Question Problem 7.1
For H₂S+Cl₂→2HCl+S; Fe₃O₄+Al→Fe+Al₂O₃; and Na+H₂→NaH, identify which substances are oxidised and reduced.
Steps only
Oxidation number rises = oxidation; falls = reduction.
Answer
- (i) H₂S / Cl₂;
- (ii) Al / Fe₃O₄;
- (iii) Na / H₂.
CLASS 11 · CHEMISTRY · CHAPTER 7
Redox Reactions
Simple, point-wise answers to the chapter’s ten in-text problems and all 30 end exercises on oxidation, reduction and redox reactions.
40 worked answersComplete chapter answers
40 worked answers · short steps first, explanations when you need them
In-text Problems
Textbook pages 236–251
Question Problem 7.2
Show that 2Na+H₂→2NaH is a redox reaction.
Steps only
2Na→2Na⁺+2e⁻; H₂+2e⁻→2H⁻.
Answer
Na is oxidised; H₂ is reduced.
Question Problem 7.3
Write the Stock names of HAuCl₄, Tl₂O, FeO, Fe₂O₃, CuI, CuO, MnO and MnO₂.
Steps only
Balance the charges to determine each metal oxidation number.
Answer
Hydrogen tetrachloroaurate(III), thallium(I) oxide, iron(II) oxide, iron(III) oxide, copper(I) iodide, copper(II) oxide, manganese(II) oxide, manganese(IV) oxide.
Question Problem 7.4
For 2Cu₂O+Cu₂S→6Cu+SO₂, identify the oxidised and reduced species and the oxidising and reducing agents.
Steps only
S: −2→+4; Cu: +1→0.
Answer
Cu₂S sulfur is oxidised/reductant; Cu(I) from Cu₂O is reduced/oxidant.
Question Problem 7.5
Which of ClO⁻, ClO₂⁻, ClO₃⁻ and ClO₄⁻ cannot disproportionate? Write balanced disproportionation reactions for the others.
Steps only
ClO₄⁻ has chlorine at its maximum oxidation number, +7.
Answer
ClO₄⁻ cannot disproportionate; the other three can.
Question Problem 7.6
Classify: (a) N₂+O₂→2NO; (b) Pb(NO₃)₂ decomposition to PbO, NO₂ and O₂; (c) NaH+H₂O→NaOH+H₂; (d) 2NO₂+2OH⁻→NO₂⁻+NO₃⁻+H₂O.
Steps only
Identify whether the reaction combines, breaks apart, displaces, or disproportionates.
Answer
(a) Combination; (b) decomposition; (c) displacement; (d) disproportionation.
Question Problem 7.7
Why does Pb₃O₄ react with HCl to form PbCl₂ and Cl₂, but with HNO₃ to form Pb(NO₃)₂ and PbO₂?
Steps only
Cl⁻ reduces Pb(IV) and becomes Cl₂; nitrate does not, so PbO₂ remains.
Answer
The difference is the reducing ability of chloride versus nitrate.
Question Problem 7.8
Write the net ionic equation for acidified dichromate reacting with sulfite to form Cr³⁺ and sulfate.
Steps only
Cr₂O₇²⁻+3SO₃²⁻+8H⁺→2Cr³⁺+3SO₄²⁻+4H₂O.
Answer
Cr₂O₇²⁻+3SO₃²⁻+8H⁺→2Cr³⁺+3SO₄²⁻+4H₂O.
Question Problem 7.9
In basic solution, permanganate oxidises bromide to bromate and forms MnO₂. Write the balanced ionic equation.
Steps only
2MnO₄⁻+Br⁻+H₂O→2MnO₂+BrO₃⁻+2OH⁻.
Answer
2MnO₄⁻+Br⁻+H₂O→2MnO₂+BrO₃⁻+2OH⁻.
Question Problem 7.10
In basic solution, permanganate oxidises iodide to iodine and forms MnO₂. Write the balanced ionic equation.
Steps only
2MnO₄⁻+6I⁻+4H₂O→2MnO₂+3I₂+8OH⁻.
