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Questions on variable oxidation states: vanadium, chromium and manganate(VII) titrations

The four oxidation states of vanadium and their colours, reduction by zinc in acid and the electrode potentials that govern it, chromium(VI), (III) and (II), the effect of pH and ligand on redox stability including chromium(III) to chromate(VI) with hydrogen peroxide in alkali, iron(II) hydroxide oxidised by air, and Tollens' reagent, chromium(III) with sodium hydroxide and ammonia, and manganate(VII) titrations of iron(II) and ethanedioate with full calculations.

20 original questions · 67 marks · the variable oxidation states: vanadium, chromium and manganate(VII) titrations notes · Transition metals · Flashcards for this unit

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• 6 marks · Calculate · VO-Q21

  Calculate the value of x in hydrated iron(II) sulfate, FeSO₄·xH₂O. A student dissolves 3.475 g of the crystals in dilute sulfuric acid and makes the solution up to 250.0 cm³ in a volumetric flask. 25.0 cm³ portions are titrated with 0.0180 mol dm⁻³ potassium manganate(VII), and the titres are given in the table. Ar: Fe 55.8, S 32.1, O 16.0, H 1.0.

  Titration	Rough	1	2	3
  Titre / cm³	14.30	14.15	13.85	13.95

  Mark scheme

  1. M1 the concordant titres are 13.85 and 13.95 cm³, which agree within 0.10 cm³; the rough titre and 14.15 cm³ are left out, giving a mean of 13.90 cm³
  2. M1 moles of MnO₄⁻ in the mean titre = 0.0180 × 13.90 ÷ 1000 = 2.502 × 10⁻⁴ mol, using the concentration given
  3. M1 moles of Fe²⁺ in each 25.0 cm³ portion = 5 × 2.502 × 10⁻⁴ = 1.251 × 10⁻³ mol, from the 1 : 5 ratio in the equation
  4. M1 moles of FeSO₄·xH₂O in the 250.0 cm³ flask, and so in the 3.475 g weighed out, = 1.251 × 10⁻² mol
  5. M1 Mr of the hydrate = 3.475 ÷ 1.251 × 10⁻² = 277.8, of which FeSO₄ contributes 55.8 + 32.1 + 64.0 = 151.9
  6. A1 water contributes 277.8 − 151.9 = 125.9, and 125.9 ÷ 18.0 = 6.99, so x = 7; allow error carried forward from an incorrect mean titre

• 5 marks · Calculate · VO-Q9

  Calculate the percentage purity of a sample of sodium ethanedioate, Na₂C₂O₄ (Mr = 134.0). 2.00 g of the impure solid is dissolved and made up to 250.0 cm³; 25.0 cm³ portions are acidified, warmed and titrated with 0.0200 mol dm⁻³ potassium manganate(VII), and the mean titre is 23.50 cm³.

  Mark scheme

  1. M1 moles of MnO₄⁻ = 0.0200 × 23.50 ÷ 1000 = 4.70 × 10⁻⁴ mol
  2. M1 moles of C₂O₄²⁻ in 25.0 cm³ = 4.70 × 10⁻⁴ × 5 ÷ 2 = 1.175 × 10⁻³ mol
  3. M1 moles in 250.0 cm³ = 1.175 × 10⁻² mol
  4. M1 mass of Na₂C₂O₄ = 1.175 × 10⁻² × 134.0 = 1.5745 g
  5. A1 percentage purity = 1.5745 ÷ 2.00 × 100 = 78.7%, to three significant figures

• 5 marks · Suggest · VO-Q10

  Suggest what goes wrong when a manganate(VII) titration of Fe²⁺ is acidified with hydrochloric acid, and when it is acidified with ethanoic acid, instead of with dilute sulfuric acid.

  Mark scheme

  1. B1 dilute sulfuric acid supplies the H⁺ that the reduction of MnO₄⁻ to Mn²⁺ needs, and is not itself oxidised
  2. B1 manganate(VII) oxidises the chloride ions of hydrochloric acid to chlorine
  3. B1 so extra manganate(VII) is used and the titre is too large
  4. B1 ethanoic acid is a weak acid and does not supply enough H⁺ ions
  5. B1 MnO₄⁻ is then reduced only to manganese(IV) oxide, a brown precipitate, so the 1 : 5 ratio no longer holds

• 4 marks · Give · VO-Q1 AQA · Edexcel

  Give the formula and the colour in solution of the vanadium species in the +5, +4, +3 and +2 oxidation states.

  Mark scheme

  1. B1 +5 is VO₂⁺, a yellow solution
  2. B1 +4 is VO²⁺, a blue solution
  3. B1 +3 is V³⁺, a green solution
  4. B1 +2 is V²⁺, a violet solution

• 4 marks · Complete · VO-Q2 AQA · Edexcel

  Complete and balance the equation for the reduction of VO²⁺ by zinc in acid, Zn + VO²⁺ + H⁺ →, and explain why a solution of VO₂⁺ turns green before it turns blue when zinc and acid are added.

