# What indicators can be used when substituting iodine with other halogens in the Winkler method?

I'm a chemistry student who is doing an investigation with respect to the Winkler method. The Winkler method is one which is used to find the oxygen content in water. Primarily, the halide which is used is iodine. However, I want to make use of other halogens such as fluorine, chlorine and bromine, together with their respective halides. However, I am stuck on how I will actually do the titration, since the indicator used for the case of iodine is a starch indicator. I don't exactly know which indicator I should use for the other halides.

My question is this: What indicators should I use in the titrations involving the other halides?

• Note that the iodine is formed in solution from the iodide reagent. I'm pretty sure you don't want to go there with the other halides. The acidification step liberates iodine, and if other halogens acted analogously, you would then form the more volatile bromine, chlorine or fluorine gases. No thanks. I doubt any of those would work except for possibly bromide anyway. What's wrong with iodide? – airhuff Jan 20 '17 at 7:13
• – Klaus-Dieter Warzecha Jan 20 '17 at 8:00
• There's nothing much wrong with it, it's just that I'm just doing an investigation for other halides within the method. Like, just trying to see how other halides might affect the results obtained. – Abhijeet Vats Jan 20 '17 at 9:23

It will likely not work with other halides. As far as I understood the winkle method, the oxygen is first fixated with $\ce{Mn^2+}$ in basic conditions: \begin{align} \ce{Mn^2+ (aq) + 2 OH- (aq)&-> Mn(OH)2 (aq)}\tag{1}\label{alkalinemanganese}\\ \ce{2Mn(OH)2 (aq) + O2 (aq) &-> 2 MnO(OH)2 (s) v}\tag{2}\label{bindingoxygen}\\ \end{align}

The sample is acidified and iodide is added. This is oxidised to $\ce{I2}$, which forms a complex with remaining iodide. \begin{align} \ce{MnO(OH)2 (s) + H+ (aq) &-> Mn(OH)3 (aq)}\tag{3}\label{acidicmanganese}\\ \ce{2 Mn(OH)3 (aq) + 2 I- (aq) + 6 H+ (aq) &-> 2 Mn^2+ (aq) + I2 (aq) + 6 H2O}\tag4\label{iodide-ox}\\ \ce{I2 (aq) + I- (aq) &<=> I3- (aq)}\tag5\label{triiodide}\\ \end{align}

The $\ce{I3-}$ forms the indication complex with the starch, which can then be titrated with thiosulfate solution. When the colour is gone you have successfully converted all of the prepared $\ce{MnO(OH)2}$ back to manganese(II) ions. $$\ce{I3- (aq) + 2 S2O3^2- (aq) -> 3 I- (aq) + S4O6^2- (aq)}\tag6\label{triiodide-reduction}$$

The reason why this works is because you have the equilibrium $\eqref{triiodide}$ which produces the triiodide ion, which is relatively stable in aqueous solution. Iod by itself does not dissolve in water (much). This complex again forms a very colourful complex with starch, the switch can easily be detected by the human eye.

Polyatomic ions of other halogens are not unheard of (ref. Trichloride ion (c.f. triiodide ion) and therein), but they are a lot less stable. I am not aware that they form a complex with starch so they won't be that easily detected.

Another aspect, I don't want to go much deeper here, is that the redox reactions necessary $(\ref{iodide-ox},\ref{triiodide-reduction})$ for this procedure to work might not work at all with other halides.

The original method was proposed in the 19th century and is still applied today. It is a quite accurate and robust method. There is not really a reason to tamper with it.

• But suppose that I did want to try doing this experiment anyways, what kind of indicator should i use for each halogen? The thing here is that there may be no point in tampering with it but I'm still quite curious how this may work with other halogens and really want to do this investigation. – Abhijeet Vats Jan 20 '17 at 9:25
• @AbhijeetVats If the species necessary to detect does not form, there is no way you can use any other molecule to detect it. It simply is not possible. – Martin - マーチン Jan 20 '17 at 9:29