Answer
2MnO₄⁻+6I⁻+4H₂O→2MnO₂+3I₂+8OH⁻.
Chapter 7 Exercises
Textbook pages 252–255
Question 7.1
Find the oxidation numbers of the marked central or variable element in NaH₂PO₄, NaHSO₄, H₄P₂O₇, K₂MnO₄, CaO₂, NaBH₄, H₂S₂O₇ and KAl(SO₄)₂·12H₂O.
Steps only
Apply the oxidation-number sum rule; use O=−1 in peroxide and H=−1 in NaBH₄.
Answer
+5, +6, +5, +6, −1, +3, +6, +6, respectively.
Question 7.2
Find and explain the oxidation numbers in KI₃, H₂S₄O₆, Fe₃O₄, ethanol and acetic acid.
Steps only
Use the charge-sum rule and report individual values where atoms have different bonding.
Answer
KI₃: average I −1/3; H₂S₄O₆: average S +2.5; Fe₃O₄: Fe²⁺/Fe³⁺; ethanol C: −3, −1; acetic-acid C: −3, +3.
Question 7.3
Show that each reaction is redox: (a) CuO+H₂→Cu+H₂O; (b) Fe₂O₃+3CO→2Fe+3CO₂; (c) 4BCl₃+3LiAlH₄→2B₂H₆+3LiCl+3AlCl₃; (d) 2K+F₂→2KF; (e) 4NH₃+5O₂→4NO+6H₂O.
Steps only
Track oxidation numbers on both sides; increases and decreases identify redox.
Answer
All five are redox. In (c), the exercise assigns B +3→−3 and H −1→+1.
Question 7.4
Fluorine reacts with ice: H₂O(s)+F₂(g)→HF(g)+HOF(g). Explain why this is a redox reaction.
Steps only
O: −2→0; F: 0→−1.
Answer
Oxygen is oxidised and fluorine is reduced, so it is redox.
Question 7.5
Find the oxidation numbers of S in H₂SO₅, Cr in Cr₂O₇²⁻ and N in NO₃⁻. Suggest their structures and check the apparent oxidation-number issue in H₂SO₅.
Steps only
H₂SO₅: H–O–S(=O)₂–O–O–H (peroxide O atoms are −1); Cr₂O₇²⁻: [O₃Cr–O–CrO₃]²⁻; NO₃⁻: trigonal planar with three equivalent resonance bonds.
Answer
- H₂SO₅: H–O–S(=O)₂–O–O–H. The O–O pair is a peroxide bond; those two O atoms are −1, the other three are −2, so S is +6.
- Cr₂O₇²⁻: [O₃Cr–O–CrO₃]²⁻, with one oxygen bridging the two chromium groups; each Cr is +6.
- NO₃⁻: trigonal planar, with three equivalent resonance N–O bonds; N is +5. The peroxide bond explains why assigning −2 to every O in H₂SO₅ would be incorrect.
Question 7.6
Write formulas for mercury(II) chloride, nickel(II) sulfate, tin(IV) oxide, thallium(I) sulfate, iron(III) sulfate and chromium(III) oxide.
Steps only
Balance the positive and negative ionic charges in each formula.
Answer
HgCl₂, NiSO₄, SnO₂, Tl₂SO₄, Fe₂(SO₄)₃, Cr₂O₃.
Question 7.7
Give examples in which carbon shows oxidation states from −4 to +4 and nitrogen from −3 to +5.
Steps only
Use the total oxidation-number sum to check each example.
Answer
Examples range from CH₄ (C −4) to CO₂ (C +4), and NH₃ (N −3) to HNO₃ (N +5).
Question 7.8
Why can SO₂ and H₂O₂ act as oxidising and reducing agents, while ozone and nitric acid act only as oxidants?
Steps only
Intermediate oxidation states may move up or down; maximum states can only move down.
Answer
SO₂ and H₂O₂ can act both ways; O₃ and HNO₃ act as oxidants.
