JEE Advance - Chemistry (2015 - Paper 2 Offline - No. 1)
In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by $$MnO_4^-$$. For this reaction,
the ratio of the rate of change of [H+] to the rate of change of $$[MnO_4^-]$$ is
Answer
8
Explanation
In complex, $$\mathop {{{[Fe{{({C_2}{O_4})}_2}{{({H_2}O)}_2}]}^{2 - }}}\limits_{Diaquodioxalatoferrate\,(II)} $$,
Fe is in +2 oxidation state.
In acidic medium, $$KMn{O_4}$$ oxidises $$F{e^{2 + }}$$ to $$F{e^{3 + }}$$,
$$2MnO_4^ - + 16{H^ + } + 10F{e^{2 + }} \to 2M{n^{2 + }} + 8{H_2}O + 10F{e^{3 + }}$$
or $$MnO_4^ - + 8{H^ + } + 5F{e^{2 + }} \to M{n^{2 + }} + 4{H_2}O + 5F{e^{3 + }}$$
$${{Rate\,of\,change\,of\,[{H^ + }]} \over {Rate\,of\,change\,of\,[MnO_4^ - ]}} = {8 \over 1} = 8$$
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