Worked examples — Balancing redox equations — ion-electron (half-reaction) method, oxidation-number method

Before we start, one promise: I will not use a single symbol you have not already met. Everything here rests on three ideas from the parent note, and only these three:

Recall The only three laws we ever use

Mass: every kind of atom appears equally on both sides. Charge: total electric charge (the little $+$ and $-$ superscripts, added up) is equal on both sides. Electrons: the number of electrons lost by the atom being oxidized equals the number gained by the atom being reduced. Nothing drops on the floor.

A "$\mathrm{half}\text{-}\mathrm{reaction}$" is one side of the story: only the losing, or only the gaining, with the electrons $e^{-}$ written out loud. See Oxidation states / oxidation number rules and Conservation of mass and charge.

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The scenario matrix

Every redox balancing problem you will ever be handed falls into one of these cells. The examples below are labelled with the cell they cover — together they fill the whole grid.

#	Case class	What is tricky about it	Covered by
A	Acidic medium, two different elements change	supply O and H from ${\mathrm{H}}_2\mathrm{O}$ / ${\mathrm{H}}^{+}$	Ex 1
B	Basic medium	must neutralize ${\mathrm{H}}^{+}$ into water at the end	Ex 2
C	Disproportionation — one element is both oxidized and reduced	you split the same species into two halves	Ex 3
D	Comproportionation — two forms of one element meet in the middle	halves merge rather than split	Ex 4
E	Spectator trap — some atoms of a changing element don't actually change	count only the atoms that move	Ex 5
F	Peroxide / unusual oxidation state	the $-1$ oxygen breaks the usual rule	Ex 6
G	Real-world word problem (titration, moles)	translate grams → the balanced ratio	Ex 7
H	Exam twist — degenerate skeleton where "no reaction" is a live answer	recognise zero net electron transfer	Ex 8

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Case A — Acidic, two elements change

1. Split into halves. Reduction: $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}\to \mathrm{C}{\mathrm{r}}^{3+}$. Oxidation: $\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}$. Why this step? One species clearly gains (Cr goes down in oxidation number, so it grabs electrons), the other clearly loses; separating them lets us count electrons cleanly.
2. Balance the changing atom first: $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}\to 2C{\mathrm{r}}^{3+}$. Why? There are 2 Cr on the left; mass conservation forces 2 on the right before we touch O or H.
3. Balance O with water: left has 7 oxygen, so $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$. Why? Oxygen isn't written in the skeleton because it comes from the solvent — the only oxygen source we're allowed is ${\mathrm{H}}_2\mathrm{O}$.
4. Balance H with ${\mathrm{H}}^{+}$: the 7 waters bring 14 hydrogen, so add 14 ${\mathrm{H}}^{+}$ on the left: $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$. Why? Acidic solution has free ${\mathrm{H}}^{+}$ available (see Acid–base neutralization (for the OH⁻ step)); it's the legal hydrogen source.
5. Balance charge with electrons. Left charge: $-2+14=+12$. Right charge: $2\times (+3)=+6$. Add 6 electrons to the left (the more positive side): $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$. Why? $+12+6(-1)=+6$. ✔ And notice: 2 Cr each dropping $6\to 3$ is $2\times 3=6$ electrons — the arithmetic agrees with the chemistry.
6. Oxidation half: $\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}$ (loses one). Equalize: the reduction eats 6 electrons, the oxidation gives 1, so multiply the iron half by 6. Why? Electron law: 6 lost must equal 6 gained. The figure below shows this as two "rivers" of electrons — one flowing out of the iron (amber), one into the dichromate (cyan) — that must carry the same number, 6, or electrons would pile up or vanish.

Figure 1 — Ex 1. The cyan arrow is the reduction half swallowing 6 electrons; the amber arrow is the oxidation half releasing 6. They meet at the "6 electrons transferred" node: the flows are equal, which is exactly the electron law that fixes the $6:1$ scaling of the iron half.

