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I am taking a lot of iron properties this semester, but I have found some contradicting piece of information about the reaction of iron with nonmetals illustrated with two cases
\begin{align}\ce{ **First Case** }\\\end{align} If we react the iron with chlorine, only FeCl3 will result as following:

$$\begin{align}\ce{2Fe +3Cl2 &->[\space \Delta ]2FeCl3}\tag{1}\\\end{align}$$

No FeCl2 has produced because iron prefers to have oxidation state at +3 as sublevel 3d will be half-filled.

One case that FeCl2 will result when H2 exists as $$\begin{align}\ce{Fe +2HCl &->FeCl2 +H2}\tag{2}\\\end{align}$$ In this situatuon if any Fe2Cl3 formed, H2 will reduce it to FeCl2 \begin{align}\ce{ **Second Case** }\\\end{align} Although, Iron favor +3 oxidation state (as illustrated in 1), If we react iron with sulfur; it will actually favor +2 oxidation state as follows: $$\begin{align}\ce{Fe +S &->[\space \Delta ]FeS}\tag{3}\\\end{align}$$ I tried to find a convenient way that explains how this happen (as the way illustrated at 2), but

I failed to do so. I am really confused:

Does the iron actually favor +3 state or not?

If yes, Why 3 happens? Why there isn't any Fe2S3 in the products?

If no, Why 1 happens?

Is this has to do anything with some hidden conditions or unobvious reaction mechanism?

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    $\begingroup$ Iron is not the only party to decide how many electrons will be transferred. $\endgroup$ – Ivan Neretin Sep 13 '18 at 20:09
  • $\begingroup$ Better stop thinking about "preferences" for ox. states or "stability" - while such thinking might have some use, it's leading you astray. Fe can "be made" to have ox. state as high as +6, or lower then 0. It's just a matter of right reagents. $\endgroup$ – Mithoron Sep 13 '18 at 22:43
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Not an answer per se (I lack sufficient reputation to comment): are these reactions in aqueous solution, or dry gases? Just a suggestion, but by omitting the charges on these species and not writing them as half-reactions you are obscuring the electron exchange in each. Thus in your first example (iron reacting with chlorine gas), you take metallic Fe, whose oxidation half-reaction can be written as $$\ce{Fe^0 -> Fe^3+ + 3e-}$$ The equivalent for reduction of chlorine would be $$\ce{2e- + Cl2_{(g)} -> 2Cl-}$$ Normalizing half reactions to electrons transferred ($6$), adding and simplifying stoichiometry gives

$$\ce{2Fe^0 + 3Cl2 -> 2FeCl3}$$

Thus chlorine is the electron acceptor (strong oxidizer) and iron the donor. In your reaction (2), iron is again a donor (this time of only $2$ electrons), hydrogen is the acceptor, with chloride ion just along for the ride:

$$\ce{Fe^0 -> Fe^2+ + 2e-}$$ $$\ce{e- + H+ -> \tfrac{1}{2}H2} $$

In contrast to these simple reactions, those involving sulfur are exceeding complex because of the diversity of species that can form. In your reaction to form iron monosulfide, one possible way for Fe to be oxidized but native sulfur (zero-valent S actually present as stable ring, $\mathrm{S}_8$) to be both reduced and oxidized is via a disproportionation reaction, with water reduction to hydrogen consuming additional electrons in an anoxic environment, thus yielding both FeS and iron sulfate: $$\ce{Fe^0 + \tfrac{1}{8} S8 + 2H2O -> \tfrac{1}{2}FeSO4 + \tfrac{1}{2}FeS + 2H2}$$

Amended answer: If you're heating metallic iron and S to significant temperatures, the elemental sulfur will form a gas, $\mathrm{S}_{n}(g)$ ($n$ indicates which T-dependent allotrope is present). To synthesize FeS however, I think you would need an oxygen-free system. If $\mathrm{O_2}$ is initially present, consumption could be accomplished by sacrificing some of the reactants via in situ oxidation of the metal, or production of $\mathrm{SO_2}$. Then this can proceed directly, using the remaining $\mathrm{S}_{n}(g)$, e.g.: $$ \tfrac{1}{2}\mathrm{S_2}(g) + \mathrm{Fe}(s) = \mathrm{FeS}(s) $$ I believe the ratio of S/Fe of the product will increase with temperature, and at sufficiently high temperature reach 2 ($\mathrm{FeS_2}$, pyrite).

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  • $\begingroup$ Thanks for clarification, but my textbook just mentions that If we heated sulfur and iron (without any water involved), only Fes will result. Also, I saw the full procedure on a video and there is no water in the process. $\endgroup$ – Amer Esmail Sep 14 '18 at 17:16
  • $\begingroup$ Amer, sorry for the confusion. I thought for some reason you were referring to room temperature reactions in solution (my bias). I've amended the answer, above. Hope this clarifies. The reaction you're describing is quite commonly used as a showy lab demo of an exothermic reaction (remembering from many years ago). $\endgroup$ – Rolf Sep 14 '18 at 19:58
  • $\begingroup$ @Amer: Also, I forgot to answer your question re why Fe2S3 is not present in the products: probably due its poor stability at temperature. In their solution syntheses of ferric sulfide by addition of ferric chloride with sodium sulfide nonahydrate, Stansberry et al. (1993) stated that Fe2S3 disproportionated to mixtures of pyrite (FeS2), pyrrhotite, and elemental S above temperatures of 10C. Apparently there was significant interest in Fe2S3 as a catalyst for coal liquefaction. $\endgroup$ – Rolf Sep 15 '18 at 17:13

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