Types of reactions — combination, decomposition, displacement, double displacement, redox: Visual walkthrough

Before we start, the only fact we borrow: atoms are made of a positive nucleus (the heavy centre) surrounded by tiny negative electrons. That's it. We build everything else.

Step 1 — The one thing that moves: an electron

WHAT. Picture two atoms sitting side by side. Between them we draw one small red dot — a single electron (charge $- 1$ ). An atom that is electrically neutral has exactly as many electrons as the positive charges in its nucleus, so its net charge is $0$ .

WHY start here? Because the parent note's whole claim — "redox = electron transfer" — only makes sense once we can count electrons on each atom. If we can't count them, we can't say one was lost or gained.

PICTURE. Two neutral atoms, each with a balanced ledger of $+$ and $-$ . The red electron is the piece we will follow.

Figure — Types of reactions — combination, decomposition, displacement, double displacement, redox

Step 2 — Move ONE electron: charge is born

WHAT. Slide the red electron from the left atom to the right atom. Now count again:

Left atom lost one $- 1$ ⇒ it is left with a net charge of $+ 1$ .
Right atom gained one $- 1$ ⇒ net charge $- 1$ .

WHY this move? This is the entire physical event of a redox reaction, stripped to its bones. Everything else on this page is just bookkeeping for this one transfer.

PICTURE. The arrow shows the electron leaving left, arriving right. Watch the two charge tags flip from $0$ to $+ 1$ and $- 1$ .

$net 0 A \to net + 1 A^{+} + the mover e^{-}$

$A$ — the left atom, starting neutral.
$A^{+}$ — same atom after losing the electron; the little $+$ is the leftover positive charge.
$e^{-}$ — the electron that left, carrying charge $- 1$ away.

Step 3 — Naming the count: the oxidation number

WHAT. We invent one integer per atom, the oxidation number, that answers: "how much charge would this atom carry if we handed every shared electron to the greedier partner?" For the clean transfer in Step 2 it is simply the net charge: left $= + 1$ , right $= - 1$ .

WHY invent this? In real molecules electrons are shared, not fully given. We need a fair rule for who "owns" a shared pair so we can still assign a whole number. The rule: the more electron-hungry atom (higher electronegativity) is pretended to own the pair. That pretence turns fuzzy sharing into countable ownership.

PICTURE. A shared electron pair sitting on the bond, then a dashed "ownership arrow" pulling both electrons toward the greedier (yellow) atom. The number that pops out on each atom is its oxidation number.

See Oxidation Number Rules for the full rule ladder.

Step 4 — The test: did any number change?

WHAT. Take any reaction. Write the oxidation number over every atom, before and after. If at least one number changed, an electron moved ⇒ the reaction is redox. If no number changed, no electron moved ⇒ not redox.

WHY is this the whole test? Because a changed oxidation number is defined as charge appearing/disappearing on an atom, and Step 2 showed that only happens when an electron crosses. Change ⇔ transfer. Nothing else can move the number.

PICTURE. Two side-by-side ledgers. Green ✓ where a number changed (redox), a flat grey line where every number is unchanged (not redox).

Step 5 — Apply the test to the star reaction: iron in copper sulfate

WHAT. Assign numbers to $F e + C u S O_{4} \to F e S O_{4} + C u$ .

$F e$ starts as a free element ⇒ $0$ . In $F e S O_{4}$ it is $+ 2$ (the sulfate $S O_{4}$ carries $- 2$ , and the whole salt is neutral, so Fe must be $+ 2$ ). So $F e : 0 \to + 2$ .
$C u$ in $C u S O_{4}$ is $+ 2$ (same balancing logic). As metal deposited on the nail it is a free element ⇒ $0$ . So $C u : + 2 \to 0$ .
The sulfate block $S O_{4}^{2 -}$ is a spectator — its atoms never change number.

WHY these values? Because " $\sum$ oxidation numbers $=$ overall charge" forces them. $S O_{4}$ is locked at $- 2$ , the salts are neutral, so the metal must supply the balance.

PICTURE. Fe's number climbs the ladder $0 \to + 2$ (going up = oxidation); Cu's number slides $+ 2 \to 0$ (going down = reduction). Two red electrons hop from Fe to Cu.

