Redox Reactions ISC Class 11: Oxidation Numbers and Balancing

What Redox Reactions is about

Weightage: Redox Reactions belongs to the unit Inorganic Chemistry: Redox Reactions, which CISCE weights at 6 of the 70 marks in the ISC Class XI Chemistry Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Oxidation and reduction: oxygen, hydrogen and electrons

Classical definitions: oxidation is addition of oxygen (or an electronegative element) or removal of hydrogen (or an electropositive element); reduction is the reverse. Electronic definitions: oxidation is loss of electrons and reduction is gain of electrons (OIL RIG). Oxidation and reduction always occur together. The oxidising agent accepts electrons (is reduced); the reducing agent donates electrons (is oxidised).

Worked example

Identify the substance oxidised and the oxidising agent in (i) $2Mg+O_2\to 2MgO$ (ii) $H_{2}S+C{l}_2\to 2HCl+S$ (iii) $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$.

Try it first, then show the solutionSolution

1. (i) Mg gains oxygen (loses electrons): Mg oxidised; O₂ is the oxidising agent
2. (ii) H₂S loses hydrogen: H₂S oxidised; Cl₂ is the oxidising agent
3. (iii) Zn loses 2 electrons: Zn oxidised; Cu²⁺ gains them and is the oxidising agent

Answer: Mg, O₂; H₂S, Cl₂; Zn, Cu²⁺

Rules for oxidation numbers

The oxidation number (ON) is the charge an atom would have if all its bonds were ionic. Rules: (1) an element in the free state has ON 0 (Na, O₂, P₄, S₈); (2) a monatomic ion's ON equals its charge; (3) O is −2, except in peroxides (−1), superoxides (−½) and OF₂ (+2); (4) H is +1, except in metal hydrides (−1); (5) F is always −1; alkali metals +1 and alkaline earth metals +2 in compounds; (6) the sum of ONs is 0 in a neutral molecule and equals the charge in an ion. Other halogens are −1 unless bonded to O or a more electronegative halogen.

Worked example

Find the oxidation numbers of O in H₂O₂, KO₂ and OF₂, and of H in NaH.

Try it first, then show the solutionSolution

1. H₂O₂: 2(+1) + 2x = 0 ⇒ x = −1 (peroxide)
2. KO₂: +1 + 2x = 0 ⇒ x = −½ (superoxide); OF₂: x + 2(−1) = 0 ⇒ x = +2
3. NaH: +1 + x = 0 ⇒ x = −1 (hydride)

Answer: −1; −½; +2; −1

Oxidation numbers in molecules and ions such as K₂Cr₂O₇ and S₂O₃²⁻

Set the unknown ON as x and use the sum rule. K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 ⇒ x = +6. S₂O₃²⁻: 2x + 3(−2) = −2 ⇒ x = +2 (an average, because the two S atoms are in different environments). Fe₃O₄: 3x = +8 ⇒ x = +8/3, an average of Fe²⁺ and two Fe³⁺. S₄O₆²⁻: 4x − 12 = −2 ⇒ x = +2.5 (average of two +5 and two 0). When the formula hides a peroxide group (CrO₅, H₂SO₅), find the structure first, because peroxide O is −1.

Worked example

Find the oxidation number of Cr in K₂Cr₂O₇ and of S in S₂O₃²⁻, and of Fe in Fe₃O₄.

Try it first, then show the solutionSolution

1. K₂Cr₂O₇: 2 + 2x − 14 = 0 ⇒ x = +6
2. S₂O₃²⁻: 2x − 6 = −2 ⇒ x = +2 (average)
3. Fe₃O₄: 3x − 8 = 0 ⇒ x = +8/3 (average of +2, +3, +3)

Answer: +6; +2; +8/3

Oxidation and reduction as change in oxidation number

Oxidation = increase in ON; reduction = decrease in ON. In any redox reaction, find the elements whose ON changes. The element whose ON increases belongs to the reducing agent; the one whose ON decreases belongs to the oxidising agent. Reactions in which no ON changes (neutralisation, precipitation) are not redox.

