PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC/AB short answer · 24 marksAP 风格选择题 + 安/卑/阿省考短答 · 共 24 分
Section A · Short ResponseA 部分 · 短答题
For MCQs, circle the letter and show enough working in the margin for a marker to verify. For short-answer items, write full chemical equations where asked and state units. Use standard reduction potentials where given in the question. No calculator on Q1–Q3; calculator permitted on Q4–Q5.选择题请圈出字母并在空白处写下足以让阅卷人核对的过程。短答题在要求时写出完整化学方程式并注明单位。在题目给出标准还原电势的地方使用它。Q1–Q3 不可使用计算器；Q4–Q5 可用计算器。
(a)Identify which species is oxidised and state the change in its oxidation number.确定哪种物质被氧化，并写出其氧化数的变化。[2]
(b)Identify the oxidising agent and the reducing agent.确定氧化剂和还原剂。[2]
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]
How many electrons are transferred per $\text{MnO}_4^-$ ion in its reduction to $\text{Mn}^{2+}$ in acidic solution?在酸性溶液中，每个 $\text{MnO}_4^-$ 离子还原为 $\text{Mn}^{2+}$ 时转移多少个电子？
(A)22
(B)33
(C)55
(D)77
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]
Use the activity series to answer the following. Activity order (most to least active): $\text{Mg} > \text{Zn} > \text{Fe} > \text{Cu} > \text{Ag}$.利用活动性顺序回答以下问题。活动性由强到弱：$\text{Mg} > \text{Zn} > \text{Fe} > \text{Cu} > \text{Ag}$。
(a)Predict whether a reaction occurs when zinc metal is placed in a solution of silver nitrate, $\text{AgNO}_3(aq)$. Explain your reasoning using the activity series.预测将锌片放入硝酸银溶液 $\text{AgNO}_3(aq)$ 中是否发生反应。用活动性顺序解释你的推理。[3]
(b)Write the balanced ionic equation for the reaction in part (a), or write "NR" if no reaction occurs.为 (a) 中的反应写出配平的离子方程式，若无反应则写"NR"。[2]
(c)State one observable sign that a reaction has occurred.写出一个可观察到的反应发生的证据。[1]
A galvanic cell is constructed with a zinc electrode in $\text{Zn}^{2+}(aq)$ and a copper electrode in $\text{Cu}^{2+}(aq)$, connected by a salt bridge and an external wire.一个原电池由锌电极（浸于 $\text{Zn}^{2+}(aq)$ 中）和铜电极（浸于 $\text{Cu}^{2+}(aq)$ 中）组成，用盐桥和外部导线连接。
(a)Identify the anode and cathode, and state at which electrode oxidation occurs.确定阳极和阴极，并写出氧化反应发生在哪个电极。[2]
(b)Write the half-reaction at each electrode.写出每个电极处的半反应。[2]
(c)State the direction of electron flow in the external circuit.写出外部电路中电子流动的方向。[2]
(d)Describe the role of the salt bridge and state the direction that negative ions (anions) move through it.描述盐桥的作用，并说明负离子（阴离子）在其中的移动方向。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + Honors · 33 marksAP 衔接简答题 + 荣誉级 · 共 33 分
Section B · Extended ResponseB 部分 · 简答题
Show every step. Write unbalanced half-reactions before balancing. State oxidation-number changes explicitly. When asked for a cell EMF, use $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ with all potentials as standard reduction potentials. Calculator permitted on Q6–Q9.每一步都要写出。配平前先写出未配平的半反应。明确写出氧化数变化。计算电池电动势时使用 $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$，所有电势均为标准还原电势。Q6–Q9 可用计算器。
The following unbalanced reaction takes place in acidic aqueous solution: $\text{MnO}_4^-(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{Fe}^{3+}(aq)$以下未配平的反应在酸性水溶液中进行：$\text{MnO}_4^-(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{Fe}^{3+}(aq)$
(a)Write the unbalanced half-reaction for the reduction of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$. State the change in oxidation number of manganese.写出 $\text{MnO}_4^-$ 还原为 $\text{Mn}^{2+}$ 的未配平半反应。写出锰的氧化数变化。[2]
(b)Balance the reduction half-reaction by (i) balancing Mn, then O using $\text{H}_2\text{O}$, then H using $\text{H}^+$, then charge using electrons. Show each sub-step.按以下步骤配平还原半反应：(i) 配平 Mn，用 $\text{H}_2\text{O}$ 配平 O，用 $\text{H}^+$ 配平 H，最后用电子配平电荷。写出每个子步骤。[3]
(c)Write and balance the oxidation half-reaction for $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$.写出并配平 $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$ 的氧化半反应。[1]
(d)Combine the two half-reactions to produce the overall balanced ionic equation. State the mole ratio of $\text{MnO}_4^-$ to $\text{Fe}^{2+}$.合并两个半反应，写出总配平离子方程式。写出 $\text{MnO}_4^-$ 与 $\text{Fe}^{2+}$ 的摩尔比。[3]
