因此在配平电子数时绝不能乘以 $E^\circ$ 值。
Answer:答案：(a) $E^\circ_\text{cell} = +3.17\ \text{V}$, spontaneous · (b) $E^\circ_\text{cell} = -0.47\ \text{V}$, not spontaneous; swap to make Pb anode, Cu cathode for $+0.47\ \text{V}$ · (c)more-negative $E^\circ$ means stronger tendency to lose electrons (be oxidised), so it is easier to oxidise and acts as anode$E^\circ$ 更负意味着更强的失去电子（被氧化）倾向，因此更容易被氧化，作为阳极
(a) Mg/Mg$^{2+}$ (anode) and Ag$^+$/Ag (cathode)Mg/Mg$^{2+}$（阳极）和 Ag$^+$/Ag（阴极）M1·A1·A1
Mg has the more negative $E^\circ$, so Mg is the anode (oxidised); Ag$^+$/Ag is the cathode (reduced).Mg 的 $E^\circ$ 更负，故 Mg 为阳极（被氧化）；Ag$^+$/Ag 为阴极（被还原）。
$$ E^\circ_\text{cell} \;=\; E^\circ(\text{Ag}^+/\text{Ag}) - E^\circ(\text{Mg}^{2+}/\text{Mg}) \;=\; (+0.80) - (-2.37) \;=\; +3.17\ \text{V} $$
$E^\circ_\text{cell} = +3.17\ \text{V} > 0$: the cell is spontaneous.$E^\circ_\text{cell} = +3.17\ \text{V} > 0$：电池是自发的。
(b) Cu anode / Pb cathode: calculation and correctionCu 阳极 / Pb 阴极：计算与纠正M1·A1·A1
With Cu as anode and Pb as cathode as stated:按题目假设 Cu 为阳极、Pb 为阴极：
$$ E^\circ_\text{cell} \;=\; E^\circ(\text{Pb}^{2+}/\text{Pb}) - E^\circ(\text{Cu}^{2+}/\text{Cu}) \;=\; (-0.13) - (+0.34) \;=\; -0.47\ \text{V} $$
$E^\circ_\text{cell} = -0.47\ \text{V} < 0$: this arrangement is not spontaneous. For a spontaneous cell, the metal with the more negative $E^\circ$ must be the anode. Since $E^\circ(\text{Pb}) = -0.13\ \text{V} < E^\circ(\text{Cu}) = +0.34\ \text{V}$, Pb should be the anode and Cu the cathode:$E^\circ_\text{cell} = -0.47\ \text{V} < 0$：该配置不自发。要使电池自发，$E^\circ$ 更负的金属必须作为阳极。由于 $E^\circ(\text{Pb}) = -0.13\ \text{V} < E^\circ(\text{Cu}) = +0.34\ \text{V}$，应将 Pb 作为阳极，Cu 作为阴极：
$$ E^\circ_\text{cell} \;=\; (+0.34) - (-0.13) \;=\; +0.47\ \text{V} $$
(c) Why more-negative $E^\circ$ favours the anode role为何更负的 $E^\circ$ 倾向于作阳极A1·A1
A more negative standard reduction potential means the species has a weaker tendency to be reduced and a stronger tendency to be oxidised (to lose electrons). In a galvanic cell, the electrode that is most easily oxidised becomes the anode. Therefore, the metal with the more negative $E^\circ$ acts as the anode because oxidation is thermodynamically favoured for it.标准还原电势更负意味着该物质被还原的倾向较弱，而被氧化（失去电子）的倾向更强。在原电池中，最容易被氧化的电极成为阳极。因此，$E^\circ$ 更负的金属作为阳极，因为氧化在热力学上对其更有利。
