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Question

Question: Calculate the equivalent weight of \(KMn{{O}_{4}}\) in acidic medium....

Calculate the equivalent weight of KMnO4KMn{{O}_{4}} in acidic medium.

Explanation

Solution

Hint: We know that different species tend to behave differently in different mediums, so based on the medium KMnO4KMn{{O}_{4}} 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 KMnO4KMn{{O}_{4}} 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=MolecularweightNumberofelectronsgainedbyonemolecule= MolecularWeightChangeinOxidationNumber \begin{aligned} & Equivalent\,weight\,of\,oxidi\sin g\,agent=\frac{Molecular\,weight}{Number\,of\,electrons\,gained\,by\,one\,molecule}= \\\ & \frac{Molecular\,Weight}{Change\,in\,Oxidation\,Number} \\\ \end{aligned}

EquivalentweightofReducingagent=MolecularweightNumberofelectronslostbyonemolecule= MolecularWeightChangeinOxidationNumber \begin{aligned} & Equivalent\,weight\,of\,\operatorname{Re}ducing\,agent=\frac{Molecular\,weight}{Number\,of\,electrons\,lost\,by\,one\,molecule}= \\\ & \frac{Molecular\,Weight}{Change\,in\,Oxidation\,Number} \\\ \end{aligned}

In acidic medium
MnO4 + 8H+ + 5e = Mn2++4H2OMn{{O}_{4}}^{-}\text{ }+\text{ }8{{H}^{+}}\text{ }+\text{ }5{{e}^{-}}\text{ }=\text{ }M{{n}^{2+}}+4{{H}_{2}}O
From this reaction we can deduce that Mn7+M{{n}^{7+}} reduced to Mn2+M{{n}^{2+}}so number of electrons gained for this reaction is 5
Equivalentweight = molecular weight  n factorEquivalent\,weight\text{ }=\text{ }\frac{molecular\text{ }weight\text{ }}{\text{ }n\text{ }factor}
So, for KMnO4KMn{{O}_{4}} molecular weight is 158 gram
Equivalentweight = 158 5=31.6perequivalentEquivalent\,weight\text{ }=\text{ }\frac{158}{\text{ 5}}=31.6\,per\,equivalent
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.