$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6F{\mathrm{e}}^{2+}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}+6F{\mathrm{e}}^{3+}$

Verify: charge left $=-2+14+6(+2)=+24$; right $=2(+3)+6(+3)=+24$. ✔ Atoms: Cr $2=2$, O $7=7$, H $14=14$, Fe $6=6$. ✔

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Case B — Basic medium

1. Reduction half, balanced as if acidic: $\mathrm{Mn}{\mathrm{O}}_4^{-}\to \mathrm{Mn}{\mathrm{O}}_2$. O: add 2 water right. H: add 4 ${\mathrm{H}}^{+}$ left. Charge: left $-1+4=+3$, right $0$, so add 3 electrons left. $\mathrm{Mn}{\mathrm{O}}_4^{-}+4{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\to \mathrm{Mn}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$ Why balance in acid first? Acidic bookkeeping is cleaner — one hydrogen source (${\mathrm{H}}^{+}$). We fix the "no free ${\mathrm{H}}^{+}$ in base" problem at the very end.
2. Oxidation half, as if acidic: $\mathrm{S}{\mathrm{O}}_3^{2-}\to \mathrm{S}{\mathrm{O}}_4^{2-}$. Add 1 water left (to supply the extra O), 2 ${\mathrm{H}}^{+}$ right, charge: left $-2$, right $-2+2=0$, so add 2 electrons right. $\mathrm{S}{\mathrm{O}}_3^{2-}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{S}{\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$ Why? S climbs $+4\to +6$ = loses 2 electrons; the arithmetic must show 2 electrons produced.
3. Equalize electrons. LCM (least common multiple) of 3 and 2 is 6: multiply reduction by 2, oxidation by 3. Why 6? We need the electron count in both halves to become the same number so they cancel; 6 is the smallest number both 3 and 2 divide into, so $2\times 3=6$ gained and $3\times 2=6$ lost. $2Mn{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+3S{\mathrm{O}}_3^{2-}+3{\mathrm{H}}_2\mathrm{O}\to 2Mn{\mathrm{O}}_2+4{\mathrm{H}}_2\mathrm{O}+3S{\mathrm{O}}_4^{2-}+6{\mathrm{H}}^{+}$
4. Tidy the acidic form. Cancel: $8{\mathrm{H}}^{+}$ left vs $6{\mathrm{H}}^{+}$ right → net $2{\mathrm{H}}^{+}$ left. $3{\mathrm{H}}_2\mathrm{O}$ left vs $4{\mathrm{H}}_2\mathrm{O}$ right → net $1{\mathrm{H}}_2\mathrm{O}$ right. $2Mn{\mathrm{O}}_4^{-}+2{\mathrm{H}}^{+}+3S{\mathrm{O}}_3^{2-}\to 2Mn{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+3S{\mathrm{O}}_4^{2-}$
5. Neutralize to base — done slowly. There are $2{\mathrm{H}}^{+}$ on the left. To kill them, add $2\mathrm{O}{\mathrm{H}}^{-}$ to both sides (adding the same thing to both sides never changes an equation): $2Mn{\mathrm{O}}_4^{-}+2{\mathrm{H}}^{+}+2O{\mathrm{H}}^{-}+3S{\mathrm{O}}_3^{2-}\to 2Mn{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+3S{\mathrm{O}}_4^{2-}+2O{\mathrm{H}}^{-}$
   • Substep 5a: on the left, $2{\mathrm{H}}^{+}+2\mathrm{O}{\mathrm{H}}^{-}$ combine into $2{\mathrm{H}}_2\mathrm{O}$ (acid + base → water). The left now reads $2Mn{\mathrm{O}}_4^{-}+2{\mathrm{H}}_2\mathrm{O}+3S{\mathrm{O}}_3^{2-}$.
   • Substep 5b: count waters on each side. Left has $2{\mathrm{H}}_2\mathrm{O}$; right has $1{\mathrm{H}}_2\mathrm{O}$ (from step 4). Cancel the smaller amount, $1{\mathrm{H}}_2\mathrm{O}$, from both: left keeps $1{\mathrm{H}}_2\mathrm{O}$, right keeps $0$.
   • Substep 5c: the $2\mathrm{O}{\mathrm{H}}^{-}$ we added on the right stay as $2\mathrm{O}{\mathrm{H}}^{-}$. Why add OH⁻ at all? In base there is no free ${\mathrm{H}}^{+}$ floating around; converting each ${\mathrm{H}}^{+}$ into water with an $\mathrm{O}{\mathrm{H}}^{-}$ gives a physically honest equation. This is exactly the acid + base → water idea from Acid–base neutralization (for the OH⁻ step). $2Mn{\mathrm{O}}_4^{-}+3S{\mathrm{O}}_3^{2-}+{\mathrm{H}}_2\mathrm{O}\to 2Mn{\mathrm{O}}_2+3S{\mathrm{O}}_4^{2-}+2O{\mathrm{H}}^{-}$

