Question

${{\mathrm{SO}}_4}^{-2}\to {\mathrm{S}}_2{{\mathrm{O}}_8}^{-2}+2{\mathrm{e}}^{-}(\mathrm{ox}.)$How it is possible as there is no change in the oxidation state of $\mathrm{S}$ from sulphate ion to thionate i.e. +6 to +6

Answer 1

Oxidation state of $\mathrm{S}$ from sulphate ion to thionate:

1. So in ${\mathrm{S}}_2{{\mathrm{O}}_8}^{-2}$ ion, there is the presence of peroxy bond and so if we take the oxidation state of the Oxygen to be (-1) and as there is presence of two Oxygen atoms the charge would going to be (-2)
2. Hence, there would be a change in the oxidation state of the Oxygen atoms and not the Sulphur atoms
3. So, there would be the increase in the oxidation state from -2 to -1.
4. So, on calculating the oxidation state of the Sulphur in both ${\mathrm{S}}_2{{\mathrm{O}}_8}^{-2}$ and ${{\mathrm{SO}}_4}^{-2}$ would be equal to +6.
5. To find the oxidation state of Sulphur in ${\mathrm{S}}_2{{\mathrm{O}}_8}^{-2}$ , let the assumption be the oxidation state of Sulphur be X .and the total of eight Oxygen atoms are there.
6. Among these the two will involve the peroxide linkage with the oxidation state of the (-1) and that of the remaining six Oxygen atoms will have the oxidation state as (-2).
7. Hence, $$\begin{gathered} 2\times \mathrm{X}+2\times (-1)+6\times (-2)=-2 \\ \mathrm{X}=+6 \end{gathered}$$
8. Hence, the oxidation state of the Sulphur in the given ions would be +6.
9. Structure for Sulphate and Peroxodisulphateion ion respectively are:

and

Therefore, The change in oxidation is not due to the Sulphur atom but due to the Oxygen atom present in them. Sulphur oxidation state is +6 in both ${\mathrm{S}}_2{{\mathrm{O}}_8}^{-2}$ and ${{\mathrm{SO}}_4}^{-2}$