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Questions on oxidation states and the rules for assigning them

Oxidation and reduction as loss and gain of electrons and as a rise and fall in oxidation state, the rules for assigning oxidation states in order of priority including peroxides, OF₂ and metal hydrides, oxidation states in polyatomic ions and in compounds such as K₂Cr₂O₇, NH₄NO₃, Fe₃O₄ and Na₂S₂O₃, the metal in complex ions and oxo-ions, Group 2 metals with chlorine, metals with dilute acids, oxidation number as the other boards' name, oxidising and reducing agents, and Roman numerals in names.

19 original questions · 65 marks · the oxidation states and the rules for assigning them notes · Redox · Flashcards for this unit

To print: the questions (PDF, 80 KB) · the mark schemes (PDF, 86 KB)

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• 6 marks · Explain · OSR-Q21

  Explain, using oxidation states, which species are oxidised and reduced in the reaction 2KMnO₄(s) + 16HCl(aq) → 2KCl(aq) + 2MnCl₂(aq) + 5Cl₂(g) + 8H₂O(l), and why hydrochloric acid acts both as a reducing agent and as an acid in it.

  Mark scheme

  Indicative content

  1. manganese falls from +7 in KMnO₄ to +2 in MnCl₂, a gain of five electrons per manganese atom, so the manganate(VII) is reduced
  2. ten of the sixteen chlorine atoms rise from −1 in HCl to 0 in Cl₂, losing one electron each, so those chloride ions are oxidised
  3. the electrons balance: two manganese atoms gain ten electrons and ten chlorine atoms lose ten electrons
  4. the other six chlorine atoms stay at −1 in KCl and MnCl₂, so that part of the HCl supplies H⁺ and chloride ions as an acid would
  5. potassium stays at +1, hydrogen at +1 and oxygen at −2 throughout, so none of them takes part in the electron transfer
  6. KMnO₄ is therefore the oxidising agent, accepting electrons, and HCl is the reducing agent, donating them, as well as acting as the acid

  Levels

  • Level 3 (5–6 marks) all the oxidation states are assigned correctly, the reduction of manganese and the oxidation of chlorine are identified with the electron count balanced, the six chlorine atoms that stay at −1 are used to show the acid role, and the oxidising and reducing agents are named
  • Level 2 (3–4 marks) manganese is shown to go from +7 to +2 and chlorine from −1 to 0; the electrons are not balanced, the six chlorines that stay at −1 are not used, or the agents are not named
  • Level 1 (1–2 marks) one oxidation state change is right; HCl is often called the oxidising agent, and its role as an acid is not explained

• 4 marks · Give · OSR-Q2

  Give the oxidation state of sulfur in H₂SO₄, of nitrogen in NO₂, of chlorine in ClO₃⁻ and of manganese in MnO₄²⁻.

  Mark scheme

  1. B1 sulfur in H₂SO₄ is +6, since 2(+1) + S + 4(−2) = 0
  2. B1 nitrogen in NO₂ is +4, since N + 2(−2) = 0
  3. B1 chlorine in ClO₃⁻ is +5, since Cl + 3(−2) = −1
  4. B1 manganese in MnO₄²⁻ is +6, since Mn + 4(−2) = −2

• 4 marks · Explain · OSR-Q6

  Explain, using the definitions of oxidation and reduction in terms of electrons, why the reaction Mg(s) + Cl₂(g) → MgCl₂(s) is a redox reaction even though it involves no oxygen.

  Mark scheme

  1. B1 oxidation is the loss of electrons and reduction is the gain of electrons; reject definitions in terms of oxygen alone
  2. B1 each magnesium atom loses two electrons to form Mg²⁺, so magnesium is oxidised from 0 to +2
  3. B1 each chlorine atom gains one electron to form Cl⁻, so chlorine is reduced from 0 to −1
  4. B1 oxidation and reduction both occur, with electrons passing from magnesium to chlorine, so the reaction is redox

• 4 marks · Name · OSR-Q9

  Name the compounds Cu₂O and NaClO₃, using Roman numerals for oxidation states, and give the formulae of iron(III) sulfate and of potassium manganate(VII).

  Mark scheme

  1. B1 Cu₂O is copper(I) oxide, since each copper is +1
  2. B1 NaClO₃ is sodium chlorate(V), since chlorine is +5
  3. B1 iron(III) sulfate is Fe₂(SO₄)₃
  4. B1 potassium manganate(VII) is KMnO₄

• 4 marks · Deduce · OSR-Q16

  Deduce the oxidation state of phosphorus in H₄P₂O₇, of nitrogen in hydrazine, N₂H₄, of chlorine in Cl₂O₇ and of hydrogen in calcium hydride, CaH₂.

