the number of moles of kmn o4 that will be needed to react with one mole of sulphite ion in acidic solution is?
please give me some tricks to solve this kind of problem BECOZ I'm getting confused.
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Answers
Hey it's too simple dear!
Basically in whole syllabus you just read three main equation of KMnO4 : -
(1)KMnO4 (Mn in 7+) ----> Mn(2+)
in acidic medium
(2)KMnO4 (Mn in 7+) ---> MnO2 (Mn in +2)
in aqueous medium
(3)KMnO4 (Mn in 7+) ---> MnO2 ( Mn in +2)
in weakly basic medium
(4)KMnO4 (Mn in 7+) ----> MnO4(2-)
In basic medium
Now sulphite ion undergoes oxidation to form sulphate ion in presence of strong oxidizing agents like KMnO4 or K2Cr2O7, etc
SO3(2-) (S in 4+) ---> SO4(2-) (S in 6+)
Now ,
Reactions involved are : -
SO3(2-)(aq) + 2H(+)(aq) ---> SO4(2-)(aq) + 2e(-)(aq) + H2O(l)
MnO4(-)(aq) + 5e(-)(aq) + 8H(+)(aq) ----> Mn(2+)(aq) + 4H2O(l)
When 1 mole sulphite ion convert into sulphate ion , 2 moles e- are released
When 1 mole MnO4- converts into Mn2+ , 5 mole e- are needed
So ,
5 mole e- ---> 1 mole MnO4-
2 mole e- ---> 2/5 mole MnO4-
Answered by an IIT JEE ASPIRANT and all India mathematics OLYMPIAD TOPPER in class 10th