Redox and Equivalent ConceptMedium
Question
Reduction of the metal centre in aqueous permanganate ion involves
Options
A.3 electrons in neutral medium
B.5 electrons in neutral medium
C.3 electrons in alkaline medium
D.5 electrons in acidic medium
Solution
In acidic medium
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
In neutral medium
MnO4- + 2H2O + 3e- → MnO2 + 4OH+
Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
In neutral medium
MnO4- + 2H2O + 3e- → MnO2 + 4OH+
Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.
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