A galvanic cell is set up using a zinc half-cell ($\text{Zn}^{2+}/\text{Zn}$) and a copper half-cell ($\text{Cu}^{2+}/\text{Cu}$). Use the standard reduction potentials: $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$; $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}$.用锌半电池（$\text{Zn}^{2+}/\text{Zn}$）和铜半电池（$\text{Cu}^{2+}/\text{Cu}$）建立原电池。使用以下标准还原电势：$E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$；$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}$。
(a)Identify which half-cell acts as the cathode. Justify your answer.确定哪个半电池作为阴极。说明理由。[2]
(b)Calculate the standard cell EMF, $E^\circ_\text{cell}$, using $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$.用 $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ 计算标准电池电动势 $E^\circ_\text{cell}$。[2]
(c)State whether the overall cell reaction is spontaneous under standard conditions. Explain the link between the sign of $E^\circ_\text{cell}$ and spontaneity.说明在标准条件下总电池反应是否自发。解释 $E^\circ_\text{cell}$ 的符号与自发性之间的关系。[2]
(d)Write the overall balanced equation for the cell reaction.写出电池总反应的配平方程式。[2]
(a)Consider a cell made from $\text{Mg}(s)/\text{Mg}^{2+}(aq)$ and $\text{Ag}^+(aq)/\text{Ag}(s)$. Calculate $E^\circ_\text{cell}$ and state whether this cell is spontaneous.考虑由 $\text{Mg}(s)/\text{Mg}^{2+}(aq)$ 和 $\text{Ag}^+(aq)/\text{Ag}(s)$ 组成的电池。计算 $E^\circ_\text{cell}$ 并说明该电池是否自发。[3]
(b)Consider a cell made from $\text{Cu}(s)/\text{Cu}^{2+}(aq)$ and $\text{Pb}^{2+}(aq)/\text{Pb}(s)$. Calculate $E^\circ_\text{cell}$ assuming $\text{Cu}$ is the anode and $\text{Pb}$ is the cathode. Is this arrangement spontaneous? If not, identify which metal should be the anode for a spontaneous cell.考虑由 $\text{Cu}(s)/\text{Cu}^{2+}(aq)$ 和 $\text{Pb}^{2+}(aq)/\text{Pb}(s)$ 组成的电池。假设 $\text{Cu}$ 为阳极、$\text{Pb}$ 为阴极，计算 $E^\circ_\text{cell}$。该配置是否自发？若否，确定哪种金属应作为阳极使电池自发。[3]
(c)Explain in one sentence why a metal with a more negative standard reduction potential tends to act as the anode in a galvanic cell.用一句话解释为什么标准还原电势更负的金属在原电池中倾向于作为阳极。[2]
Molten sodium chloride (NaCl) is electrolysed using inert carbon electrodes. The ions present are $\text{Na}^+$ and $\text{Cl}^-$.用惰性碳电极电解熔融氯化钠（NaCl）。体系中存在的离子为 $\text{Na}^+$ 和 $\text{Cl}^-$。
(a)In electrolysis, which electrode is connected to the positive terminal of the power supply? Is it the anode or the cathode?在电解中，哪个电极连接到电源正极？它是阳极还是阴极？[1]
(b)Identify the product formed at the cathode. Write the half-reaction and explain which ion is attracted to the cathode and why.确定阴极处产生的产物。写出半反应，并解释哪种离子被吸引到阴极以及原因。[3]
(c)Identify the product formed at the anode. Write the half-reaction and explain which ion is attracted to the anode and why.确定阳极处产生的产物。写出半反应，并解释哪种离子被吸引到阳极以及原因。[3]
(d)Compare electrolysis with a galvanic cell: in which device does electrical energy drive a non-spontaneous chemical reaction?将电解与原电池进行比较：在哪种装置中，电能驱动非自发化学反应？[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 26 marks阿省毕业考 + 通用题型 · 共 26 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all symbols before substituting. Write the formula used before substituting numbers. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.代入数值前先定义所有符号。先写出所用公式，再代入数值。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
A technician is designing cells for energy storage. She considers the following half-cells and their standard reduction potentials:
$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$, $E^\circ = +0.80\ \text{V}$
$\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)$, $E^\circ = -0.76\ \text{V}$
$\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s)$, $E^\circ = -0.25\ \text{V}$一名技术员正在设计储能电池。她考虑以下几个半电池及其标准还原电势：
$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$，$E^\circ = +0.80\ \text{V}$
$\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s)$，$E^\circ = -0.76\ \text{V}$
$\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s)$，$E^\circ = -0.25\ \text{V}$
