Balanced Equations

Spec Mapping — OCR H432 Module 2.1.3 — Amount of substance / chemical equations, content statements covering the construction of balanced chemical equations with state symbols; ionic equations including the cancellation of spectator ions; balanced half-equations for oxidation and reduction; and the combination of half-equations to give overall redox equations (refer to the official OCR H432 specification document for exact wording). Balanced equations are the master language of stoichiometry; the moles, ideal-gas and yield lessons that follow all assume the equations they cite are correctly balanced for atoms and charge.

A balanced chemical equation is more than a symbolic shorthand — it is a quantitative claim about how many particles of one species reorganise into how many particles of another. Every subsequent stoichiometry calculation, every titration arithmetic, every limiting-reagent and yield problem in this course feeds off a correctly balanced equation. Antoine Lavoisier's late-eighteenth-century work on combustion in sealed vessels established the law of conservation of mass — matter is neither created nor destroyed in chemical reactions, only rearranged — and that conservation principle underwrites every coefficient you will balance in this lesson. We cover three categories of equation: full molecular equations with state symbols; ionic equations with spectator ions cancelled; and half-equations for redox processes that can be combined to give the overall ionic equation. Charge as well as atoms must balance in ionic equations — a constraint that becomes the central tool when we meet electrochemistry and standard electrode potentials in Module 5.

Key Rule: A balanced chemical equation conserves both atoms of each element and total charge on each side. Coefficients (whole numbers placed before formulae) are the only quantities you may adjust; formulae themselves must never be changed to "make the maths work".

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Why Equations Must Balance

Chemical reactions obey the law of conservation of mass: matter is neither created nor destroyed, only rearranged. A balanced equation must satisfy three constraints:

• Equal numbers of each type of atom on both sides
• Equal total charge on both sides (especially important for ionic equations)
• Correct state symbols where required

Lavoisier established conservation of mass through precise measurements of combustion reactions in sealed containers in the 1770s and 1780s. Before this work many chemists believed (following the now-discredited phlogiston theory) that burning destroyed matter or released a weightless fluid. Lavoisier's quantitative sealed-vessel experiments demonstrated that the total mass before and after combustion was unchanged — laying the foundation for the entire modern quantitative-chemistry tradition.

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State Symbols

OCR expects state symbols in almost every equation answer. They are:

Symbol	Meaning
(s)	Solid
(l)	Pure liquid
(g)	Gas
(aq)	Aqueous (dissolved in water)

Tip: Missing state symbols are a common reason for losing 1 mark on a 2-mark equation question. They are not decorative; they encode the physical state and therefore the chemistry.

Common state-symbol rules:

• Water produced in a neutralisation is (l), not (aq)
• NaOH dissolved in water is (aq); solid NaOH from a bottle is (s); molten NaOH (electrolysis) is (l)
• Precipitates in solution reactions are (s), even though the rest of the solution is (aq)
• Gases evolved in test-tube reactions are (g)

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A Systematic Method for Balancing

1. Write the correct formulae for reactants and products (never change these).
2. Count atoms of each element on each side.
3. Add whole-number coefficients in front of formulae (avoid fractions in final answers unless the question specifies per-mole energetics).
4. Balance one element at a time; leave H and O until last for inorganic equations.
5. Do a final atom-count check.
6. Add state symbols.

Worked Example 1 — Combustion of methane

Unbalanced: $\text{CH}_4(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O(l)}$CH4​(g)+O2​(g)→CO2​(g)+H2​O(l)

• C: 1 = 1 ✓
• H: 4 on left, 2 on right → coefficient 2 before $\text{H}_2\text{O}$H2​O
• O: 2 on left, $2 + 2 = 4$2+2=4 on right → coefficient 2 before $\text{O}_2$O2​
• Final: $\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O(l)}$CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(l) ✓

Worked Example 2 — Combustion of propane

Unbalanced: $\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$C3​H8​+O2​→CO2​+H2​O

• C: 3 → $3\text{CO}_2$3CO2​. H: 8 → $4\text{H}_2\text{O}$4H2​O. O: $6 + 4 = 10$6+4=10 on right → $5\text{O}_2$5O2​.
• $\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O(l)}$C3​H8​(g)+5O2​(g)→3CO2​(g)+4H2​O(l)

Worked Example 3 — Aluminium with oxygen

Unbalanced: $\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3$Al+O2​→Al2​O3​

