RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Chemistry · 10 questions · 24 marks
Rusting iron, a burning candle, the cell in your phone and the reaction that digests your food are all the same kind of change: electrons moving from one species to another. This chapter gives you a bookkeeping tool called the oxidation number, then two reliable methods for balancing even the messiest redox equation in acidic or basic medium.
The oxidation number of sulphur in the tetrathionate ion S₄O₆²⁻ is:
Answer
+2.5 is the oxidation number of sulphur in S₄O₆²⁻. Let the oxidation number of each sulphur be x. Oxygen is −2, and the total must equal the charge on the ion, which is −2. So 4x + 6(−2) = −2, giving 4x − 12 = −2, then 4x = +10 and x = +2.5. A fraction does not mean a fractional charge on any atom; it is an average, because in the real structure two of the sulphur atoms are joined to each other in an S−S bridge and are in a different environment from the other two. The oxidation number method is only a bookkeeping device, and averages are perfectly acceptable within it.
Which of the following changes represents a disproportionation reaction?
Answer
2H₂O₂(l) → 2H₂O(l) + O₂(g) is a disproportionation. In disproportionation one element in a single oxidation state is simultaneously oxidised and reduced. Oxygen in hydrogen peroxide is in the −1 state; in water it falls to −2 (reduction) and in O₂ it rises to 0 (oxidation), so the same element goes both ways from one starting state. The first option is a displacement reaction, where two different elements change state. The third is a straightforward combination in which magnesium is oxidised and oxygen reduced. The fourth is thermal decomposition with no change in any oxidation number at all, so it is not even a redox reaction.
In the reaction 3Br₂ + 6NaOH → 5NaBr + NaBrO₃ + 3H₂O, the element bromine:
Answer
Bromine is both oxidised and reduced. In Br₂ the oxidation number is 0. In NaBr the bromine is −1, which is a decrease of 1, so that portion has been reduced. In NaBrO₃ the bromine is +5, since +1 + x + 3(−2) = 0 gives x = +5, an increase of 5, so that portion has been oxidised. The electron balance also checks out: five bromine atoms each gain one electron, a total of 5, while one bromine atom loses 5 electrons. Sodium and oxygen keep their oxidation numbers throughout, so bromine alone carries the redox change and this is a disproportionation in alkaline medium.
The number of moles of electrons required to reduce one mole of dichromate ion to Cr³⁺ in acidic medium is:
Answer
6 moles of electrons are required. In Cr₂O₇²⁻ each chromium is +6, since 2x + 7(−2) = −2 gives 2x = +12 and x = +6. The product Cr³⁺ has chromium at +3. Each chromium atom therefore gains 3 electrons, and the ion contains two chromium atoms, so the total is 2 × 3 = 6 electrons per dichromate ion. The balanced half-reaction confirms this: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, where the charges are −2 + 14 − 6 = +6 on the left and 2 × (+3) = +6 on the right. This n-factor of 6 is the key number in every dichromate titration calculation.
Assertion (A): Fluorine is the strongest oxidising agent among the halogens. Reason (R): Fluorine has the highest electron gain enthalpy in magnitude of all the halogens.
Answer
A is true but R is false. Fluorine really is the strongest oxidising agent of the halogens; its standard electrode potential of about +2.87 V is the highest of any common aqueous oxidant, and it can oxidise water itself. The reason, however, states a common misconception. Chlorine, not fluorine, has the more negative electron gain enthalpy. The fluorine atom is so small that the incoming electron enters a compact 2p shell already crowded with electrons, and the resulting repulsion reduces the energy released. Fluorine's supremacy as an oxidant comes instead from its very low bond dissociation enthalpy (the weak F−F bond breaks easily) combined with the very high hydration enthalpy of the small F⁻ ion, which together outweigh the electron gain enthalpy term in the overall energy cycle.
Determine the oxidation number of the underlined element in each: (i) nitrogen in NH₄NO₃, (ii) chlorine in HClO₄, (iii) manganese in K₂MnO₄, (iv) carbon in CH₃OH.
Answer
(i) NH₄NO₃ is a salt of two different ions, so the two nitrogens must be treated separately. In NH₄⁺: x + 4(+1) = +1, so x = −3. In NO₃⁻: x + 3(−2) = −1, so x = +5. Quoting a single average of +1 would hide the real chemistry. (ii) HClO₄: (+1) + x + 4(−2) = 0, so x − 7 = 0 and x = +7, the maximum available to chlorine. (iii) K₂MnO₄: 2(+1) + x + 4(−2) = 0, so x − 6 = 0 and x = +6. This is the manganate ion, not permanganate, in which manganese would be +7. (iv) CH₃OH: x + 3(+1) + (−2) + (+1) = 0, so x + 2 = 0 and x = −2.
