MCAT content library / Chemistry & physics / Redox bookkeeping: balance atoms, charge, and electrons
A valid redox equation conserves charge as well as atoms. Use separate electron accounts to build the complete reaction and its mole relationships.
About 22 minutes, including the worked examples and checks.
Useful foundations
On this page
- Track an electron account, not just a changed formula
- Assign states with the formula’s total charge in view
- Balance a half-reaction in a stated medium
- Worked application: dichromate accepts electrons from iodide
- Convert the bookkeeping to a basic medium
- One reactant can supply both redox roles
- Check the final equation before using its coefficients
- Check your reasoning
- Continue learning
- Sources
Track an electron account, not just a changed formula
A redox reaction changes the oxidation state of at least one element. Oxidation raises an atom’s formal oxidation state and corresponds to electron loss; reduction lowers it and corresponds to electron gain. A formula can change during precipitation or proton transfer without any oxidation-state change, so “a reaction happened” is not enough to classify it as redox. These bookkeeping skills connect AAMC categories 4E and 4C.12
An oxidation state is a formal electron-assignment convention. For a monatomic ion it equals the ion’s charge. For an atom in a covalent molecule, it need not equal a measured partial charge or an integer charge physically sitting on that atom. The oxidation states of all atoms must sum to the total charge of the species.3
Assign states with the formula’s total charge in view
An element in its uncombined form has oxidation state zero: the oxygen atoms in O₂ and the chlorine atoms in Cl₂ are each zero. Fluorine is −1 in its compounds. Oxygen is usually −2, but in a peroxide such as H₂O₂ it is −1. Hydrogen is usually +1 when bonded to nonmetals and −1 in simple metal hydrides. Use the chemical context and charge sum rather than applying “usually” as an exceptionless rule.4
For SO₃²−, let sulfur’s oxidation state be x. With oxygen at −2, x + 3(−2) = −2, giving x = +4. For SO₄²−, x + 4(−2) = −2, giving x = +6. Conversion from sulfite to sulfate therefore oxidizes sulfur by two units. Adding an oxygen does not itself define oxidation; the electron-state account does.
The reducing agent donates the electrons and is itself oxidized. The oxidizing agent accepts them and is itself reduced. The names describe what each reagent does to the other reaction partner. A statement that “the oxidizing agent is oxidized” swaps these two perspectives.
Balance a half-reaction in a stated medium
A half-reaction isolates the electron loss or gain. Balance elements other than hydrogen and oxygen first. In an acidic aqueous representation, use H₂O to balance oxygen and H⁺ to balance hydrogen, then add electrons to balance charge. These additions are accounting steps for the stated medium; they are not permission to change the stipulated main products. Finally combine oxidation and reduction so that the electron counts cancel.3
The examples below are original worked constructions using standard chemical species. Products and solution conditions are supplied. The task is to conserve atoms and charge, not to predict an unprovided product or an experimental yield.
Worked application: dichromate accepts electrons from iodide
In acidic solution, suppose Cr₂O₇²− is converted to Cr³⁺ while I⁻ is converted to I₂. Chromium starts at +6 because 2x + 7(−2) = −2. Each chromium gains three electrons on reaching +3. Two chromium atoms therefore require six electrons per dichromate ion.
The reduction half-reaction is Cr₂O₇²− + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. The left charge is −2 + 14 − 6 = +6, matching the two Cr³⁺ ions. The seven waters supply seven oxygen atoms and fourteen hydrogen atoms on the right.
Iodide oxidation is 2I⁻ → I₂ + 2e⁻. Multiply this entire half-reaction by three, giving 6I⁻ → 3I₂ + 6e⁻. Adding it to the reduction half cancels the six electrons:
Cr₂O₇²− + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂.
| Quantity | Reactant total | Product total |
|---|---|---|
| Chromium atoms | 2 | 2 |
| Iodine atoms | 6 | 6 |
| Oxygen atoms | 7 | 7 |
| Hydrogen atoms | 14 | 14 |
| Charge, elementary units | −2 + 14 − 6 = +6 | 2(+3) = +6 |
Dichromate is the oxidizing agent because it accepts electrons; iodide is the reducing agent because it supplies them. If 0.012 mol of dichromate is completely reduced by this specified reaction, it accepts 0.072 mol of electrons and consumes 0.072 mol of iodide. It produces 0.036 mol of I₂. The electron count is not the number of dichromate ions or the coefficient of iodine molecules.
Convert the bookkeeping to a basic medium
For a separate specified reduction of MnO₄⁻ to MnO₂, manganese changes from +7 to +4 and gains three electrons. An acidic intermediate balance is MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O. To express the half-reaction in basic solution, add four OH⁻ to both sides, combine H⁺ with OH⁻ into water, and cancel waters common to both sides.3
The result is MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Both sides contain one manganese, six oxygens, four hydrogens, and charge −4. Simply replacing four H⁺ with four OH⁻ on the same side would not conserve charge or atoms. This is a half-reaction; the electrons will cancel only after a compatible oxidation half is added.
The medium can also affect which chemical products are favored. Balancing MnO₂ as a supplied product does not prove that it is the product under every set of conditions. Conservation is necessary for a candidate equation, while thermodynamics and kinetics determine whether and how it occurs.
