Class 11 Chemistry Chapter 7 Redox Reactions – Extra Questions with Answers

Extra practice questions for Class 11 Chemistry Chapter 7 (Redox Reactions), beyond the textbook. These Class 11 Chemistry Chapter 7 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. What is the oxidation number of oxygen in H₂O₂ (hydrogen peroxide)?
Ans: −1.

Q2. What happens to the oxidation number of an element during oxidation?
Ans: It increases.

Q3. Where does oxidation occur in a galvanic cell?
Ans: At the anode.

Q4. What is the oxidation number of an element in its free (uncombined) state?
Ans: Zero.

Q5. Name the two common methods of balancing redox equations.
Ans: Oxidation number method and ion-electron (half-reaction) method.

Short Answer Questions (2–3 marks)

Q6. Find the oxidation number of Mn in KMnO₄.
Ans: K=+1, O=−2(×4=−8). Sum=0: 1+Mn−8=0, so Mn=+7.

Q7. Find the oxidation number of Cr in K₂Cr₂O₄.
Ans: K=+1(×2=+2), O=−2(×7=−14). Sum=0: 2+2Cr−14=0, so 2Cr=12, Cr=+6.

Q8. In the reaction Zn+CuSO₄→ZnSO₄+Cu, identify the substance oxidized and the substance reduced.
Ans: Zn is oxidized (0→+2); Cu²⁺ is reduced (+2→0).

Higher-Order Thinking / Application Questions

Q9. In the reaction Cl₂+2NaOH→NaCl+NaOCl+H₂O, explain why this is classified as a disproportionation reaction, showing the oxidation number changes of chlorine.
Ans: In Cl₂, chlorine has an oxidation number of 0 (free element). In the products, chlorine appears in two different oxidation states: in NaCl, chlorine is −1 (reduced, since it decreased from 0), and in NaOCl (sodium hypochlorite), chlorine is +1 (oxidized, since it increased from 0). Because the same starting element (chlorine, oxidation number 0) is simultaneously oxidized (to +1 in NaOCl) and reduced (to −1 in NaCl) within the same reaction, this is classified as a disproportionation reaction — a special type of redox reaction where a single species acts as both the oxidizing and reducing agent for itself.

Q10. Explain, using oxidation numbers, why the reaction 2KMnO₄+10FeSO₄+8H₂SO₄→2MnSO₄+5Fe₂(SO₄)₃+K₂SO₄+8H₂O requires 5 moles of FeSO₄ for every 1 mole of KMnO₄, connecting this to the number of electrons transferred.
Ans: In this reaction, manganese in KMnO₄ is reduced from +7 (in MnO₄⁻) to +2 (in Mn²⁺), a gain of 5 electrons per Mn atom. Meanwhile, iron in FeSO₄ is oxidized from +2 (Fe²⁺) to +3 (Fe³⁺), a loss of only 1 electron per Fe atom. For the overall electron transfer to balance (total electrons lost by the reducing agent must equal total electrons gained by the oxidizing agent), 1 Mn atom gaining 5 electrons must be balanced by 5 Fe atoms each losing 1 electron, giving the 1:5 mole ratio of KMnO₄ to FeSO₄ seen in the balanced equation — this illustrates how the ion-electron/oxidation number method fundamentally relies on equalizing total electrons lost and gained to correctly balance redox reactions.

Written by Satish

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