Worked example

In $2KMn{O}_4+16HCl\to 2KCl+2MnC{l}_2+5C{l}_2+8H_{2}O$, identify the changes in oxidation number, the oxidising agent and the reducing agent.

Try it first, then show the solutionSolution

1. Mn: +7 (KMnO₄) → +2 (MnCl₂): decrease of 5, reduced
2. Cl: −1 (HCl) → 0 (Cl₂): increase, oxidised (only 10 of the 16 Cl atoms change)
3. KMnO₄ is the oxidising agent; HCl is the reducing agent

Answer: Mn +7 → +2; Cl −1 → 0; oxidant KMnO₄, reductant HCl

Types of redox reactions

Combination: two substances combine, at least one being an element ($C+O_2\to C{O}_2$). Decomposition: a compound breaks into components, at least one an element ($2KCl{O}_3\to 2KCl+3O_2$). Displacement: an atom or ion in a compound is replaced by another element: metal displacement ($Zn+CuS{O}_4\to ZnS{O}_4+Cu$), non-metal displacement ($C{l}_2+2KBr\to 2KCl+B{r}_2$). Disproportionation: one element in one oxidation state is simultaneously oxidised and reduced ($2H_2{O}_2\to 2H_{2}O+O_2$; $C{l}_2+2NaOH\to NaCl+NaOCl+H_{2}O$). An element can disproportionate only if it has at least one higher and one lower oxidation state available.

Worked example

Show that $3HN{O}_2\to HN{O}_3+2NO+H_{2}O$ is a disproportionation, and check the electron balance.

Try it first, then show the solutionSolution

1. N in HNO₂: +1 + x − 4 = 0 ⇒ +3
2. N in HNO₃: +5 (oxidised, loses 2 e⁻); N in NO: +2 (reduced, each gains 1 e⁻)
3. One N loses 2 e⁻ and two N gain 1 e⁻ each: the same element in one state goes both up and down

Answer: N +3 → +5 and +3 → +2: disproportionation; 2 electrons lost = 2 gained

Balancing by the oxidation-number method

Steps: (1) write the skeleton equation and assign ONs; (2) find the increase and decrease in ON per formula unit; (3) multiply so that the total increase equals the total decrease; (4) balance all atoms other than O and H; (5) in acidic solution balance O by adding H₂O and H by adding H⁺ (in basic solution, add OH⁻ and H₂O); (6) check that atoms and charges balance.

Worked example

Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution by the oxidation-number method.

Try it first, then show the solutionSolution

1. Mn: +7 → +2 (decrease 5); Fe: +2 → +3 (increase 1); so 1 MnO₄⁻ : 5 Fe²⁺
2. MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺; balance O with 4H₂O on the right and H with 8H⁺ on the left
3. Charge check: −1 + 10 + 8 = 17 on the left; 2 + 15 = 17 on the right ✓

Answer: $Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$

Balancing by the ion-electron method (acidic medium)

Steps: (1) split into oxidation and reduction half-reactions; (2) balance atoms other than O and H; (3) balance O with H₂O; (4) balance H with H⁺; (5) balance charge with electrons; (6) multiply the half-reactions so that electrons lost = electrons gained, add, and cancel common species; (7) check atoms and charge.

Worked example

Balance Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ in acidic solution.

Try it first, then show the solutionSolution

1. Reduction: $C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$
2. Oxidation: $2I^{-}\to I_2+2e^{-}$; multiply by 3
3. Add: charge check −2 + 14 − 6 = +6 = 2 × (+3) ✓

Answer: $C{r}_2{O}_7^{2-}+14H^{+}+6I^{-}\to 2C{r}^{3+}+3I_2+7H_{2}O$

Balancing in basic medium

Balance as for acidic solution, then add OH⁻ to both sides equal in number to the H⁺ present; combine H⁺ + OH⁻ into H₂O and cancel water appearing on both sides. Alternatively, balance O and H directly using OH⁻ and H₂O. The final equation must contain OH⁻ (not H⁺).

Worked example

Balance MnO₄⁻ + I⁻ → MnO₂ + IO₃⁻ in basic solution.

