Worked example 1
Assign the oxidation number of chromium in Cr₂O₇²⁻.
Try it first: Write the known oxygen value first and let chromium be the unknown.
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What you'll be able to do: Assign oxidation numbers, identify what is oxidized and reduced, and balance redox equations by the half-reaction method.
Every battery you have ever used runs on electrons moving from one substance to another. Before you can analyze a cell, you need a reliable way to track those electrons, and that starts with oxidation numbers.
These are recommended, not required. You can start this lesson at any time.
An oxidation number is the charge an atom would have if every bond in the species were fully ionic. It is a bookkeeping device, not a real charge: the carbon in CH₄ is assigned -4 even though it carries nothing close to four units of negative charge. The value is still useful because a change in oxidation number reliably signals electron transfer.
Free elements are 0. A monatomic ion equals its charge. Group 1 is +1 and group 2 is +2 in compounds. Fluorine is always -1. Hydrogen is +1 except in metal hydrides, where it is -1. Oxygen is -2 except in peroxides (-1) and when bonded to fluorine. Everything else is solved for so the oxidation numbers sum to the overall charge of the species.
sum of oxidation numbers = overall charge
The species whose oxidation number rises is oxidized and it donates electrons, so it is the reducing agent. The species whose oxidation number falls is reduced and it accepts electrons, so it is the oxidizing agent. Students mix up these two labels more often than any other pair in this unit; the agent name describes what the species does to its partner, not to itself.
Split the reaction into an oxidation half and a reduction half. Balance all atoms except O and H. Add H₂O to balance oxygen, then H⁺ to balance hydrogen. Add electrons to whichever side is needed so the net charge matches on both sides. Finally, scale the halves so the electron counts are equal, add them, and cancel anything that appears on both sides.
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Balance the equation in acid first. Then add as many OH⁻ to both sides as there are H⁺, combine H⁺ with OH⁻ into water on the side where they meet, and cancel water that now appears on both sides. Trying to balance directly in base without this trick almost always fails.
H⁺ + OH⁻ → H₂O
Two checks catch nearly every error: count each element on both sides, then sum the charges on both sides. If the charges do not match, the electron count is wrong. Since electrons are never left over in the overall equation, seeing a stray e⁻ in a final answer is an immediate signal to go back.
The charge an atom would carry if all its bonds were fully ionic.
An increase in oxidation number; electrons are lost.
A decrease in oxidation number; electrons are gained.
The oxidized species is the reducing agent; the reduced species is the oxidizing agent.
Other atoms, then O with H₂O, then H with H⁺, then charge with electrons.
sum of oxidation numbers = charge of the species
add OH⁻ to both sides equal to the number of H⁺
Assign the oxidation number of chromium in Cr₂O₇²⁻.
Try it first: Write the known oxygen value first and let chromium be the unknown.
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Balance in acidic solution: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.
Try it first: Decide which species is oxidized before writing anything else.
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Convert the balanced acidic equation ClO⁻ + 2 H⁺ + 2 e⁻ → Cl⁻ + H₂O into basic solution.
Try it first: Count the H⁺ that must disappear.
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Why it's wrong: The oxidizing agent oxidizes something else, so it must itself be reduced.
Check instead: Name the agent by what it does to its partner.
Why it's wrong: Peroxides have an O-O bond, so each oxygen is -1.
Check instead: Check for a peroxide linkage before applying the -2 rule.
Why it's wrong: Adding water to fix oxygen changes the hydrogen count, so hydrogen must come second.
Check instead: Follow the order: other atoms, O, H, then charge.
Why it's wrong: Electrons are transferred, not produced; they must cancel completely.
Check instead: Scale the half-reactions until the electron counts match.
Why it's wrong: A redox equation can have every atom balanced and still be wrong if charge is not.
Check instead: Sum the charges on both sides as a final check.
No practice questions are available for this topic yet. You can still practice the whole unit.
Oxidation numbers are a bookkeeping tool that tracks how electrons are distributed in a compound. An increase in oxidation number is oxidation (loss of electrons) and a decrease is reduction (gain of electrons); the species that is oxidized is the reducing agent, and the species that is reduced is the oxidizing agent. To balance a redox reaction, split it into two half-reactions, balance atoms other than O and H first, add water to balance oxygen, add H⁺ to balance hydrogen, then add electrons to balance charge. Multiply the half-reactions so the electrons cancel, add them, and simplify. In basic solution, finish by adding OH⁻ to both sides to neutralize every H⁺ and cancel the water that forms. A correctly balanced redox equation has both atoms and total charge balanced, and the electrons must cancel completely.
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