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

The unusual oxidation number of oxygen in hydrogen peroxide, −1 rather than the usual −2, opens this set of redox questions. Later ones cover where oxidation occurs in a galvanic cell and the two standard methods for balancing redox equations.

Last Updated: September 23, 2026

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.

Frequently Asked Questions

In the reaction Zn + Cu2+ to Zn2+ + Cu, how would you identify the oxidizing and reducing agents?
Zinc loses electrons and is oxidized, acting as the reducing agent, while copper ion gains electrons and is reduced, acting as the oxidizing agent.

How would you assign the oxidation number of chromium in K2Cr2O7?
Since potassium is +1 and oxygen is -2, balancing the compound gives chromium an oxidation number of +6.

Written by Satish

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