Balance the following equation in basic medium by the ion–electron method: MnO₄⁻(aq) + I⁻(aq) → MnO₂(s) + I₂(s)
Answer
Step 1 — write the half-reactions. Oxidation: 2I⁻ → I₂ + 2e⁻ (iodine goes from −1 to 0, losing one electron per atom). Reduction: MnO₄⁻ → MnO₂ (manganese falls from +7 to +4, a gain of 3 electrons). Step 2 — balance the reduction half in acidic form first. Oxygen: MnO₄⁻ → MnO₂ + 2H₂O. Hydrogen: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O. Charge: left is −1 + 4 = +3, right is 0, so add 3 electrons on the left: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O. Step 3 — convert to basic medium by adding 4OH⁻ to both sides. The 4H⁺ and 4OH⁻ on the left combine to 4H₂O: MnO₄⁻ + 4H₂O + 3e⁻ → MnO₂ + 2H₂O + 4OH⁻, which simplifies to MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Step 4 — equalise electrons. Multiply the oxidation half by 3 and the reduction half by 2, giving 6 electrons each: 6I⁻ → 3I₂ + 6e⁻ and 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻. Step 5 — add and cancel the electrons: 2MnO₄⁻(aq) + 6I⁻(aq) + 4H₂O(l) → 2MnO₂(s) + 3I₂(s) + 8OH⁻(aq). Check: charge is −2 − 6 = −8 on the left and −8 on the right, and every atom balances.
(i) Explain the difference between an oxidising agent and a reducing agent in terms of both electron transfer and oxidation number. (ii) Balance the following in acidic medium by the ion–electron method: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺. (iii) Identify the oxidising and reducing agents in your balanced equation. (iv) In a titration, 25.0 mL of an iron(II) sulphate solution needed 20.0 mL of 0.020 M KMnO₄ for complete reaction in dilute sulphuric acid. Calculate the molarity of the iron(II) sulphate solution. (v) Explain why dilute sulphuric acid, and not hydrochloric acid, is used to acidify such a titration.
Answer
(i) An oxidising agent is the species that takes electrons from something else; in doing so it is itself reduced and the oxidation number of its key atom falls. A reducing agent supplies electrons; it is itself oxidised and the oxidation number of its key atom rises. The two labels always describe the same reaction from opposite sides, so no substance can act as an oxidising agent unless something else is simultaneously acting as a reducing agent. (ii) Oxidation half: Fe²⁺ → Fe³⁺ + e⁻ (iron rises from +2 to +3). Reduction half: manganese falls from +7 in MnO₄⁻ to +2 in Mn²⁺, a gain of 5 electrons. Balance oxygen with water: MnO₄⁻ → Mn²⁺ + 4H₂O. Balance hydrogen with H⁺: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O. Balance charge: left is −1 + 8 = +7, right is +2, so add 5 electrons on the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Multiply the oxidation half by 5 and add: MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l). Check the charge: −1 + 10 + 8 = +17 on the left, and +2 + 15 = +17 on the right. (iii) The permanganate ion MnO₄⁻ is the oxidising agent, since manganese is reduced from +7 to +2. Fe²⁺ is the reducing agent, since iron is oxidised from +2 to +3. (iv) Moles of KMnO₄ used = 0.020 mol/L × 0.0200 L = 4.0 × 10⁻⁴ mol. From the balanced equation, 1 mol of MnO₄⁻ reacts with 5 mol of Fe²⁺. Moles of Fe²⁺ = 5 × 4.0 × 10⁻⁴ = 2.0 × 10⁻³ mol. This was contained in 25.0 mL = 0.0250 L, so molarity = (2.0 × 10⁻³) ÷ 0.0250 = 0.080 mol/L. The iron(II) sulphate solution is therefore 0.080 M. (v) Hydrochloric acid cannot be used because chloride ion is itself oxidisable. Permanganate in acid is powerful enough to oxidise Cl⁻ to chlorine, so some of the titrant would be consumed by the acid rather than by the iron, and the burette reading would be falsely high. Nitric acid is equally unsuitable, since it is an oxidising agent in its own right and would oxidise Fe²⁺ before the titration even began. Dilute sulphuric acid supplies the H⁺ the half-reaction needs while the sulphate ion stays completely inert to both oxidation and reduction under these conditions.
(i) State the steps of the oxidation number method of balancing redox equations. (ii) Use that method to balance: Cr₂O₇²⁻ + SO₃²⁻ + H⁺ → Cr³⁺ + SO₄²⁻ + H₂O. (iii) Calculate the volume of 0.050 M potassium dichromate solution needed to react completely with 30.0 mL of 0.10 M sodium sulphite solution. (iv) Predict, with reasoning from the electrochemical series, whether copper metal will displace zinc from zinc sulphate solution, given E° values of Cu²⁺/Cu = +0.34 V and Zn²⁺/Zn = −0.76 V. (v) Give one everyday consequence of the answer to part (iv).