To predict which metal is anode: the one with the lower (more negative) $E^\circ$ is always the anode in a spontaneous galvanic cell.预测哪种金属为阳极：在自发原电池中，$E^\circ$ 更低（更负）的金属始终为阳极。This is a direct consequence of the sign convention: $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$. For $E^\circ_\text{cell} > 0$ (spontaneous), we need $E^\circ_\text{cathode} > E^\circ_\text{anode}$, i.e., the cathode has the higher reduction potential. Equivalently, the anode has the lower $E^\circ$. The Mg/Ag cell in part (a) illustrates this powerfully: Mg, with its very negative $E^\circ = -2.37\ \text{V}$, is one of the strongest reducing agents in the table, generating a large positive $E^\circ_\text{cell}$ when paired with a strong oxidiser like Ag$^+$.这是符号约定的直接推论：$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$。要使 $E^\circ_\text{cell} > 0$（自发），需要 $E^\circ_\text{cathode} > E^\circ_\text{anode}$，即阴极的还原电势更高。等价地，阳极的 $E^\circ$ 更低。(a) 中的 Mg/Ag 电池很好地说明了这一点：Mg 的 $E^\circ = -2.37\ \text{V}$ 极为负，是表中最强的还原剂之一，与 Ag$^+$ 这样的强氧化剂配对时能产生很大的正 $E^\circ_\text{cell}$。
Molten NaCl electrolysed with inert carbon electrodes: (a) which electrode is the anode; (b) cathode product and half-reaction; (c) anode product and half-reaction; (d) electrolysis vs galvanic cell.用惰性碳电极电解熔融 NaCl：(a) 哪个电极为阳极；(b) 阴极产物与半反应；(c) 阳极产物与半反应；(d) 电解与原电池的比较。
Answer:答案：(a)Anode is connected to positive terminal of power supply阳极连接电源正极 · (b)Cathode: Na metal; $\text{Na}^+(l) + e^- \to \text{Na}(l)$阴极：Na 金属；$\text{Na}^+(l) + e^- \to \text{Na}(l)$ · (c)Anode: Cl$_2$ gas; $2\text{Cl}^-(l) \to \text{Cl}_2(g) + 2e^-$阳极：Cl$_2$ 气体；$2\text{Cl}^-(l) \to \text{Cl}_2(g) + 2e^-$ · (d)Electrolysis drives non-spontaneous reactions using electrical energy电解利用电能驱动非自发化学反应
(a) Electrode connected to positive terminal连接正极的电极A1
In electrolysis, the anode is connected to the positive terminal of the external power supply. Oxidation still occurs at the anode; the power supply forces electrons out of the anode and into the cathode (the reverse of a galvanic cell).在电解中，阳极连接外部电源的正极。氧化仍在阳极发生；电源强制将电子从阳极抽出并注入阴极（与原电池方向相反）。
(b) Cathode product: sodium metal阴极产物：钠金属A1·A1·A1
The cathode is connected to the negative terminal and attracts cations. In molten NaCl, the only cation is $\text{Na}^+$. Sodium ions migrate to the cathode and are reduced to sodium metal:阴极连接电源负极，吸引阳离子。熔融 NaCl 中唯一的阳离子是 $\text{Na}^+$。钠离子迁移至阴极，被还原为钠金属：
$$ \text{Na}^+(l) + e^- \;\rightarrow\; \text{Na}(l) \quad \text{(cathode / 阴极)} $$
Sodium is produced as a liquid (the melting point of NaCl is above that of Na).钠以液态生成（NaCl 的熔点高于 Na 的熔点）。
(c) Anode product: chlorine gas阳极产物：氯气A1·A1·A1
The anode is connected to the positive terminal and attracts anions. In molten NaCl, the only anion is $\text{Cl}^-$. Chloride ions migrate to the anode and are oxidised to chlorine gas:阳极连接电源正极，吸引阴离子。熔融 NaCl 中唯一的阴离子是 $\text{Cl}^-$。氯离子迁移至阳极，被氧化为氯气：
$$ 2\text{Cl}^-(l) \;\rightarrow\; \text{Cl}_2(g) + 2e^- \quad \text{(anode / 阳极)} $$
(d) Electrolysis vs galvanic cell电解与原电池的比较A1
