Chemistry 2e · Electrochemistry

Review of Redox Chemistry

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Many of the most important reactions in chemistry and biology involve the transfer of electrons. Redox (–) reactions include combustion, rusting, photosynthesis, respiration, battery discharge, and the metabolism of drugs. In every redox reaction, one species loses electrons — it is oxidized — while another species gains them — it is reduced. The two processes are inseparable: oxidation cannot occur without a simultaneous reduction, which is why we call them half-reactions that must be combined.

Because electrons are invisible, chemists use oxidation numbers as a bookkeeping tool to track electron transfer, and they split balanced equations into half-reactions to see exactly where electrons go. This topic reviews how to assign oxidation numbers, identify oxidizing and reducing agents, and balance redox equations in acidic and basic solution. It is the foundation for everything else in this chapter: galvanic cells, cell potentials, batteries, corrosion, and electrolysis all assume you can dissect a reaction into its half-reactions.

Why this matters

Every battery and fuel cell works by separating the oxidation and reduction half-reactions so electrons must travel through a wire.
Corrosion of metals (rusting) is a redox process; understanding it leads to prevention strategies covered later in this chapter.
Biology depends on it: cellular respiration oxidizes glucose and reduces oxygen; photosynthesis does the reverse. Electron-transfer chains move energy through living systems.
Industrial chemistry: extraction of metals from ores, electroplating, and water treatment (e.g., chlorine disinfection) are all redox processes.
Exams: oxidation-number assignment and balancing are guaranteed skills on any electrochemistry assessment — get them solid now.

The college version

Core Concepts

Oxidation and reduction

Oxidation = loss of electrons (an increase in ).
Reduction = gain of electrons (a decrease in oxidation number).
Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain.

Electrons are conserved: the total number lost by the oxidized species equals the total gained by the reduced species. That is the constraint that makes balancing redox equations work.

Oxidation numbers: the bookkeeping rules

Oxidation numbers are assigned by a small set of priority rules:

Free elements (uncombined) have oxidation number 0.
Monatomic ions have oxidation number equal to their charge.
In compounds, fluorine is always -1.
Hydrogen is +1 except in metal hydrides (e.g., NaH), where it is -1.
Oxygen is -2 except in peroxides (-1, e.g., H₂O₂) and in OF₂ (+2).
The sum of oxidation numbers equals the charge of the species: 0 for neutral molecules, the ion charge for polyatomic ions.

Oxidizing and reducing agents

The is the species that gets reduced (it accepts electrons and causes the other species to be oxidized).
The is the species that gets oxidized (it donates electrons).
The agent is the reactant species, not the product. In Zn + Cu²⁺ → Zn²⁺ + Cu, Cu²⁺ is the oxidizing agent and Zn is the reducing agent.

Half-reactions

A half-reaction shows only the oxidation or only the reduction part, including electrons as explicit participants. For example:

$Zn \to Zn 2+ + 2e -$

$Cu 2+ + 2e - \to Cu$

Adding the two half-reactions (electrons cancel) gives the overall balanced equation. Half-reactions are also the units used to describe electrodes in cells.

Balancing redox equations in acidic solution

The half-reaction method, in acid:

Split the skeleton equation into oxidation and reduction half-reactions.
Balance atoms other than O and H in each half-reaction.
Balance O by adding H₂O.
Balance H by adding H⁺.
Balance charge by adding electrons.
Multiply each half-reaction so electrons cancel; add and simplify.

Balancing redox equations in basic solution

Balance as if acidic, then neutralize: add OH^- to both sides to convert each H⁺ into H₂O, combine and cancel water molecules.

How It Works / Step-by-Step Process

Assign oxidation numbers to every atom; identify which species changes state.
Write the two half-reactions (oxidation and reduction) with electrons on the appropriate sides.
Balance atoms: non-O/H first, then O with H₂O, then H with H⁺ (acidic) or convert later (basic).
Balance charge with electrons; equalize electron counts by multiplying half-reactions.
Add the half-reactions, cancel electrons and any species appearing on both sides, and verify atoms and charge balance in the final equation.

Common Confusions

Do Not Confuse	With	Difference
Oxidizing agent	Species that is oxidized	The oxidizing agent is reduced (it takes electrons); the reducing agent is oxidized.
Oxidation number	Ionic charge	Oxidation numbers are bookkeeping assignments and can be fractional or larger than typical charges.
O in all compounds	O = -2 everywhere	Peroxides (H₂O₂) have O = -1; OF₂ has O = +2.
H in all compounds	H = +1 everywhere	In metal hydrides (NaH, CaH₂), H = -1.
Balancing by inspection	Half-reaction balancing	Redox equations with polyatomic ions usually need the half-reaction method to balance charge properly.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Some reactions trade electrons like trading cards. When an atom gives away electrons, we say it got oxidized; when an atom takes electrons, it got reduced. Oxidizing agents are "electron stealers," and reducing agents are "electron givers." Chemists keep score with oxidation numbers, just like keeping score in a game, so they can see exactly who gave and who took.