Question 7.9
Why are expanded forms of the photosynthesis and ozone–hydrogen peroxide equations useful, and how could the reaction pathway be investigated?
Steps only
Label an oxygen source with ¹⁸O and check where it appears using mass spectrometry.
Answer
Use oxygen-isotope labelling and mass spectrometry.
Question 7.10
Why would AgF₂, although unstable, act as a strong oxidising agent if formed?
Steps only
Ag²⁺ tends to gain an electron and become the more stable Ag⁺.
Answer
AgF₂ is a strong oxidant because Ag(II) is easily reduced to Ag(I).
Question 7.11
Explain with three examples why excess reducing agent gives a lower oxidation-state product, while excess oxidising agent gives a higher one.
Steps only
A greater supply of oxidant accepts more electrons and produces a higher oxidation state.
Answer
Examples include PCl₃/PCl₅ and Fe²⁺/Fe³⁺ formation as oxidant supply changes.
Question 7.12
(a) Explain the use of alcoholic KMnO₄ to oxidise toluene to benzoic acid and give a balanced equation. (b) Why does concentrated H₂SO₄ give HCl with chloride but Br₂ with bromide?
Steps only
Permanganate oxidises toluene’s side chain; bromide, unlike chloride, also reduces concentrated sulfuric acid.
Answer
Toluene is oxidised to benzoate/benzoic acid; Br⁻ is oxidised to Br₂, while Cl⁻ forms HCl.
Question 7.13
For each of the five listed reactions (silver bromide with hydroquinone; formaldehyde with Tollens’ or Cu²⁺ reagent; hydrazine with H₂O₂; and lead with PbO₂), identify the oxidised and reduced substances and the agents.
Steps only
The species whose oxidation number rises is oxidised; the species whose value falls is reduced.
Answer
(a) hydroquinone / Ag(I); (b,c) HCHO / Ag(I) or Cu(II); (d) N₂H₄ / H₂O₂; (e) Pb / PbO₂.
Question 7.14
Why does thiosulfate form tetrathionate with iodine but sulfate with bromine?
Steps only
Bromine is the stronger oxidant and carries sulfur to a higher oxidation state.
Answer
I₂ gives tetrathionate; stronger Br₂ gives sulfate.
Question 7.15
Show that fluorine is the strongest halogen oxidant and HI is the strongest hydrohalic-acid reductant.
Steps only
F₂ is most readily reduced; I⁻ is most readily oxidised.
Answer
F₂ is the strongest oxidant; HI is the strongest reductant.
Question 7.16
Explain why XeO₆⁴⁻ reacts with F⁻ in acid to form XeO₃ and F₂. What does this show about Na₄XeO₆?
Steps only
Xe: +8→+6; F: −1→0.
Answer
Na₄XeO₆ is an oxidising agent.
Question 7.17
Use the four reactions with H₃PO₂ and benzaldehyde to compare the oxidising strengths of Ag⁺ and Cu²⁺.
Steps only
Ag⁺ reacts with benzaldehyde but Cu²⁺ does not, so Ag⁺ is the stronger oxidant.
Answer
Ag⁺ is a stronger oxidant than Cu²⁺.
Question 7.18
Balance four ionic redox reactions: permanganate with iodide in base; permanganate with SO₂ in acid; H₂O₂ with Fe²⁺ in acid; and dichromate with SO₂ in acid.
Steps only
Balance atoms and charge with half-reactions; use H₂O/H⁺ in acid and OH⁻ in base.
Answer
(a) 2MnO₄⁻+6I⁻+4H₂O→2MnO₂+3I₂+8OH⁻; (b) 2MnO₄⁻+5SO₂+2H₂O→2Mn²⁺+5SO₄²⁻+4H⁺; (c) H₂O₂+2Fe²⁺+2H⁺→2Fe³⁺+2H₂O; (d) Cr₂O₇²⁻+3SO₂+2H⁺→2Cr³⁺+3SO₄²⁻+H₂O.
