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Redox Reactions: questions with answers

Class 11 Chemistry, Chapter 7 (NCERT). 15 questions by concept. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Ion–electron balancing in acidic and basic media

Question 1

Which net ionic equation results when Fe²⁺ is oxidised by MnO₄⁻ in acid?

  1. (a)5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
  2. (b)Correct-looking coefficients remove the need for charge checking
  3. (c)Cancel unequal electron counts directly
  4. (d)Convert to basic medium by adding OH− to one side
Show answer

(a) 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Five Fe²⁺ ions release the five electrons consumed by one permanganate reduction half-reaction.

Question 2

A proposed final net ionic equation still contains 6e⁻ on the reactant side. What is wrong?

  1. (a)Correct-looking coefficients remove the need for charge checking
  2. (b)The half-reaction electron counts were not cancelled completely; electrons cannot remain in the overall chemical equation.
  3. (c)Cancel unequal electron counts directly
  4. (d)Convert to basic medium by adding OH− to one side
Show answer

(b) The half-reaction electron counts were not cancelled completely; electrons cannot remain in the overall chemical equation.

Electrons are bookkeeping intermediates exchanged between the two half-reactions.

Average, fractional, and atom-specific oxidation states

Question 3

In S₄O₆²⁻, the average sulfur oxidation number is +2.5. What atom-specific assignment is supported by the source structure?

  1. (a)A molecular formula fixes every atom-specific oxidation state
  2. (b)A fractional average means a fraction of an electron was transferred
  3. (c)All atoms of one element in a formula have the same oxidation state
  4. (d)The two terminal sulfur atoms are +5 and the two central sulfur atoms are 0.
Show answer

(d) The two terminal sulfur atoms are +5 and the two central sulfur atoms are 0.

(+5 + 0 + 0 + 5)/4 = +2.5, so the fraction is an average over inequivalent sites.

Question 4

Why can the formula C₃O₂ give an average carbon oxidation number but not by itself identify every carbon site's value?

  1. (a)A molecular formula fixes every atom-specific oxidation state
  2. (b)A fractional average means a fraction of an electron was transferred
  3. (c)Charge balance fixes the total carbon contribution, but we need to know how the atoms are joined to share it among carbon atoms that differ.
  4. (d)All atoms of one element in a formula have the same oxidation state
Show answer

(c) Charge balance fixes the total carbon contribution, but we need to know how the atoms are joined to share it among carbon atoms that differ.

Atom-specific oxidation states depend on which atoms are bonded to more electronegative partners.

Classical, electron-transfer, and oxidation-number views of redox

Question 5

In Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), which agent assignment is correct?

  1. (a)Cu²⁺ is the oxidising agent and Zn is the reducing agent.
  2. (b)Any reactant containing oxygen is a redox agent
  3. (c)Oxidation always requires oxygen
  4. (d)The oxidised species is the oxidising agent
Show answer

(a) Cu²⁺ is the oxidising agent and Zn is the reducing agent.

Cu²⁺ accepts two electrons and is reduced; Zn donates them and is oxidised.

Question 6

In the net ionic reaction Zn + Cu²⁺ → Zn²⁺ + Cu, nitrate ions from the salts do not appear. Why are they not redox agents?

  1. (a)The oxidised species is the oxidising agent
  2. (b)Oxidation always requires oxygen
  3. (c)Their composition and oxidation states do not change. They are spectator ions.
  4. (d)Any reactant containing oxygen is a redox agent
Show answer

(c) Their composition and oxidation states do not change. They are spectator ions.

A species is not a redox agent merely because it is present in a reagent formula or contains oxygen.

Oxidants, reductants, and oxidation-state limits

Question 7

Why does thiosulfate form tetrathionate with I₂ but is oxidised to sulfate by Br₂ under the stated conditions?

  1. (a)An intermediate oxidation state has only one redox role
  2. (b)A reagent has the same observed role with every partner
  3. (c)Br₂ is the stronger oxidising agent and drives sulfur to a higher oxidation state than I₂ does.
  4. (d)Reagent excess cannot change a redox product
Show answer

(c) Br₂ is the stronger oxidising agent and drives sulfur to a higher oxidation state than I₂ does.

Different oxidant strengths can give different oxidation products from the same reductant.

Question 8

Why can oxidant excess yield a higher oxidation-state product than reductant excess?

