Oxidation number method

Redox equations rarely balance by inspection — count the atoms of KMnO4+FeSO4+H2SO4→K2SO4+MnSO4+Fe2(SO4)3+H2O\text{KMnO}_4 + \text{FeSO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{MnSO}_4 + \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2\text{O} by trial and you will be guessing coefficients forever. The oxidation number method replaces guesswork with a single bookkeeping rule: every electron an oxidised atom loses must be picked up by a reduced atom, so the total rise in oxidation number on one side must exactly equal the total fall on the other. Fix that ratio first, then mop up the remaining atoms — oxygen and hydrogen — with water and, in acid, H+^+ (or with OH−^- if the reaction runs in base). This is one of the two prescribed balancing techniques in the ISC Class 11 Redox Reactions unit, alongside the ion-electron (half-reaction) method, and it leans on the oxidation-number rules taught just before it in the same unit. Balanced redox equations feed directly into redox titration calculations later in the course, where the mole ratio of oxidant to reductant comes straight from the coefficients you find here.

The core insight

Balance a redox equation by first choosing coefficients so the total increase in oxidation number (oxidation) equals the total decrease (reduction), then balancing the remaining atoms with H2_2O and H+^+ in acid, or H2_2O and OH−^- in base.

The core idea

Every redox reaction is, underneath the formulas, an electron transfer. The atom being oxidised loses electrons and its oxidation number climbs; the atom being reduced gains electrons and its oxidation number falls. Because electrons are neither created nor destroyed in the reaction, the electrons lost by the oxidised species must equal the electrons gained by the reduced species. Oxidation number is just a bookkeeping stand-in for electron count, so this conservation law reads:

Δ(ON)oxidised, total=Δ(ON)reduced, total\Delta(\text{ON})_{\text{oxidised, total}} = \Delta(\text{ON})_{\text{reduced, total}}

where: Δ(ON)oxidised, total\Delta(\text{ON})_{\text{oxidised, total}} is the oxidation-number rise per atom multiplied by the number of such atoms in the formula unit (for example two Cr atoms in Cr2_2O72−_7^{2-}), and Δ(ON)reduced, total\Delta(\text{ON})_{\text{reduced, total}} is the corresponding fall for the reduced species, counted the same way. Getting the per-formula-unit total right — not just the per-atom change — is the step that actually decides the coefficients. Once that ratio of oxidant to reductant is fixed, the rest of the equation is finished by ordinary atom counting, in a fixed sequence:

  • Balance every atom except O and H by inspection, using the coefficients already fixed by the electron-transfer step.
  • In acidic medium: balance O by adding H2_2O to the side short of oxygen, then balance the resulting H with H+^+.
  • In basic medium: balance exactly as if acidic first (H2_2O and H+^+), then add an equal number of OH−^- to both sides to neutralise every H+^+, combine H++OH−→H2O^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}, and cancel any water common to both sides.
  • Finish with a check: total charge on the left must equal total charge on the right, in addition to every atom balancing.
Figure
Electron transfer balance showing five Fe atoms oxidised for every one Mn atom reduced
Total decrease must equal total increase reduction (Mn) Mn +7 to +2 fall of 5 1 Mn atom total fall = 5 = oxidation (Fe) +1 +1 +1 +1 +1 5 Fe atoms total rise = 5
Derivation AFinding the coefficient ratio from oxidation-number changes

The goal is to turn the raw oxidation-number changes of the oxidised and reduced species into the smallest whole-number coefficients that make total electron loss equal total electron gain, which is the one calculation every application of this method depends on, whether the species are simple ions or polyatomic formula units with several atoms changing state at once.

Step 1. For the species being reduced, find the fall in oxidation number per atom, call it frf_r, and the number of atoms of that element in one formula unit, nrn_r. The total decrease contributed by one formula unit is nrfrn_r f_r. For KMnO4\text{KMnO}_4 reducing to Mn2+\text{Mn}^{2+}, nr=1n_r = 1 and fr=5f_r = 5, so the total decrease per formula unit is 5.

