Classifying Chemical Reactions¶
Syllabus mapping
CH-11-03 — Chemical Reactions. Covers: using balanced equations to describe and predict the products of synthesis, decomposition, combustion, precipitation and neutralisation reactions (including laboratory experiments); the removal of toxicity in Aboriginal and Torres Strait Islander Peoples' preparation of nardoo as an example of a decomposition reaction (explicit, examinable content integration point); full and net ionic equations for precipitation, predicted using solubility rules; metal displacement reactions; oxidation and reduction as electron transfer and change in oxidation state; half-equations; comparing metal reactivity with oxygen, water, dilute acid and metal ions to build an activity series; linking reactivity to ionisation energy, atomic radius and electronegativity; a secondary-source investigation of a chemical reaction's impact in medicine or agriculture.
Wording above is paraphrased for teaching use — check the official syllabus before treating any line as verbatim NESA text.
You need to know: how to balance an equation, how to recognise and predict the products of the five main reaction types, how to use solubility rules to predict a precipitate and write a net ionic equation, what oxidation and reduction mean, and how to use an activity series to predict whether a metal displacement reaction will happen.
This is a long page. It's in three parts — take them one at a time:
- Balancing equations and the five reaction types
- Precipitation reactions and ionic equations
- Redox, displacement and the activity series
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Writing formulas and naming compounds, including polyatomic ions
- Working out an ion's charge
- Conservation of mass — why equations have to balance
- The reaction types and acids, bases and pH from Year 10
Part 1: Balancing equations and the five reaction types¶
Balancing equations¶
A balanced equation has the same number of each type of atom on both sides. You balance it by changing the coefficients (the big numbers in front) — never the small subscripts, because that would change what the substance is.
State symbols show the state of each substance: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).
Method:
- Write the correct formulas for every reactant and product.
- Count the atoms of each element on each side.
- Balance one element at a time by changing coefficients. Leave elements that appear on their own (like \(\ce{O2}\) or a metal) until last.
- Re-count everything at the end.
Worked example
Balance: \(\ce{C3H8 + O2 -> CO2 + H2O}\) (propane burning)
Carbon: 3 on the left, so put 3 in front of \(\ce{CO2}\): \(\ce{C3H8 + O2 -> 3CO2 + H2O}\)
Hydrogen: 8 on the left, so put 4 in front of \(\ce{H2O}\): \(\ce{C3H8 + O2 -> 3CO2 + 4H2O}\)
Oxygen (last): the right now has \(3 \times 2 + 4 \times 1 = 10\) O atoms, so put 5 in front of \(\ce{O2}\):
Check: C 3 = 3 ✓, H 8 = 8 ✓, O 10 = 10 ✓
The five reaction types¶
| Type | Pattern | Example |
|---|---|---|
| Synthesis | Two or more substances join to make one product: A + B → AB | \(\ce{2Mg(s) + O2(g) -> 2MgO(s)}\) |
| Decomposition | One substance breaks down into two or more: AB → A + B | \(\ce{CaCO3(s) -> CaO(s) + CO2(g)}\) (when heated) |
| Combustion | A fuel reacts with oxygen, releasing energy. Hydrocarbon fuels make \(\ce{CO2}\) and \(\ce{H2O}\) | \(\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}\) |
| Precipitation | Two solutions mix and form an insoluble solid (a precipitate) | \(\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}\) |
| Neutralisation | An acid reacts with a base to make a salt and water | \(\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}\) |
Predicting products — some reliable patterns:
- Metal + oxygen (synthesis/combustion) → metal oxide
- Metal carbonate, heated (decomposition) → metal oxide + carbon dioxide
- Hydrocarbon + oxygen (complete combustion) → carbon dioxide + water
- Acid + base (neutralisation) → salt + water
- Acid + carbonate → salt + water + carbon dioxide (you'll see fizzing)
The "salt" is the ionic compound made from the metal of the base and the non-metal part of the acid: hydrochloric acid gives chlorides, sulfuric acid gives sulfates, nitric acid gives nitrates.
Worked example — predict and balance
Predict the products and write a balanced equation for sulfuric acid reacting with potassium hydroxide.
Type: acid + base → salt + water (neutralisation)
Salt: potassium (from the base) + sulfate (from sulfuric acid) → \(\ce{K+}\) and \(\ce{SO4^{2-}}\) → \(\ce{K2SO4}\)
Unbalanced: \(\ce{H2SO4 + KOH -> K2SO4 + H2O}\)
Balance: 2 K needed on the left → \(\ce{2KOH}\); that gives 4 H on the left (2 + 2), so \(\ce{2H2O}\):
Aboriginal and Torres Strait Islander content — placeholder, not yet written
NESA lists the removal of toxicity in the preparation of nardoo by Aboriginal and Torres Strait Islander Peoples, as an example of a decomposition reaction, as explicit, examinable syllabus content — not an optional cultural add-on. The specific practices, and whose Knowledge they are, need proper research — ideally with input from your school's Aboriginal Education team, local Community, or a community-verified source — rather than a generic paraphrase.
