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QJEEAdv15Chbd8a: JEE Advanced 2015 Chemistry - Class 12 JEEAdv15Chbd8a Solution: In dilute aqueous H 2 SO 4 , the complex…

In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by MnO4−MnO_4^- . For this reaction, the ratio of the rate of change of [H+] to the rate of change of [MnO4−][MnO_4^-] is

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In complex, [Fe(C2O4)2(H2O)2]2−Diaquodioxalatoferrate (II)\mathop {{{[Fe{{({C_2}{O_4})}_2}{{({H_2}O)}_2}]}^{2 - }}}\limits_{Diaquodioxalatoferrate\,(II)} ,

Fe is in +2 oxidation state.

In acidic medium, KMnO4KMn{O_4} oxidises Fe2+F{e^{2 + }} to Fe3+F{e^{3 + }} ,

2MnO4−+16H++10Fe2+→2Mn2++8H2O+10Fe3+2MnO_4^ - + 16{H^ + } + 10F{e^{2 + }} \to 2M{n^{2 + }} + 8{H_2}O + 10F{e^{3 + }}

or MnO4−+8H++5Fe2+→Mn2++4H2O+5Fe3+MnO_4^ - + 8{H^ + } + 5F{e^{2 + }} \to M{n^{2 + }} + 4{H_2}O + 5F{e^{3 + }}

Rate of change of [H+]Rate of change of [MnO4−]=81=8{{Rate\,of\,change\,of\,[{H^ + }]} \over {Rate\,of\,change\,of\,[MnO_4^ - ]}} = {8 \over 1} = 8

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