If $\ce{FeSO4.7H2O}$ is dissolved in an acidic environment, for example in 0.05 M aqueous solutions of methanoic or ethanoic acid, would that prevent it from getting oxidized if an oxidizing agent like bromine water was added to it? If so, how? Since this is not examining oxidation with respect to air or oxygen.

  • $\begingroup$ Diluted solutions of $\ce{FeSO4}$ are partially hydrolyzed, producing $\ce{[Fe(OH)(H2O)5]+}$ ion. In the presence of an acid in aqueous solution, the main ion containing iron is $\ce{[Fe(H2O)6]^2+}.$ This ion must be more resistant to oxidation than the partially hydrolyzed ion. $\endgroup$
    – Maurice
    Apr 3 at 10:08
  • $\begingroup$ @Maurice can you please provide the equations for the formation of the respective ions? $\endgroup$
    – user107975
    Apr 3 at 11:28
  • 1
    $\begingroup$ @user109975 You should be able to do it yourself, shouldn't you ? Ferric ions have much greater hydrolysis tendency than ferrous ions, what leads to decreasing ratio of ferric/ferrous ions and decreasing redox potential, making it some what less vulnerable to oxidation in acidic solutions. But is is not sufficient not to be oxidized by bromine. Carboxylic acids presence brings another factor, forming complexes with iron. See en.wikipedia.org/wiki/Iron%28III%29_acetate and en.wikipedia.org/wiki/Iron%28II%29_acetate $\endgroup$
    – Poutnik
    Apr 3 at 12:38

No, adding acid to a ferrous sulfate solution will not prevent its being oxidized by bromine water.

Fresh FeSO$_4$ solutions are light blue-green, almost colorless, but pick up O$_2$ readily:

Fe$^3$$^+$ + e$^-$ --> Fe$^2$$^+$ +0.770 V and

O$_2$ + 2 H$_2$O + 4e$^-$ --> 4 OH$^-$ +0.401 V

(The half cell voltages are only indicative of what's going on, because we are not at standard concentrations of 1 M.)

The light-colored solution quickly becomes yellowish and consumes all the oxygen in the vessel, and then sucks in more air and consumes that oxygen, and then things slow down a little, but diffusion still lets a bit more oxygen get in, and the "ferrous" becomes somewhat ferric, and the liquid becomes a muddy color and you wish you could preserve the nice blue-green color somehow.

You could preserve the ferrous if you could somehow complex it and keep it from oxidizing to ferric. But nature prefers to complex the ferric ion: smaller ion, bigger charge. Typical ferric stability constants are larger than ferrous by ten orders of magnitude.

It is instinctive to suppose that metals will oxidize in the presence of O$_2$, and if higher valences are possible, well, then, the metal will go as far as possible. But the oxidation of ferrous to ferric is unfavored (+0.770 V) and Fe$^2$$^+$ needs help from O$_2$. The reduction of water is also unfavored, but is driven forward by the consumption of OH$^-$ ions by precipitation by Fe$^+$$^3$, which also keeps the Fe$^+$$^3$ low.

LeChatelier predicted as much: by increasing the acidity, you reduce the availability of electrons from OH$^-$, and O$_2$ is more able to take electrons from Fe$^+$$^2$.

So you could say that adding acid stabilizes the ferrous ion, but what it really does is inhibit the oxygen molecule. Bromine is totally unaffected.

  • $\begingroup$ How about just adding metallic iron to the ferrous sulfate solution and, of course, keeping the solution in a sealed bottle to reduce oxygen issues? Ferric ions will certainly oxidize iron, with ferrous ions being the result. $\endgroup$
    – Ed V
    Apr 3 at 18:35
  • $\begingroup$ @Ed V: I like that approach to keeping the ferrous as Fe+2. Why have I never heard of it before? We put desiccants in solvents to keep them dry. Sodium in benzene. Iron in ferrous sulfate solution sounds great! $\endgroup$ Apr 3 at 19:53
  • $\begingroup$ We used to do it in labs I was in. One little problem: iron powder is readily available, and I even used to have two 5 pound bottles of it. But high purity iron is a bit harder to find. We used small pieces of high purity iron wire: just drop a clean piece in the solution. $\endgroup$
    – Ed V
    Apr 3 at 19:58

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.