Worked example 1
Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.
Try it first: Write the two half-reactions before attempting any coefficient.
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What you'll be able to do: Balance a redox equation in acidic or basic solution using the half-reaction method.
Redox equations resist ordinary trial-and-error balancing because water, hydrogen ions and electrons are all available to be added. The half-reaction method solves this by handling oxidation and reduction separately, then combining them.
These are recommended, not required. You can start this lesson at any time.
In aqueous redox chemistry the solvent participates. Water supplies oxygen, hydrogen ions supply hydrogen and electrons carry the charge, so an equation can be atom-balanced while remaining charge-unbalanced. The half-reaction method forces both to be handled explicitly.
Work in this exact order and the method never fails. Balance the element being oxidised or reduced. Balance oxygen by adding H₂O. Balance hydrogen by adding H⁺. Balance charge by adding electrons to the more positive side. Doing oxygen before hydrogen is essential, because adding water changes the hydrogen count.
Electrons must cancel completely. Find the lowest common multiple of the electron counts and multiply each entire half-reaction by the appropriate factor, including every species in it. Then add the halves and cancel anything appearing on both sides, typically water and hydrogen ions.
5 × (Fe²⁺ → Fe³⁺ + e⁻) plus MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Balance the equation for acidic solution first. Then add as many OH⁻ to both sides as there are H⁺, combine each H⁺ with an OH⁻ to make water on one side, and cancel any water that now appears on both sides. This preserves the balance because equal amounts were added to each side.
A finished redox equation must satisfy three tests: every element balances, the total charge is equal on both sides, and no free electrons remain. Verifying charge catches the majority of errors, so make it a habit rather than an afterthought.
Identifying redox reactionsOne side of a redox process written on its own, with electrons shown explicitly.
Main element, then oxygen with water, then hydrogen with H⁺, then charge with electrons.
Scale each half so the electrons lost exactly equal the electrons gained, then cancel them.
Add equal OH⁻ to both sides to convert every H⁺ into water, then cancel duplicate water.
A correct answer balances atoms, balances charge and contains no free electrons.
Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.
Try it first: Write the two half-reactions before attempting any coefficient.
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Balance Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ in acidic solution.
Try it first: Notice that both chromium and iodine need atom balancing before anything else.
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Convert MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O into basic solution.
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Why it's wrong: Adding water to fix oxygen changes the hydrogen count, undoing the earlier work.
Check instead: Always do oxygen with water first, then hydrogen with H⁺.
Why it's wrong: Electrons carry no mass in this bookkeeping, they only adjust charge.
Check instead: Add electrons only in the final step of each half-reaction.
Why it's wrong: The factor applies to every species in that half, otherwise the half is no longer balanced.
Check instead: Rewrite the whole scaled half-reaction rather than editing coefficients in place.
Why it's wrong: Hydrogen ions are not present at appreciable concentration in basic solution.
Check instead: Add matching OH⁻ to both sides and convert every H⁺ to water.
No practice questions are available for this topic yet. You can still practice the whole unit.
Split the reaction into an oxidation half and a reduction half. In each half, balance the main element, then oxygen with water, then hydrogen with H⁺, then charge with electrons. Multiply the halves so the electrons cancel, add them, and simplify. For basic solution, add hydroxide to both sides at the end to neutralise every H⁺ into water.
This lesson is original Chem Help content. No external sources were adapted.