One reactant can supply both redox roles
In disproportionation, atoms of the same element in one initial oxidation state end in both higher and lower states. For the specified basic conversion Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O, chlorine begins at zero. It reaches −1 in chloride and +1 in hypochlorite. The chlorine reagent therefore supplies both the reduced and oxidized products.45
The equation is balanced: two chlorine atoms, two oxygen atoms, two hydrogens, and charge −2 occur on each side. There is no need to invent a second elemental reagent to provide the opposite electron change. Nor does the presence of oxygen on one side mean that oxygen itself has changed oxidation state; here its state remains −2.
Check the final equation before using its coefficients
Three separate checks are useful: every element balances, total charge balances, and electrons cancel in the complete reaction. Once all three hold, the coefficients determine mole relationships for that specified reaction. An atom-balanced equation with unequal charges can still give a plausible-looking but incorrect reagent ratio.
Oxidation states and a balanced equation do not by themselves supply cell voltage, spontaneity, rate, or percent yield. They establish the conserved electron account on which those later calculations depend.
ACTIVE PRACTICE · REACTION BOOKKEEPING
Balance Electron Transfer Before Scaling the Reaction
Use the lesson’s acidic-solution reaction: dichromate is reduced to Cr³⁺ while I⁻ is oxidized to I₂. Each chromium changes from oxidation state +6 to +3. The two chromium atoms in dichromate therefore accept six electrons in total.
Electrons accepted: mol e⁻. Iodide consumed: mol I⁻. I₂ formed: mol.
Original balanced equation:
Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂
Reveal Atom, Charge, and Electron Audit
Cr: 2 = 2; I: 6 = 6; O: 7 = 7; H: 14 = 14. Left total charge: −2 + 14 − 6 = +6. Right total charge: 2(+3) = +6. The reduction half-reaction takes six electrons; three iodide oxidation half-reactions supply six electrons in total.
This is stoichiometry for the specifically supplied complete reaction in acidic solution, not a prediction of reaction rate, yield, or spontaneity.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
In an acidic reaction, bromate BrO₃⁻ is reduced to Br⁻. Oxygen is assigned −2 in bromate. If 0.025 mol of bromate is fully converted, how many moles of electrons are accepted?
A. 0.025 mol.
Review this choice. This counts one electron per bromate. Bromine changes by six oxidation-state units, not one.
B. 0.050 mol.
Review this choice. Bromine goes from +5 in BrO₃⁻ to −1 in Br⁻, requiring six electrons per bromate.
C. 0.150 mol.
Correct. x + 3(−2) = −1 gives x = +5. Reduction to −1 gains six electrons; 6 × 0.025 = 0.150 mol.
D. 0.300 mol.
Review this choice. This doubles the six-electron requirement. Each bromate contains one bromine atom, not two.
The principle to carry forward
Determine the change per atom, multiply by atoms per species, and then apply the amount reacted.
Check 2
The supplied half-reactions are Sn²⁺ → Sn⁴⁺ + 2e⁻ and Ag⁺ + e⁻ → Ag. Which combined equation conserves atoms and charge with electrons canceled?
A. Sn²⁺ + Ag⁺ → Sn⁴⁺ + Ag.
Review this choice. Only one electron is accepted for the two released. Charge is +3 on the left and +4 on the right.
B. Sn²⁺ + 2Ag⁺ → Sn⁴⁺ + 2Ag.
Correct. Two copies of the silver reduction accept the two electrons released by tin oxidation. Both sides have net charge +4.
C. 2Sn²⁺ + Ag⁺ → 2Sn⁴⁺ + Ag.
Review this choice. Four electrons would be released but only one accepted. Multiplying the tin half worsens the mismatch.
D. Sn⁴⁺ + 2Ag⁺ → Sn²⁺ + 2Ag.
Review this choice. Both tin and silver would be reduced in this direction. The supplied oxidation half has been reversed without an electron donor.
The principle to carry forward
Multiply a complete half-reaction until electron loss and gain match; do not change its direction arbitrarily.
Check 3
A stipulated acidic reaction is 2CrO₄²⁻ + 3SO₃²⁻ + 10H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 5H₂O. Assign oxygen −2. Which reagent is reduced?
A. Chromate, because chromium changes from +6 to +3.
Correct. The chromium oxidation state falls by three. Chromate accepts electrons and is the oxidizing agent.
B. Chromate, because chromium changes from +3 to +6.
Review this choice. This reverses the states in the supplied equation. The reagent contains Cr at +6 and the product at +3.
C. Sulfite, because sulfur changes from +4 to +6.
Review this choice. Those sulfur states are correct, but an increase is oxidation. Sulfite is the reducing agent.
D. Hydrogen ion, because hydrogen changes from +1 to 0.
Review this choice. Hydrogen remains +1 in water. No elemental hydrogen product is supplied.
The principle to carry forward
The species reduced accepts electrons; its agent name is oxidant, even though its own oxidation state falls.
Continue learning
Related lessons
- Moles, formulas, and yield: conserve the reacting material
- Periodic trends: explain the pattern from electron structure
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Sources
1. AAMC: C/P content category 4E
2. AAMC: C/P content category 4C
3. OpenStax Chemistry 2e: Review of Redox Chemistry
4. OpenStax Chemistry 2e: Classifying Chemical Reactions
5. OpenStax Chemistry 2e: Occurrence, Preparation, and Compounds of Oxygen