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Questions on half-equations and building redox equations

Half-equations for oxidation and reduction, the four-step method for balancing a half-equation in acidic solution, half-equations in alkaline solution such as chromium(III) to chromate(VI), combining two half-equations by matching electrons, balancing from oxidation-number changes, the manganate(VII) with iron(II) and dichromate(VI) with iodide equations, identifying the species oxidised and reduced, spectator ions, and disproportionation of chlorine in water and in cold sodium hydroxide, and in hot concentrated sodium hydroxide to chlorate(V).

20 original questions · 68 marks · the half-equations and building redox equations notes · Redox · Flashcards for this unit

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• 6 marks · Explain · HEB-Q17

  Explain, using half-equations and oxidation states, how the overall equation is built for the oxidation of ethanol, CH₃CH₂OH, to ethanal, CH₃CHO, by acidified potassium dichromate(VI), and state what is seen.

  Mark scheme

  Indicative content

  1. the dichromate(VI) half-equation is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, with each chromium falling from +6 to +3
  2. the ethanol half-equation is CH₃CH₂OH → CH₃CHO + 2H⁺ + 2e⁻, balanced for hydrogen with H⁺ and for charge with electrons
  3. the mean oxidation state of carbon rises from −2 in ethanol to −1 in ethanal, two carbons giving two electrons in all, so ethanol is oxidised
  4. the ethanol half-equation is multiplied by three so that both halves carry six electrons, which then cancel when the two are added
  5. after cancelling 6H⁺ from each side the overall equation is Cr₂O₇²⁻ + 8H⁺ + 3CH₃CH₂OH → 2Cr³⁺ + 3CH₃CHO + 7H₂O, with a charge of +6 on each side
  6. the solution turns from orange, the colour of Cr₂O₇²⁻, to green, the colour of Cr³⁺, as the dichromate(VI) is reduced; reject 'yellow to green'

  Levels

  • Level 3 (5–6 marks) both half-equations are correct with the oxidation state changes, the electrons are balanced by multiplying the ethanol half-equation by three, the overall equation is given with the H⁺ cancelled and the charge balanced, and the colour change from orange to green is stated
  • Level 2 (3–4 marks) both half-equations are correct and the overall equation is attempted; the electrons are left unbalanced, H⁺ is not cancelled, or the oxidation states or colour change are absent
  • Level 1 (1–2 marks) one half-equation is right or the colour change is stated, often the wrong way round as green to orange, with no overall equation

• 4 marks · Give · HEB-Q3

  Give the half-equations for the oxidation of iodide ions to iodine and for the reduction of hydrogen peroxide to water in acid, and combine them into an overall ionic equation.

  Mark scheme

  1. B1 2I⁻(aq) → I₂(aq) + 2e⁻
  2. B1 H₂O₂(aq) + 2H⁺(aq) + 2e⁻ → 2H₂O(l)
  3. B1 both halves carry two electrons, so they are added as they stand and the electrons cancel
  4. B1 H₂O₂(aq) + 2H⁺(aq) + 2I⁻(aq) → I₂(aq) + 2H₂O(l); ignore state symbols

• 4 marks · Explain · HEB-Q7

  Explain why the reaction of chlorine with cold, dilute sodium hydroxide is described as disproportionation. Include an equation.

  Mark scheme

  1. B1 Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)
  2. B1 chlorine falls from 0 in Cl₂ to −1 in NaCl, which is reduction
  3. B1 chlorine rises from 0 in Cl₂ to +1 in NaClO, which is oxidation
  4. B1 the same element in the same species, Cl₂, is both oxidised and reduced, which is disproportionation

• 4 marks · Outline · HEB-Q10

  Outline the steps used to construct a half-equation for a reaction in acidic solution.

  Mark scheme

  1. B1 balance the atoms of every element other than oxygen and hydrogen
  2. B1 balance oxygen by adding H₂O molecules to the side short of oxygen
  3. B1 balance hydrogen by adding H⁺ ions to the side short of hydrogen
  4. B1 balance the charge by adding electrons to the more positive side

• 4 marks · Compare · HEB-Q14

  Compare the reaction of chlorine with water with the reaction of chlorine with cold, dilute sodium hydroxide, and explain why both are described as disproportionation.

  Mark scheme

  1. B1 disproportionation is a redox reaction in which the same element in one species is simultaneously oxidised and reduced
  2. B1 in both reactions chlorine goes from 0 in Cl₂ to −1, which is reduction, and to +1, which is oxidation
  3. B1 with water the products are HCl and HClO; with sodium hydroxide they are NaCl and NaClO
  4. B1 the reaction with water is reversible and only a small fraction of chlorine reacts, whereas hydroxide removes the acids as they form so the reaction goes to completion

• 4 marks · Deduce · HEB-Q21

  Deduce the half-equation for the reduction of iodate(V) ions, IO₃⁻, to iodine in acidic solution, and use it to deduce the overall equation for the reaction of iodate(V) ions with iodide ions in acid.

