In neutral or faintly alkaline solution, permanganate ions oxidise a manganous salt to manganese dioxide. The presence of zinc sulphate or zinc oxide catalyses the oxidation. The reaction is represented as:
\( \ce{2MnO4^-} + x\,\ce{Mn^2+} + \ce{2H2O} \longrightarrow y\,\ce{MnO2} + \ce{4H+} \)
What are the values of x and y in the balanced equation?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → x = 3, y = 5
In this reaction manganese is present on both sides in different oxidation states. Permanganate is reduced to MnO2 and the Mn2+ ion is oxidised to MnO2. The electrons lost must equal the electrons gained.
Oxidation state of Mn: +7 in MnO4−, +2 in Mn2+ and +4 in MnO2.
Each MnO4− gains 3 electrons (+7 → +4), so 2 MnO4− gain 6 electrons.
Each Mn2+ loses 2 electrons (+2 → +4). To give 6 electrons, 3 Mn2+ ions are needed, so x = 3.
Manganese atoms on the left = 2 + 3 = 5, so 5 MnO2 are formed and y = 5.
Check: oxygen 8 + 2 = 10 = 5 × 2; hydrogen 4 = 4; charge 2(−1) + 3(+2) = +4 on both sides.
\( \displaystyle \ce{2MnO4^- + 3Mn^2+ + 2H2O} \xrightarrow{\ce{ZnSO4}} \ce{5MnO2 + 4H+} \)
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “2. In neutral or faintly alkaline solutions:” … “(c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation: 2MnO4– + 3Mn2+ + 2H2O → 5MnO2 + 4H+”