Class 11 Redox: NCERT Question Patterns
Class 11 redox questions follow a small set of patterns. The NCERT exercise and the board papers ask you to find oxidation numbers in unusual compounds, to name the species that is oxidised or reduced, to spot a disproportionation reaction, or to decide whether a reaction is a redox reaction at all. This page covers these question patterns. For the method of balancing redox equations, see our separate article on redox reactions.
The rules you need
Oxidation = increase in oxidation number (loss of electrons) | Reduction = decrease (gain of electrons)
- An element in its free state is 0.
- Oxygen is usually −2 (exceptions: peroxides −1, OF2 +2).
- Hydrogen is usually +1 (exception: metal hydrides such as NaH, where it is −1).
- Alkali metals are +1, alkaline earth metals are +2, fluorine is always −1.
Pattern 1: Oxidation number of the central atom
Step 1: Let the oxidation number of Cr be x. Two Cr atoms give 2x. Seven O atoms give 7 × (−2) = −14.
Step 2: The total must equal the charge, −2: 2x − 14 = −2.
Step 3: 2x = 12, so x = +6.
Check: 2(+6) + 7(−2) = 12 − 14 = −2 ✓.
Answer: +6.
Pattern 2: Fractional oxidation number
Step 1 (sulphur): Na is +1, so two Na give +2. Six O give −12. Let S be x: 2 + 4x − 12 = 0.
Step 2: 4x = 10, so x = +2.5. This is the average oxidation number of the four sulphur atoms. In the real structure, two S atoms are 0 and two are +5, and (0 + 0 + 5 + 5)/4 = 2.5 ✓.
Step 3 (iron): Four O give −8. Three Fe must add up to +8, so each Fe is 8/3 = +2.67 on average. In reality Fe3O4 contains Fe2+ and Fe3+ in the ratio 1 : 2, and 2 + 3 + 3 = 8 ✓.
Answer: S = +2.5 (average) and Fe = +8/3 (average).
Pattern 3: Same element, two different oxidation numbers
Step 1: The compound has two ions, NH4+ and NO3−. Work out each ion separately.
Step 2: In NH4+: x + 4(+1) = +1, so x = −3.
Step 3: In NO3−: x + 3(−2) = −1, so x = +5.
Answer: N is −3 in the ammonium ion and +5 in the nitrate ion. If you add all the N atoms together and divide, you get a meaningless average of +1. Do not do this when the structure shows two different ions.
Pattern 4: Identify the oxidising and reducing agents
The oxidising agent is the species that is reduced (its oxidation number goes down). The reducing agent is the species that is oxidised (its oxidation number goes up).
Step 1: Fe in Fe2O3: 2x + 3(−2) = 0, so x = +3. Fe in the product is 0. Fe goes from +3 to 0: reduction.
Step 2: C in CO: x + (−2) = 0, so x = +2. C in CO2: x + 2(−2) = 0, so x = +4. C goes from +2 to +4: oxidation.
Step 3: Total electrons check: 2 Fe atoms gain 2 × 3 = 6 electrons. 3 C atoms lose 3 × 2 = 6 electrons ✓.
Answer: Fe2O3 is the oxidising agent (it is reduced). CO is the reducing agent (it is oxidised).
Pattern 5: Disproportionation
In a disproportionation reaction, one element in one species is oxidised and reduced at the same time.
Step 1: Cl in Cl2 is 0.
Step 2: In Cl− it is −1 (reduced, 0 → −1).
Step 3: In ClO−: x + (−2) = −1, so x = +1 (oxidised, 0 → +1).
Step 4: One reactant, Cl2, gives a product with a lower and a product with a higher oxidation number. This is disproportionation.
Answer: Chlorine is both oxidised and reduced. Electron balance: the Cl atom that goes to −1 gains 1 electron, and the one that goes to +1 loses 1 electron ✓.
Pattern 6: Is it a redox reaction?
Write the oxidation number of every element on both sides. If no number changes, it is not a redox reaction. For example, in AgNO3 + NaCl → AgCl + NaNO3, Ag stays +1, Na stays +1, Cl stays −1 and N stays +5. Nothing changes, so it is a precipitation reaction and not redox.
- Forgetting the charge. For an ion, the sum equals the ion's charge, not zero.
- Treating every oxygen as −2. In H2O2 oxygen is −1. Check for a peroxide before you assume −2.
- Treating every hydrogen as +1. In NaH and CaH2, hydrogen is −1.
- Reading the "agent" the wrong way round. The oxidising agent is the one that gets reduced.
- Averaging across different ions. Work out NH4+ and NO3− separately.
- Counting atoms in the electron balance. Multiply the change in oxidation number by the number of atoms that change.
Exam relevance
| Question type | First step |
|---|---|
| Find the oxidation number of an element in a compound or ion | Set the sum equal to the charge, solve for x |
| Identify oxidised and reduced species | Write the oxidation numbers on both sides and compare |
| Name the oxidising and reducing agents | Oxidising agent = species reduced; reducing agent = species oxidised |
| Disproportionation | Look for one element ending at both a higher and a lower number |
Check your own syllabus and the latest sample paper to see how redox is tested in your board. Do not assume a fixed number of marks.
Check your own answers with the calculator after you have worked each one by hand. Do the working first, then use the tool to confirm.
Open the Oxidation Number Calculator →