Worked example

Example 1: Assigning oxidation numbers

Find the oxidation number of each element in MnO₄^-, H₂O₂, and Cr₂O₇^2-.

For MnO₄^-: O is -2 each (4 O atoms = -8); the ion charge is -1, so Mn must be +7:

$x + 4(-2) = -1 \Rightarrow x = +7$

For H₂O₂: H is +1 each (total +2); the molecule is neutral, so the two O atoms sum to -2, giving O = -1 each — the peroxide exception to the usual -2.

For Cr₂O₇^2-: O is -2 each (7 O atoms = -14); the ion charge is -2, so the two Cr atoms sum to +12, giving Cr = +6 each:

$2y + 7(-2) = -2 \Rightarrow 2y = +12 \Rightarrow y = +6$

Example 2: Balancing in acidic solution

Balance MnO₄^- + Fe²⁺ → Mn²⁺ + Fe³⁺ in acid.

Reduction half-reaction (Mn: +7 → +2, gains 5 e^-):

$MnO 4 - + 8 H + + 5e - \to Mn 2+ + 4 H 2 O$

Oxidation half-reaction (Fe: +2 → +3, loses 1 e^-):

$Fe 2+ \to Fe 3+ + e -$

Electrons must cancel: multiply the oxidation half-reaction by 5, then add:

$MnO 4 - + 8 H + + 5 Fe 2+ \to Mn 2+ + 4 H 2 O + 5 Fe 3+$

Check: atoms balance (1 Mn, 4 O, 8 H, 5 Fe each side) and charge balances (-1 + 8 + 10 = +17 left; +2 + 0 + 15 = +17 right). The permanganate ion is the oxidizing agent (reduced); Fe²⁺ is the reducing agent (oxidized).

Example 3: Balancing in basic solution

Balance Al + NO₃^- → Al(OH)₄^- + NH₃ in base.

Oxidation: Al + 4OH^- → Al(OH)₄^- + 3e^-

Reduction: NO₃^- + 6H₂O + 8e^- → NH₃ + 9OH^-

Balance electrons (LCM of 3 and 8 is 24): multiply oxidation by 8, reduction by 3, add and cancel:

$8 Al + 3 NO 3 - + 5 OH - + 18 H 2 O \to 8 Al(OH) 4 - + 3 NH 3$

Verify atoms and charge: left 3(-1) + 5(-1) = -8; right 8(-1) = -8. Balanced. Nitrate is the oxidizing agent; aluminum is the reducing agent.

Key takeaways

Oxidation = loss of electrons (number increases); reduction = gain (number decreases). OIL RIG.
Oxidation numbers: free element 0; O usually -2 (peroxides -1); H usually +1 (hydrides -1); sum equals species charge.
The oxidizing agent is reduced; the reducing agent is oxidized. Identify agents from reactants, not products.
Half-reactions must be balanced for atoms and charge; electrons cancel when combined.
In acidic solution: balance O with H₂O, H with H⁺, then charge with e^-.
In basic solution: balance as acidic, then add OH^- to each side to convert H⁺ to water.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

In Cu²⁺ + Zn → Cu + Zn²⁺, which species is oxidized, which is reduced, and which is the oxidizing agent?
Show answer
Zn is oxidized (0 → +2, loses electrons) and is the reducing agent; Cu²⁺ is reduced (+2 → 0, gains electrons) and is the oxidizing agent.
What is the oxidation number of S in SO₄^2-?
Show answer
S = +6: x + 4(-2) = -2 ⇒ x = +6.
Why must oxidation and reduction always occur together?
Show answer
Electrons are conserved — the electrons lost by the oxidized species must be gained by the reduced species, so the two processes are always coupled.
What is the first step in balancing a redox equation in basic solution?
Show answer
Balance the equation as if it were acidic (using H⁺ and H₂O), then add OH^- to both sides to convert H⁺ into water and simplify.
True or false: In H₂O₂, oxygen has an oxidation number of -2. Explain.
Show answer
False — H₂O₂ is a peroxide, so each O is -1 (H is +1 each, and the molecule is neutral).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

oxidation: Loss of electrons, raising an atom's oxidation number.
reduction: Gain of electrons, lowering an atom's oxidation number.
oxidation number: Bookkeeping charge assigned by rules to track electron transfer.
oxidizing agent: The reactant that is reduced (accepts electrons).
reducing agent: The reactant that is oxidized (donates electrons).
half-reaction: One side of a redox process written with electrons shown.

Sources & references

openstax.org — Chemistry 2e

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