Deliberately left unwritten here rather than filled with unverified content.
Part 2: Precipitation reactions and ionic equations¶
Solubility rules¶
When two solutions are mixed, a precipitate forms if any new combination of ions is insoluble. You predict this using solubility rules:
| These compounds are… | …soluble, except: |
|---|---|
| All compounds of Group 1 ions (\(\ce{Na+}\), \(\ce{K+}\), \(\ce{Li+}\)) and ammonium (\(\ce{NH4+}\)) | No exceptions |
| All nitrates (\(\ce{NO3-}\)) and acetates (\(\ce{CH3COO-}\)) | No exceptions |
| Chlorides, bromides, iodides | Silver (\(\ce{Ag+}\)) and lead (\(\ce{Pb^{2+}}\)) |
| Sulfates (\(\ce{SO4^{2-}}\)) | Barium, lead, calcium, strontium (and silver is only slightly soluble) |
| These compounds are… | …insoluble, except: |
|---|---|
| Hydroxides (\(\ce{OH-}\)) | Group 1, ammonium, barium (calcium hydroxide is slightly soluble) |
| Carbonates (\(\ce{CO3^{2-}}\)) and phosphates (\(\ce{PO4^{3-}}\)) | Group 1 and ammonium |
A useful shortcut: if a compound contains a Group 1 ion, ammonium or nitrate, it's soluble — so those ions are never in the precipitate.
Writing full and net ionic equations¶
When ionic compounds dissolve, they split into their ions. So a precipitation reaction can be written three ways.
Worked example — silver nitrate and sodium chloride
1. Balanced (molecular) equation:
How do we know \(\ce{AgCl}\) is the solid? The two possible new combinations are \(\ce{AgCl}\) and \(\ce{NaNO3}\). Chlorides of silver are insoluble; sodium nitrate is soluble.
2. Full ionic equation — split every (aq) compound into its ions. Leave the solid together:
3. Net ionic equation — cross out the ions that appear unchanged on both sides (the spectator ions, here \(\ce{Na+}\) and \(\ce{NO3-}\)):
The net ionic equation shows only what actually happens: silver ions and chloride ions join to make a solid.
Worked example — predict a precipitate
Solutions of lead(II) nitrate and potassium iodide are mixed. Is there a precipitate? Write the net ionic equation.
Swap partners: the new combinations are lead(II) iodide (\(\ce{PbI2}\)) and potassium nitrate (\(\ce{KNO3}\)).
Check the rules: iodides are soluble except lead → \(\ce{PbI2}\) is the precipitate (it's bright yellow). \(\ce{KNO3}\) is soluble.
Net ionic equation:
Part 3: Redox, displacement and the activity series¶
Oxidation and reduction¶
Many reactions involve electrons moving from one substance to another. These are redox (reduction–oxidation) reactions.
- Oxidation is loss of electrons.
- Reduction is gain of electrons.
A memory trick: OIL RIG — Oxidation Is Loss, Reduction Is Gain.
They always happen together: whatever loses electrons must be giving them to something that gains them.
Oxidation states¶
An oxidation state (or oxidation number) is a way of keeping track of electrons. The rules you need:
- An element on its own (\(\ce{Zn}\), \(\ce{O2}\), \(\ce{Cu}\)) has oxidation state 0.
- A simple ion has an oxidation state equal to its charge (\(\ce{Zn^{2+}}\) is +2, \(\ce{Cl-}\) is −1).
- Oxygen in a compound is usually −2; hydrogen is usually +1.
- The oxidation states in a compound add up to 0; in a polyatomic ion they add up to the charge.
Oxidation = oxidation state goes up. Reduction = oxidation state goes down.
Worked example
What is the oxidation state of sulfur in the sulfate ion, \(\ce{SO4^{2-}}\)?
