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

Jan 3, 2016

This is a somewhat complicated redox equation. You should end with $C {r}^{3 +}$ and sulfate ions.

#### Explanation:

$C r \left(V I\right)$ is reduced to $C r \left(I I I\right)$:

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \left(l\right)$ $\left(i\right)$

Thiolsulfate is oxidized to sulfate:

${S}_{2} {O}_{3}^{2 -} + 5 {H}_{2} O \left(l\right) \rightarrow 2 S {O}_{4}^{2 -} + 10 {H}^{+} + 8 {e}^{-}$ $\left(i i\right)$

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:

$4 \times \left(i\right) + 3 \times \left(i i\right)$:

$4 C {r}_{2} {O}_{7}^{2 -} + 26 {H}^{+} + 3 {S}_{2} {O}_{3}^{2 -} \rightarrow 8 C {r}^{3 +} + 6 S {O}_{4}^{2 -} + 13 {H}_{2} O \left(l\right)$

So at the end of your titration, you should have $C {r}^{3 +}$, 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 $- I I$ oxidation state, whereas the central sulfur has a $V {I}^{+}$ oxidation state, as in sulfate. The average sulfur oxidation state in thiosulfate is still $I {I}^{+}$.