Balance Al by setting $2\text{Al}_2\text{O}_3$2Al2​O3​ on the right (4 Al, 6 O), then 4 Al on left and $3\text{O}_2$3O2​ on left:

$4\text{Al(s)} + 3\text{O}_2(\text{g}) \rightarrow 2\text{Al}_2\text{O}_3(\text{s})$4Al(s)+3O2​(g)→2Al2​O3​(s)

Worked Example 4 — Sodium with water

Unbalanced: $\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$Na+H2​O→NaOH+H2​

The H count on the right is odd ($\text{NaOH} + \text{H}_2$NaOH+H2​ has odd H if coefficients are 1), so double both Na and H₂O:

$2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2(\text{g})$2Na(s)+2H2​O(l)→2NaOH(aq)+H2​(g)

Worked Example 5 — Combustion of butane

$\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$C4​H10​+O2​→CO2​+H2​O

C: 4 → $4\text{CO}_2$4CO2​. H: 10 → $5\text{H}_2\text{O}$5H2​O. O on right $8 + 5 = 13$8+5=13, so $\text{O}_2$O2​ coefficient $= 13/2$=13/2. Multiply through by 2:

$2\text{C}_4\text{H}_{10}(\text{g}) + 13\text{O}_2(\text{g}) \rightarrow 8\text{CO}_2(\text{g}) + 10\text{H}_2\text{O(l)}$2C4​H10​(g)+13O2​(g)→8CO2​(g)+10H2​O(l)

Worked Example 6 — Thermite reaction

$\text{Fe}_2\text{O}_3 + \text{Al} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}$Fe2​O3​+Al→Al2​O3​+Fe

Al: 2 on right → 2 Al on left. Fe: 2 on left → 2 Fe on right. O: 3 each side.

$\text{Fe}_2\text{O}_3(\text{s}) + 2\text{Al(s)} \rightarrow \text{Al}_2\text{O}_3(\text{s}) + 2\text{Fe(l)}$Fe2​O3​(s)+2Al(s)→Al2​O3​(s)+2Fe(l)

The Fe state is (l): the reaction is so exothermic that molten iron is produced, which is why thermite is used for welding rail tracks.

Worked Example 6b — A tricky disproportionation: chlorine in cold alkali

In cold dilute alkali, chlorine disproportionates simultaneously to chloride (Cl⁻, oxidation state −1) and chlorate(I) (ClO⁻, oxidation state +1):

$\text{Cl}_2(\text{aq}) + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}$Cl2​(aq)+2NaOH(aq)→NaCl(aq)+NaClO(aq)+H2​O(l)

Balance check: Cl 2 each side; Na 2 each side; O $2 = 1 + 1$2=1+1 ✓; H 2 each side. The challenge here is recognising that the same element (Cl) is being both oxidised and reduced — a disproportionation. This is a classic Module 3 (Group 17 chemistry) reaction whose balancing skills are developed here.

Worked Example 6c — Hot concentrated alkali

In hot concentrated alkali, the products change: the chlorate(I) further disproportionates to chloride and chlorate(V) (ClO₃⁻, oxidation state +5):

$3\text{Cl}_2(\text{aq}) + 6\text{NaOH(aq)} \rightarrow 5\text{NaCl(aq)} + \text{NaClO}_3(\text{aq}) + 3\text{H}_2\text{O(l)}$3Cl2​(aq)+6NaOH(aq)→5NaCl(aq)+NaClO3​(aq)+3H2​O(l)

The coefficients can be derived from the half-equation logic: 5 Cl atoms each gain one electron to become Cl⁻ (5 electrons total), while one Cl atom loses 5 electrons to become ClO₃⁻ (the −1 to +5 change). So the Cl atom count is 5 + 1 = 6, hence 3 Cl₂ on the left. This is a textbook example of how oxidation-number changes drive coefficient assignment.

Worked Example 6d — Industrial Haber process (with state symbols)

$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$N2​(g)+3H2​(g)⇌2NH3​(g)

Note the equilibrium arrow (⇌) rather than the single arrow. OCR examines this distinction explicitly: forward arrow for "going to completion" reactions, double arrow for equilibrium-limited reactions. The 1 : 3 : 2 mole ratio is essential for the Module 5 Haber-yield optimisation arithmetic.

