4
$\begingroup$

$\ce{Fe(OH)2 (s) <=> Fe^{2+} + 2OH-}$

The reaction will be more active in the reverse direction if .......... is added

a) $\ce{KOH}$

b) $\ce{Na2S}$

c) $\ce{Fe(OH)2}$

d) $\ce{Fe(NO3)2}$

My thoughts:

Well, adding more $\ce{KOH}$ will increase the $\ce{OH-}$ ions so the reaction will go in the reverse direction to get back to equilibrium. Adding $\ce{Fe(NO3)2}$ Will increase $\ce{Fe^{2+}}$ ions and the reaction will go in the reverse direction too.

However, the book answers states that only (d) $\ce{Fe(NO3)2}$. I searched google but all the examples involved having only choice which works.

Note: Refer me to sites if necessary. I am ready to read.

$\endgroup$
1
  • $\begingroup$ I have improved the formatting of your question using $\LaTeX$. For more information on how to do this yourself please see here. $\endgroup$ – bon May 27 '15 at 14:00
7
$\begingroup$

In fact, adding more ion hydroxide won't displace the equilibrium towards the reverse reaction. This is due to the fact that ion complex $\ce{[Fe(OH)4]^{-2}}$, which is stable in presence of excess of ion hydroxide, is formed by successive addition of ion hydroxide on ferrous hydroxide. If you want to study this more in depth, I invite you to read this webpage

It's explained at the end of the page under "numerical problems".

$\endgroup$
3
  • $\begingroup$ Need a bit more explanation , It's too difficult to my current state. $\endgroup$ – Hoyt Volker May 27 '15 at 16:36
  • 1
    $\begingroup$ @HoytVolker Essentially, if you add enough hydroxide ions, another product will be formed, the complex mentioned in the Yomen's response. If this is beyond the scope of your class, then the lack of KOH as a correct answer was likely an oversight. I wouldn't expect my high school chemistry students to know this. $\endgroup$ – Jason Patterson May 27 '15 at 17:45
  • $\begingroup$ I think it's upper my current level. "I wouldn't expect my high school chemistry students to know this" so It's not lack of effort from me ?. $\endgroup$ – Hoyt Volker May 27 '15 at 18:23
3
$\begingroup$

There are basically two processes, that make adding more hydroxide the wrong answer.

The first one is already described in Yomen's answer, but I would like to extend that a little bit further. If you dissolve iron (II) hydroxide in water, an aqueous complex will form: $$\ce{Fe(OH)2~(s) + 4H2O <=> [Fe(H2O)4(OH)2]}$$ Apart from this, due to the auto-ionisation of water and the relative stabilities of the complexes, there is a variety of complexes in solution, among others: $$\ce{[Fe(H2O)5(OH)1]- <=>C[H+] [Fe(H2O)4(OH)2] <=>C[OH-] [Fe(H2O)3(OH)3]+}$$ From these complexes, only the neutral one precipitates. From this equation it is also deducible, that the stability of the precipitate is highly pH dependent.

Upon adding more hydroxide ions, the equilibrium will shift more and more to the right, and even further, resulting in the dissolution of the precipitate $$\ce{[Fe(H2O)4(OH)2] <=>C[OH-] [Fe(H2O)3(OH)3]+ <=>C[OH-] [Fe(H2O)2(OH)4]^2+}$$

The other possibility is, that iron (II) will easily be oxidised in basic conditions and access to air. This process is often referred to as rusting.

$$\ce{4Fe(OH)2 + O2 + 2H2O <=>C[OH-] 4Fe(OH)3 \equiv 2Fe2O3(OH2)3}$$

Further reactions will form the more stable, but less hydrated iron (III) oxide: $$\ce{Fe2O3(H2O)3 <=> Fe2O3(H2O)_{$x$} + $x$ H2O} \quad (x\in[0,3])$$

$\endgroup$
0
1
$\begingroup$

I believe the question is attempting to get you to think about the stochiometry of the reaction.

The reverse reaction takes one mole of $\ce{Fe^{2+}}$ and two moles of $\ce{OH-}$. This means that if you add the same molar amount of $\ce{Fe(NO3)2}$ and $\ce{KOH}$ you would expect (roughly) half the amount of $\ce{Fe(OH)2(s)}$ formed from the $\ce{KOH}$ versus the $\ce{Fe(NO3)2}$, because it takes two (molar) units of $\ce{OH-}$ to get one unit of reaction to happen.

So both will cause the reaction to reverse because of Le Chatelier's principle, but adding $\ce{Fe(NO3)2}$ will be "more active" (on a mole/mole comparison basis) because of the stochiometry of the reaction.

I agree the question is poorly worded, though. Although using a molar comparison basis for things like this is pretty standard, it's not the only way to do it. (E.g. adding equal masses of $\ce{Fe(NO3)2}$ or $\ce{KOH}$ will result in the $\ce{KOH}$ solution having greater reverse reactions.) For completeness, how "equivalent" amounts of the various substances are determined should have been specified in the question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.