(a)Calculate $E^\circ_\text{cell}$ for a cell made from the $\text{Zn}/\text{Zn}^{2+}$ and $\text{Ag}^+/\text{Ag}$ half-cells, with $\text{Zn}$ as the anode.计算由 $\text{Zn}/\text{Zn}^{2+}$ 和 $\text{Ag}^+/\text{Ag}$ 半电池组成的电池的 $E^\circ_\text{cell}$，以 $\text{Zn}$ 为阳极。[2]
(b)Calculate $E^\circ_\text{cell}$ for a cell made from the $\text{Zn}/\text{Zn}^{2+}$ and $\text{Ni}^{2+}/\text{Ni}$ half-cells, with $\text{Zn}$ as the anode.计算由 $\text{Zn}/\text{Zn}^{2+}$ 和 $\text{Ni}^{2+}/\text{Ni}$ 半电池组成的电池的 $E^\circ_\text{cell}$，以 $\text{Zn}$ 为阳极。[2]
(c)Which cell produces the higher voltage? Identify which half-cell pairing maximises the EMF and explain why in terms of the reduction potential values.哪个电池产生更高的电压？确定哪对半电池组合可使电动势最大，并用还原电势的数值解释原因。[2]
(d)Write the balanced overall cell reaction for the higher-voltage cell from part (c).为 (c) 中电压较高的电池写出配平的总反应方程式。[2]
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]
(a)At the cathode, two reductions are possible: $\text{Cu}^{2+}$ can be reduced, or water can be reduced to produce $\text{H}_2$. Which reaction is preferred and why? (Hint: the species with the higher reduction potential is preferentially reduced.)在阴极，有两种可能的还原反应：$\text{Cu}^{2+}$ 可被还原，或水被还原产生 $\text{H}_2$。哪种反应优先发生？为什么？（提示：还原电势较高的物质优先被还原。）[3]
(b)Write the half-reaction for the product formed at the anode. Name the gas produced.写出阳极处生成产物的半反应。写出产生的气体名称。[2]
(c)Predict what would change at the cathode if all the $\text{Cu}^{2+}$ ions were consumed. What product would then form?预测若所有 $\text{Cu}^{2+}$ 离子耗尽后阴极会发生什么变化。之后会产生什么产物？[2]
(d)Industrial electroplating uses electrolysis to coat a metal object with copper. Describe which electrode the object to be plated should be connected to and explain why.工业电镀利用电解在金属制品上镀铜。描述待镀件应连接到哪个电极，并解释原因。[2]
The following reaction is proposed in acidic solution: $\text{Cr}_2\text{O}_7^{2-}(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Cr}^{3+}(aq) + \text{Fe}^{3+}(aq)$ (unbalanced). Standard reduction potentials: $E^\circ(\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}) = +1.33\ \text{V}$; $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\ \text{V}$.以下反应在酸性溶液中进行（未配平）：$\text{Cr}_2\text{O}_7^{2-}(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Cr}^{3+}(aq) + \text{Fe}^{3+}(aq)$。标准还原电势：$E^\circ(\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}) = +1.33\ \text{V}$；$E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\ \text{V}$。
(a)Assign oxidation numbers to Cr in $\text{Cr}_2\text{O}_7^{2-}$ and in $\text{Cr}^{3+}$. State whether chromium is oxidised or reduced.对 $\text{Cr}_2\text{O}_7^{2-}$ 和 $\text{Cr}^{3+}$ 中的 Cr 分配氧化数。说明铬是被氧化还是被还原。[2]
(b)Write the balanced half-reaction for the reduction of $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Cr}^{3+}$ in acidic solution. (Balance Cr, then O with $\text{H}_2\text{O}$, then H with $\text{H}^+$, then charge with electrons.)写出酸性溶液中 $\text{Cr}_2\text{O}_7^{2-}$ 还原为 $\text{Cr}^{3+}$ 的配平半反应。（依次配平 Cr、用 $\text{H}_2\text{O}$ 配平 O、用 $\text{H}^+$ 配平 H、用电子配平电荷。）[3]
(c)Calculate $E^\circ_\text{cell}$ for this reaction using $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$. State whether the forward reaction is spontaneous and justify your answer.用 $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$ 计算该反应的 $E^\circ_\text{cell}$。说明正向反应是否自发，并说明理由。[2]
(d)If the sign of $E^\circ_\text{cell}$ were negative for this reaction, what type of device would be needed to drive it forward? Name that device.若该反应的 $E^\circ_\text{cell}$ 为负值，需要什么装置才能驱动其正向进行？写出该装置的名称。[2]
🇺🇸 US NGSS美国 NGSSHS-PS1-2 (qualitative redox); electrochemistry honors-levelHS-PS1-2（定性氧化还原）；电化学为荣誉级
🇨🇦 Ontario安大略SCH4U · C2: ElectrochemistryC2：电化学
🇨🇦 British Columbia不列颠哥伦比亚Chem 12: Oxidation-Reduction化学 12：氧化还原
🇨🇦 Alberta阿尔伯塔Chem 30 · Unit D: ElectrochemistryD 单元：电化学
Full Syllabus Map in ../Study Guides/Unit_13_Redox_and_Electrochemistry.html. NGSS covers oxidation-number identification only; half-reaction balancing and cell-EMF calculations are ON/BC/AB Chem 30 and AP Chem territory.完整大纲对照见 ../Study Guides/Unit_13_Redox_and_Electrochemistry.html。NGSS 仅涵盖氧化数识别；半反应配平与电池电动势计算属于 ON/BC/AB Chem 30 及 AP 化学范围。