Redox Reactions — Spotting Oxidation and Reduction Reliably
Most students can recite "oxidation is loss of electrons". The trouble starts when the equation has no visible electrons — when you are handed MnO₄⁻ + Fe²⁺ and asked which species is oxidised. This article gives you one method that never fails: track the oxidation number. If a number rises, that atom was oxidised. If it falls, that atom was reduced. If nothing changes, the reaction is not redox at all.
The three definitions, and which one to trust
| Basis | Oxidation is… | Reduction is… | Works when? |
|---|---|---|---|
| Oxygen | gain of oxygen | loss of oxygen | Only for simple oxide reactions — the oldest and weakest definition |
| Hydrogen | loss of hydrogen | gain of hydrogen | Useful in organic chemistry, useless for ionic reactions |
| Electrons | loss of electrons | gain of electrons | Always true, but electrons are usually invisible in the equation |
| Oxidation number | increase in ON | decrease in ON | Always. This is the working tool. |
The classic mnemonic still helps: OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons).
Assigning oxidation numbers — the rules in order
Apply these from the top; the earlier rule wins any conflict.
- An atom in its free element has ON = 0 (Na, O₂, P₄, S₈, Cl₂).
- A monatomic ion has ON equal to its charge (Na⁺ = +1, S²⁻ = −2).
- Fluorine is always −1 in compounds.
- Group 1 metals are +1, Group 2 metals are +2 in their compounds.
- Hydrogen is +1, except −1 in metal hydrides (NaH, CaH₂).
- Oxygen is −2, except −1 in peroxides (H₂O₂), −½ in superoxides (KO₂) and +2 in OF₂, where fluorine's rule outranks it.
- The oxidation numbers in a neutral compound sum to 0; in a polyatomic ion they sum to the ion's charge.
The 30-second redox test
Is CaCO₃ → CaO + CO₂ a redox reaction?
Ca: +2 → +2. C: +4 → +4. O: −2 → −2.
Nothing moved, so it is not redox — it is thermal decomposition.
Neutralisation (NaOH + HCl) and precipitation (AgNO₃ + NaCl) fail the same test for the
same reason.
Is Zn + CuSO₄ → ZnSO₄ + Cu redox?
Zn: 0 → +2 (rise, oxidised). Cu: +2 → 0 (fall, reduced). Yes.
Zn is the reducing agent, Cu²⁺ is the oxidising agent.
Note the naming trap carefully: the species that is oxidised is the reducing agent, because by giving up electrons it reduces something else. The species that is reduced is the oxidising agent. Students routinely write these the wrong way round.
Balancing by the oxidation-number method — acidic medium
The recipe has five steps:
- Assign oxidation numbers and find the atoms that change.
- Work out electrons lost per atom and gained per atom.
- Multiply each species by whatever factor makes total electrons lost = total gained.
- Balance every atom except O and H. Then balance O with H₂O and H with H⁺.
- Check that the total charge on the left equals the total charge on the right. This is the step that catches mistakes.
Example 1 — permanganate oxidising iron(II) in acid.
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Mn: +7 → +2, so each Mn gains 5 electrons.
Fe: +2 → +3, so each Fe loses 1 electron.
To match, we need 5 Fe per 1 Mn:
MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺
Left has 4 O, right has none → add 4 H₂O on the right.
Right now has 8 H → add 8 H⁺ on the left.
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Charge check: left = (−1) + 5(+2) + 8(+1) = +17.
Right = (+2) + 5(+3) = +17. ✓
Atom check: Mn 1 = 1, Fe 5 = 5, O 4 = 4, H 8 = 8. ✓
Example 2 — dichromate oxidising sulphite in acid.
Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻
Cr: +6 → +3, gain of 3 electrons per Cr, and there are 2 Cr atoms → 6 electrons
gained per dichromate ion.
S: +4 → +6, loss of 2 electrons per S → we need 3 sulphite ions.
Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻
O count: left 7 + 9 = 16; right 12 → add 4 H₂O on the right (12 + 4 = 16). ✓
H count: right has 8 → add 8 H⁺ on the left.
Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
Charge check: left = (−2) + 3(−2) + 8(+1) = 0. Right = 2(+3) + 3(−2) = 0. ✓
Balancing in basic medium — two safe routes
In alkaline solution there is no free H⁺ to write, so the equation must end up with OH⁻ and H₂O only. There are two ways to get there, and they must agree.
Example 3 — permanganate oxidising iodide in base.
MnO₄⁻ + I⁻ → MnO₂ + I₂
Mn: +7 → +4, gain of 3 electrons. I: −1 → 0, loss of 1 electron each, so producing one I₂ costs 2 electrons. The lowest common multiple of 3 and 2 is 6, so take 2 MnO₄⁻ and 6 I⁻ (giving 3 I₂):
2MnO₄⁻ + 6I⁻ → 2MnO₂ + 3I₂
Route A — balance charge with OH⁻ directly. Left charge = −2 + (−6) = −8; right = 0. Add 8 OH⁻ to the right to make it −8. Left O = 8; right O = 4 + 8 = 12, so add 4 H₂O on the left. H is then 8 on each side. ✓
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻
Route B — balance in acid first, then neutralise. In acid the equation comes out
as 2MnO₄⁻ + 6I⁻ + 8H⁺ → 2MnO₂ + 3I₂ + 4H₂O. Now add 8 OH⁻ to both sides. On the
left, 8H⁺ + 8OH⁻ becomes 8H₂O:
2MnO₄⁻ + 6I⁻ + 8H₂O → 2MnO₂ + 3I₂ + 4H₂O + 8OH⁻
Cancel the 4 H₂O common to both sides and you get the identical answer. Two independent
routes agreeing is a genuine check on your working.
Example 4 — disproportionation. Cl₂ + OH⁻ → Cl⁻ + ClO⁻
Chlorine starts at 0 and ends at both −1 and +1, so the same element is oxidised
and reduced. One Cl loses an electron, one gains it, so the electron count is already
balanced within one Cl₂ molecule:
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
Charge: left −2, right (−1) + (−1) = −2 ✓; O: 2 = 1 + 1 ✓; H: 2 = 2 ✓.
Common mistakes that cost marks
- Swapping the agents. Oxidised species = reducing agent. Write it on your formula sheet until it is automatic.
- Adding H⁺ in a basic medium (or OH⁻ in acid). Read the question's medium before you start.
- Balancing atoms before electrons. The electron count fixes the coefficients; atoms come after. Doing it the other way round almost always fails.
- Skipping the charge check. An equation can be atom-balanced and still wrong. The charge check is 10 seconds and catches nearly everything.
- Panicking at fractional oxidation numbers. They are legitimate averages. In Fe₃O₄ the three Fe atoms share +8, giving an average of +8/3 ≈ +2.67 — the real solid contains one Fe²⁺ and two Fe³⁺.
- Forgetting oxygen's exceptions. In H₂O₂ oxygen is −1, and in OF₂ it is +2 because fluorine's rule outranks oxygen's.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11–12 | Identify oxidising and reducing agents; balance by the oxidation-number method |
| JEE/NEET | Balancing in acidic/basic medium; equivalent weight of oxidants; disproportionation |
| IIT-JAM / CUET-PG | Redox titrations with KMnO₄ and K₂Cr₂O₇; ion–electron half-reaction method |
| GATE / CSIR-NET | Electrode potentials, Latimer and Frost diagrams, predicting disproportionation |
Verify every oxidation number before you balance. One wrong ON at step 1 ruins the whole equation. The Oxidation Number calculator assigns oxidation states for any formula you type — including polyatomic ions and the awkward cases like Fe₃O₄, S₂O₃²⁻ and H₂O₂.
Open the Oxidation Number Calculator →Redox is the chapter that quietly decides your electrochemistry marks in Class 12. ABC Chemistry runs Class 11–12 chemistry coaching at its Gurugram centre and online classes across India — details at abcchemistry.in.