IMAT Chemistry · Redox Reactions

Balancing Redox Equations

A redox equation balances when the electrons lost by the reducing agent equal the electrons gained by the oxidising agent; fix that ratio first, then the spectator atoms.

All Redox Reactions notes

Why this matters

The 2026 paper had three balancing questions: a metal with nitric acid, an electron transfer between two metal bromides, and a disproportionation. No earlier paper asked one, so this is the newest shape in the chapter.

Concept 1 of 3: Balancing by matching electrons lost and gained

Trial and error is slow for redox equations because the coefficients are linked by the electrons. Find how many electrons each atom loses or gains, choose multipliers so the two totals match, and most of the equation falls into place. Atoms that are not oxidised or reduced are filled in at the end.

Definition

The oxidation-number method:

Find the atoms whose oxidation numbers change, and the change per atom.
Multiply each side of the transfer so that electrons lost = electrons gained. This fixes the ratio of oxidised to reduced species.
Add any extra reagent that is used without changing (for example nitrate that only supplies ions to the salt).
Balance hydrogen with $\mathrm{H_2O}$ or $\mathrm{H^+}$ , and check oxygen last.
In a multiple-choice question, check the electron ratio first: it usually eliminates most options at once.

The balancing condition

(\text{atoms oxidised}) \times (\text{rise per atom}) = (\text{atoms reduced}) \times (\text{fall per atom})

Worked example

Very dilute nitric acid oxidises zinc and is itself reduced to ammonium ions. Balance:

\mathrm{Zn + HNO_3 \rightarrow Zn(NO_3)_2 + NH_4NO_3 + H_2O}

Practice this conceptself-check · 3 quick reps

Atom balance alone can hide a wrong electron count

Some wrong options balance the metal atoms but not the electrons or the spectator ions. Check three things: the electron ratio, every element, and (for ionic equations) the total charge. An equation is balanced only when all three agree.

Concept 2 of 3: Half-equations in acidic solution

Splitting a reaction into its two halves lets you balance each one separately, using only water,

\mathrm{H^+}

and electrons as fillers. Then you scale the halves so the electrons cancel and add them together. This is the standard way to handle ions such as permanganate and dichromate.

Definition

For each half-equation, in acid:

Balance the atoms other than O and H.
Add $\mathrm{H_2O}$ to balance oxygen.
Add $\mathrm{H^+}$ to balance hydrogen.
Add electrons to the more positive side to balance the charge.

Then multiply the halves so the electrons are equal, add them, and cancel anything that appears on both sides.

Two common reduction half-equations in acid

\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O} \qquad \mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}

Worked example

Write the overall equation for acidified permanganate oxidising iron(II) ions to iron(III).

Practice this conceptself-check · 3 quick reps

Electrons go on the side that is more positive

After adding water and

\mathrm{H^+}

, compare the total charges. Electrons are added to the side with the larger positive charge, enough to make the two sides equal. In a reduction they end up on the left; in an oxidation, on the right.

Concept 3 of 3: Disproportionation and comproportionation

Sometimes one element in a single species is both the giver and the taker of electrons: some of its atoms go up and the rest go down. That is disproportionation. The balancing idea is unchanged: the electrons lost by the atoms going up must equal those gained by the atoms going down, so you split the same species between the two products.

Definition

Disproportionation: the same element, from one oxidation state, is simultaneously oxidised and reduced. The species is both the oxidising and the reducing agent.
Examples: $\mathrm{Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O}$ (cold alkali); $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$ ; $\mathrm{2Cu^+ \rightarrow Cu + Cu^{2+}}$ .
Comproportionation is the reverse: two oxidation states of one element meet at a middle value, as in $\mathrm{2H_2S + SO_2 \rightarrow 3S + 2H_2O}$ .
A species can disproportionate only if its element sits in an intermediate state, with room both above and below.

Splitting one species between two products

(\text{atoms going up}) \times (\text{rise}) = (\text{atoms going down}) \times (\text{fall})

Worked example

Chlorine reacts with hot sodium hydroxide to give sodium chloride and sodium chlorate(V). Balance:

\mathrm{Cl_2 + NaOH \rightarrow NaCl + NaClO_3 + H_2O}

Practice this conceptself-check · 4 quick reps

In disproportionation one species is both agents

\mathrm{Cl_2 + H_2O \rightarrow HCl + HClO}

, chlorine is both the oxidising agent and the reducing agent: one Cl atom falls to −1 while the other rises to +1. An option saying the reaction is not redox because only one element changes is wrong.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Balancing by matching electrons lost and gained
The balancing condition
$(\text{atoms oxidised}) \times (\text{rise per atom}) = (\text{atoms reduced}) \times (\text{fall per atom})$
Half-equations in acidic solution
Two common reduction half-equations in acid
$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O} \qquad \mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$
Disproportionation and comproportionation
Splitting one species between two products
$(\text{atoms going up}) \times (\text{rise}) = (\text{atoms going down}) \times (\text{fall})$

Watch out for (3)

Atom balance alone can hide a wrong electron count
→ Balancing by matching electrons lost and gained
Electrons go on the side that is more positive
→ Half-equations in acidic solution
In disproportionation one species is both agents
→ Disproportionation and comproportionation

Drill the past-year questions on this topic

Open the bank with this topic pre-filtered.

Open in the bank