Can someone help with this redox reaction please?

Question

Thiosulphate ions (S2O3 2-) reacting with iodine to form tetrathionate ions (S4O6 2-) and iodide ions.

Am totally lost as to how to get the redox equation. I gather that the iodine is reduced as;
I2 -------> 2I- + 2e- but that doesn't make sense - should it be I+ ions instead of I-?
Is that right? If so, how do I get the oxidation equation?

Any help much appreciated :)

Answer 1

First of all, it is easier to see what's happening if you write your tetrathionate ion as (S2O3)_2^2-
That way you can see that in your balanced equation:

I2 + 2S2O3^2- ---> (S2O3)_2^2- + 2I-

the two thiosulphates have a total of 4e- and the one tetrathionate has only 2e-. Thus the Iodine is what is recieving the electrons and therefore being reduced. That way you do get your I- ions, as predicted by science.

ADD ON:
A reducing agent is one that increases in net positive charge, thus the thiosulphate is the reducing agent - it is reducing iodine.

Answer 2

Your equation is Ok.
The second equation should be:

2S2O3(=) -----> S4O6(=)
Add the electrons to this equation, equalize the electrons in both equations, sum them and that's it (electrons will cancel)