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Chemical Kinetics and Nuclear Chemistry question

2015 · Shift 2 · Q1
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Chemical Kinetics and Nuclear Chemistry question

2015 · Shift 2 · Q1

JEE AdvancedChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
In dilute aqueous H2SO4H_2SO_4H2​SO4​, the complex diaquodioxalatoferrate(II) is oxidized by MnO4−MnO_4^-MnO4−​. For this reaction, the ratio of the rate of change of [H+] to the rate of change of [MnO4−][MnO_4^-][MnO4−​] is
Numerical answer
View written solutionFree

Correct answer: 8

  1. Identify the reacting species

The complex diaquodioxalatoferrate(II) is: [Fe(C2O4)2(H2O)2]2−[Fe(C_2O_4)_2(H_2O)_2]^{2-}[Fe(C2​O4​)2​(H2​O)2​]2−

Here, iron is in the +2+2+2 oxidation state and is oxidized to iron(III).

Permanganate ion MnO4−MnO_4^-MnO4−​ in acidic medium is reduced to Mn2+Mn^{2+}Mn2+.

  1. Write the oxidation half-reaction

Only the oxidation state of iron changes from +2+2+2 to +3+3+3; the oxalate ligands remain coordinated.

So, [Fe(C2O4)2(H2O)2]2−→[Fe(C2O4)2(H2O)2]−+e−[Fe(C_2O_4)_2(H_2O)_2]^{2-} \rightarrow [Fe(C_2O_4)_2(H_2O)_2]^- + e^-[Fe(C2​O4​)2​(H2​O)2​]2−→[Fe(C2​O4​)2​(H2​O)2​]−+e−

  1. Write the reduction half-reaction for permanganate in acidic medium

MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2OMnO4−​+8H++5e−→Mn2++4H2​O

  1. Balance electrons

Multiply the oxidation half-reaction by 555:

5[Fe(C2O4)2(H2O)2]2−→5[Fe(C2O4)2(H2O)2]−+5e−5[Fe(C_2O_4)_2(H_2O)_2]^{2-} \rightarrow 5[Fe(C_2O_4)_2(H_2O)_2]^- + 5e^-5[Fe(C2​O4​)2​(H2​O)2​]2−→5[Fe(C2​O4​)2​(H2​O)2​]−+5e−

Now add with the reduction half-reaction:

MnO4−+8H++5[Fe(C2O4)2(H2O)2]2−→Mn2++4H2O+5[Fe(C2O4)2(H2O)2]−MnO_4^- + 8H^+ + 5[Fe(C_2O_4)_2(H_2O)_2]^{2-} \rightarrow Mn^{2+} + 4H_2O + 5[Fe(C_2O_4)_2(H_2O)_2]^-MnO4−​+8H++5[Fe(C2​O4​)2​(H2​O)2​]2−→Mn2++4H2​O+5[Fe(C2​O4​)2​(H2​O)2​]−

  1. Extract the stoichiometric rate relation

From the balanced equation:

  • 888 moles of H+H^+H+ are consumed
  • 111 mole of MnO4−MnO_4^-MnO4−​ is consumed

Hence, −d[H+]/dt−d[MnO4−]/dt=81=8\frac{-d[H^+]/dt}{-d[MnO_4^-]/dt} = \frac{8}{1} = 8−d[MnO4−​]/dt−d[H+]/dt​=18​=8

Therefore, the required ratio of the rate of change of [H+][H^+][H+] to the rate of change of [MnO4−][MnO_4^-][MnO4−​] is:

888

  1. Compare with stored correct answer

Stored correct answer = 888

This matches the derived answer.

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