Electrolysis uses electrical energy (from an external power supply) to drive a non-spontaneous chemical reaction. A galvanic cell converts chemical energy from a spontaneous reaction into electrical energy. The electrolysis of molten NaCl is non-spontaneous under standard conditions.电解利用电能（来自外部电源）驱动非自发化学反应。原电池则将自发反应的化学能转化为电能。熔融 NaCl 的电解在标准条件下是非自发的。
Molten-salt electrolysis eliminates water as a competing reactant, giving pure metal and halogen products.熔盐电解排除了水作为竞争反应物，从而得到纯金属和卤素产物。In aqueous NaCl electrolysis, water competes at both electrodes: $\text{H}_2$ forms at the cathode (water is reduced in preference to Na$^+$ because Na$^+$ has a very negative reduction potential) and $\text{Cl}_2$ or $\text{O}_2$ can form at the anode depending on concentration. Using molten (fused) NaCl removes this competition, ensuring Na metal at the cathode and $\text{Cl}_2$ at the anode. This is the industrial Downs process for producing sodium. The concept that the species with the higher reduction potential is preferentially reduced applies to competing cathode reactions, and the species most easily oxidised (lowest $E^\circ$) is preferentially oxidised at the anode.在 NaCl 水溶液电解中，水在两个电极处竞争：阴极优先产生 $\text{H}_2$（水的还原优先于 Na$^+$，因为 Na$^+$ 的还原电势极负），阳极根据浓度不同可产生 $\text{Cl}_2$ 或 $\text{O}_2$。使用熔融 NaCl 消除了这种竞争，确保阴极产生 Na 金属、阳极产生 $\text{Cl}_2$。这就是工业上生产钠的道恩斯法。"还原电势较高的物质优先被还原"这一概念适用于竞争性阴极反应；而"最容易被氧化（$E^\circ$ 最低）的物质在阳极优先被氧化"则适用于竞争性阳极反应。
(c) Which pair gives higher voltage哪对组合产生更高电压A1·A1
The Zn/Ag pair gives the higher voltage ($+1.56$ V vs $+0.51$ V). The cell EMF equals the difference between the two standard reduction potentials. The larger the gap between the cathode and anode $E^\circ$ values, the higher the cell voltage. Ag has the most positive $E^\circ$ ($+0.80$ V) and Zn the most negative $E^\circ$ ($-0.76$ V) of the three, maximising the difference.Zn/Ag 组合产生更高电压（$+1.56$ V 对比 $+0.51$ V）。电池电动势等于两个标准还原电势之差。阴极与阳极 $E^\circ$ 之差越大，电池电压越高。三者中 Ag 的 $E^\circ$ 最正（$+0.80$ V），Zn 的 $E^\circ$ 最负（$-0.76$ V），两者之差最大。
(d) Balanced overall equation for Zn/Ag cellZn/Ag 电池的总配平方程式M1·A1
To maximise cell voltage, pair the most negative $E^\circ$ (anode) with the most positive $E^\circ$ (cathode) available.要使电池电压最大，应将最负的 $E^\circ$（阳极）与最正的 $E^\circ$（阴极）配对。This principle underlies battery engineering: the larger the electrochemical potential difference, the more energy per coulomb of charge transferred. However, high-voltage cells also tend to use reactive metals like Zn or Li as the anode, raising safety and shelf-life concerns. Note again that $E^\circ$ values are not multiplied by stoichiometric factors: $E^\circ(\text{Ag}^+/\text{Ag})$ remains $+0.80$ V regardless of whether we write the half-reaction as $\text{Ag}^+ + e^- \to \text{Ag}$ or $2\text{Ag}^+ + 2e^- \to 2\text{Ag}$.这一原理是电池工程设计的基础：电化学电势差越大，每库仑电荷传输的能量越多。然而，高电压电池往往使用 Zn 或 Li 等活泼金属作阳极，带来安全性和保质期方面的问题。再次注意：$E^\circ$ 值不随化学计量系数而改变：无论将半反应写成 $\text{Ag}^+ + e^- \to \text{Ag}$ 还是 $2\text{Ag}^+ + 2e^- \to 2\text{Ag}$，$E^\circ(\text{Ag}^+/\text{Ag})$ 始终为 $+0.80$ V。