Step 2. For the species being oxidised, find the rise per atom, fof_o, and the number of such atoms per formula unit, non_o, giving a total increase of nofon_o f_o per formula unit. For FeSO4\text{FeSO}_4 oxidising to Fe3+^{3+}, no=1n_o = 1 and fo=1f_o = 1, so the total increase per formula unit is 1.

Step 3. The coefficients aa (for the reduced species) and bb (for the oxidised species) must satisfy a(nrfr)=b(nofo)a(n_r f_r) = b(n_o f_o), since total electrons gained must equal total electrons lost. The smallest whole numbers satisfying this are a=nofogcd⁡(nrfr, nofo)a = \dfrac{n_o f_o}{\gcd(n_r f_r,\, n_o f_o)} and b=nrfrgcd⁡(nrfr, nofo)b = \dfrac{n_r f_r}{\gcd(n_r f_r,\, n_o f_o)}, found in practice by taking the ratio nrfr:nofon_r f_r : n_o f_o and reducing it, or by using the LCM when the two totals share no common factor.

Step 4. Apply it to the KMnO4_4/FeSO4_4 case: nrfr=5n_r f_r = 5, nofo=1n_o f_o = 1, so a:b=1:5a : b = 1 : 5 — one KMnO4_4 for every five FeSO4_4, exactly the ratio used in the worked example that follows. For the Cr2_2O72−_7^{2-}-to-Cr3+^{3+} case, nr=2n_r = 2, fr=3f_r = 3 gives nrfr=6n_r f_r = 6; against a one-electron oxidant like Fe2+^{2+} (nofo=1n_o f_o = 1), the ratio is 1:61 : 6, showing why forgetting the factor of 2 for Cr2_2O72−_7^{2-} silently halves the reductant coefficient and breaks the final balance.

Worked example

Question: Balance KMnO4+FeSO4+H2SO4→K2SO4+MnSO4+Fe2(SO4)3+H2O\text{KMnO}_4 + \text{FeSO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{MnSO}_4 + \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2\text{O} (acidic medium) by the oxidation number method, and confirm the balanced coefficients.

Step 1. Assign oxidation numbers and spot the changes. Mn runs from +7+7 in KMnO4\text{KMnO}_4 to +2+2 in MnSO4\text{MnSO}_4 — a fall of 5 (reduction). Fe runs from +2+2 in FeSO4\text{FeSO}_4 to +3+3 in Fe2(SO4)3\text{Fe}_2(\text{SO}_4)_3 — a rise of 1 per atom (oxidation).

Step 2. Equalise total change. Each Mn atom accounts for a fall of 5; each Fe atom for a rise of 1. To make total rise equal total fall, 5 Fe atoms must be oxidised for every 1 Mn atom reduced, giving the ratio 1 KMnO4:5 FeSO41\,\text{KMnO}_4 : 5\,\text{FeSO}_4. Doubling to clear the fraction that would otherwise appear on K2_2SO4_4 gives 2KMnO4+10FeSO4→K2SO4+2MnSO4+5Fe2(SO4)32\text{KMnO}_4 + 10\text{FeSO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3.

Step 3. Balance sulphate by inspection to fix H2_2SO4_4. The right side needs 11 (K2_2SO4_4) +2+ 2 (MnSO4_4) +15+ 15 (5×\times3 in Fe2_2(SO4_4)3_3) =18= 18 sulphate groups; FeSO4_4 already supplies 10 on the left, so H2_2SO4_4 must supply the remaining 18−10=818 - 10 = 8.

Step 4. Balance H and O with water. 8 H2_2SO4_4 carries 16 H atoms, so 8 H2_2O must form on the right to absorb them.