Try it first, then show the solutionSolution

1. Reduction (acid form): MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O; oxidation: I⁻ + 3H₂O → IO₃⁻ + 6H⁺ + 6e⁻
2. Double the reduction and add: 2MnO₄⁻ + 8H⁺ + I⁻ + 3H₂O → 2MnO₂ + 4H₂O + IO₃⁻ + 6H⁺, i.e. 2MnO₄⁻ + 2H⁺ + I⁻ → 2MnO₂ + IO₃⁻ + H₂O
3. Add 2OH⁻ to both sides: 2H⁺ + 2OH⁻ = 2H₂O; cancel one H₂O

Answer: $2Mn{O}_4^{-}+I^{-}+H_{2}O\to 2Mn{O}_2+I{O}_3^{-}+2O{H}^{-}$

Applications of redox reactions

Titrations: KMnO₄ (purple → colourless in acid, a self-indicator) and K₂Cr₂O₇ are used to estimate Fe²⁺, oxalates and H₂O₂; the n-factor (electrons per formula unit) gives the equivalent weight (KMnO₄ in acid 158/5 = 31.6; K₂Cr₂O₇ 294/6 = 49). Metallurgy: ores are reduced to metals (by carbon, CO or electrolysis). Cells and batteries: separating the half-reactions lets electrons flow through a wire. Biology: respiration oxidises glucose; photosynthesis reduces CO₂. Corrosion: rusting is oxidation of iron; galvanising protects it because zinc is oxidised first.

Worked example

Find the equivalent weights of K₂Cr₂O₇ (M = 294) and FeSO₄ (M = 152) when K₂Cr₂O₇ oxidises Fe²⁺ in acid.

Try it first, then show the solutionSolution

1. Cr: +6 → +3, two Cr: n = 6 ⇒ 294/6 = 49
2. Fe: +2 → +3: n = 1 ⇒ 152/1 = 152

Answer: 49; 152

Chemistry in real life: Redox Reactions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Balancing Redox Equations and Redox Titrations
• Oxidation Numbers and Types of Redox Reactions

Redox Reactions formula sheet

Electronic definitions

$\text{oxidation}=\text{loss of }{e}^{-};\text{reduction}=\text{gain of }{e}^{-}$

OIL RIG.

Sum rule

$\sum \text{ON}=0(\text{molecule})=\text{charge}(\text{ion})$

O −2, H +1, F −1 by default.

Permanganate in acid

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

n-factor 5.

Dichromate in acid

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

n-factor 6.

Electron balance

$\text{electrons lost}=\text{electrons gained}$

Basis of every balancing method.

Equivalent weight of a redox agent

$\text{EW}=\frac{M}{\text{electrons transferred per formula unit}}$

Depends on the reaction.

Redox Reactions: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Redox Reactions

The question patterns to expect, and the method that works for each.

Tricks to remember Redox Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

• OIL RIG

  Oxidation Is Loss, Reduction Is Gain (of electrons).

  Use it when: Electronic definitions.

• Up is oxidation

  ON goes up: oxidised; goes down: reduced.

  Use it when: Identifying changes.

• Agents are opposite

  The oxidising agent is reduced; the reducing agent is oxidised.

  Use it when: Naming agents.

• Water for O, H⁺ for H, e⁻ for charge

  Balance in this order in acidic solution.

  Use it when: Half-reactions.

• OH⁻ follows H⁺

  In basic solution, add as many OH⁻ as there are H⁺ to both sides.

  Use it when: Basic medium.

Redox Reactions: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Redox Reactions

The slips examiners see most often, and what to write instead.

• ✗Mistake: Naming the agent backwards

  Students call the substance oxidised the oxidising agent.

  Wrong working

  In Zn + Cu²⁺, Zn is the oxidising agent because it is oxidised.

  Right working

  Zn is oxidised, so it is the reducing agent.

  ✓Do this: The agent does to others what is not done to itself.

  Usually costs: 1 mark.

• ✗Mistake: Taking O as −2 in peroxides

  The default rule is applied blindly.

  Wrong working

  ON of Cr in CrO₅ = +10.

  Right working

  CrO₅ has two peroxide groups: x − 2 − 4(1) = 0 ⇒ +6.