Let sulfur be \(x\). Four oxygens at −2 each, and the total must equal the ion's charge (−2):
Metal displacement reactions¶
Put a piece of zinc into blue copper(II) sulfate solution, and after a while:
- the zinc becomes coated in pink-brown copper
- the blue colour fades
- the solution gets warm
The zinc has displaced the copper from solution:
The sulfate ion is a spectator, so the net ionic equation is:
This is a redox reaction. Split it into two half-equations that show where the electrons go:
| Half-equation | What happens | Oxidation state |
|---|---|---|
| \(\ce{Zn(s) -> Zn^{2+}(aq) + 2e-}\) | Zinc loses 2 electrons → oxidation | 0 → +2 (up) |
| \(\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}\) | Copper ions gain 2 electrons → reduction | +2 → 0 (down) |
The electrons lost by zinc are exactly the electrons gained by the copper ions. Adding the two half-equations (the electrons cancel) gives back the net ionic equation.
The activity series¶
A metal will only displace another metal from solution if it is more reactive (it gives up its electrons more easily). Ranking metals by reactivity gives the activity series:
Most reactive at the left, least reactive at the right.
The rule: a metal displaces the ions of any metal to its right in the series — but not to its left.
Worked example — will it react?
(a) Will copper metal react with silver nitrate solution? (b) Will copper metal react with zinc sulfate solution?
(a) Copper is left of silver → copper is more reactive → yes.
Half-equations: \(\ce{Cu -> Cu^{2+} + 2e-}\) (oxidation) and \(\ce{Ag+ + e- -> Ag}\) (reduction).
Copper loses 2 electrons but each silver ion only takes 1, so double the silver half-equation to balance the electrons:
(b) Copper is right of zinc → less reactive → no reaction.
How metals react with oxygen, water and acid¶
The activity series isn't just a list to memorise — you can see it in how strongly metals react:
| Metal | With oxygen (air) | With water | With dilute acid |
|---|---|---|---|
| Potassium, sodium | Tarnish instantly; stored under oil | React violently with cold water, making hydrogen gas | Dangerously violent — never done in school |
| Calcium | Tarnishes quickly | Reacts steadily with cold water | Vigorous |
| Magnesium | Burns with a brilliant white flame | Very slowly with cold water; quickly with steam | Fast fizzing |
| Aluminium, zinc, iron | Slowly form an oxide layer (iron rusts) | Little or no reaction | Steady fizzing (aluminium slow at first) |
| Copper | Turns black (copper oxide) when heated | No reaction | No reaction with dilute HCl |
| Silver, gold | No reaction | No reaction | No reaction |
When a metal reacts with acid, it makes a salt and hydrogen gas — another redox reaction: \(\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}\). Hydrogen sits in the activity series because metals to its left can displace it from acids; metals to its right (like copper) can't.
Why some metals are more reactive: linking to periodic trends¶
Metals react by losing electrons. A metal is more reactive if its outer electrons are easier to remove. That links straight back to the periodic trends:
| Property | More reactive metals have… | Because… |
|---|---|---|
| First ionisation energy | Lower ionisation energy | Less energy needed to remove the outer electron(s) |
| Atomic radius | Larger atoms | Outer electrons further from the nucleus, held less tightly |
| Electronegativity | Lower electronegativity | Weaker pull on electrons |
That's why reactivity increases down a group (potassium is more reactive than sodium, which is more reactive than lithium) and decreases across a period (sodium > magnesium > aluminium).
Secondary-source investigation: a reaction's impact in medicine or agriculture¶
Your task: choose one chemical reaction used in medicine or agriculture, and evaluate its impact — the benefits and the problems. Some starting points:
- Agriculture: making ammonia for fertilisers (\(\ce{N2 + 3H2 -> 2NH3}\), the Haber process) — it feeds billions of people, but fertiliser run-off damages rivers and the process uses a lot of energy.
- Medicine: antacids neutralising excess stomach acid (e.g. \(\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}\)); or the synthesis of a common drug like aspirin.
An "evaluate" answer needs a judgement backed by evidence from both sides — see the command verb glossary and Communicating for referencing.
Practical¶
The syllabus lists several pracs for this sub-topic. Typical versions:
- Reaction types: burn magnesium ribbon (synthesis/combustion); heat green copper(II) carbonate until it turns black and the gas turns limewater milky (decomposition); burn a candle or ethanol (combustion); neutralise an acid with a base using an indicator (neutralisation). Write a balanced equation for each.
- Precipitation: mix pairs of solutions in a spotting tile, record which pairs form a precipitate and its colour, and check against your predictions from the solubility rules.
- Metal displacement: place small pieces of different metals in solutions of different metal ions. Record which combinations react.
- Reactivity of metals: compare how a set of metals react with water and with dilute acid (e.g. measuring the time to produce a set volume of gas). Use the displacement and reactivity results to construct your own activity series, then compare it with the one above.
Safety: magnesium burns very brightly — don't look directly at it. Never test sodium or potassium with acid.
