Redox in General Chemistry: balancing with half-reactions

First-semester lectures began in mid-September — at the Politecnico di Milano on 14 September 2026 — and in General Chemistry the syllabus usually follows the same order: stoichiometry and nomenclature first, then redox reactions between October and November. Every university has its own calendar, but if you have made it this far the sequence is probably this one. Where they are scheduled, the midterm tests fall in late October–November — at the Politecnico from 29 October to 2 November 2026 — and the winter exam session starts on 7 January 2027 at universities with an early calendar such as the Politecnico.

In the written General Chemistry exam — theoretical-numerical problems, multiple-choice questions, identifying and balancing reactions, nomenclature — stoichiometry, redox reactions and pH calculation are the three topics that trip almost everyone up at the start. Redox reactions in particular have a specific problem: the concepts are not hard, but balancing them “by eye”, the way you do with a simple reaction, stops working, and without a systematic method you waste precious time in a timed exam.

Oxidation numbers: the rules and the exceptions

Before you can recognise a redox reaction, you need to be able to assign an oxidation number (ox. no.) to every atom, using these rules in order of priority:

1. An element in its free state has ox. no. 0 — $\mathrm{Fe}$, ${\mathrm{O}}_2$, $\mathrm{C}{\mathrm{l}}_2$.
2. In a monatomic ion, the ox. no. equals the charge — ${\mathrm{Fe}}^{3+}$ has ox. no. $+3$.
3. Oxygen almost always has ox. no. $-2$, except in peroxides (${\mathrm{H}}_2{\mathrm{O}}_2$, $\mathrm{N}{\mathrm{a}}_2{\mathrm{O}}_2$), where it is $-1$, and in the compound with fluorine $\mathrm{O}{\mathrm{F}}_2$, where it is $+2$ (a rare case, but you should know it exists).
4. Hydrogen almost always has ox. no. $+1$, except in metal hydrides ($\mathrm{NaH}$, $\mathrm{Ca}{\mathrm{H}}_2$), where it is $-1$.
5. Fluorine always has ox. no. $-1$, in every compound, with no exceptions: it is the most electronegative element in the periodic table and never gives up that position.
6. The sum of the oxidation numbers is 0 in a neutral compound, and equals the overall charge in a polyatomic ion.

Order matters: apply the exceptions first (peroxides, hydrides, fluorine), then the general rules, and finally find the ox. no. of the “unknown” atom by difference — usually the central one, such as manganese in $\mathrm{Mn}{\mathrm{O}}_4^{-}$ or chromium in $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}$.

Recognising oxidation and reduction

A reaction is a redox reaction when at least one atom changes oxidation number between reactants and products. The species that is oxidised increases its ox. no. (loses electrons); the one that is reduced decreases it (gains electrons). As a result:

• The oxidising agent is the species that is reduced — it is the one that “oxidises” the other, taking electrons from it.
• The reducing agent is the species that is oxidised — it is the one that “reduces” the other, giving it electrons.

It is a symmetrical, compulsory exchange: if one species loses electrons, another must gain the same number, in the same reaction. This electron balance is exactly the pivot of the method that follows.

The ion-electron method in acidic solution

The most reliable way to balance a redox reaction is to split it into two half-reactions — one oxidation, one reduction — balance each separately, and then add them. In acidic solution, the steps are always the same:

1. Write the two half-reactions, working out from the oxidation numbers which species is oxidised and which is reduced.
2. Balance the atom that changes ox. no. (usually all you need is a coefficient).
3. Balance the oxygen by adding ${\mathrm{H}}_2\mathrm{O}$ molecules on the side that has fewer.
4. Balance the hydrogen by adding ${\mathrm{H}}^{+}$ ions on the side that has fewer.
5. Balance the charge by adding electrons ($e^{-}$) on the side with the higher charge, as many as needed to equalise it.
6. Equalise the number of electrons in the two half-reactions by multiplying them by suitable coefficients, then add them and cancel whatever appears identically on both sides.

Worked exercise 1 — $\mathrm{Mn}{\mathrm{O}}_4^{-}$ + $\mathrm{F}{\mathrm{e}}^{2+}$ in acidic solution

Permanganate oxidises iron(II) to iron(III), and is itself reduced to manganese(II).