  ✓Do this: Check whether the formula contains O–O bonds.

  Usually costs: 1 mark.

• ✗Mistake: Unbalanced charge

  Students balance atoms only.

  Wrong working

  MnO₄⁻ + Fe²⁺ + 8H⁺ → Mn²⁺ + Fe³⁺ + 4H₂O.

  Right working

  Five Fe²⁺ are needed: charge 17 on both sides.

  ✓Do this: Always add up charges on both sides at the end.

  Usually costs: 1-2 marks.

• ✗Mistake: Using H⁺ in a basic medium answer

  The acidic answer is not converted.

  Wrong working

  2MnO₄⁻ + 2H⁺ + I⁻ → 2MnO₂ + IO₃⁻ + H₂O (in base).

  Right working

  Add 2OH⁻ to both sides: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻.

  ✓Do this: No H⁺ may remain in a basic-medium equation.

  Usually costs: 1 mark.

• ✗Mistake: Counting all chlorine atoms as oxidised

  Some HCl only supplies chloride to the products.

  Wrong working

  16 Cl atoms are oxidised in the KMnO₄–HCl reaction.

  Right working

  Only 10 Cl (forming 5Cl₂) change; 6 stay as Cl⁻ in KCl and MnCl₂.

  ✓Do this: Compare each product's ON separately.

  Usually costs: 1 mark.

Practice questions on Redox Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The oxidation number of oxygen in OF₂ is:
   (a) −2
   (b) −1
   (c) +1
   (d) +2

   Hint

   F is always −1.

   Answer

   (d)

2. Q2 · 1 mark · MCQExam level

   In Zn + CuSO₄ → ZnSO₄ + Cu, the reducing agent is:
   (a) Zn
   (b) Cu²⁺
   (c) SO₄²⁻
   (d) Cu

   Hint

   Which is oxidised?

   Answer

   (a)

3. Q3 · 1 mark · Very short answerExam level

   Find the oxidation number of Fe in Fe₃O₄.

   Hint

   3x = 8.

   Answer

   +8/3

4. Q4 · 1 mark · Very short answerExam level

   What is the oxidation number of H in NaH?

   Hint

   Metal hydride.

   Answer

   −1

5. Q5 · 2 marks · Short answerExam level

   Classify as combination, decomposition or displacement: (i) $2KCl{O}_3\to 2KCl+3O_2$ (ii) $C{l}_2+2KBr\to 2KCl+B{r}_2$.

   Hint

   What happens to the compound?

   Answer

   (i) Decomposition (ii) Non-metal displacement

6. Q6 · 2 marks · Short answerHarder exam level

   Find the oxidation number of Cr in CrO₅ (which contains two peroxide groups) and of S in H₂SO₅ (one peroxide group).

   Hint

   Peroxide O is −1.

   Answer

   CrO₅: one O at −2, four at −1: x − 2 − 4 = 0 ⇒ +6. H₂SO₅: 2 + x − 6 − 2 = 0 ⇒ +6.

7. Q7 · 2 marks · Short answerHarder exam level

   Write the half-reactions for $C{l}_2+2B{r}^{-}\to 2C{l}^{-}+B{r}_2$.

   Hint

   Who loses electrons?

   Answer

   Oxidation: 2Br⁻ → Br₂ + 2e⁻; reduction: Cl₂ + 2e⁻ → 2Cl⁻

8. Q8 · 3 marks · Short answerHarder exam level

   Balance MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O by the oxidation-number method.

   Hint

   Mn +4 → +2; Cl −1 → 0.

   Answer

   Decrease 2 for Mn; increase 1 per Cl, so 2Cl oxidised; plus 2Cl⁻ in MnCl₂: $Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$

9. Q9 · 3 marks · Short answerHarder exam level

   Balance H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O in acidic solution.

   Hint

   O in H₂O₂ goes −1 → −2.

   Answer

   $H_2{O}_2+2F{e}^{2+}+2H^{+}\to 2F{e}^{3+}+2H_{2}O$

10. Q10 · 3 marks · Short answerHarder exam level

    Balance Cr(OH)₃ + IO₃⁻ → CrO₄²⁻ + I⁻ in basic solution.