Step 5. Write the balanced equation and check it: 2KMnO4+10FeSO4+8H2SO4→K2SO4+2MnSO4+5Fe2(SO4)3+8H2O2\text{KMnO}_4 + 10\text{FeSO}_4 + 8\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3 + 8\text{H}_2\text{O}. K: 2=22=2. Mn: 2=22=2. Fe: 10=1010=10. Sulphate: right 1+2+15=181+2+15=18, left 10+8=1810+8=18. H: 16=1616=16 on both sides, and O balances along with it. The balanced equation is 2KMnO4+10FeSO4+8H2SO4→K2SO4+2MnSO4+5Fe2(SO4)3+8H2O2\text{KMnO}_4 + 10\text{FeSO}_4 + 8\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3 + 8\text{H}_2\text{O}.

Competitive stretch

For a basic-medium case, take MnO4−+I−→MnO2+I2\text{MnO}_4^- + \text{I}^- \rightarrow \text{MnO}_2 + \text{I}_2. Mn falls from +7+7 to +4+4 (decrease 3); I rises from −1-1 to 00 (increase 1 per atom). The LCM of 3 and 2 (I2_2 pairs iodine atoms) fixes the ratio at 2MnO4−+6I−→2MnO2+3I22\text{MnO}_4^- + 6\text{I}^- \rightarrow 2\text{MnO}_2 + 3\text{I}_2. Balancing O first as if acidic needs 4 H2_2O on the right and 8 H+^+ on the left; adding 8 OH−^- to both sides and combining H++OH−→H2O^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}, then cancelling water common to both sides, gives 2MnO4−+6I−+4H2O→2MnO2+3I2+8OH−2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^-. Charge check: left 2(−1)+6(−1)=−82(-1) + 6(-1) = -8; right 8(−1)=−88(-1) = -8. Balanced.

Common traps

  • Using the oxidation-number change per atom directly as the coefficient, instead of first multiplying by the number of such atoms in the formula unit (two Cr in Cr2_2O72−_7^{2-}, two I in I2_2).
  • Declaring the equation balanced once major atoms match, without running the final charge-balance check — oxygen, hydrogen, or overall charge can still be off.
  • Applying the acidic-medium O/H procedure to a reaction stated to be in basic medium and skipping the OH−^- neutralisation step entirely.
  • Leaving fractional coefficients uncleared instead of multiplying the whole equation through by an integer to make them whole numbers.

Ask the tutor: Try balancing a fresh polyatomic redox equation, such as one involving Cr2_2O72−_7^{2-} or MnO4−_4^- in a medium you haven't practised, and ask your AI tutor to check your oxidation-number-change table and final charge balance step by step.

Why do we multiply the oxidation number change by the number of atoms in the formula, not just use the per-atom value?

The conservation law is about total electrons transferred, not about a single atom. If two atoms of an element each change oxidation number by the same amount inside one formula unit — like both Cr atoms in Cr2_2O72−_7^{2-} falling by 3 — the formula unit as a whole accounts for a fall of 6, because both atoms are reduced together every time that ion reacts. Using only the per-atom value of 3 would understate the electrons actually transferred and give the wrong coefficient ratio, so the balancing would fail the final atom or charge check even though it looked plausible early on.

How do I know whether to balance in acidic or basic medium?

The question tells you directly, either by naming the medium or by showing H+^+/H3_3O+^+ (acidic) or OH−^- (basic) among the reactants or products. If nothing is stated and the species involved are typical of acidic conditions — for example permanganate reacting with Fe2+^{2+} in dilute sulphuric acid — treat it as acidic. There is no way to guess correctly from the skeletal equation alone if the medium genuinely is not indicated, so always balance using H2_2O and H+^+ first; only add the OH−^- conversion step if you are explicitly told the reaction happens in base.

What if the same equation can be balanced by both the oxidation number method and the ion-electron method — do I get different coefficients?

No, both methods must give the same final balanced equation for the same reaction, because both are enforcing the identical physical law: electrons lost equal electrons gained. The oxidation number method tracks that conservation through whole-formula oxidation-number changes, while the ion-electron method splits the reaction into separate oxidation and reduction half-equations and balances electrons explicitly in each. They are two bookkeeping routes to the same coefficients, so if you get different answers from the two methods on the same equation, one of the two attempts has an arithmetic or medium-handling error.

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Where this leads