Assign the oxidation number of sulfur in SO_4^2-. [1 point]

In Mg + Cl_2 → MgCl_2, identify the reducing agent. [1 point]

United StatesChemistrySyllabus dot point

How do we assign oxidation numbers, identify what is oxidized and reduced, and balance a redox reaction using half-reactions?

Topic 4.9 Oxidation-Reduction (Redox) Reactions: assign oxidation numbers, identify the species oxidized and reduced and the oxidizing and reducing agents, and balance redox reactions using half-reactions.

A focused answer to AP Chemistry Topic 4.9, covering oxidation-number rules, identifying oxidation and reduction, oxidizing and reducing agents, and balancing redox reactions by half-reactions including electron and charge balance, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this topic is asking
Oxidation numbers
Oxidation, reduction and the agents
Balancing with half-reactions
Try this

What this topic is asking

The College Board (Topic 4.9) wants you to analyze oxidation-reduction (redox) reactions: assign oxidation numbers, identify which species is oxidized and which is reduced, name the oxidizing and reducing agents, and balance the reaction using half-reactions. Redox is the third major reaction type (Topic 4.7) and underpins electrochemistry in Unit 9, so the bookkeeping of electrons and oxidation states must be secure.

Oxidation numbers

Oxidation numbers are the bookkeeping device that makes electron transfer visible. By comparing an atom's oxidation number before and after a reaction, you can see whether it lost electrons (number went up) or gained them (number went down), even when the electrons are not obviously transferred as in a simple ion exchange.

Oxidation, reduction and the agents

A memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The agents are easy to mix up: the species that is oxidized is the reducing agent (it causes reduction in its partner), and the species that is reduced is the oxidizing agent. The electrons lost in oxidation are exactly the electrons gained in reduction, which is why the two halves balance.

Balancing with half-reactions

The cleanest way to balance a redox reaction is to split it into two half-reactions: an oxidation half (showing electrons as products) and a reduction half (showing electrons as reactants). In each half, balance the atoms first, then balance charge by adding electrons. Finally, multiply the halves by whole numbers so that the electrons lost equal the electrons gained, and add them; the electrons cancel, leaving the balanced overall equation. This method guarantees both atom balance and charge balance.

Analyzing and balancing a redox reaction

For $\text{Fe}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s)$ , assign oxidation numbers, identify oxidation and reduction, and write the balanced half-reactions.

step 1 Assign oxidation numbers

Iron: $0$ (element) to $+2$ (in $\text{Fe}^{2+}$ ). Copper: $+2$ (in $\text{Cu}^{2+}$ ) to $0$ (element).

step 2 Identify oxidation and reduction

Iron's number increases ( $0 \to +2$ ), so iron is oxidized. Copper's number decreases ( $+2 \to 0$ ), so copper is reduced.

step 3 Write the half-reactions

Oxidation: $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$ . Reduction: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ .

step 4 Combine

Both halves involve two electrons, so they balance directly; adding them cancels the electrons to give $\text{Fe} + \text{Cu}^{2+} \rightarrow \text{Fe}^{2+} + \text{Cu}$ . Iron is the reducing agent and $\text{Cu}^{2+}$ is the oxidizing agent.

Try this

Q1. Assign the oxidation number of sulfur in $\text{SO}_4^{2-}$ . [1 point]

Cue. Oxygen is $-2$ (four give $-8$ ); the ion is $-2$ , so sulfur is $+6$ .

Q2. In $\text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2$ , identify the reducing agent. [1 point]

Cue. Magnesium (it is oxidized from $0$ to $+2$ , donating electrons, so it is the reducing agent).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (short FRQ). Consider the reaction

\text{Zn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s)

. (a) Assign oxidation numbers to zinc and silver before and after. (b) Identify the species oxidized and the species reduced. (c) Identify the oxidizing agent and the reducing agent. (d) Write the two half-reactions and show that electrons balance.

Show worked answer →

A 4-point FRQ on redox analysis.

(a) Oxidation numbers (1 point): zinc goes from $0$ (element) to $+2$ ; silver goes from $+1$ (in $\text{Ag}^+$ ) to $0$ (element).
(b) Oxidized and reduced (1 point): zinc is oxidized (its oxidation number increases, it loses electrons); silver is reduced (its oxidation number decreases, it gains electrons).
(c) Agents (1 point): the oxidizing agent is $\text{Ag}^+$ (it is reduced and causes oxidation); the reducing agent is $\text{Zn}$ (it is oxidized and causes reduction).
(d) Half-reactions (1 point): oxidation $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ ; reduction $2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}$ . Both involve two electrons, so electrons balance and cancel.

Markers reward correct oxidation numbers, the oxidized/reduced identification, the agents, and balanced half-reactions whose electrons cancel.

AP 2022 (style)1 marksSection I (multiple choice). In the reaction

\text{Cu}(s) + 2\text{H}^+ + ... \rightarrow \text{Cu}^{2+} + ...

, copper is (A) reduced and acts as the oxidizing agent (B) oxidized and acts as the reducing agent (C) reduced and acts as the reducing agent (D) oxidized and acts as the oxidizing agent. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

Copper goes from $0$ to $+2$ , so it loses electrons and is oxidized. The species that is oxidized is the reducing agent (it donates electrons, causing the reduction of something else). So copper is oxidized and acts as the reducing agent.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)