Q11MEDIUM中🇨🇦 ON安🇨🇦 BC卑AP-feeder FRQAP 衔接简答题§4 + §7 Activity series · electrolysis of aqueous solution活动性顺序 · 水溶液电解[9 marks][9 分]
Electrolysis of aqueous $\text{CuSO}_4$ with Pt electrodes. $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V; $E^\circ(\text{H}_2\text{O}/\text{H}_2) = -0.83$ V; $E^\circ_\text{anode}(\text{H}_2\text{O}/\text{O}_2) = -1.23$ V. (a) preferred cathode reaction; (b) anode half-reaction and gas; (c) what happens when Cu$^{2+}$ is exhausted; (d) electroplating connection.用 Pt 电极电解 $\text{CuSO}_4$ 水溶液。$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V；$E^\circ(\text{H}_2\text{O}/\text{H}_2) = -0.83$ V；$E^\circ_\text{anode}(\text{H}_2\text{O}/\text{O}_2) = -1.23$ V。(a) 优先阴极反应；(b) 阳极半反应和气体；(c) Cu$^{2+}$ 耗尽后的变化；(d) 电镀连接。
Answer:答案：(a)Cu$^{2+}$ reduction preferred ($E^\circ = +0.34$ V $>$ $-0.83$ V)Cu$^{2+}$ 优先被还原（$+0.34$ V $>$ $-0.83$ V） · (b) $2\text{H}_2\text{O}(l) \to \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$; oxygen gas氧气 · (c)Water is reduced; H$_2$ gas forms at cathode水被还原；阴极产生 H$_2$ 气体 · (d)Object connected to cathode; Cu$^{2+}$ is reduced and deposits on it待镀件连接阴极；Cu$^{2+}$ 在其上被还原沉积
(a) Preferred cathode reaction优先阴极反应A1·A1·A1
At the cathode, two reductions compete: $\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$, $E^\circ = +0.34$ V, and $2\text{H}_2\text{O}(l) + 2e^- \to \text{H}_2(g) + 2\text{OH}^-(aq)$, $E^\circ = -0.83$ V. The species with the higher reduction potential is preferentially reduced. Since $+0.34\ \text{V} > -0.83\ \text{V}$, Cu$^{2+}$ is preferentially reduced and copper metal is deposited at the cathode.阴极处两种还原反应竞争：$\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$，$E^\circ = +0.34$ V；以及 $2\text{H}_2\text{O}(l) + 2e^- \to \text{H}_2(g) + 2\text{OH}^-(aq)$，$E^\circ = -0.83$ V。还原电势较高的物质优先被还原。由于 $+0.34\ \text{V} > -0.83\ \text{V}$，Cu$^{2+}$ 优先被还原，铜金属沉积在阴极。
(b) Anode half-reaction and gas produced阳极半反应与产生的气体A1·A1
At the anode, water is oxidised (sulfate ion is too stable to be oxidised under these conditions):在阳极，水被氧化（硫酸根离子在此条件下过于稳定，不被氧化）：
$$ 2\text{H}_2\text{O}(l) \;\rightarrow\; \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- $$
The gas produced is oxygen (O$_2$).产生的气体是氧气（O$_2$）。
(c) When Cu$^{2+}$ is exhaustedCu$^{2+}$ 耗尽后A1·A1