  1. (a)Reagent excess cannot change a redox product
  2. (b)After the first electron-transfer step, remaining oxidant can continue oxidising an intermediate, whereas excess reductant favours stopping at a lower state.
  3. (c)An intermediate oxidation state has only one redox role
  4. (d)A reagent has the same observed role with every partner
Show answer

(b) After the first electron-transfer step, remaining oxidant can continue oxidising an intermediate, whereas excess reductant favours stopping at a lower state.

Stoichiometric conditions can determine how far a multistep redox sequence proceeds.

Oxidation-number rules, exceptions, and Stock notation

Question 9

What is the oxidation number of oxygen in H₂O₂?

  1. (a)Hydrogen is always +1
  2. (b)−1
  3. (c)Oxidation numbers always sum to zero
  4. (d)Oxygen is always −2
Show answer

(b) −1

Hydrogen peroxide contains an O–O peroxide linkage, so each oxygen is −1 rather than the usual −2.

Question 10

What is the oxidation number of hydrogen in NaH?

  1. (a)Oxidation numbers always sum to zero
  2. (b)Hydrogen is always +1
  3. (c)−1
  4. (d)Oxygen is always −2
Show answer

(c) −1

NaH is a binary metal hydride; Na is +1 and hydrogen is −1.

Balancing redox equations by oxidation-number change

Question 11

Which is the balanced acidic equation for dichromate oxidising sulfite to sulfate?

  1. (a)Use H+ as a final balancing species in basic medium
  2. (b)Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
  3. (c)Atom balance alone proves a redox equation is balanced
  4. (d)Oxidation and reduction electron changes need not match
Show answer

(b) Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

The equation conserves Cr, S, O, H and has net charge zero on both sides.

Combination, decomposition, displacement, and disproportionation

Question 12

Why is CaCO₃ → CaO + CO₂ not a redox reaction although it is a decomposition?

  1. (a)Every combination reaction is redox
  2. (b)Either metal can displace the other
  3. (c)Every decomposition reaction is redox
  4. (d)Ca remains +2, C remains +4, and O remains −2, so no oxidation numbers change.
Show answer

(d) Ca remains +2, C remains +4, and O remains −2, so no oxidation numbers change.

Decomposition alone does not prove electron transfer.

Redox titrations, indicators, endpoints, and stoichiometry

Question 13

What does the first persistent faint pink colour mean in a permanganate titration of Fe²⁺?

  1. (a)All Fe²⁺ has been consumed and a slight excess of MnO₄⁻ now persists, marking the endpoint just beyond equivalence.
  2. (b)Endpoint and equivalence point are identical by definition
  3. (c)Dichromate gives its own sharp endpoint
  4. (d)Permanganate always needs an external indicator
Show answer

(a) All Fe²⁺ has been consumed and a slight excess of MnO₄⁻ now persists, marking the endpoint just beyond equivalence.

Permanganate is its own indicator; the observed endpoint is close to but not identical by definition with the stoichiometric equivalence point.

Redox couples, the Daniell cell, and charge transport

Question 14

What happens at the zinc electrode of an operating Daniell cell?

  1. (a)Electrons flow from copper to zinc
  2. (b)Zn(s) → Zn²⁺(aq) + 2e⁻; it is the negative anode.
  3. (c)Conventional current follows electron flow
  4. (d)The anode is always positive
Show answer

(b) Zn(s) → Zn²⁺(aq) + 2e⁻; it is the negative anode.

Zinc oxidation supplies electrons to the external circuit.

Standard reduction potentials, redox strength, and feasibility

Question 15

Using E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V, what is E° for Cu + 2Ag⁺ → Cu²⁺ + 2Ag?

  1. (a)Scaling a half-reaction scales its E°
  2. (b)Positive E° guarantees reaction under every condition and rate
  3. (c)Cell potential is always the sum of two listed reduction potentials
  4. (d)+0.46 V
Show answer

(d) +0.46 V

Silver ion is reduced at the cathode and copper is oxidised at the anode, so 0.80 − 0.34 = +0.46 V.

Common misconception in Redox Reactions

Open full misconception guide →

Can something be oxidised without oxygen?

Wrong

Sodium cannot be oxidised in a reaction with chlorine because there is no oxygen.

Why it feels right. I recognise the oxygen in the word oxidation.

Right

Oxidation involves loss of electrons or an increase in oxidation number.

Worked example

Sodium
Na · oxidation number 0→Loses one electron per atom→Na⁺ · oxidation number +1
The sodium oxidation number rises even though the reaction uses chlorine and no oxygen.
  1. In 2Na + Cl₂ → 2NaCl, each sodium atom loses an electron and its oxidation number rises from 0 to +1.
  2. This is oxidation in an oxygen-free reaction.