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.
0 of 3 steps revealed.
ElectrochemistryCourse level: AP Chemistry
Oxidation Numbers and Balancing Redox Reactions
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.
Introduction
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.
Learning objectives
- Assign oxidation numbers using the standard rules
- Identify the species oxidized, reduced, and the two agents
- Balance redox half-reactions in acidic solution
- Convert a balanced acidic equation to basic solution
What oxidation numbers mean
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.
OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.
The assignment rules, in priority order
Free elements are 0. A monatomic ion equals its charge. Group 1 is +1 and group 2 is +2 in compounds. Bonded fluorine is −1; elemental F₂ has oxidation number zero. 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.
For Cr₂O₇²⁻, write 2x + 7(−2) = −2 and solve x = +6 per chromium atom. For H₂O₂, two H at +1 require the two O to total −2, so each is −1. Elemental O₂ is zero and superoxide O₂⁻ has average O = −1/2. Apply the free-element and known-ion rules before ordinary O/H defaults. These oxidation numbers differ from Lewis formal charges and need not equal measured partial charges.
sum of oxidation numbers = overall charge
Oxidized, reduced, and the agents
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.
In 2 Al + 3 Cu²⁺ → 2 Al³⁺ + 3 Cu, two Al atoms lose 2 × 3 = 6 electrons while three Cu²⁺ gain 3 × 2 = 6. Aluminum is oxidized and is the reducing agent; the supplied Cu²⁺ is reduced and is the oxidizing agent. The product Cu(s) is not the agent originally accepting electrons. Count atoms as well as per-atom oxidation-state changes when determining the electron-transfer number for the overall reaction.
The reducing agent is the one that gets oxidized.
Half-reaction method in acid
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.
For MnO₄⁻ → Mn²⁺, add 4 H₂O to the product side and 8 H⁺ to the reactant side. The charge is then +7 left and +2 right, requiring 5 e⁻ on the left. Combining this with five Fe²⁺ → Fe³⁺ + e⁻ gives MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺. Net charge is +17 on each side. In an overall equation no electrons remain; in a half-reaction they must remain to show charge transfer.
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Finishing in base
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.
Direct balancing in base is possible, but the acid-then-OH⁻ route is a reliable systematic method.
H⁺ + OH⁻ → H₂O
Checking your work
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 reaction medium determines the balanced half-reaction
The same oxidant may yield different reduction products depending on acidity and other conditions. When a product is specified, balance that product in the stated medium; do not copy an acidic permanganate half-reaction ending in Mn²⁺ into a basic question ending in MnO₂. Audit atoms, charge, and electron direction.
Balance the specified products rather than predicting them from a memorized half-reaction. MnO₄⁻ → MnO₂ changes Mn +7 → +4, so it needs three electrons, not the five required for Mn²⁺. Its basic half is MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻: both sides contain one Mn, six O, four H and charge −4. Algebraic conversion of a hypothetical acidic equation to base preserves balance but does not prove those products are stable in that medium. Product identity and balancing are separate chemical questions.
Read both the product and the solution medium before balancing.
Key ideas
Definition
Oxidation number
The charge an atom would carry if all its bonds were fully ionic.
Rule
Oxidation
An increase in oxidation number; electrons are lost.
Rule
Reduction
A decrease in oxidation number; electrons are gained.
Rule
Agents
The oxidized species is the reducing agent; the reduced species is the oxidizing agent.
Rule
Balancing order
Other atoms, then O with H₂O, then H with H⁺, then charge with electrons.
Equation
Oxidation number sum
sum of oxidation numbers = charge of the species
- charge = 0 for a neutral compound, the ion charge otherwise
Equation
Basic conversion
add OH⁻ to both sides equal to the number of H⁺
- OH⁻ = hydroxide added to neutralize H⁺
Worked examples
Worked example 2
Balance in acidic solution: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.
Try it first: Decide which species is oxidized before writing anything else.
0 of 5 steps revealed.
Worked example 3
Convert the balanced acidic equation ClO⁻ + 2 H⁺ + 2 e⁻ → Cl⁻ + H₂O into basic solution.
Try it first: Count the H⁺ that must disappear.
0 of 3 steps revealed.
Worked example 4
Balance MnO₄⁻ → MnO₂ as a reduction half-reaction in basic solution.
Try it first: Mn goes from +7 to +4 and gains three electrons.
0 of 3 steps revealed.
Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Calling the oxidizing agent the species that is oxidized. | The oxidizing agent oxidizes something else, so it must itself be reduced. | Name the agent by what it does to its partner. |
| Assigning oxygen -2 in H₂O₂. | Peroxides have an O-O bond, so each oxygen is -1. | Check for a peroxide linkage before applying the -2 rule. |
| Balancing hydrogen before oxygen. | Adding water to fix oxygen changes the hydrogen count, so hydrogen must come second. | Follow the order: other atoms, O, H, then charge. |
| Leaving electrons in the final overall equation. | Electrons are transferred, not produced; they must cancel completely. | Scale the half-reactions until the electron counts match. |
| Balancing only atoms and not charge. | A redox equation can have every atom balanced and still be wrong if charge is not. | Sum the charges on both sides as a final check. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
32 topic questions across 3 difficulty levels.
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What you should now know
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.
- Oxidation numbers track electron bookkeeping, not real charges
- Oxidation raises the oxidation number, reduction lowers it
- The reducing agent is oxidized; the oxidizing agent is reduced
- Balance in acid using the order: other atoms, O with water, H with H⁺, charge with electrons
- Convert to base by adding OH⁻ to both sides and cancelling water
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Sources and further reading
- Chemistry 2e, Section 17.1: Review of Redox ChemistryOpenStax · Rice University · Chapter 17.1View source
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