Answer
(i) The steps are: write the skeletal equation and assign oxidation numbers to every atom; identify which atoms change and calculate the total increase per formula unit for the oxidised species and the total decrease for the reduced species; multiply the two species by the smallest whole numbers that make the total increase equal the total decrease; balance all remaining atoms except hydrogen and oxygen; balance oxygen using H₂O and hydrogen using H⁺ in acidic medium; and finally verify that both the atom count and the net charge match on the two sides. (ii) Chromium falls from +6 to +3, a decrease of 3 per atom and 6 per dichromate ion. Sulphur rises from +4 in SO₃²⁻ to +6 in SO₄²⁻, an increase of 2 per ion. To equalise, take 1 dichromate (decrease 6) against 3 sulphite ions (increase 3 × 2 = 6). This gives Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻. Oxygen count: left has 7 + 9 = 16, right has 12, so add 4H₂O to the right. Hydrogen then needs 8H⁺ on the left. Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. Charge check: −2 − 6 + 8 = 0 on the left; +6 − 6 = 0 on the right. Balanced. (iii) Moles of SO₃²⁻ = 0.10 mol/L × 0.0300 L = 3.0 × 10⁻³ mol. The balanced equation needs 1 mol of Cr₂O₇²⁻ for every 3 mol of SO₃²⁻. Moles of Cr₂O₇²⁻ required = (3.0 × 10⁻³) ÷ 3 = 1.0 × 10⁻³ mol. Volume = moles ÷ molarity = (1.0 × 10⁻³) ÷ 0.050 = 0.020 L = 20.0 mL. (iv) No, copper will not displace zinc. A metal can displace another from its salt only if the metal is the better reducing agent, which means its reduction potential must be the more negative. Zinc at −0.76 V is far more negative than copper at +0.34 V, so zinc is the stronger reducing agent and the spontaneous direction is Zn + Cu²⁺ → Zn²⁺ + Cu. The cell potential for the reaction as proposed, Cu + Zn²⁺ → Cu²⁺ + Zn, is E°cell = E°cathode − E°anode = (−0.76) − (+0.34) = −1.10 V, and a negative cell potential means the change is non-spontaneous. (v) Because zinc is the more reactive metal, it corrodes in preference to less reactive metals it is connected to. This is used deliberately in galvanising, where a coating of zinc on iron sheets protects the iron even when the coating is scratched, since the zinc dissolves first and the exposed iron is left as the protected cathode. The same idea explains why sacrificial zinc blocks are bolted to ship hulls and underground pipelines.
A pharmacy quality-control chemist checks an iron tonic that is claimed to contain iron(II) sulphate. She dissolves one 0.850 g tablet in dilute sulphuric acid, makes the solution up to 100.0 mL, and titrates 20.0 mL portions of it against 0.0100 M KMnO₄, which is added from a burette until a faint pink colour persists. The average titre is 24.0 mL. Answer: (a) Write the balanced ionic equation and explain why no separate indicator is needed. (b) Calculate the moles of Fe²⁺ in the 20.0 mL portion and hence in the whole tablet. (c) Calculate the percentage by mass of iron in the tablet (molar mass of Fe = 56 g/mol). (d) The chemist finds a slightly lower result on a tablet left open on the bench for a week. Explain this observation and state one precaution that would prevent it.
Answer
(a) The balanced equation is MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l). No indicator is needed because potassium permanganate is self-indicating. The MnO₄⁻ ion is intensely purple while the Mn²⁺ product is almost colourless, so every drop added before the end point is decolourised instantly by the Fe²⁺ still in the flask. The moment the last of the Fe²⁺ has reacted, the very next drop has nothing to reduce it and its own colour survives, giving a faint permanent pink that marks the end point sharply. (b) Moles of MnO₄⁻ used = 0.0100 mol/L × 0.0240 L = 2.40 × 10⁻⁴ mol. The mole ratio is 1 MnO₄⁻ to 5 Fe²⁺, so moles of Fe²⁺ in the 20.0 mL portion = 5 × 2.40 × 10⁻⁴ = 1.20 × 10⁻³ mol. The portion is one-fifth of the 100.0 mL made up from a single tablet, so moles of Fe²⁺ in the whole tablet = 5 × 1.20 × 10⁻³ = 6.00 × 10⁻³ mol. (c) Mass of iron = 6.00 × 10⁻³ mol × 56 g/mol = 0.336 g. Percentage by mass = (0.336 ÷ 0.850) × 100 = 39.5%. (d) Iron(II) is a reasonably good reducing agent and is slowly oxidised to iron(III) by atmospheric oxygen, especially once the tablet is exposed to damp air: 4Fe²⁺ + O₂ + 4H⁺ → 4Fe³⁺ + 2H₂O. Iron(III) does not react with permanganate at all, so less titrant is needed and the calculated iron(II) content comes out low, even though the total iron present is unchanged. Storing the tablets in a sealed, dry container away from air, and dissolving each sample in freshly boiled and cooled acid immediately before titrating, prevents the loss; adding the sulphuric acid at once also helps, because oxidation of Fe²⁺ by air is much slower in strongly acidic solution.
RowQ generates fresh questions on Redox Reactions, marks your answers, and explains every step.
Start free