  Mark scheme

  1. B1 2IO₃⁻ → I₂ needs 6H₂O on the right for the oxygen and then 12H⁺ on the left for the hydrogen
  2. B1 the charge is +10 on the left and 0 on the right, so 2IO₃⁻(aq) + 12H⁺(aq) + 10e⁻ → I₂(aq) + 6H₂O(l); iodine falls from +5 to 0 in each of two atoms
  3. B1 the iodide half-equation, 2I⁻ → I₂ + 2e⁻, is multiplied by five so that ten electrons cancel
  4. B1 IO₃⁻(aq) + 5I⁻(aq) + 6H⁺(aq) → 3I₂(aq) + 3H₂O(l), after dividing through by two; accept the doubled equation

• 4 marks · Show · HEB-Q22

  Show that one mole of dichromate(VI) ions oxidises six moles of iron(II) ions in acidic solution, by constructing the two half-equations and combining them.

  Mark scheme

  1. M1 Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l), balanced for oxygen with water, for hydrogen with H⁺ and for charge with electrons
  2. M1 Fe²⁺(aq) → Fe³⁺(aq) + e⁻, one electron lost by each iron(II) ion
  3. M1 the iron half-equation is multiplied by six so that the electrons gained and lost are equal, and the two are added
  4. A1 Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6Fe²⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l) + 6Fe³⁺(aq), so the ratio is 1 : 6; the check from oxidation states is two chromiums falling by three each, six electrons

• 3 marks · Complete · HEB-Q2

  Complete the half-equation for the reduction of dichromate(VI) ions to chromium(III) ions in acid, Cr₂O₇²⁻(aq) → 2Cr³⁺(aq), and check the number of electrons against the change in oxidation state.

  Mark scheme

  1. B1 7H₂O(l) on the right balances the oxygen, and then 14H⁺(aq) on the left balances the hydrogen
  2. B1 6e⁻ on the left balance the charge: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)
  3. B1 each chromium falls from +6 to +3, and two chromium atoms change, so six electrons are needed

• 3 marks · Calculate · HEB-Q4

  Calculate the mass of iodine formed when 25.0 cm³ of 0.0500 mol dm⁻³ acidified potassium dichromate(VI) reacts with excess potassium iodide: Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O. Mr of I₂ is 253.8.

  Mark scheme

  1. M1 moles of Cr₂O₇²⁻ = 0.0500 × 25.0/1000 = 1.25 × 10⁻³ mol
  2. M1 moles of I₂ = 3 × 1.25 × 10⁻³ = 3.75 × 10⁻³ mol, from the 1 : 3 ratio
  3. A1 mass = 3.75 × 10⁻³ × 253.8 = 0.952 g, to three significant figures

• 3 marks · Calculate · HEB-Q5

  Calculate the final oxidation state of a metal X when 2.40 × 10⁻³ mol of X²⁺ ions are oxidised by exactly 9.60 × 10⁻⁴ mol of MnO₄⁻ ions in acid, the MnO₄⁻ being reduced to Mn²⁺.

  Mark scheme

  1. M1 each MnO₄⁻ gains five electrons, so electrons transferred = 5 × 9.60 × 10⁻⁴ = 4.80 × 10⁻³ mol
  2. M1 electrons lost per X ion = 4.80 × 10⁻³ ÷ 2.40 × 10⁻³ = 2
  3. A1 the oxidation state of X rises from +2 to +4, an integer with no unit

• 3 marks · Explain · HEB-Q6

  Explain why two half-equations must contain the same number of electrons before they are added to give an overall redox equation.

  Mark scheme

  1. B1 every electron released in the oxidation is taken up in the reduction; electrons are not created or left over
  2. B1 matching the numbers lets the electrons cancel, so none appears in the overall equation
  3. B1 the multipliers used fix the mole ratio of oxidising agent to reducing agent, such as 1 MnO₄⁻ to 5 Fe²⁺

• 3 marks · Explain · HEB-Q8

  Explain why chlorine is added to drinking water even though chlorine is toxic.

  Mark scheme

  1. B1 chlorine reacts with water to form chloric(I) acid, HClO, which kills bacteria
  2. B1 this prevents water-borne diseases such as cholera and typhoid, and chlorine left in the water protects it in the pipes
  3. B1 it is used at a low concentration, at which the benefit outweighs the risk of toxicity and of forming chlorinated hydrocarbons

• 3 marks · Suggest · HEB-Q11

  Suggest a half-equation for the reduction of VO₂⁺ ions to VO²⁺ ions in acidic solution.

