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Chemistry: Atoms First 2e · Electrochemistry

Review of Redox Chemistry

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Want it in plain words first? Jump to Eli explains — the same idea, no jargon.
On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Redox chemistry is the study of reactions in which electrons are transferred from one species to another. The word redox fuses reduction and oxidation, two processes that can never occur alone: every electron one atom loses must be accepted by a different atom. Iron rusting and a battery lighting a flashlight are the same kind of event — electron transfer. This topic builds the vocabulary and bookkeeping tools — states, half-reactions, oxidizing and reducing agents — used throughout the chapter.

Why this matters

Redox reactions power modern life: they run the batteries in phones and cars, drive the corrosion that destroys bridges and pipelines, and underlie bleaching, metal refining, and chlorine and aluminum production. Photosynthesis and cellular respiration are elaborate redox sequences, and glucose test strips detect sugar in blood through a tiny redox reaction. Every topic that follows — galvanic cells, electrode potentials, batteries — is this electron-transfer chemistry organized and measured, so mastering oxidation states and balancing now pays off later.

The college version

Core Concepts

Oxidation and reduction are two halves of one event

Oxidation is the loss of electrons; is the gain of electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain). Electrons cannot float free in solution, so an oxidation must always pair with a reduction — together they form a redox reaction. The older oxygen-based definitions describe many real reactions, but the electron definition is the general one used in this course.

Oxidation states: the bookkeeping tool

An is a bookkeeping charge assigned to each atom in a formula, as if the more electronegative atom owned the shared electrons — a counting device, not a real charge. The key rules: free elements are 0; monatomic ions equal their charge; fluorine is always −1; hydrogen is usually +1 but −1 in metal hydrides (NaH); oxygen is usually −2 but −1 in peroxides; and the states in a species sum to its charge (0 for a neutral molecule). Atoms whose oxidation state rises are oxidized; those whose state falls are reduced.

Oxidizing and reducing agents

The (oxidant) is the species that gets reduced — it accepts electrons and oxidizes something else. The (reductant) is the species that gets oxidized — it donates electrons. In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is the reducing agent and Cu²⁺ is the oxidizing agent. Exams love to flip this pair: the agent does to the other species what its name says — the oxidizing agent oxidizes, the reducing agent reduces.

Balancing redox equations by half-reactions

A shows one half of the electron transfer with electrons written explicitly. The half-reaction method balances redox by treating oxidation and reduction separately. In acidic solution: (1) split the skeleton into two half-reactions; (2) balance atoms other than O and H; (3) balance O with H₂O; (4) balance H with H⁺; (5) balance charge with electrons; (6) multiply so electrons cancel; (7) add the halves. In basic solution, balance as if acidic, then neutralize each H⁺ by adding OH⁻ to both sides.

Redox stoichiometry: counting electrons

The balanced half-reactions give the electron bookkeeping for calculations: one mole of MnO₄⁻ in acid accepts 5 moles of electrons (Mn +7 → +2), while one mole of Fe²⁺ donates 1 (Fe +2 → +3). These ratios are the mole bridge of titration-style problems.

How It Works / Step-by-Step Process

  1. Assign oxidation states to every atom and circle the ones that change.
  2. Split the skeleton into oxidation and reduction half-reactions.
  3. Balance each half-reaction — atoms (H₂O, then H⁺ in acid), then charge with electrons.
  4. Multiply so electrons cancel, add the halves, and verify atoms and charge balance.

Common Confusions

Do Not ConfuseWithDifference
OxidationReductionOxidation is electron loss; reduction is electron gain.
Oxidizing agentReducing agentThe oxidizing agent is reduced (accepts e⁻); the reducing agent is oxidized (donates e⁻).
Oxidation stateActual chargeOxidation states are bookkeeping assignments; only monatomic ions carry that real charge.
Oxygen in peroxidesOxygen in ordinary compoundsIn H₂O₂, O is −1; in H₂O or CO₂ it is −2.
Hydrogen in metal hydridesHydrogen in acidsH is −1 in NaH and other hydrides; +1 with nonmetals.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of electrons as marbles. Some atoms want marbles and some want to give them away. When an atom gives a marble away, that's oxidation; when it grabs one, that's reduction — and every giveaway must be a grab by somebody else. Chemists keep score with a pretend charge number on each atom, making sure every marble that leaves one atom arrives at another.