Verify: charge left $=2(-1)+3(-2)+0=-8$; right $=0+3(-2)+2(-1)=-8$. ✔ Atoms: Mn $2=2$, S $3=3$, O: left $8+9+1=18$, right $4+12+2=18$. ✔ H: left $2$, right $2$. ✔

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Case C — Disproportionation (one element does both jobs)

1. Split — but both halves start from $\mathrm{C}{\mathrm{l}}_2$. Reduction: $\mathrm{C}{\mathrm{l}}_2\to 2C{\mathrm{l}}^{-}$. Oxidation: $\mathrm{C}{\mathrm{l}}_2\to 2Cl{\mathrm{O}}^{-}$. Why? Disproportionation means one substance is both the oxidiser and the reducer. We treat each fate as its own half-reaction. The figure below shows this: a single $\mathrm{C}{\mathrm{l}}_2$ box splits, one arrow going down to $\mathrm{C}{\mathrm{l}}^{-}$ (gaining electrons, reduced) and one going up to $\mathrm{Cl}{\mathrm{O}}^{-}$ (losing electrons, oxidised).

Figure 2 — Ex 3. One species, two fates. The cyan arrow (upper) carries chlorine down to oxidation number $-1$ (reduction); the amber arrow (lower) carries chlorine up to $+1$ (oxidation). Recognising this split is the whole trick of disproportionation — there is no second reactant to look for.

2. Reduction half. $\mathrm{C}{\mathrm{l}}_2+2{\mathrm{e}}^{-}\to 2C{\mathrm{l}}^{-}$. Charge: left $0-2=-2$, right $-2$. ✔ (Each Cl gained 1 electron, $0\to -1$.)
3. Oxidation half (acidic bookkeeping first). $\mathrm{C}{\mathrm{l}}_2\to 2Cl{\mathrm{O}}^{-}$ needs 2 O: add 2 water left, 4 ${\mathrm{H}}^{+}$ right. Charge: left $0$, right $2(-1)+4=+2$, so add 2 electrons right. $\mathrm{C}{\mathrm{l}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2Cl{\mathrm{O}}^{-}+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$ Why? Each Cl climbs $0\to +1$ = loses 1 electron; two Cl = 2 electrons out. ✔
4. Equalize. Both halves already use 2 electrons — add directly. $2C{\mathrm{l}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2C{\mathrm{l}}^{-}+2Cl{\mathrm{O}}^{-}+4{\mathrm{H}}^{+}$ (the two $\mathrm{C}{\mathrm{l}}_2$ come from adding the halves — one is reduced, one is oxidised.)
5. Base conversion. Add $4\mathrm{O}{\mathrm{H}}^{-}$ both sides. Left $4{\mathrm{H}}^{+}+4\mathrm{O}{\mathrm{H}}^{-}=4{\mathrm{H}}_2\mathrm{O}$; right gains $4\mathrm{O}{\mathrm{H}}^{-}$. Cancel: $4{\mathrm{H}}_2\mathrm{O}$ left vs $2{\mathrm{H}}_2\mathrm{O}$ left originally → net $2{\mathrm{H}}_2\mathrm{O}$ left. Finally divide everything by 2: $\mathrm{C}{\mathrm{l}}_2+2O{\mathrm{H}}^{-}\to \mathrm{C}{\mathrm{l}}^{-}+\mathrm{Cl}{\mathrm{O}}^{-}+{\mathrm{H}}_2\mathrm{O}$

Verify: charge left $=0+2(-1)=-2$; right $=-1-1+0=-2$. ✔ Atoms: Cl $2=2$, O $2=1+1$, H $2=2$. ✔

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Case D — Comproportionation (two forms meet in the middle)

