Half-reaction method

A redox equation like dichromate oxidising iron(II) looks intimidating until you split it in two. The half-reaction (ion-electron) method takes one messy ionic equation and separates it into an oxidation half and a reduction half, balances each on its own for atoms and then for charge, and finally scales the two so that every electron released by the reducing agent is absorbed by the oxidising agent. Nothing is guessed — each step follows a fixed order: atoms other than O and H first, then oxygen with H2_2O, then hydrogen with H+^+, then charge with electrons. The technique is the second balancing tool in the ISC Class 11 Redox Reactions unit, alongside the oxidation number method, and it matters beyond balancing on paper: the electron count it exposes is exactly what fixes the equivalent ratio used later in redox titrations.

The core insight

Split a redox equation into oxidation and reduction half-reactions, balance each independently for atoms (non-O/H, then O with H2_2O, then H with H+^+) and then for charge with electrons, then scale and add so the electrons cancel exactly.

The core idea

Two half-reactions, one electron currency. Every redox change is really two things happening together: something loses electrons (oxidation) and something gains the same number of electrons (reduction). The half-reaction method makes this literal — it writes each process as its own equation, complete with electrons as a term, and only combines them once the electron count matches.

The fixed balancing order. Within a single half-reaction, balance in this sequence: atoms other than oxygen and hydrogen first (get the skeleton right), then oxygen by adding water to the side short of oxygen, then hydrogen by adding H+^+ to the side short of hydrogen, and only then charge, by adding electrons to whichever side carries the more positive total. Doing charge before atoms is the single most common source of errors.

Electron matching is the join. If the oxidation half releases aa electrons per formula unit and the reduction half absorbs bb electrons per formula unit, the two half-reactions cannot simply be added — one side would have leftover electrons. Multiplying the oxidation half by bb and the reduction half by aa makes both sides supply exactly abab electrons, so adding them and cancelling the electron terms leaves a fully balanced net-ionic equation with no free electrons anywhere.

nox elost per unit−=nred egained per unit−n_{\text{ox}}\,e^-_{\text{lost per unit}} = n_{\text{red}}\,e^-_{\text{gained per unit}}

where: noxn_{\text{ox}} and nredn_{\text{red}} are the integer multipliers applied to the oxidation and reduction half-reactions respectively, and elost per unit−e^-_{\text{lost per unit}}, egained per unit−e^-_{\text{gained per unit}} are the electrons shown in each balanced half-reaction before scaling. The multipliers are chosen so the two sides of this equation are equal — usually by taking the LCM of the two electron counts.

Acidic versus basic medium. In acidic solution the atom-balancing tools are H2_2O and H+^+, as above. In basic solution, balance exactly as if the medium were acidic first, then add OH−^- to both sides equal to the number of H+^+ present, combine H++OH−→H2O^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} wherever they meet, and cancel any water molecules that now appear on both sides — the same conversion trick used after the oxidation number method.

Figure
Two half reactions balanced separately then added so their electrons cancel
Ion-electron method — balance separately, then join on electrons Oxidation half Fe2+ to Fe3+ + e- loses electrons, charge balanced with e- on right Reduction half Cr2O7 2- + 14H+ + 6e- to 2Cr3+ + 7H2O gains electrons, atoms then charge balanced multiply by 6 multiply by 1 Cr2O7 2- + 14H+ + 6Fe2+ to 2Cr3+ + 7H2O + 6Fe3+ electrons cancel exactly, atoms and charge match both sides
Derivation AWhy the electron multipliers must be b and a

The goal is to justify, from the plain requirement that no free electrons can survive in a net-ionic equation, why an oxidation half-reaction releasing aa electrons per formula unit must be multiplied by bb and a reduction half-reaction absorbing bb electrons per formula unit must be multiplied by aa before the two half-reactions are added, so that the resulting combined equation is balanced for both atoms and charge with the electron terms cancelling completely.

Step 1. Write the two half-reactions in their already atom-and-charge-balanced forms, oxidation releasing electrons on the product side and reduction absorbing electrons on the reactant side: Red→Ox+a e−\text{Red} \rightarrow \text{Ox} + a\,e^- and Ox′+b e−→Red′\text{Ox}' + b\,e^- \rightarrow \text{Red}', where aa and bb are the whole-number electron counts already fixed by the charge-balancing step.

Step 2. Adding these two half-reactions as they stand only makes sense chemically if the electrons produced by the first exactly equal the electrons consumed by the second — otherwise the sum would still contain bare e−e^- terms, which cannot appear in a real chemical equation between neutral or ionic species. So the requirement is aa (from however many times the oxidation half is used) equal to bb (from however many times the reduction half is used).

Step 3. Multiply the oxidation half-reaction throughout by bb and the reduction half-reaction throughout by aa: scaling oxidation by bb raises its electron output from aa to abab, and scaling reduction by aa raises its electron intake from bb to abab. Both half-reactions now involve the identical electron count, abab.