Worked Example 6e — Methanol carbonylation (Cativa process)

A modern industrial reaction:

$\text{CH}_3\text{OH(l)} + \text{CO(g)} \rightarrow \text{CH}_3\text{COOH(l)}$CH3​OH(l)+CO(g)→CH3​COOH(l)

Atom count: C $1 + 1 = 2$1+1=2 on each side; H 4 each side; O $1 + 1 = 2$1+1=2 each side. Single-step addition with 100 % atom economy — the iridium-catalysed Cativa process replaced the older Monsanto rhodium-based process in the early 2000s for industrial methanol-to-ethanoic-acid conversion. The equation is trivially balanced because of the addition nature, but the catalyst identity matters: the catalyst does not appear in the balanced equation (it is regenerated each cycle) and so does not affect the stoichiometry.

Decision tree — balancing strategy

flowchart TD
    A[Write correct formulae] --> B[Count atoms of each element]
    B --> C{Combustion of CxHy or CxHyOz?}
    C -->|Yes| D[Balance C first, then H, then O last]
    C -->|No| E[Balance the rarest element first]
    D --> F{Any fractional coefficient?}
    E --> F
    F -->|Yes| G[Multiply through to clear fractions]
    F -->|No| H[Final atom count check]
    G --> H
    H --> I[Add state symbols]

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Ionic Equations

For reactions in aqueous solution, many ions are spectators — they appear on both sides unchanged. An ionic equation strips them out, showing only the species that actually react.

Procedure

1. Write the full balanced equation with state symbols.
2. Expand aqueous ionic compounds into their constituent ions.
3. Cancel any ions appearing identically on both sides.
4. Keep solids, liquids, gases and covalent molecules intact.

Worked Example 7 — Precipitation of silver chloride

Full: $\text{AgNO}_3(\text{aq}) + \text{NaCl(aq)} \rightarrow \text{AgCl(s)} + \text{NaNO}_3(\text{aq})$AgNO3​(aq)+NaCl(aq)→AgCl(s)+NaNO3​(aq)

Expanded: $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq}) + \text{Na}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl(s)} + \text{Na}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$Ag+(aq)+NO3−​(aq)+Na+(aq)+Cl−(aq)→AgCl(s)+Na+(aq)+NO3−​(aq)

Cancel NO₃⁻ and Na⁺ (spectators):

$\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl(s)}$Ag+(aq)+Cl−(aq)→AgCl(s)

This compact ionic equation represents any silver-halide precipitation regardless of the original spectator counter-ions.

Worked Example 8 — Neutralisation

Full: $\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$HCl(aq)+NaOH(aq)→NaCl(aq)+H2​O(l)

After cancelling Na⁺ and Cl⁻:

$\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O(l)}$H+(aq)+OH−(aq)→H2​O(l)

This is the universal strong-acid / strong-alkali ionic equation — the same equation describes HNO₃ + KOH, HBr + LiOH and every other strong-strong neutralisation.

Worked Example 9 — Acid + metal carbonate

Full: $2\text{HCl(aq)} + \text{CaCO}_3(\text{s}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O(l)} + \text{CO}_2(\text{g})$2HCl(aq)+CaCO3​(s)→CaCl2​(aq)+H2​O(l)+CO2​(g)

CaCO₃ is (s) and is not expanded. After cancelling Cl⁻:

$2\text{H}^+(\text{aq}) + \text{CaCO}_3(\text{s}) \rightarrow \text{Ca}^{2+}(\text{aq}) + \text{H}_2\text{O(l)} + \text{CO}_2(\text{g})$2H+(aq)+CaCO3​(s)→Ca2+(aq)+H2​O(l)+CO2​(g)

Worked Example 10 — Redox displacement

Full: $\text{Zn(s)} + \text{CuSO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{Cu(s)}$Zn(s)+CuSO4​(aq)→ZnSO4​(aq)+Cu(s)

Ionic after cancelling SO₄²⁻: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)

Charge check: left $= 0 + (+2) = +2$=0+(+2)=+2; right $= (+2) + 0 = +2$=(+2)+0=+2 ✓.

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Charge balance in ionic equations

In addition to atom balance, ionic equations must satisfy charge balance: the total electrical charge must be the same on both sides. Many candidates verify atom balance carefully but skip the charge check, losing the marks that distinguish a grade C from a grade A.

