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$$\ce{Mn^2+ Solution ->[reagent 'A'] MnO4^-}$$ The reagent '$\ce{A}$' can be
(A) $\ce{Pb2O3 + \text{conc.} HNO3}$
(B) $\ce{K2S2O8}$.
(C) $\ce{(NH4)2S2O8}$
(D) $\ce{K2SO4}$

All of the options can be reduced to their corresponding compounds so, Maybe all of them should be the answer. But the answer given is (A), (B), (C). Why is (D) not the answer? $\ce{SO4^2-}$ Can be reduced to $\ce{SO2}$ right?

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    $\begingroup$ D can be reduced, but you would need much stronger reducing agens than $\ce{Mn^2+}$ . And even if SO2 had been created somehow, it would be immmediately oxidized by $\ce{MnO4-}$. $\endgroup$ – Poutnik Sep 10 at 12:18
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    $\begingroup$ Everything can be reduced. That's not what the question asks, though. $\endgroup$ – Ivan Neretin Sep 10 at 12:19
  • $\begingroup$ On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ – Martin - マーチン Sep 10 at 12:21
  • $\begingroup$ $\ce{Pb2O3}$ does not exist. It should be replaced by $\ce{Pb3O4}$ which a useful substance for protecting iron pieces against rust. $\endgroup$ – Maurice Sep 10 at 12:27
  • $\begingroup$ @Maurice Hmm, without external reference, but: Less common lead oxides are: Lead(II,IV) oxide, Pb2O3, lead sesquioxide (reddish yellow), Pb12O19 (monoclinic, dark-brown or black crystals) $\endgroup$ – Poutnik Sep 10 at 12:34
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Looking at the standard electrode potentials answers your question (Note that conc. $\ce{HNO3}$ oxidizes $\ce{Pb2O3}$ to $\ce{PbO2}$) : $$\begin{align} \ce{PbO2 + 4H+ + 2e- &-> Pb^2+ + 2H2O } & E^0 = \pu{1.46 V} \\ \ce{S2O8^2- + 2e- &-> 2SO4- } & E^0 = \pu{2.01 V} \\ \ce{SO4^2- + 4H+ + 2e- &-> SO2 + 2H2O } & E^0 = \pu{0.17 V} \end{align}$$

I couldn't find a reference for the couple $\ce{MnO4^2- | Mn^2+}$, but I'm guessing it would be slightly less than that of $\ce{MnO4- | Mn^2+}$, which is $\pu{1.51 V}$ (seeing that $\ce{PbO2}$ oxidizes it).

The standard electrode potential for the reduction of sulphur is less than that required to oxidize $\ce{Mn^2+}$, and hence the reaction does not proceed in the forward direction (net potential of cell is negative, which means that $\Delta G$ would be positive and cell reaction would be non-spontaneous)

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