The product obtained from the electrolytic oxidation of acidified sulphate solutions, is:

We start with an aqueous, strongly acidified sulphate solution. In such a medium the dominant anion is the bisulphate ion, written as $$HSO_4^-$$.

At the anode of the electrolytic cell this bisulphate ion undergoes oxidation. First we write the oxidation half-reaction that is always quoted in standard texts:

$$2\,HSO_4^- \;\longrightarrow\; S_2O_8^{2-} + 2\,e^- + 2\,H^+$$

This tells us that two $$HSO_4^-$$ ions lose a total of two electrons and combine to give the peroxodisulphate ion $$S_2O_8^{2-}$$, while also releasing two protons. The key feature is the formation of the peroxy O-O linkage between the two $$SO_4$$ groups.

Because the solution is already acidified, the $$S_2O_8^{2-}$$ ions will immediately pick up the two available protons $$2\,H^+$$ (produced in the same reaction and present from the external acid) to give the free acid form. We therefore write:

$$S_2O_8^{2-} + 2\,H^+ \;\longrightarrow\; H_2S_2O_8$$

The molecular formula $$H_2S_2O_8$$ corresponds to peroxydisulphuric (Marshall’s) acid. To make its structure absolutely clear, we expand it, showing each sulphur atom bonded to three terminal oxygen atoms and sharing a central peroxy oxygen-oxygen bridge:

$$H-O_3S-O-O-SO_3-H$$

Written more compactly, but preserving the connectivity, this becomes $$HO_3SOOSO_3H$$.

Now we match this structural formula with the options given in the question. Option B is exactly $$HO_3SOOSO_3H$$, whereas the other options represent different species (bisulphate ion, sulphurous/peroxomonosulphuric variants, etc.).

Hence, the correct answer is Option B.