  Mark scheme

  1. B1 phosphorus is +5: 4(+1) + 2P + 7(−2) = 0, so 2P = +10
  2. B1 nitrogen is −2: 2N + 4(+1) = 0
  3. B1 chlorine is +7: 2Cl + 7(−2) = 0, chlorine bonded to oxygen being positive
  4. B1 hydrogen is −1: calcium is +2 by the Group 2 rule, which outranks the rule for hydrogen in a metal hydride

• 4 marks · Explain · OSR-Q17

  Explain, using oxidation states, why the reaction of concentrated sulfuric acid with sodium bromide to give bromine and sulfur dioxide is a redox reaction, and identify the oxidising agent.

  Mark scheme

  1. B1 bromine rises from −1 in Br⁻ to 0 in Br₂, so bromide is oxidised
  2. B1 sulfur falls from +6 in H₂SO₄ to +4 in SO₂, so the acid is reduced
  3. B1 one element is oxidised and another reduced in the same reaction, so it is redox; sodium, hydrogen and oxygen do not change
  4. B1 sulfuric acid accepts the electrons, so it is the oxidising agent

• 3 marks · Deduce · OSR-Q3

  Deduce the mean oxidation state of iron in the mixed oxide Fe₃O₄, and explain how the oxidation states of the iron ions present account for it.

  Mark scheme

  1. M1 four oxygen atoms at −2 give −8, so the three iron atoms together are +8
  2. A1 mean oxidation state = +8/3, which is +2.67 to three significant figures; accept +2⅔
  3. B1 the oxide holds one Fe²⁺ for every two Fe³⁺, and 2 + 3 + 3 = 8, so the fraction is the average of two real oxidation states

• 3 marks · Calculate · OSR-Q4

  Calculate the final oxidation state of vanadium when 25.0 cm³ of a 0.100 mol dm⁻³ solution of vanadium in the +5 oxidation state is reduced by zinc in acid and accepts 5.00 × 10⁻³ mol of electrons.

  Mark scheme

  1. M1 moles of vanadium = 0.100 × 25.0/1000 = 2.50 × 10⁻³ mol
  2. M1 electrons per vanadium = 5.00 × 10⁻³ ÷ 2.50 × 10⁻³ = 2
  3. A1 the oxidation state falls by 2, from +5 to +3; it is an integer with no unit

• 3 marks · Explain · OSR-Q5

  Explain why oxygen has an oxidation state of +2 in OF₂ but −2 in H₂O.

  Mark scheme

  1. B1 fluorine is the most electronegative element and is always −1 in its compounds, a rule of higher priority than the rule for oxygen
  2. B1 in OF₂ two fluorines contribute −2, so by the sum rule oxygen must be +2
  3. B1 in H₂O oxygen is more electronegative than hydrogen, hydrogen is +1, so the two hydrogens give +2 and oxygen is −2

• 3 marks · Explain · OSR-Q7

  Explain why it is wrong to give the oxidation state of nitrogen in ammonium nitrate, NH₄NO₃, as +1.

  Mark scheme

  1. B1 +1 is the mean found by treating NH₄NO₃ as one unit: 2N + 4(+1) + 3(−2) = 0
  2. B1 the two nitrogen atoms are in different ions, NH₄⁺ and NO₃⁻, and have different oxidation states
  3. B1 in NH₄⁺ nitrogen is −3 and in NO₃⁻ it is +5, so no nitrogen atom is at +1

• 3 marks · Identify · OSR-Q8

  Identify the oxidising agent and the reducing agent in the thermite reaction, 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(l), giving the change in oxidation state of each element.

  Mark scheme

  1. B1 Fe₂O₃ is the oxidising agent: iron falls from +3 to 0, so it is reduced; accept iron(III) oxide, reject 'iron'
  2. B1 Al is the reducing agent: aluminium rises from 0 to +3, so it is oxidised
  3. B1 oxygen stays at −2 in Fe₂O₃ and in Al₂O₃, so it takes no part in the electron transfer

• 3 marks · Suggest · OSR-Q11

  Suggest the oxidation state of hydrogen in lithium aluminium hydride, LiAlH₄, and suggest why this compound is a strong reducing agent.