Once all $\text{Cu}^{2+}$ ions are consumed, there is no longer a high-$E^\circ$ species available for reduction at the cathode. The next-best reduction reaction is water: $2\text{H}_2\text{O}(l) + 2e^- \to \text{H}_2(g) + 2\text{OH}^-(aq)$. Hydrogen gas begins to form at the cathode.一旦所有 $\text{Cu}^{2+}$ 离子耗尽，阴极处不再有高 $E^\circ$ 的物质可供还原。次优还原反应是水：$2\text{H}_2\text{O}(l) + 2e^- \to \text{H}_2(g) + 2\text{OH}^-(aq)$。阴极开始产生氢气。
(d) Electroplating: which electrode to connect电镀：连接哪个电极A1·A1
The object to be plated should be connected to the cathode (negative terminal). At the cathode, $\text{Cu}^{2+}$ ions are reduced and deposit as copper metal on the surface of the object. If connected to the anode, the object would be oxidised (corroded) rather than plated.待镀件应连接到阴极（负极）。在阴极，$\text{Cu}^{2+}$ 离子被还原，以铜金属形式沉积在待镀件表面。若连接到阳极，待镀件将被氧化（腐蚀），而非镀层。
In aqueous electrolysis, the species with the highest $E^\circ$ is reduced first at the cathode; the species most easily oxidised reacts first at the anode.在水溶液电解中，阴极处 $E^\circ$ 最高的物质优先被还原；阳极处最容易被氧化的物质优先反应。This problem illustrates the industrial significance of selective electrode reactions. In the copper refining industry, impure copper anodes dissolve while pure copper plates out at the cathode from a $\text{CuSO}_4$ electrolyte, selectively purifying the metal. The "competitive reduction" concept is fundamental to predicting electrolysis products whenever more than one ionic species is present. A common mistake is to assume that all ions discharge simultaneously; in practice, the most favoured reaction dominates until that reactant is depleted.这道题说明了选择性电极反应的工业意义。在铜精炼工业中，粗铜阳极溶解，纯铜从 $\text{CuSO}_4$ 电解质中在阴极沉积，从而选择性地纯化金属。"竞争性还原"概念对于预测存在多种离子时的电解产物至关重要。常见错误是认为所有离子同时放电；实际上，最有利的反应占主导地位，直至该反应物耗尽。
(a) Oxidation numbers and identification氧化数与识别A1·A1
In $\text{Cr}_2\text{O}_7^{2-}$: $2x + 7(-2) = -2 \Rightarrow x = +6$. In $\text{Cr}^{3+}$: $x = +3$. Change $+6 \to +3$: Cr is reduced (gains electrons). In $\text{Fe}^{2+}$: $x = +2$; in $\text{Fe}^{3+}$: $x = +3$. Change $+2 \to +3$: Fe is oxidised (loses electrons).在 $\text{Cr}_2\text{O}_7^{2-}$ 中：$2x + 7(-2) = -2 \Rightarrow x = +6$。在 $\text{Cr}^{3+}$ 中：$x = +3$。变化 $+6 \to +3$：Cr 被还原（得到电子）。在 $\text{Fe}^{2+}$ 中：$x = +2$；在 $\text{Fe}^{3+}$ 中：$x = +3$。变化 $+2 \to +3$：Fe 被氧化（失去电子）。
(b) Balance the Cr$_2$O$_7^{2-}$ reduction half-reaction配平 Cr$_2$O$_7^{2-}$ 还原半反应A1·A1·A1
Step 1 (balance Cr): 2 Cr on each side:步骤 1（配平 Cr）：两边各 2 个 Cr：
$$ \text{Cr}_2\text{O}_7^{2-} \;\rightarrow\; 2\text{Cr}^{3+} $$
Step 2 (balance O with H$_2$O): 7 O on left, add 7 H$_2$O on right:步骤 2（用 H$_2$O 配平 O）：左边 7 个 O，右边加 7 个 H$_2$O：
$$ \text{Cr}_2\text{O}_7^{2-} \;\rightarrow\; 2\text{Cr}^{3+} + 7\text{H}_2\text{O} $$
Step 3 (balance H with H$^+$): 14 H on right, add 14 H$^+$ on left:步骤 3（用 H$^+$ 配平 H）：右边 14 个 H，左边加 14 个 H$^+$：