Question 7.19
Balance three redox reactions in basic solution and identify the oxidising and reducing agents: P₄ with OH⁻; N₂H₄ with ClO₃⁻; and Cl₂O₇ with H₂O₂.
Steps only
Balance in basic medium and identify the species whose oxidation number rises or falls.
Answer
(a) P₄ is both agents; (b) N₂H₄ reduces ClO₃⁻; (c) H₂O₂ reduces Cl₂O₇.
Question 7.20
What can be inferred from (CN)₂+2OH⁻→CN⁻+CNO⁻+H₂O?
Steps only
One reactant is both oxidised and reduced.
Answer
It is a disproportionation reaction.
Question 7.21
Write the balanced ionic equation for Mn³⁺ forming Mn²⁺, MnO₂ and H⁺ in water.
Steps only
Balance atoms and charge: 2Mn³⁺+2H₂O→Mn²⁺+MnO₂+4H⁺.
Answer
2Mn³⁺+2H₂O→Mn²⁺+MnO₂+4H⁺.
Question 7.22
For Cs, Ne, I and F, identify which (a) shows only negative oxidation state, (b) only positive, (c) both positive and negative, and (d) neither.
Steps only
F is negative; Cs positive; iodine can be positive or negative; neon is generally zero.
Answer
(a) F; (b) Cs; (c) I; (d) Ne.
Question 7.23
Write a balanced reaction for removing excess chlorine from drinking water with sulfur dioxide.
Steps only
Cl₂+SO₂+2H₂O→2Cl⁻+SO₄²⁻+4H⁺.
Answer
Cl₂+SO₂+2H₂O→2Cl⁻+SO₄²⁻+4H⁺.
Question 7.24
Using the periodic table, name non-metals and three metals that can show disproportionation.
Steps only
Choose elements with multiple accessible oxidation states.
Answer
Examples: Cl, Br, I, S, P; metals Cu, Mn and Hg.
Question 7.25
In the first step of the Ostwald process, find the maximum NO mass from 10.00 g NH₃ and 20.00 g O₂.
Steps only
4NH₃+5O₂→4NO+6H₂O; 20.00/32.00×4/5×30.00=15.0 g NO.
Answer
15.0 g NO.
Question 7.26
Use standard electrode potentials to decide whether Fe³⁺/I⁻, Ag⁺/Cu, Fe³⁺/Cu, Ag/Fe³⁺ and Br₂/Fe²⁺ react spontaneously.
Steps only
Positive E°cell means feasible: (a), (b), (c), and (e) are feasible; (d) is not.
Answer
Feasible: (a), (b), (c), (e). Not feasible as written: (d).
Question 7.27
Predict the electrode products for aqueous AgNO₃ with silver electrodes, aqueous AgNO₃ with platinum, dilute H₂SO₄ with platinum, and aqueous CuCl₂ with platinum.
Steps only
Reduction at cathode, oxidation at anode; Ag is deposited, water gives H₂/O₂, and chloride can give Cl₂.
Answer
- (i) Ag/Ag⁺;
- (ii) Ag and O₂;
- (iii) H₂ and O₂;
- (iv) Cu and Cl₂.
Question 7.28
Arrange Al, Cu, Fe, Mg and Zn by their ability to displace one another from salt solutions.
Steps only
Compare the metals’ reducing strengths.
Answer
Mg > Al > Zn > Fe > Cu.
Question 7.29
Use the listed E° values for K, Ag, Hg, Mg and Cr to arrange the metals in increasing reducing power.
Steps only
More negative E° means stronger metal reducing power.
Answer
Ag < Hg < Cr < Mg < K.
Question 7.30
Draw or write the cell notation for Zn(s)+2Ag⁺(aq)→Zn²⁺(aq)+2Ag(s). Identify the negative electrode, charge carriers and electrode reactions.
Steps only
Zn is oxidised at the negative anode; Ag⁺ is reduced at the cathode.
Answer
Zn|Zn²⁺||Ag⁺|Ag; e⁻ flow Zn→Ag; Zn→Zn²⁺+2e⁻; 2Ag⁺+2e⁻→2Ag.