1. Reduction half: $\mathrm{I}{\mathrm{O}}_3^{-}\to \frac{1}{2}{\mathrm{I}}_2$; clear the fraction: $2I{\mathrm{O}}_3^{-}\to {\mathrm{I}}_2$. O: add 6 water right, 12 ${\mathrm{H}}^{+}$ left. Charge: left $2(-1)+12=+10$, right $0$, add 10 electrons left. $2I{\mathrm{O}}_3^{-}+12{\mathrm{H}}^{+}+10{\mathrm{e}}^{-}\to {\mathrm{I}}_2+6{\mathrm{H}}_2\mathrm{O}$ Why 10? Two iodine atoms each dropping $+5\to 0$ is $2\times 5=10$ electrons. ✔
2. Oxidation half: $2{\mathrm{I}}^{-}\to {\mathrm{I}}_2+2{\mathrm{e}}^{-}$ (each loses 1).
3. Equalize. LCM of 10 and 2 is 10: multiply oxidation half by 5. $2I{\mathrm{O}}_3^{-}+12{\mathrm{H}}^{+}+10{\mathrm{I}}^{-}\to {\mathrm{I}}_2+6{\mathrm{H}}_2\mathrm{O}+5{\mathrm{I}}_2$ Combine the iodine: $I_2+5I_2=6I_2$. $2I{\mathrm{O}}_3^{-}+12{\mathrm{H}}^{+}+10{\mathrm{I}}^{-}\to 6{\mathrm{I}}_2+6{\mathrm{H}}_2\mathrm{O}$ Why does the product merge? Both halves make the same substance, ${\mathrm{I}}_2$ — the reverse of disproportionation.

Verify: charge left $=2(-1)+12+10(-1)=0$; right $=0$. ✔ Atoms: I: left $2+10=12$, right $6\times 2=12$. ✔ O: $6=6$, H: $12=12$. ✔

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Case E — The spectator trap

1. Oxidation half: $\mathrm{Cu}\to \mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}$ ($0\to +2$, loses 2).
2. Reduction half: $\mathrm{N}{\mathrm{O}}_3^{-}\to \mathrm{NO}$. N goes $+5\to +2$. O: add 2 water right, 4 ${\mathrm{H}}^{+}$ left. Charge: left $-1+4=+3$, right $0$, add 3 electrons left. $\mathrm{N}{\mathrm{O}}_3^{-}+4{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\to \mathrm{NO}+2{\mathrm{H}}_2\mathrm{O}$ Why only these nitrates? Only the N atoms changing oxidation state belong in a half-reaction. Spectator ions (see Oxidizing and reducing agents) are invisible here — this is the whole trap.
3. Equalize. LCM of 2 and 3 is 6: ×3 copper half, ×2 nitrate half. $3Cu+2N{\mathrm{O}}_3^{-}+8{\mathrm{H}}^{+}\to 3C{\mathrm{u}}^{2+}+2NO+4{\mathrm{H}}_2\mathrm{O}$

Verify: charge left $=0+2(-1)+8=+6$; right $=3(+2)+0+0=+6$. ✔ Atoms: Cu $3=3$, N $2=2$, O: left $6$, right $2+4=6$, H $8=8$. ✔

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Case F — Peroxide oddity

1. Reduction half (unchanged from parent Ex 1): $\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to \mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$ Mn: $+7\to +2$, gains 5.
2. Oxidation half: ${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{O}}_2$. Each O rises $-1\to 0$, and there are 2 of them → 2 electrons lost. H: add 2 ${\mathrm{H}}^{+}$ right. Charge check: left $0$, right $2(+1)-2=0$. ✔ ${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$ Why is peroxide oxygen $-1$? In ${\mathrm{H}}_2{\mathrm{O}}_2$ the two oxygens share a bond with each other; the special rule from Oxidation states / oxidation number rules assigns them $-1$ each. Miss this and you'll balance the wrong element.
3. Equalize. LCM of 5 and 2 is 10: ×2 reduction, ×5 oxidation. $2Mn{\mathrm{O}}_4^{-}+16{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{O}}_2\to 2M{\mathrm{n}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2+10{\mathrm{H}}^{+}$ Cancel $10{\mathrm{H}}^{+}$: $16-10=6$ left. $2Mn{\mathrm{O}}_4^{-}+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{O}}_2\to 2M{\mathrm{n}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2$