$F e \to F e^{2 +} + 2 e^{-} (oxidation, numbers up)$ $C u^{2 +} + 2 e^{-} \to C u (reduction, numbers down)$

$F e^{2 +}$ — iron after donating two electrons.
$2 e^{-}$ — the two electrons that leave Fe and are exactly the two that arrive at Cu.
$C u^{2 +}$ — the copper ion in solution, waiting to collect electrons.

This is a displacement and a redox reaction — two classification axes, one event.

Step 6 — Why the coefficients must match: electron conservation

WHAT. Fe releases 2 electrons; Cu needs exactly 2. They match one-to-one, so the coefficients in $F e + C u S O_{4} \to F e S O_{4} + C u$ are all $1$ . If a metal released $3$ and its partner needed $2$ , you'd scale to a common multiple ( $6$ ) to make lost $=$ gained.

WHY must they match? Electrons can't pile up in mid-air or vanish. Electrons lost $=$ electrons gained is charge conservation — the electrical twin of "atoms are conserved," which is why we balance in the first place.

PICTURE. A balance scale: left pan holds the electrons Fe lost, right pan holds the electrons Cu gained. The scale sits level only when the two counts are equal.

Step 7 — Naming the roles: agents

WHAT. The atom that got reduced ( $C u^{2 +} \to C u$ ) is the one that pulled electrons off Fe — so it caused Fe to oxidise. We call it the oxidising agent. The atom that got oxidised ( $F e \to F e^{2 +}$ ) handed electrons over — it caused Cu to reduce — so it is the reducing agent.

WHY the crossover names? Each agent does to itself the opposite of what it does to the other. The reducing agent gets oxidised; the oxidising agent gets reduced. It feels backwards until you see it as "the giver enables the taking."

PICTURE. Two labelled arrows crossing: "does the reducing" points at Cu (which itself is reduced), "does the oxidising" points at Fe (which itself is oxidised). The role label and the fate label sit on opposite ends of each arrow.

The one-picture summary

Every idea on this page is one red electron leaving a nucleus and joining another: that departure raises one oxidation number (oxidation), the arrival lowers another (reduction), the counts must match (conservation), and the roles are named by cause, not by fate (agents).

Recall Feynman retelling — say it in plain words

Imagine two kids, Iron and Copper-in-water. Iron is generous and hands over two marbles (electrons). The moment those marbles leave, Iron's "charge score" ticks up from $0$ to $+ 2$ — we say Iron got oxidised. Copper catches exactly two marbles, so its score drops from $+ 2$ down to $0$ — Copper got reduced. The number of marbles thrown must equal the number caught (none fall on the floor) — that's why the equation balances. And here's the twist in the naming: because Iron supplied the marbles that let Copper change, we call Iron the reducing agent even though Iron itself got oxidised. To test any reaction for this game, just write a score over every atom before and after; if even one score changes, marbles moved and it's redox; if every score stays put (like calcium in slaking lime), nobody threw anything and it's not redox, no matter how busy the reaction looks.

Recall

Two atoms start neutral — what is each atom's oxidation number? ::: $0$ (free element, electrons balance the nucleus) After one electron moves left→right, the two net charges are? ::: Left $+ 1$ (lost $e^{-}$ ), right $- 1$ (gained $e^{-}$ ) The single test for "is it redox?" ::: Did any atom's oxidation number change? Yes ⇒ redox; no change ⇒ not redox In $F e + C u S O_{4}$ , how does Fe's oxidation number move? ::: $0 \to + 2$ (oxidation, up) In $F e + C u S O_{4}$ , how does Cu's oxidation number move? ::: $+ 2 \to 0$ (reduction, down) Why are all coefficients $1$ in $F e + C u S O_{4} \to F e S O_{4} + C u$ ? ::: Fe loses 2 electrons, Cu gains 2 — they match, so no scaling needed Which is the oxidising agent here, and what happens to it? ::: $C u^{2 +}$ — it gets reduced while causing Fe to oxidise

Parent: Types of reactions (topic note) · Related: Oxidation Number Rules, Reactivity Series of Metals, Balancing Chemical Equations, Electrolysis, Acids, Bases and Salts, Exothermic and Endothermic Reactions