When Manganese (II) reacts with peroxodisulphate, the respective products obtained are:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → MnO4− and SO42−
The peroxodisulphate ion, \( \ce{S2O8^{2-}} \), contains a peroxide (O–O) link and is one of the strongest oxidising agents. When it oxidises something, the O–O link breaks and each ion takes up two electrons to give two sulphate ions:
\( \displaystyle \ce{S2O8^2- + 2e^- -> 2SO4^2-} \)
It is strong enough to take manganese from +2 all the way up to +7, so \( \ce{Mn^{2+}} \) is oxidised to the permanganate ion:
\( \displaystyle \ce{Mn^2+ + 4H2O -> MnO4^- + 8H^+ + 5e^-} \)
Balancing the electrons (10 given, 10 taken) gives the full balanced equation:
\( \displaystyle \ce{2Mn^2+ + 5S2O8^2- + 8H2O -> 2MnO4^- + 10SO4^2- + 16H^+} \)
So the products are MnO4− and SO42−.
Why the other options are incorrect
Option (1) MnO42− only — Incorrect. The oxidising agent must itself be reduced, so sulphate always forms as well. Manganate is also not the product in acidic solution.
Option (3) MnO2 and S2O42− — Incorrect. Peroxodisulphate is reduced to sulphate (S stays +6), not to dithionite (S +3), and \( \ce{MnO2} \) (+4) is not the end point of such a strong oxidant.
Option (4) MnO42− and SO42− — Incorrect. Manganate(VI) is stable only in alkaline solution; in acid it disproportionates into \( \ce{MnO4^{-}} \) and \( \ce{MnO2} \).
NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – …
