In Analytical Chemistry, I was asked to determine the iron content of a screw.

I thought it could be solved in the following way:

  1. Treat the screw with a solution of sulfuric acid or hydrochloric acid.

  2. As the iron in solution can be found in the form of $\ce{Fe^{II}}$ or $\ce{Fe^{III}}$, I would reduce the possible $\ce{Fe^{III}}$ ions with $\ce{SnCl2}$ or Walden's reducer.

  3. Titrate the solution with potassium permanganate.

  4. From the volume spent, the amount of iron present could be determined.

I don't know if it will be OK, it is the first time I have faced a problem of this type.

  • $\begingroup$ 0. Weigh the screw. $\endgroup$ – imalipusram Apr 24 '20 at 12:34
  • $\begingroup$ Yes, you're right on that! @imalipusram $\endgroup$ – aprendiendo-a-programar Apr 24 '20 at 12:47
  • 1
    $\begingroup$ First, use sulfuric acid, to avoid Cl- ions. Second, If you titrate the solution within one hour after the end of the acidic dissolution, the amount of ferric ions formed will be negligible. And you can immediately titrate with $\ce{KMnO_4}$ without using Walden's reducer. I have done it in my classes for years without trouble about ferric oxidation. $\endgroup$ – Maurice Apr 24 '20 at 15:38
  • $\begingroup$ Weigh the screw and deduct 0.7 % for Mn, Si, C, and few tramp elements. $\endgroup$ – blacksmith37 Apr 24 '20 at 19:59

There is another problem with this.

Potassium permanganate is a very strong oxidizing agent, which is capable of oxidizing Chloride ions to Chlorine. This will result in additional consumption of $\ce{KMnO4}$ solution, which makes you think there is more iron in your screw than there actually is.

To solve this, you should look up the titration of Iron using the Reinhardt-Zimmermann solution. It contains sulfuric acid, Manganese (II) sulfate and phosphoric acid.

The $\ce{MnSO4}$ changes the redox potential so the acidic $\ce{KMnO4}$ solution can no longer oxidize Chloride ions. Because Iron(III) gives an intense yellow solution, the addition of Phosphoric acid is necessary since it will react with Iron(III) to colorless Iron phosphates. That makes it easier to detect the end point.

So the way I did this back in my first semester was to make sure we have all Iron as Fe(II) by adding an excess of $\ce{SnCl2}$. The excessive $\ce{SnCl2}$ was quenched using $\ce{HgCl2}$. Then the Reinhard-Zimmermann solution was added and titration with $\ce{KMnO4}$ could start.

  • $\begingroup$ Thank you very much. I really like analytical chemistry. I am surprised by the amount of knowledge you have, would you recommend me any book to deepen? Because, for example, in the same problem I am also asked to explain how to determine the lactose content of a commercial milk using titrations or gravimetries. $\endgroup$ – aprendiendo-a-programar Apr 24 '20 at 15:26
  • $\begingroup$ Nice method, which textbook did you use? Today, it will generate a lot of fuss if someone teaches to use mercury salts in titrations. $\endgroup$ – M. Farooq Apr 24 '20 at 17:12
  • $\begingroup$ You almost have the chemistry right, you miss the part about radical formations. For those interested in the likely actual reaction mechanics and not century-old chemistry, see my now edited answer on the suggested underlying chemistry below. $\endgroup$ – AJKOER Apr 25 '20 at 13:30

A simple, safe and effective electrochemical based path (absence strong acids which, acting by themselves, are generally inert on stainless steel at intermediate temperatures) to gather surface metal salts for testing is, per a source, to quote:

The iodine uptake, and corrosion of the stainless steel surfaces was extensive. The observed time-dependent behaviour suggests that the iodine adsorption on the stainless steel surface occurred in two stages, a slow induction period followed by a fast linear adsorption rate region. Iodine deposited on the stainless steel was desorbed during purging with air, but not with N2. These observations could be explained by the formation of metal-oxy-iodide scale, in which iodine is initially incorporated into but subsequently replaced by oxygen. This paper presents the preliminary results of the study, and their potential implication for "calibration" of iodine transmission through stainless steel sampling lines.

Some hot I2 vapor based corrosion on a stainless steel screw in an atmosphere of oxygen is a path to gathering surface formed oxy-iodides. The latter can be used to assess metal content of the alloy which is mainly Fe, but significant alloyed elements of Ni, Cr, Cu,..., also possible.

Also, in time, even very dilute iodized water is problematic on stainless steel, to quote NASA:

The effects of stainless steel exposure to iodinated water is a concern in developing the Integrated Water System (IWS) for Space Station Freedom.

Interestingly, the rate of corrosion may be reduced (or, even possibly accelerated) in the presence of chromium and nickel, to quote a source:

Iodine is capable of reacting with iron in the steel, as well as with nickel and chromium. This serves to complicate the situation. As well, there may exist synergistic relationships between these metals. This means that the presence of one metal may inhibit or enhance the reactions of iodine with one of the other component metals. Hellmann and Funke [19] performed a literature review of relevant reactions on this topic.

[EDIT] I found a good albeit dated (like 1913) reference of on the Zimmermann-Reinhardt (ZM) method here. I must correct, however, my misperception above that, somehow, a combination of H2SO4 and 85% H3PO4 in the presence of ferrous or stannous constitutes a 'strong' acid mix that attacks stainless steel. The less appealing truth, having an advanced knowledge of the reaction mechanics, KMnO4 (a substitute for H2O2), fosters powerful acidic radical formations. This is the actual basis of ZM method (which some may cite as historical, and others, as inefficient, reagent exhausting, and possibly dangerous route to the sulfate radical anion ($\ce{.SO4-}$) and likely the phosphate radical anion as well).

Now while, I have proclaimed elsewhere that the nitrate radical anion (see comments and source here), for example, exceeds the nitrating ability of concentrated Nitric acid, this was not well received. Here, the sulfate radical anion apparently assists in the attack of stainless steel and its creation, for the record, does not require strong H2SO4 or HgCl2 or KMnO4, just a source of hydroxyl radicals via a Fenton/Fenton-like reaction with transition metals in their lower valence states at an appropriate pH, or a photocatalyst creating an electron-hole in the presence of water, introducing the powerful hydroxyl radical ($\ce{.OH}$), which furthers reacts with the nitrate or sulfate or chloride ion as follows:

$\ce{ .OH + SO4(2-) -> OH- + .SO4- }$

$\ce{ .OH + NO3- -> OH- + .NO3 }$

$\ce{ .OH + Cl- -> OH- + .Cl }$

$\ce{ .Cl + .Cl -> Cl2 (g) }$

However, apparently couched in the veil of nostalgia ZM is applauded, congratulations.

For those actually interested in modern chemistry, I quote some revealing passages from the cited ZM source:

If dilute permanganate is allowed to run into a cold dilute solution of ferrous chloride containing hydrochloric acid, the permanganate is decolorized and the iron is oxidized, but there is a noticeable evolution of chlorine...If, however, permanganate is run into cold dilute hydrochloric acid in the absence of ferrous chloride, there is no evolution of chlorine.

So, an acidic promoted Fenton/Fenton-like reaction with ferrous (or stannous, also mentioned in this source work):

$\ce{ Fe(2+) + H2O2 (or KMnO4) + H+ -> Fe(3+) + .OH + H2O }$

and the creation of chlorine as detailed above. The sulfate radical itself is likely a stabilized transport for the powerful, but transient, hydroxyl radical, to quote source:

It has also been suggested that the sulfate radical anion may be converted to the hydroxyl radical in aqueous solution.

So, I speculate that the attack of steel may be due to an aqueous sulfate radical assisted release of the hydroxyl radical on the iron surface:

$\ce{Fe + 2 .OH -> Fe(OH)2}$


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