Question
Question: Calculate the equivalent weight of \(KMn{{O}_{4}}\) in acidic medium....
Calculate the equivalent weight of KMnO4 in acidic medium.
Solution
Hint: We know that different species tend to behave differently in different mediums, so based on the medium KMnO4 act as an oxidiser in an acidic medium. Equivalent weight is the weight per equivalent that reacts. An equivalent is a reacting packet. It is an amount that will react.
Complete step by step answer:
We know that KMnO4 is a strong oxidising agent, meaning it will get reduced to oxidise another species.
Equivalent weight of a substance (oxidant or reductant) is equal to the molecular weight divided by the number of electrons lost or gained by one molecule of the substance in redox reaction.
Equivalentweightofoxidisingagent=NumberofelectronsgainedbyonemoleculeMolecularweight=ChangeinOxidationNumberMolecularWeight
EquivalentweightofReducingagent=NumberofelectronslostbyonemoleculeMolecularweight=ChangeinOxidationNumberMolecularWeight
In acidic medium
MnO4− + 8H+ + 5e− = Mn2++4H2O
From this reaction we can deduce that Mn7+ reduced to Mn2+so number of electrons gained for this reaction is 5
Equivalentweight = n factormolecular weight
So, for KMnO4 molecular weight is 158 gram
Equivalentweight = 5158=31.6perequivalent
So, our answer is 31.6 per equivalent.
Note: We should always keep a check about n factor, which is decided by the factor from number of electrons gained or lost during reaction in acidic and basic medium. We should revise concepts about determining n factors of various chemical species.
We should know about Potassium permanganate. When we observe potassium permanganate, it appears as a purplish colour crystalline solid. It is non-combustible. We should be careful that it doesn’t come in contact with sulphuric acid, as result in spontaneous ignition.