Verify: charge left $=2(-1)+6+0=+4$; right $=2(+2)+0+0=+4$. ✔ Atoms: Mn $2=2$, O: left $8+10=18$, right $8+10=18$, H: left $6+10=16$, right $16$. ✔

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Case G — Real-world word problem

1. Moles of permanganate used. $n=\text{concentration}\times \text{volume}=0.0200\times 0.0250\mathrm{L}=5.00\times 10^{-4}\mathrm{mol}$. Why? Concentration (mol per litre) times volume (litres) gives moles — the definition of molarity from Stoichiometry & mole ratios. Note $25.0\mathrm{mL}=0.0250\mathrm{L}$.
2. Apply the balanced ratio. From $\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5F{\mathrm{e}}^{2+}\to \dots$, every 1 permanganate consumes 5 iron(II): $n(\mathrm{F}{\mathrm{e}}^{2+})=5\times 5.00\times 10^{-4}=2.50\times 10^{-3}\mathrm{mol}.$ Why 5? Because 1 Mn accepts 5 electrons and each Fe supplies 1 — the electron law fixes the mole ratio. Balancing is the stoichiometry.

Verify: electrons balance: permanganate accepts $5.00\times 10^{-4}\times 5=2.50\times 10^{-3}\mathrm{mol}{e}^{-}$; iron supplies $2.50\times 10^{-3}\times 1=2.50\times 10^{-3}\mathrm{mol}{e}^{-}$. Equal. ✔ Units check: $\mathrm{mol}{\mathrm{L}}^{-1}\times \mathrm{L}=\mathrm{mol}$, a pure amount, as required. ✔ Answer: $2.50\times 10^{-3}\mathrm{mol}\mathrm{F}{\mathrm{e}}^{2+}$ (i.e. $2.50\mathrm{mmol}$).

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Case H — The exam twist: "no reaction"

1. Assign oxidation numbers to both sides. Fe is $+3$ on both sides; Cu is $+2$ on both sides. Why start here? A redox reaction requires at least one element to change oxidation state. If none changes, there is no electron transfer to balance. (See Oxidation states / oxidation number rules.)
2. Try to build half-reactions. Oxidation half: needs a species losing electrons — none exists. Reduction half: needs a species gaining electrons — none exists. Why does this matter? $e_{\text{lost}}^{-}=e_{\text{gained}}^{-}=0$. The "machine" runs but transfers zero electrons: this is not a redox reaction at all (in fact not a reaction — both sides are identical).
3. Answer: there is nothing to balance beyond writing it as-is; report "no redox change; not a redox reaction." Exam lesson: always check oxidation numbers first. If nothing moves, don't force electrons that aren't there — a "no reaction" verdict is a valid, examinable answer.

Verify: total oxidation-number change $=0$. Electrons transferred $=0$. Charge already equal on both sides ($+5=+5$). ✔ (a consistent — if trivial — equation.)

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Recall Self-test

In $\mathrm{C}{\mathrm{l}}_2\to \mathrm{C}{\mathrm{l}}^{-}+\mathrm{Cl}{\mathrm{O}}^{-}$, which element is oxidised and which reduced? ::: Chlorine is both — it disproportionates ($0\to -1$ reduced, $0\to +1$ oxidised). In ${\mathrm{H}}_2{\mathrm{O}}_2$, what is the oxidation number of oxygen? ::: $-1$ (peroxide), not the usual $-2$. In Ex 7, why is the iron:permanganate mole ratio $5:1$? ::: Each $\mathrm{Mn}{\mathrm{O}}_4^{-}$ accepts 5 electrons; each $\mathrm{F}{\mathrm{e}}^{2+}$ gives 1. How do you spot a "no reaction" trap? ::: Assign oxidation numbers first; if nothing changes, no electrons transfer.

Connections

• 1.3.07 Balancing redox equations — ion-electron (half-reaction) method, oxidation-number method (Hinglish)
• Oxidation states / oxidation number rules
• Oxidizing and reducing agents
• Electrochemical cells & standard electrode potentials
• Stoichiometry & mole ratios
• Acid–base neutralization (for the OH⁻ step)
• Conservation of mass and charge

Concept Map