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When $\ce{Na2S2O3}$ reacts with $\ce{I2}$, it undergoes a redox reaction as
$$\ce{2 Na2S2O3 + I2 -> Na2S4O6 + 2 NaI}$$
Sulfur is being oxidized from a $+2$ state to a $+2.5$ state. Why isn't it getting oxidized any further, say to $3+$, to form a compound like $\ce{Na2HSO3}$?
I do notice that there's no hydrogen involved in the equation, but maybe the reaction occurs in an acidic medium?
Could someone please tell me whether the above scenario is possible or not and why?
Whether compound $\ce{X}$ is going to be reduced/oxidized and to what degree is mostly dictated by the corresponding redox potential for the given medium. The thiosulfate anion $\ce{S2O3^2-}$ is used in quantitative analysis and in iodometric titration in particular since it is a reasonably strong reducing agent [1 pp. 714-717]:
$$\ce{S4O6^2- + 2 e- <=> 2 S2O3^2-} \qquad E^\circ = \pu{0.169 V}$$ $$\ce{2 S2O3^2- + I2 -> S4O6^2- + 2 I-}$$
However, stronger oxidants such as chlorine are capable of oxidizing thiosulfate further down to sulfur(VI), e.g. forming a sulfate instead:
$$\ce{S2O3^2- + 4 Cl2 + 5 H2O -> 2 HSO4^- + 8 H+ + 8 Cl-}$$
Bromine, being intermediate between iodine and chlorine, can cause $\ce{S2O3^2-}$ to act either as a 1-electron or an 8-electron reducer according to conditions. For example, in an amusing and instructive experiment, if concentrated aqueous solutions of $\ce{S2O3^2-}$ and $\ce{Br2}$ are titrated, and the titration is then repeated after having diluted both the $\ce{S2O3^2-}$ and $\ce{Br2}$ solutions 100-fold, then the titre will be found to have increased by a factor of exactly 8.
Note, however, that it is not entirely correct to state that sulfur in thiosulfate has oxidation state $+2$. In fact, there are two nonequivalent $\ce{S}$ atoms with oxidation states -1 and +5. Same goes for tetrathionate anion $\ce{S4O6^2−}$ where two central $\ce{S}$ atoms have oxidation state $0$ and the two terminal ones $+5$.