Advanced
Incomplete combustion. With too little oxygen, hydrocarbons burn to make carbon monoxide (\(\ce{CO}\), toxic) or soot (\(\ce{C}\)) instead of \(\ce{CO2}\): \(\ce{2CH4 + 3O2 -> 2CO + 4H2O}\). This is why gas heaters need ventilation.
Why aluminium seems less reactive than it is. Aluminium is high in the activity series, but in practice it reacts slowly with water and acid at first. Its surface is covered by a thin, tough layer of aluminium oxide that protects the metal underneath. Scratch or remove the layer and it reacts much faster.
Spotting redox using oxidation states. In \(\ce{2Mg + O2 -> 2MgO}\), magnesium goes 0 → +2 (oxidised) and oxygen goes 0 → −2 (reduced) — so this synthesis reaction is also a redox reaction, even though no ions are in solution. Neutralisation and precipitation reactions are not redox: no oxidation states change.
Net ionic equation for neutralisation. For any strong acid with a strong base, the spectator ions cancel and you're left with \(\ce{H+(aq) + OH^-(aq) -> H2O(l)}\). That's why the energy released by these neutralisations is almost the same whichever acid and base you use (see Energy Changes).
Extension
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
From an activity series to electricity. If you separate the two half-equations of a displacement reaction into two different beakers and connect them with a wire, the electrons travel through the wire — that's a galvanic cell (a battery). The further apart two metals are in the activity series, the bigger the voltage. Chemists measure this precisely as standard electrode potentials, which you'll meet in Year 12.
📎 Depth study idea
Build a series of simple cells from pairs of different metals (e.g. zinc, copper, iron, magnesium) in salt solution or a lemon, measure the voltage of each pair, and test whether the voltages follow the order of your activity series.
Video/visual resources¶
- 🖼️ Diagram needed — a particle-level picture of the zinc + copper(II) sulfate reaction: a zinc surface with Zn atoms leaving as Zn²⁺ ions, electrons passing to Cu²⁺ ions which deposit as Cu atoms, and sulfate ions shown unchanged (spectators). TODO — placeholder until sourced/drawn.
- 🎥 TODO: source a video showing the alkali metals reacting with water (lithium, sodium, potassium side by side) — too dangerous to do in full in a school lab.
- 🎥 TODO: source a precipitation reactions video showing the colours (e.g. yellow lead iodide, white silver chloride, blue copper hydroxide).
- 🎥 Khan Academy — TODO: source an oxidation states / redox explainer.
- 🖥️ PhET Simulation — Balancing Chemical Equations — practise balancing, with a game mode at three difficulty levels
Check yourself¶
-
Balance: \(\ce{Al + O2 -> Al2O3}\)
Answer
\(\ce{4Al + 3O2 -> 2Al2O3}\). Check: Al 4 = 4 ✓, O 6 = 6 ✓
-
Classify each reaction: (a) \(\ce{2H2O2 -> 2H2O + O2}\), (b) \(\ce{HNO3 + NaOH -> NaNO3 + H2O}\), (c) \(\ce{Fe + S -> FeS}\).
Answer
(a) Decomposition — one reactant breaks into two products. (b) Neutralisation — acid + base → salt + water. (c) Synthesis — two elements join to make one compound.
-
Predict the products and write a balanced equation for calcium carbonate reacting with hydrochloric acid.
Answer
Acid + carbonate → salt + water + carbon dioxide: \(\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)}\)
-
Solutions of barium chloride and sodium sulfate are mixed. Identify the precipitate and write the net ionic equation.
Answer
The new combinations are barium sulfate and sodium chloride. Barium sulfate is insoluble (sulfates are soluble except barium), so it's the precipitate: \(\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4(s)}\). Sodium and chloride ions are spectators.
-
In \(\ce{Mg(s) + Cu^{2+}(aq) -> Mg^{2+}(aq) + Cu(s)}\), which substance is oxidised and which is reduced? Write the two half-equations.
Answer
Magnesium is oxidised (loses electrons, 0 → +2): \(\ce{Mg -> Mg^{2+} + 2e-}\)
Copper(II) ions are reduced (gain electrons, +2 → 0): \(\ce{Cu^{2+} + 2e- -> Cu}\)
-
Will iron react with (a) copper(II) sulfate solution, (b) magnesium sulfate solution? Explain using the activity series.
Answer
(a) Yes — iron is more reactive than copper (left of it in the series), so it displaces copper. (b) No — iron is less reactive than magnesium, so it can't displace it.
-
Explain why potassium is more reactive than sodium.
Answer
Potassium's outer electron is in a higher energy level, further from the nucleus and more shielded by inner electrons. It's held less tightly (lower ionisation energy), so potassium loses it more easily — and reacting means losing that electron.