  Mark scheme

  1. B1 vanadium balances; one H₂O on the right balances the oxygen
  2. B1 two H⁺ on the left balance the hydrogen
  3. B1 one electron on the left balances the charge, matching vanadium falling from +5 to +4: VO₂⁺(aq) + 2H⁺(aq) + e⁻ → VO²⁺(aq) + H₂O(l)

• 3 marks · Suggest · HEB-Q12

  Suggest the two half-equations and the overall equation for the reaction in which concentrated sulfuric acid oxidises bromide ions to bromine and is reduced to sulfur dioxide.

  Mark scheme

  1. B1 oxidation: 2Br⁻ → Br₂ + 2e⁻
  2. B1 reduction: H₂SO₄ + 2H⁺ + 2e⁻ → SO₂ + 2H₂O
  3. B1 overall, with two electrons cancelling: 2Br⁻ + H₂SO₄ + 2H⁺ → Br₂ + SO₂ + 2H₂O

• 3 marks · Evaluate · HEB-Q15

  Evaluate the half-equation MnO₄⁻(aq) + 8H⁺(aq) + 3e⁻ → Mn²⁺(aq) + 4H₂O(l), written by a student for the reduction of manganate(VII) in acid.

  Mark scheme

  1. B1 the atoms balance: one Mn, four O and eight H on each side
  2. B1 the charge does not balance: −1 + 8 − 3 = +4 on the left against +2 on the right
  3. B1 manganese falls from +7 to +2, a gain of five electrons, so the equation needs 5e⁻, not 3e⁻

• 3 marks · Describe · HEB-Q16

  Describe what is seen when zinc powder is stirred into copper(II) sulfate solution, and give the two half-equations and the ionic equation for the reaction.

  Mark scheme

  1. B1 the blue colour of the solution fades towards colourless and a brown solid, copper, forms as the zinc dissolves
  2. B1 Zn(s) → Zn²⁺(aq) + 2e⁻ and Cu²⁺(aq) + 2e⁻ → Cu(s)
  3. B1 Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with sulfate ions left out as spectator ions

• 3 marks · Explain · HEB-Q18 Edexcel · CAIE

  Explain, using oxidation states, why the reaction of chlorine with hot, concentrated sodium hydroxide gives five moles of sodium chloride for every mole of sodium chlorate(V): 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.

  Mark scheme

  1. B1 chlorine starts at 0 in Cl₂, falls to −1 in NaCl and rises to +5 in NaClO₃, so this is disproportionation
  2. B1 each chlorine atom oxidised to +5 releases five electrons, while each chlorine atom reduced to −1 gains one
  3. B1 electrons lost must equal electrons gained, so five chlorine atoms are reduced for every one oxidised, a 5 : 1 ratio of chloride to chlorate(V)

• 3 marks · Suggest · HEB-Q19 OCR A · Edexcel

  Suggest a half-equation for the oxidation of sulfite ions, SO₃²⁻, to sulfate ions, SO₄²⁻, in alkaline solution.

  Mark scheme

  1. B1 balanced as in acid first: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻, sulfur rising from +4 to +6
  2. B1 two OH⁻ are added to each side, and the 2H⁺ and 2OH⁻ on the right form 2H₂O, one of which cancels the H₂O on the left
  3. B1 SO₃²⁻(aq) + 2OH⁻(aq) → SO₄²⁻(aq) + H₂O(l) + 2e⁻, with a charge of −4 on each side

• 3 marks · Complete · HEB-Q20

  Complete and balance, using changes in oxidation number, the equation for the reduction of concentrated sulfuric acid to hydrogen sulfide by iodide ions: H₂SO₄ + H⁺ + I⁻ → H₂S + I₂ + H₂O.

  Mark scheme

  1. B1 sulfur falls by eight, from +6 to −2, and each iodine rises by one, from −1 to 0, so eight I⁻ react with each H₂SO₄, giving four I₂
  2. B1 four H₂O on the right balance the oxygen, and eight H⁺ on the left balance the hydrogen
  3. B1 H₂SO₄ + 8H⁺ + 8I⁻ → H₂S + 4I₂ + 4H₂O, with a total charge of 0 on each side

• 2 marks · Identify · HEB-Q13

  Identify the species oxidised and the species reduced in the reaction 3Cu(s) + 8H⁺(aq) + 2NO₃⁻(aq) → 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l), giving the change in oxidation state for each.

  Mark scheme

  1. B1 copper is oxidised, from 0 in Cu to +2 in Cu²⁺
  2. B1 the nitrate ion is reduced, nitrogen falling from +5 in NO₃⁻ to +2 in NO

Practise half-equations and building redox equations one question at a time