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$\ce{3Ca(OH)2 + 2FeCl3 -> 2Fe(OH)3 + 3CaCl2}$

What factors influence the resulting oxidation state in the iron hydroxide molecule in this reaction? Since iron's oxidation state ranges from +2 to +6, and there exists iron hydroxide (II), how do we know what the oxidation state will be in this reaction?

Would this equation yield $\ce{Fe(OH)2}$ under some conditions?

I googled and found that in order to obtain $\ce{Fe(OH)2}$ we react $\ce{FeCl2}$ and $\ce{KOH}$ with no air present:

$\ce{FeCl2 + 2KOH -> 2KCl + Fe(OH)2}$

Would my initial equation yield $\ce{Fe(OH)2}$ in the absence of air? If yes, why?


The oxidation state of $\ce{Fe}$ before the reaction is +3.

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    $\begingroup$ First of all, figure out what oxidation state do you have initially (that is, before the reaction). Then see whether it can change. Is redox reaction possible? Are there any strong reducing agents? $\endgroup$ – Ivan Neretin Feb 1 '16 at 8:12
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    $\begingroup$ I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance. $\endgroup$ – bon Feb 1 '16 at 10:21
  • $\begingroup$ @bon - thank you! I'll add that to my bookmarks. $\endgroup$ – CowperKettle Feb 1 '16 at 10:21
  • $\begingroup$ @IvanNeretin - thank you! I'll read up on redox reactions. Ferrum's pre-reaction oxidation state is +3, of course. Fe seems to be the most potent reducing agent here, since it's able to "donate" up to 6 electrons, according to my copy of the Periodic Table. $\endgroup$ – CowperKettle Feb 1 '16 at 10:23
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    $\begingroup$ Strength of a reducing agent is determined not by the number of electrons it may donate, but by its "willingness" to do so, which in case of $\ce{Fe+^3\to Fe+^6}$ is quite low. But that's irrelevant. See, you have $\ce{Fe+^3}$ and you want $\ce{Fe+^2}$; how's that possible? Some other element must reduce iron. What element that might be, if any? $\endgroup$ – Ivan Neretin Feb 1 '16 at 10:34

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