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A-LevelChemistryElectrochemistry9701/41 · May/June 2020 · Q97 Marks

(a) Manganese(IV) oxide, MnO2, catalyses the decomposition of hydrogen peroxide, H2O2, as shown. MnO2 2H2O2(aq) → 2H2O(l) + O2(g) The mechanism involves the formation of the intermediate species, Mn2+, in the first step which is subsequently used up in the second step. State and use relevant electrode potentials, Eº, to construct two equations to show how MnO2 can catalyse this reaction. equation 1 equation 2

  1. (a)Manganese(IV) oxide, MnO2, catalyses the decomposition of hydrogen peroxide, H2O2, as shown. MnO2 2H2O2(aq) → 2H2O(l) + O2(g) The mechanism involves the formation of the intermediate species, Mn2+, in the first step which is subsequently used up in the second step. State and use relevant electrode potentials, Eº, to construct two equations to show how MnO2 can catalyse this reaction. equation 1 equation 2[3]
  2. (b)The equation for the decomposition of hydrogen peroxide without a catalyst is shown. 2H2O2(aq) → 2H2O(l) + O2(g) Under certain conditions this reaction is found to be first order with respect to hydrogen peroxide, with a rate constant, k, of 2.0 × 10-6s-1 at 298 K. Calculate the initial rate of decomposition of a 0.75moldm³ hydrogen peroxide solution at 298 K. initial rate = .................................................. moldm-³s-1[1]
  3. (c)(i)A four-step mechanism is suggested for the reaction between hydrogen peroxide and iodide ions in an acidic solution. step 1 H2O2 + I¯ → IO¯ + H2O step 2 H+ + IO¯ → HIO step 3 HIO + I¯ → I2 + OH- step 4 OH- + H+ → H2O Step 1 is the rate-determining step. State what is meant by the term rate-determining step.[1]
  4. (c)(ii)Use this mechanism to construct a balanced equation for this reaction.[1]
  5. (c)(iii)Deduce the order of reaction with respect to each of the following. H2O2 = I- = H+ =[1]

Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

  1. [3] M1 data seen H2O2/H2O +1.77V and MnO2/Mn2+ +1.23 V and O2/H2O2 +0.68 V OR Ecell= 0.55 V (first step) and 0.54 V (second step)
  2. [3] M2 MnO2 + H2O2 + 2H+ → Mn2+ + O2 + 2H2O
  3. [3] M3 Mn2+ + H2O2 → MnO2 + 2H+
  4. [1] rate = 2.0 x 10-6 x 0.75 = 1.5 x 10-6
  5. [1] slowest step in overall reaction
  6. [1] H2O2 + 2H+ + 2I- → I2 + 2H2O OR H2O2 + 2HI → I2 + 2H2O
  7. [1] H2O2 = 1 AND I- = 1 AND H+ = 0

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About this question

Topic
Electrochemistry, A-Level Chemistry (syllabus 9701). Structured question, 7 marks.
Source
A-Level Chemistry, May/June 2020 Paper 4 Variant 1, question 9.

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