Question

How do we quantitatively represent the oxidation of sodium thiosulfate, `S_2O_3^(2-)`, by dichromate ion, `Cr_2O_7^(2-)`?

Answer 1

This is a somewhat complicated redox equation. You should end with `Cr^(3+)` and sulfate ions.

Explanation:

`Cr(VI)` is reduced to `Cr(III)`:

`Cr_2O_7^(2-) +14H^(+)+ 6e^(-)rarr 2Cr^(3+) + 7H_2O(l)` `(i)`

Thiolsulfate is oxidized to sulfate:

`S_2O_3^(2-) + 5H_2O(l) rarr 2SO_4^(2-) + 10H^(+) + 8e^-` `(ii)`

Both equations (I think) are balanced with respect to mass and charge (as they must be!).

So I cross multiply to give the overall redox equation (I wish to remove the electrons!) and cancel common reagents:

`4xx(i)+3xx(ii)`:

`4Cr_2O_7^(2-) +26H^(+)+ 3S_2O_3^(2-) rarr 8Cr^(3+) + 6SO_4^(2-) + 13H_2O(l)`

So at the end of your titration, you should have `Cr^(3+)`, and sulfate ions; acidium ions will be in excess with these reactions. What should signal the endpoint in this redox reaction?

Note that in thiosulfate, I have always found it useful to regard the terminal sulfur as the precise analogue of oxygen, which has the `-II` oxidation state, whereas the central sulfur has a `VI^+` oxidation state, as in sulfate. The average sulfur oxidation state in thiosulfate is still `II^+`.