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How do we represent the oxidation of thiosulfate anion, #S_2O_3^(2-)# to tetrathionite, #S_4O_6^(2-)#...?

Answer 1

#"Thiosulfate to tetrathionite..........."#

#2S_2O_3^(2-) rarr S_4O_6^(2-)+2e^(-)# #;stackrel(II)Srarrstackrel(II*1/2)S#

I always considered #"thiosulfate dianion"#, #S_2O_3^(2-)#, to be the same as sulfate dianion, #SO_4^(2-)#, EXCEPT that one of the oxygen atoms has been replaced by congeneric sulfur. And this sulfur has PRECISELY the same oxidation state as the oxygen it replaces, i.e. #stackrel(-II)S#. The AVERAGE oxidation states is still.................................

#(S(-II)+S(+VI))/2=S(+II)#

All right, let us attempt to equalize the beast.

#2S_2O_3^(2-) rarr stackrel()S_4O_6^(2-)+2e^(-)# #;stackrel(II)Srarrstackrel(II*1/2)S#

Do you also think that its mass and charge are balanced?

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Answer 2

Cooper Adams

The oxidation of thiosulfate anion (S2O3^2-) to tetrathionate (S4O6^2-) can be represented by the following balanced chemical equation:

2 S2O3^2- (aq) + I2 (aq) + 2 H2O (l) → S4O6^2- (aq) + 2 I^- (aq) + 4 OH^- (aq)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.