Step 4. Add the two scaled half-reactions. The ab e−ab\,e^- term appears on the product side of the scaled oxidation half and on the reactant side of the scaled reduction half, so it cancels identically, leaving a single combined equation with no electrons, balanced for charge because each half-reaction was separately charge-balanced before scaling (multiplying a balanced equation by an integer preserves the balance), and balanced for atoms for the same reason. Verified: for the dichromate–iron pair, a=1a = 1 (Fe2+→^{2+} \rightarrow Fe3++e−^{3+} + e^-) and b=6b = 6 (Cr2_2O72−_7^{2-} half needs 6e−6e^-), so the oxidation half is multiplied by 66 and the reduction half by 11, giving ab=6ab = 6 electrons cancelling — matching the worked example below.

Worked example

Question. Balance Cr2O72−+Fe2+→Cr3++Fe3+\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+} in acidic medium using the ion-electron method.

Step 1. Balance atoms other than O and H. The reduction half needs 2 Cr on the right: Cr2O72−→2Cr3+\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}. The oxidation half already has one Fe on each side: Fe2+→Fe3+\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}.

Step 2. Balance oxygen with H2_2O, then hydrogen with H+^+, in the reduction half, which has 7 oxygen atoms on the left: Cr2O72−+14H+→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}.

Step 3. Balance charge with electrons. On the reduction half, left charge is −2+14=+12-2 + 14 = +12 and right charge is 2(+3)=+62(+3) = +6, so add 6 electrons to the left: Cr2O72−+14H++6e−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}. On the oxidation half, left charge +2+2 and right charge +3+3, so add 1 electron to the right: Fe2+→Fe3++e−\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-.

Step 4. Equalise electrons and add. The reduction half needs 6 electrons; the oxidation half supplies only 1 per Fe, so multiply the oxidation half by 6, giving 6Fe2+→6Fe3++6e−6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-, then add the two half-reactions and cancel the 6e−6e^- on each side:

Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}

Step 5. Check both atoms and charge. Charge left: −2+14+12=+24-2 + 14 + 12 = +24; charge right: 6+18=+246 + 18 = +24. Atoms: Cr (2=22 = 2), Fe (6=66 = 6), O (7=77 = 7), H (14=1414 = 14). The balanced equation is Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}.

Competitive stretch

The same method handles basic medium with one extra move. For MnO4−+SO32−→MnO2+SO42−\text{MnO}_4^- + \text{SO}_3^{2-} \rightarrow \text{MnO}_2 + \text{SO}_4^{2-} in basic medium, first balance both halves as if acidic: the reduction half becomes MnO4−+4H++3e−→MnO2+2H2O\text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} and the oxidation half becomes SO32−+H2O→SO42−+2H++2e−\text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 2\text{H}^+ + 2e^-. Here a=2a = 2 and b=3b = 3, so multiply reduction by 2 and oxidation by 3 (LCM =6= 6), add, and cancel duplicate H2_2O and H+^+, which leaves 2H+2\text{H}^+ net on the left. Adding 2OH−2\text{OH}^- to both sides converts that 2H+2\text{H}^+ to 2H2O2\text{H}_2\text{O}, and cancelling one duplicate water gives the final basic-medium equation 2MnO4−+3SO32−+H2O→2MnO2+3SO42−+2OH−2\text{MnO}_4^- + 3\text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{SO}_4^{2-} + 2\text{OH}^-, which checks at −8-8 charge on both sides.

Common traps

  • Balancing charge with electrons before atoms are settled — always finish non-O/H atoms, then O, then H, and only then add electrons.
  • Adding H2_2O or H+^+ to a fixed side out of habit instead of comparing atom counts first and adding to whichever side is actually deficient.
  • Multiplying only the electron term when scaling a half-reaction to match electron counts, instead of scaling every species in that half-reaction.
  • Forgetting the final cleanup after adding the two half-reactions — duplicate H2_2O, H+^+, or OH−^- appearing on both sides must be cancelled down to simplest form.

Ask the tutor: Want to practise another half-reaction balancing problem, or see the ion-electron method placed side by side with the oxidation number method on the same equation?

Why not just balance the whole equation at once instead of splitting into two halves?

Splitting keeps the electron bookkeeping explicit. In the combined equation it is easy to lose track of how many electrons the oxidising agent actually needs versus how many the reducing agent supplies, especially when the numbers differ, as with dichromate needing 6 and iron supplying only 1 per ion. Balancing each half separately for atoms and then for charge guarantees the electron count in each half is correct before you ever add them, so the final combination is mechanical rather than a guess.

How do I know whether to add H2O or H+ first when balancing a half-reaction?

Always oxygen before hydrogen. Compare oxygen atom counts on the two sides and add H2_2O to whichever side is short of oxygen; only after oxygen matches do you compare hydrogen counts and add H+^+ to the side short of hydrogen. Doing it in the other order, or adding H+^+ before checking oxygen, is the single most common way this method goes wrong on a test.

Why does the basic-medium conversion add OH- equal to the H+ count instead of just removing the H+

You cannot simply delete H+^+ from an equation because that would remove charge and atoms without accounting for them chemically. Adding OH−^- to both sides is a legitimate step because it adds the same species to each side equally, and it lets the H+^+ combine with the added OH−^- to form neutral water, which is the species that actually exists in a basic solution. That is why the H+^+ "disappears" only after this conversion, never before it.

Before this

Where this leads