For the silver-chloride precipitation equation $\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl(s)}$Ag+(aq)+Cl−(aq)→AgCl(s):

• Left-hand side: $(+1) + (-1) = 0$(+1)+(−1)=0
• Right-hand side: $0$0
• Balance: ✓

For the redox displacement $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s):

• Left-hand side: $0 + (+2) = +2$0+(+2)=+2
• Right-hand side: $(+2) + 0 = +2$(+2)+0=+2
• Balance: ✓

For Worked Example 11 (Al + acid) above: $2\text{Al(s)} + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 3\text{H}_2(\text{g})$2Al(s)+6H+(aq)→2Al3+(aq)+3H2​(g):

• Left-hand side: $0 + 6(+1) = +6$0+6(+1)=+6
• Right-hand side: $2(+3) + 0 = +6$2(+3)+0=+6
• Balance: ✓

For Worked Example 14 (permanganate-iron): $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$MnO4−​+8H++5Fe2+→Mn2++4H2​O+5Fe3+:

• Left-hand side: $(-1) + (+8) + 5(+2) = +17$(−1)+(+8)+5(+2)=+17
• Right-hand side: $(+2) + 0 + 5(+3) = +17$(+2)+0+5(+3)=+17
• Balance: ✓

The discipline is to write the charge check explicitly — "$+17 = +17$+17=+17" rather than waving at the calculation. Examiners flag missing charge checks as M0 on the charge-balance mark of an ionic-equation question.

Half-Equations and Combining Them

In redox reactions, oxidation (electron loss) and reduction (electron gain) happen simultaneously. Each half can be written as a half-equation with explicit electrons:

• Oxidation: $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$Zn(s)→Zn2+(aq)+2e−
• Reduction: $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$Cu2+(aq)+2e−→Cu(s)

The electron counts are already equal (both 2), so adding and cancelling electrons gives the same equation as in Worked Example 10.

Worked Example 11 — Unequal electron counts

• Al → Al³⁺ + 3e⁻ (oxidation)
• 2H⁺ + 2e⁻ → H₂ (reduction)

To combine, scale so that both halves have 6e⁻: multiply the Al half by 2 and the H half by 3.

• $2\text{Al} \rightarrow 2\text{Al}^{3+} + 6e^-$2Al→2Al3++6e−
• $6\text{H}^+ + 6e^- \rightarrow 3\text{H}_2$6H++6e−→3H2​

Add and cancel:

$2\text{Al(s)} + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 3\text{H}_2(\text{g})$2Al(s)+6H+(aq)→2Al3+(aq)+3H2​(g)

Charge check: $0 + 6(+1) = +6$0+6(+1)=+6 on left; $2(+3) + 0 = +6$2(+3)+0=+6 on right ✓.

Worked Example 12 — Magnesium and acid

• $\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$Mg→Mg2++2e−
• $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$2H++2e−→H2​

Add: $\text{Mg(s)} + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})$Mg(s)+2H+(aq)→Mg2+(aq)+H2​(g)

Worked Example 13 — Phosphoric acid with calcium hydroxide

Unbalanced: $\text{Ca(OH)}_2(\text{aq}) + \text{H}_3\text{PO}_4(\text{aq}) \rightarrow \text{Ca}_3(\text{PO}_4)_2(\text{s}) + \text{H}_2\text{O(l)}$Ca(OH)2​(aq)+H3​PO4​(aq)→Ca3​(PO4​)2​(s)+H2​O(l)

Ca: $1 \rightarrow 3$1→3 on right, so $3\text{Ca(OH)}_2$3Ca(OH)2​. P: $1 \rightarrow 2$1→2, so $2\text{H}_3\text{PO}_4$2H3​PO4​. Now H: $6 + 6 = 12$6+6=12 → $6\text{H}_2\text{O}$6H2​O. O: $6 + 8 = 14$6+8=14 on left; $8 + 6 = 14$8+6=14 on right ✓.

$3\text{Ca(OH)}_2(\text{aq}) + 2\text{H}_3\text{PO}_4(\text{aq}) \rightarrow \text{Ca}_3(\text{PO}_4)_2(\text{s}) + 6\text{H}_2\text{O(l)}$3Ca(OH)2​(aq)+2H3​PO4​(aq)→Ca3​(PO4​)2​(s)+6H2​O(l)

Worked Example 14 — Permanganate-iron redox combination

The textbook Module 5 redox titration, balanced from half-equations:

• Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$MnO4−​+8H++5e−→Mn2++4H2​O
• Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$Fe2+→Fe3++e−

Scaling the Fe half by 5 to match the 5 e⁻ in the Mn half:

• $5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$5Fe2+→5Fe3++5e−
• Combined: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$MnO4−​+8H++5Fe2+→Mn2++4H2​O+5Fe3+