    Hint

    Cr +3 → +6; I +5 → −1.

    Answer

    $2Cr(OH{)}_3+I{O}_3^{-}+4O{H}^{-}\to 2Cr{O}_4^{2-}+I^{-}+5H_{2}O$ (charge −5 on both sides)

11. Q11 · 2 marks · Short answerHarder exam level

    Find the equivalent weights of oxalic acid crystals, H₂C₂O₄·2H₂O (M = 126), as a reducing agent towards acidified KMnO₄, and of H₂O₂ (M = 34) when it is oxidised to O₂.

    Hint

    Electrons lost per formula unit.

    Answer

    Oxalate: two C, each +3 → +4, n = 2: 126/2 = 63. H₂O₂: two O, each −1 → 0, n = 2: 34/2 = 17.

12. Q12 · 2 marks · Short answerHarder exam level

    Is $C{l}_2+H_{2}O\to HCl+HOCl$ a redox reaction? Explain with oxidation numbers.

    Hint

    Cl in each product.

    Answer

    Yes: Cl goes 0 → −1 (HCl) and 0 → +1 (HOCl), so it is a disproportionation.

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Find the oxidation number of N in NH₃, N₂H₄, NH₂OH, N₂, N₂O, NO, NO₂ and HNO₃. [2]
    (b) Balance Zn + NO₃⁻ → Zn²⁺ + NH₄⁺ in acidic solution. [3]

    Hint

    N goes from +5 to −3 in (b).

    Answer

    (a) −3, −2, −1, 0, +1, +2, +4, +5. (b) Zn loses 2e⁻, N gains 8e⁻: 4Zn : 1NO₃⁻; balance O with 3H₂O and H with 10H⁺: $4Zn+N{O}_3^{-}+10H^{+}\to 4Z{n}^{2+}+N{H}_4^{+}+3H_{2}O$ (charge 9 = 9).

14. Q14 · 3 marks · Short answerHOTS / competency

    Balance the disproportionation of white phosphorus in alkali: P₄ → PH₃ + H₂PO₂⁻ (basic medium).

    Hint

    P: 0 → −3 and 0 → +1.

    Answer

    One P gains 3e⁻ and three P lose 1e⁻ each: $P_4+3O{H}^{-}+3H_{2}O\to P{H}_3+3H_{2}P{O}_2^{-}$

15. Q15 · 3 marks · Short answerHOTS / competency

    What volume of 0.10 M KMnO₄ (acidified) is needed to oxidise 1.52 g of FeSO₄ (M = 152)?

    Hint

    1 MnO₄⁻ : 5 Fe²⁺.

    Answer

    n(Fe²⁺) = 0.010 mol; n(MnO₄⁻) = 0.0020 mol; V = 0.0020/0.10 = 0.020 L = 20 mL

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Oxidation in terms of electrons?Show answer

   Loss of electrons.

2. 2.ON of O in H₂O₂?Show answer

   −1.

3. 3.ON of S in S₂O₃²⁻?Show answer

   +2 (average).

4. 4.What is disproportionation?Show answer

   One element in one state is both oxidised and reduced.

5. 5.n-factor of KMnO₄ in acid?Show answer

   5.

6. 6.n-factor of K₂Cr₂O₇ in acid?Show answer

   6.

7. 7.Order for balancing a half-reaction in acid?Show answer

   Other atoms, O with H₂O, H with H⁺, charge with e⁻.

8. 8.How to convert an acidic-medium answer to basic?Show answer

   Add OH⁻ equal to the H⁺ on both sides and simplify.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you define oxidation in three ways?
• ☐Can you find ONs including fractional and peroxide cases?
• ☐Can you name the oxidising and reducing agents in any reaction?
• ☐Can you show a reaction is a disproportionation?
• ☐Can you balance an equation by the oxidation-number method?
• ☐Can you balance by half-reactions in acidic solution?
• ☐Can you convert to basic medium correctly?
• ☐Can you use n-factors in a KMnO₄ or K₂Cr₂O₇ titration?

Redox Reactions: frequently asked questions