Reduction half-reaction (manganese goes from $+7$ to $+2$):

I balance the oxygen with 4 ${\mathrm{H}}_2\mathrm{O}$ on the right, then the hydrogen with 8 ${\mathrm{H}}^{+}$ on the left:

Charge on the left: $-1+8=+7$; charge on the right: $+2$. I need to equalise by adding $5e^{-}$ on the left:

Oxidation half-reaction (iron goes from $+2$ to $+3$, with no need to balance O or H):

I equalise the electrons: the reduction consumes 5, the oxidation produces 1 — I multiply the second by 5:

I add the two half-reactions and cancel the 5 electrons, which appear on both sides:

Check: atoms of Mn (1 = 1), Fe (5 = 5), O (4 = 4, in the water), H (8 = 8, in the water); charge on the left $-1+5(+2)+8(+1)=17$, on the right $+2+5(+3)=17$. Balanced in both atoms and charge.

Worked exercise 2 — $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}$ + ${\mathrm{I}}^{-}$ in acidic solution

Dichromate oxidises iodide to molecular iodine, and is itself reduced to chromium(III).

Reduction (chromium goes from $+6$ to $+3$; since there are 2 Cr atoms in dichromate, the coefficient 2 goes on the right straight away):

I balance O with 7 ${\mathrm{H}}_2\mathrm{O}$ on the right, then H with 14 ${\mathrm{H}}^{+}$ on the left:

Charge on the left: $-2+14=+12$; on the right: $2(+3)=+6$. I add $6e^{-}$ on the left:

Oxidation (iodine goes from $-1$ to $0$, and you need 2 iodide ions to form ${\mathrm{I}}_2$):

I equalise the electrons: the reduction consumes 6, the oxidation produces 2 — I multiply the second by 3:

I add and cancel the $6e^{-}$:

Check: Cr (2 = 2), I (6 = 6), O (7 = 7, in the water), H (14 = 14, in the water); charge on the left $-2+6(-1)+14(+1)=6$, on the right $2(+3)+0=6$.

The method in basic solution

In basic solution there are no free ${\mathrm{H}}^{+}$ ions in significant quantity: the most direct method is to balance first as if it were in acidic solution, and only at the end convert, adding as many $\mathrm{O}{\mathrm{H}}^{-}$ ions to each side as there are ${\mathrm{H}}^{+}$ present, so as to turn them into water (${\mathrm{H}}^{+}+\mathrm{O}{\mathrm{H}}^{-}\to {\mathrm{H}}_2\mathrm{O}$) and then cancel the water molecules that appear on both sides. It is exactly the same procedure as before, with a single extra step at the end: always use this one, without mixing it with other methods, so you don’t lose count.

Worked exercise 3 — $\mathrm{Mn}{\mathrm{O}}_4^{-}$ + $\mathrm{S}{\mathrm{O}}_3^{2-}$ in basic solution

Permanganate oxidises sulfite to sulfate, and is itself reduced to manganese dioxide ($\mathrm{Mn}{\mathrm{O}}_2$) — a typical case of redox in basic solution, because in a basic medium permanganate is typically reduced only as far as $+4$, not to $+2$ as in acid.

Reduction (Mn from $+7$ to $+4$):

Balancing O: 4 on the left, 2 on the right — I add 2 ${\mathrm{H}}_2\mathrm{O}$ on the right:

I balance H with 4 ${\mathrm{H}}^{+}$ on the left (provisional, as in acid):

Charge on the left: $-1+4=+3$; on the right: $0$. I add $3e^{-}$ on the left:

Oxidation (S from $+4$ to $+6$): I start from $\mathrm{S}{\mathrm{O}}_3^{2-}\to \mathrm{S}{\mathrm{O}}_4^{2-}$, balance O with 1 ${\mathrm{H}}_2\mathrm{O}$ on the left and H with 2 ${\mathrm{H}}^{+}$ on the right (provisional):

(charge on the left $-2$, on the right $-2+2-2=-2$: it works with 2 electrons given up).