  Mark scheme

  1. B1 the products are Zn²⁺, V³⁺ and H₂O
  2. B1 balanced as Zn + 2VO²⁺ + 4H⁺ → Zn²⁺ + 2V³⁺ + 2H₂O; allow multiples
  3. B1 the first green appears while some yellow VO₂⁺ remains and some blue VO²⁺ has formed
  4. B1 yellow and blue together look green, so this green is a mixture and not V³⁺

• 4 marks · Deduce · VO-Q13 AQA · Edexcel

  Deduce the final vanadium species when an excess of each of Fe²⁺ ions, lead metal and magnesium metal is added to separate acidified solutions of VO₂⁺. The E° values needed are given in the table.

  Half-equation	E° / V
  VO₂⁺(aq) + 2H⁺(aq) + e⁻ → VO²⁺(aq) + H₂O(l)	+1.00
  Fe³⁺(aq) + e⁻ → Fe²⁺(aq)	+0.77
  VO²⁺(aq) + 2H⁺(aq) + e⁻ → V³⁺(aq) + H₂O(l)	+0.34
  Pb²⁺(aq) + 2e⁻ → Pb(s)	−0.13
  V³⁺(aq) + e⁻ → V²⁺(aq)	−0.26
  Mg²⁺(aq) + 2e⁻ → Mg(s)	−2.37

  Mark scheme

  1. B1 a step happens when E° of the vanadium couple minus E° of the reducing agent's couple is positive
  2. B1 Fe²⁺: VO₂⁺ to VO²⁺ gives +1.00 − 0.77 = +0.23 V but VO²⁺ to V³⁺ gives +0.34 − 0.77 = −0.43 V, so the final species is VO²⁺
  3. B1 lead: the first two steps are positive, +1.13 V and +0.47 V, but V³⁺ to V²⁺ gives −0.26 − (−0.13) = −0.13 V, so it stops at V³⁺
  4. B1 magnesium: all three steps are positive, the last being −0.26 − (−2.37) = +2.11 V, so the vanadium is reduced to V²⁺

• 4 marks · Show · VO-Q22

  Show that two moles of manganate(VII) ions react with five moles of ethanedioate ions, C₂O₄²⁻, in acidic solution, and give the overall ionic equation.

  Mark scheme

  1. M1 MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l), manganese falling from +7 to +2
  2. M1 C₂O₄²⁻(aq) → 2CO₂(g) + 2e⁻, each carbon rising from +3 to +4, two electrons for each ion
  3. M1 ten electrons is the lowest common multiple, so the manganate half-equation is doubled and the ethanedioate one multiplied by five, which gives the 2 : 5 ratio
  4. A1 2MnO₄⁻(aq) + 16H⁺(aq) + 5C₂O₄²⁻(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g); ignore state symbols

• 3 marks · Explain · VO-Q4 AQA · Edexcel

  Explain why tin reduces VO²⁺ to V³⁺ but does not reduce V³⁺ to V²⁺. E°(Sn²⁺/Sn) = −0.14 V, E°(VO²⁺/V³⁺) = +0.34 V and E°(V³⁺/V²⁺) = −0.26 V.

  Mark scheme

  1. B1 for VO²⁺ to V³⁺, E°cell = +0.34 − (−0.14) = +0.48 V, which is positive, so the reduction is feasible
  2. B1 for V³⁺ to V²⁺, E°cell = −0.26 − (−0.14) = −0.12 V
  3. B1 this is negative, because the V³⁺/V²⁺ couple is more negative than tin's, so the reduction is not feasible

• 3 marks · Describe · VO-Q6 OCR A · Edexcel · CAIE

  Describe how chromium(III) ions in solution can be oxidised to chromate(VI) ions, giving the colours.

  Mark scheme

  1. B1 excess sodium hydroxide is added to give [Cr(OH)₆]³⁻, a green solution
  2. B1 hydrogen peroxide is added and the mixture is warmed
  3. B1 a yellow solution of CrO₄²⁻ forms

• 3 marks · Explain · VO-Q7 OCR A · Edexcel · CAIE

  Explain why chromium(III) is oxidised to chromium(VI) more easily in alkaline solution than in acidic solution.

  Mark scheme

  1. B1 the half-equation for reducing Cr₂O₇²⁻ to Cr³⁺ uses fourteen H⁺ ions
  2. B1 in acid the high H⁺ concentration makes chromium(VI) a strong oxidising agent, so chromium(III) is the stable state and is hard to oxidise
  3. B1 in alkali there is little H⁺, so chromium(VI) is a weak oxidising agent and chromium(III) is easily oxidised

• 3 marks · Calculate · VO-Q8

  Calculate the concentration of Fe²⁺ in an acidified solution of iron(II) sulfate, given that 25.0 cm³ of it needs 23.20 cm³ of 0.0180 mol dm⁻³ potassium manganate(VII) to reach the end point.

  Mark scheme

  1. M1 moles of MnO₄⁻ = 0.0180 × 23.20 ÷ 1000 = 4.176 × 10⁻⁴ mol
  2. M1 moles of Fe²⁺ = 5 × 4.176 × 10⁻⁴ = 2.088 × 10⁻³ mol, in 0.0250 dm³
  3. A1 concentration = 0.0835 mol dm⁻³, to three significant figures

• 3 marks · Compare · VO-Q14 OCR A · Edexcel · CAIE

  Compare the reduction of dichromate(VI) by zinc in acid with the reduction of VO₂⁺ by zinc in acid.

  Mark scheme

  1. B1 in both, zinc reduces the metal one stage at a time to its +2 state
  2. B1 chromium goes from orange Cr₂O₇²⁻ (+6) to green Cr³⁺ (+3) to blue Cr²⁺ (+2); vanadium goes from yellow (+5) through blue (+4) and green (+3) to violet (+2)
  3. B1 both Cr²⁺ and V²⁺ are oxidised by air, so the flask must be closed or plugged to keep air out

• 3 marks · Explain · VO-Q15

  Explain why the green precipitate formed when sodium hydroxide is added to Fe²⁺ turns orange-brown on standing in air, whereas acidified Fe²⁺ solution changes only slowly.

  Mark scheme

  1. B1 the green precipitate is iron(II) hydroxide, and the orange-brown solid is iron(III) hydroxide
  2. B1 oxygen in the air oxidises iron(II) to iron(III)
  3. B1 the electrode potential for iron(III)/iron(II) is much less positive in alkaline conditions, so iron(II) is oxidised more easily at high pH

• 3 marks · Outline · VO-Q16

  Outline how Tollens' reagent distinguishes an aldehyde from a ketone, in terms of the oxidation state of silver.

  Mark scheme

  1. B1 Tollens' reagent contains [Ag(NH₃)₂]⁺, with silver in the +1 oxidation state
  2. B1 an aldehyde reduces silver(I) to silver metal, oxidation state 0, seen as a silver mirror, and is itself oxidised
  3. B1 a ketone is not oxidised, so the silver stays as +1 and no mirror forms

• 3 marks · Describe · VO-Q17 OCR A · Edexcel · CAIE

  Describe what is seen when sodium hydroxide solution is added drop by drop until in excess to a solution of [Cr(H₂O)₆]³⁺, and give an equation for the first change.

  Mark scheme

  1. B1 a grey-green precipitate of chromium(III) hydroxide forms
  2. B1 [Cr(H₂O)₆]³⁺ + 3OH⁻ → [Cr(H₂O)₃(OH)₃] + 3H₂O; allow Cr(OH)₃ as the product
  3. B1 in excess the precipitate dissolves to a dark green solution of [Cr(OH)₆]³⁻

• 3 marks · Suggest · VO-Q19 OCR A

  Suggest how sodium hydroxide solution could be used to tell apart solutions containing Mn²⁺ ions and Cr³⁺ ions.

  Mark scheme

  1. B1 Mn²⁺ gives a light brown precipitate (off-white at first) that darkens on standing in air
  2. B1 Cr³⁺ gives a grey-green precipitate
  3. B1 in excess sodium hydroxide the chromium precipitate dissolves to a dark green solution while the manganese precipitate does not

• 2 marks · Calculate · VO-Q3 AQA · Edexcel

  Calculate E°cell for the reduction of V³⁺ to V²⁺ by zinc. E°(V³⁺/V²⁺) = −0.26 V and E°(Zn²⁺/Zn) = −0.76 V.

  Mark scheme

  1. M1 E°cell = E°(couple reduced) − E°(couple oxidised) = −0.26 − (−0.76)
  2. A1 E°cell = +0.50 V, to two significant figures

• 2 marks · Explain · VO-Q12

  Explain why no indicator is added in a manganate(VII) titration.

  Mark scheme

  1. B1 MnO₄⁻ is deep purple but the Mn²⁺ it forms is almost colourless
  2. B1 after the end point the first excess drop of MnO₄⁻ stays unreacted and gives a permanent pale pink colour

• 2 marks · Calculate · VO-Q18

  Calculate E°cell for the oxidation of iron(II) hydroxide by oxygen in alkaline solution. E° for O₂ + 2H₂O + 4e⁻ → 4OH⁻ is +0.40 V and E° for Fe(OH)₃ + e⁻ → Fe(OH)₂ + OH⁻ is −0.56 V.

  Mark scheme

  1. M1 E°cell = E°(couple reduced) − E°(couple oxidised) = +0.40 − (−0.56)
  2. A1 E°cell = +0.96 V, to two decimal places as the data

• 2 marks · Explain · VO-Q20 OCR A · Edexcel · CAIE

  Explain why chromium(III) hydroxide dissolves in excess sodium hydroxide but iron(III) hydroxide does not.

  Mark scheme

  1. B1 chromium(III) hydroxide is amphoteric, so it reacts with hydroxide ions as an acid, giving soluble [Cr(OH)₆]³⁻
  2. B1 iron(III) hydroxide is not amphoteric; it reacts only as a base, so no soluble complex forms with excess hydroxide

Practise variable oxidation states: vanadium, chromium and manganate(VII) titrations one question at a time