  Mark scheme

  1. B1 lithium is +1 and aluminium is +3, so the four hydrogens share −4 and each is −1
  2. B1 this is a metal hydride: hydrogen is bonded to less electronegative metals, so the metal rules take priority over the rule for hydrogen
  3. B1 hydrogen at −1 is readily oxidised to 0 or +1, donating electrons to the species it reacts with, so the compound is a reducing agent

• 3 marks · Suggest · OSR-Q12

  Suggest, using oxidation states, why thiosulfate ions, S₂O₃²⁻, act as a reducing agent when they react with iodine to form tetrathionate ions, S₄O₆²⁻.

  Mark scheme

  1. B1 in S₂O₃²⁻ three oxygens give −6, so the two sulfurs share +4 and the mean is +2
  2. B1 in S₄O₆²⁻ six oxygens give −12, so the four sulfurs share +10 and the mean is +2.5
  3. B1 the mean oxidation state of sulfur rises, so thiosulfate loses electrons to iodine, which is reduced from 0 to −1: thiosulfate is the electron donor, the reducing agent

• 3 marks · Compare · OSR-Q13

  Compare the definition of oxidation as gain of oxygen with the definitions in terms of electrons and of oxidation state.

  Mark scheme

  1. B1 all three agree for a reaction such as magnesium burning in oxygen, where magnesium gains oxygen, loses electrons and rises from 0 to +2
  2. B1 the oxygen definition fails for reactions with no oxygen, such as magnesium with chlorine, which the electron definition covers
  3. B1 the oxidation state definition extends the electron definition to covalent species such as SO₂, where electrons are shared rather than fully transferred

• 3 marks · Evaluate · OSR-Q14

  Evaluate a student's claim that hydrogen is reduced in the reaction H₂O₂(aq) + 2H⁺(aq) + 2I⁻(aq) → I₂(aq) + 2H₂O(l).

  Mark scheme

  1. B1 the claim is wrong: hydrogen is +1 in H₂O₂, in H⁺ and in H₂O, so its oxidation state does not change
  2. B1 oxygen falls from −1 in the peroxide to −2 in water, so hydrogen peroxide is the species reduced
  3. B1 iodine rises from −1 to 0, so iodide is oxidised, and hydrogen peroxide is the oxidising agent

• 3 marks · Describe · OSR-Q15

  Describe the changes in oxidation state when hydrogen sulfide burns in oxygen, 2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(l), and identify the reducing agent.

  Mark scheme

  1. B1 sulfur rises from −2 in H₂S to +4 in SO₂, a change of six, so sulfur is oxidised
  2. B1 oxygen falls from 0 in O₂ to −2 in SO₂ and in H₂O, so oxygen is reduced; hydrogen stays at +1
  3. B1 hydrogen sulfide donates the electrons, so H₂S is the reducing agent

• 3 marks · Explain · OSR-Q18

  Explain why the reaction of magnesium with dilute sulfuric acid is a redox reaction, but the reaction of magnesium oxide with dilute sulfuric acid is not.

  Mark scheme

  1. B1 with magnesium, Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g): magnesium rises from 0 to +2 and is oxidised
  2. B1 hydrogen falls from +1 in H⁺ to 0 in H₂ and is reduced, so electrons pass from magnesium to hydrogen ions
  3. B1 with magnesium oxide, magnesium stays at +2, oxygen at −2 and hydrogen at +1, so no oxidation state changes and it is an acid–base reaction

• 3 marks · Deduce · OSR-Q19

  Deduce the oxidation state of the metal in each of the complex ions [Cr(NH₃)₄Cl₂]⁺, [Co(NH₃)₆]³⁺ and [Ni(CN)₄]²⁻.

  Mark scheme

  1. B1 chromium is +3: ammonia ligands count 0 and each chloride −1, so Cr + 2(−1) = +1
  2. B1 cobalt is +3: six neutral ammonia ligands add nothing, so the whole 3+ charge belongs to the cobalt
  3. B1 nickel is +2: four cyanide ligands give −4, so Ni − 4 = −2; the sign is required for each mark

• 3 marks · Suggest · OSR-Q20 Edexcel

  Suggest an equation for the reaction of barium with chlorine, and suggest how its vigour compares with that of magnesium with chlorine, giving a reason.

  Mark scheme

  1. B1 Ba(s) + Cl₂(g) → BaCl₂(s), with barium oxidised from 0 to +2 and chlorine reduced from 0 to −1
  2. B1 barium reacts more vigorously than magnesium
  3. B1 barium's outer electrons are further from the nucleus and more shielded, so its first and second ionisation energies are lower and it loses electrons more readily

Practise oxidation states and the rules for assigning them one question at a time