$$ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \;\rightarrow\; 2\text{Cr}^{3+} + 7\text{H}_2\text{O} $$
Step 4 (balance charge): Left charge $= -2 + 14(+1) = +12$; Right charge $= 2(+3) = +6$. Add $6e^-$ to left:步骤 4（配平电荷）：左边电荷 $= -2 + 14(+1) = +12$；右边电荷 $= 2(+3) = +6$。左边加 $6e^-$：
$$ \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \;\rightarrow\; 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) $$
(c) Calculate $E^\circ_\text{cell}$ and state spontaneity计算 $E^\circ_\text{cell}$ 并说明自发性M1·A1
Dichromate is the cathode (reduction); Fe$^{2+}$/Fe$^{3+}$ is the anode (oxidation):重铬酸根为阴极（还原）；Fe$^{2+}$/Fe$^{3+}$ 为阳极（氧化）：
$$ E^\circ_\text{cell} \;=\; E^\circ_\text{cathode} - E^\circ_\text{anode} \;=\; (+1.33) - (+0.77) \;=\; +0.56\ \text{V} $$
$E^\circ_\text{cell} = +0.56\ \text{V} > 0$: the forward reaction is spontaneous under standard conditions. The dichromate ion oxidises Fe$^{2+}$ to Fe$^{3+}$ in acidic solution, as predicted.$E^\circ_\text{cell} = +0.56\ \text{V} > 0$：正向反应在标准条件下是自发的。重铬酸根离子在酸性溶液中将 Fe$^{2+}$ 氧化为 Fe$^{3+}$，符合预测。
(d) Device needed if $E^\circ_\text{cell}$ were negative若 $E^\circ_\text{cell}$ 为负所需的装置A1·A1
A negative $E^\circ_\text{cell}$ means the reaction is non-spontaneous in the forward direction. An external source of electrical energy would be needed to drive it. The device is an electrolytic cell (electrolyser), which uses a power supply to force a non-spontaneous redox reaction.$E^\circ_\text{cell}$ 为负意味着正向反应非自发。需要外部电能来驱动它。所需装置是电解槽（电解器），利用电源迫使非自发氧化还原反应进行。
The integrated question tests all of: oxidation number assignment, half-reaction balancing, $E^\circ_\text{cell}$ calculation, and the galvanic-vs-electrolytic distinction.综合题考查了全部内容：氧化数分配、半反应配平、$E^\circ_\text{cell}$ 计算以及原电池与电解池的区别。The Cr$_2$O$_7^{2-}$/Cr$^{3+}$ half-reaction is the benchmark "hard" half-reaction at the AP and IB level because it involves 2 Cr atoms, 7 O atoms, 14 H$^+$, and 6 electrons per formula unit. Memorising the final form is useful; being able to derive it step-by-step is essential. The $E^\circ_\text{cell}$ arithmetic here is straightforward because both potentials are given as reduction potentials and the formula is applied directly. The part (d) answer ties the whole unit together: galvanic cells (positive $E^\circ_\text{cell}$) release energy spontaneously; electrolytic cells (input of electrical energy) force non-spontaneous reactions. This distinction is the conceptual core of electrochemistry.$\text{Cr}_2\text{O}_7^{2-}$/Cr$^{3+}$ 半反应是 AP 和 IB 水平的标志性"难"半反应，因为每个化学式涉及 2 个 Cr 原子、7 个 O 原子、14 个 H$^+$ 和 6 个电子。记住最终形式固然有用，但能够逐步推导才是关键。此处的 $E^\circ_\text{cell}$ 运算很直接，因为两个电势均以还原电势给出，直接套公式即可。(d) 的答案将整个单元串联起来：原电池（$E^\circ_\text{cell} > 0$）自发释放能量；电解池（输入电能）迫使非自发反应进行。这一区别是电化学的概念核心。