I equalise the electrons (reduction: 3, oxidation: 2 → lowest common multiple 6, I multiply the first by 2 and the second by 3):

I add and cancel the electrons, then simplify the ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{H}}^{+}$ common to both sides ($8{\mathrm{H}}^{+}-6{\mathrm{H}}^{+}=2{\mathrm{H}}^{+}$ on the left; $4{\mathrm{H}}_2\mathrm{O}-3{\mathrm{H}}_2\mathrm{O}=1{\mathrm{H}}_2\mathrm{O}$ on the right):

Now I convert to basic: I add 2 $\mathrm{O}{\mathrm{H}}^{-}$ to both sides to neutralise the 2 ${\mathrm{H}}^{+}$ on the left, turning them into ${\mathrm{H}}_2\mathrm{O}$:

And I cancel the common water ($2{\mathrm{H}}_2\mathrm{O}-{\mathrm{H}}_2\mathrm{O}=1{\mathrm{H}}_2\mathrm{O}$ on the left):

Check: Mn (2 = 2), S (3 = 3), O ($2\times 4+3\times 3+1=18$ on the left; $2\times 2+3\times 4+2=18$ on the right), H ($2=2$); charge on the left $2(-1)+3(-2)+0=-8$, on the right $0+3(-2)+2(-1)=-8$.

Disproportionation: when the same species is oxidised and reduced

A special case of redox is disproportionation: one and the same element, in the same starting compound, is partly oxidised and partly reduced, giving two different products. The classic example is chlorine in basic solution:

Chlorine starts at ox. no. $0$ (free element): half is reduced to $\mathrm{C}{\mathrm{l}}^{-}$ (ox. no. $-1$), half is oxidised to $\mathrm{Cl}{\mathrm{O}}^{-}$ (ox. no. $+1$). Seeing the same element appear in two different products is not a balancing mistake: it is the very signature of a disproportionation, and recognising it saves you from hunting for “the other reactant”, which here is simply not needed.

Redox stoichiometry: from the balanced reaction to grams

Once the redox reaction is balanced, the calculations follow exactly the scheme of ordinary stoichiometry — moles, coefficients, conversion back — with the only difference that the coefficients come from the electron balance instead of from conservation of atoms alone.

Permanganate titration. A solution of $\mathrm{F}{\mathrm{e}}^{2+}$ of volume $25.00\text{mL}$ is titrated with $\mathrm{KMn}{\mathrm{O}}_4$ $0.0200\text{M}$ (the reaction of exercise 1), and the endpoint is observed after adding $18.40\text{mL}$ of titrant. What is the concentration of the $\mathrm{F}{\mathrm{e}}^{2+}$ solution?

Moles of permanganate used:

From the balanced reaction, each mole of $\mathrm{Mn}{\mathrm{O}}_4^{-}$ reacts with 5 moles of $\mathrm{F}{\mathrm{e}}^{2+}$:

Concentration of the $\mathrm{F}{\mathrm{e}}^{2+}$ solution:

The most common mistake in this kind of problem is to use a 1:1 ratio between oxidising and reducing agent, as if it were any acid-base neutralisation: in redox the stoichiometric ratio comes from the number of electrons exchanged (here 5 to 1), and is never automatically one to one.

Where marks are lost, in short

• Forgetting the exceptions on oxidation numbers (peroxides, hydrides, fluorine) before applying the general rules.
• Balancing the atoms but not the charge, or vice versa: a balanced redox equation must work on both fronts, always.
• Mixing the two methods (acidic and basic) halfway through balancing: choose one, carry it through, and convert only at the end if needed.
• Assuming a 1:1 ratio between oxidising and reducing agent in redox stoichiometry problems, instead of reading it from the coefficients of the balanced reaction.
• Not checking the charge balance as a last step: it is the quickest way to spot a wrong coefficient before you hand in the paper.

Going further

Oxidation numbers, with a more high-school slant and without half-reactions, are also the starting point of inorganic chemical nomenclature, IUPAC and traditional: useful if you arrive at university with that part not fully solid. For the rest of stoichiometry — moles, limiting reagent, yield — the reference guide is stoichiometry: the typical mistakes. If your exam also covers gases, the guide is Gas laws in General Chemistry: worked exercises; for the heat exchanged in reactions, Thermochemistry: enthalpy and Hess’s law with exercises.

If you want to train on the format of the exam, the General Chemistry practice test follows the structure of the written paper — numerical problems and multiple-choice questions: the correct answer is free, the step-by-step guided solution is unlocked with your first lesson. For the other science subjects you will find the overview on the science page, and for university support in general on the university tutoring page. If redox reactions remain your weak point after trying this method, write to me through the contact page: the first introductory call is free and lasts half an hour.