Worked example

Example 1: Assigning oxidation states

Find the oxidation state of the element of interest, showing the equation before substituting. For chromium in K₂Cr₂O₇: potassium is +1 (two K atoms, +2 total) and oxygen is −2 (seven O atoms, −14 total), so with x as the chromium state:

2(+1) + 2x + 7(-2) = 0   ⇒  2x = +12   ⇒  x = +6

For manganese in MnO₄⁻, the species charge is −1:

x + 4(-2) = -1   ⇒  x = +7

For oxygen in H₂O₂ (hydrogen is +1; two H atoms give +2):

2(+1) + 2x = 0   ⇒  x = -1

So Cr is +6, Mn is +7, and peroxide oxygen is −1 — the classic exam trap.

Example 2: Balancing a redox equation in acidic solution

Balance the permanganate–iron(II) reaction in acid: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.

Reduction half-reaction — balance O with H₂O, H with H⁺, then charge with electrons:

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Oxidation half-reaction:

Fe2+ → Fe3+ + e-

Each MnO₄⁻ needs 5 electrons but each Fe²⁺ gives only 1, so multiply the iron half-reaction by 5 and add:

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

Check the charge balance: left side −1 + 8 + 10 = +17; right side +2 + 15 = +17.

Example 3: Redox stoichiometry with dimensional analysis

How many milliliters of 0.0200 M KMnO₄ are needed to oxidize all the Fe²⁺ in 25.00 mL of 0.100 M FeSO₄? The balanced equation gives 1 mol MnO₄⁻ : 5 mol Fe²⁺. Starting from n = C × V:

n(Fe2+) = 0.100 molL × 0.02500 L = 2.50 × 10-3 mol

Convert with the stoichiometric ratio (mol Fe²⁺ cancels):

n(MnO4-) = 2.50 × 10-3 mol Fe2+ × 1 mol MnO4-5 mol Fe2+ = 5.00 × 10-4 mol

Then use V = n/C:

V = 5.00 × 10-4 mol0.0200 molL = 0.0250 L = 25.0 mL

The mol units cancel, leaving volume: 25.0 mL of KMnO₄ solution is required.

Key takeaways

  • OIL RIG: oxidation is loss of electrons, reduction is gain — they always happen together.
  • Oxidation states are bookkeeping charges; they sum to 0 for molecules and to the ion charge for ions.
  • The oxidizing agent is reduced; the reducing agent is oxidized — name the agent by what it does to the other species.
  • Oxygen is −2 normally but −1 in peroxides; hydrogen is +1 normally but −1 in metal hydrides.
  • Balance redox with H₂O for O, H⁺ for H, and e⁻ for charge; neutralize with OH⁻ in base.
  • 1 mol MnO₄⁻ (acidic) accepts 5 mol e⁻; 1 mol Cr₂O₇²⁻ accepts 6 mol e⁻; 1 mol Fe²⁺ donates 1 mol e⁻.

Check yourself

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

  1. In the reaction 2Na + Cl₂ → 2NaCl, which element is oxidized and which is reduced?

    Show answer

    Sodium is oxidized (0 → +1); chlorine is reduced (0 → −1).

  2. What is the oxidation state of sulfur in SO₄²⁻? Of chromium in Cr₂O₇²⁻?

    Show answer

    Sulfur is +6: x + 4(-2) = -2 ⇒ x = +6. Chromium is +6: 2x + 7(-2) = -2 ⇒ x = +6.

  3. In Zn + Cu²⁺ → Zn²⁺ + Cu, identify the oxidizing agent and the reducing agent.

    Show answer

    The oxidizing agent is Cu²⁺ (it is reduced to Cu); the reducing agent is Zn (it is oxidized to Zn²⁺).

  4. Why must oxidation and reduction always occur together?

    Show answer

    Electrons cannot exist free in solution; every electron lost by one species must be gained by another.

  5. How many moles of electrons does one mole of Cr₂O₇²⁻ accept when it is reduced to Cr³⁺ in acid?

    Show answer

    Six. Cr goes from +6 to +3 (a change of 3 per atom), and Cr₂O₇²⁻ contains two Cr atoms.

  6. Balance MnO₄⁻ + H₂O₂ → Mn²⁺ + O₂ in acidic solution.

    Show answer

    2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O. (Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O; oxidation: H₂O₂ → O₂ + 2H⁺ + 2e⁻; multiply by 2 and 5, then add.)

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

oxidation
Loss of electrons by a species (rise in oxidation state).
reduction
Gain of electrons by a species (fall in oxidation state).
oxidation state
Bookkeeping charge assigned to each atom in a formula.
oxidizing agent
The species that is reduced (accepts electrons).
reducing agent
The species that is oxidized (donates electrons).
half-reaction
One half of a redox reaction, written with explicit electrons.

Sources & references

  1. openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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