Question

In the reaction
$2KMn{O}_4+3H_{2}S{O}_4⟶K_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5\left(O\right)$
Equivalent weight of $KMn{O}_4$ is (when molecular weight of $KMn{O}_4=158$)

• A. $31.6$
• B. $52.7$
• C. $79$
• D. $158$

Answer 1

The correct option is A $31.6$

KMnO4 is a strong oxidising agent . In acidic medium, itself reduced to Mn2+ while in basic or neutral medium it reduced to MnO2, logic behind this is very simple , as acidic medium is also called as electron deficient medium . So there is no support from medium to reaction that why it compeleted its all deficiency of electron from other source (substrate) . While in basic medium (electron rich medium ) there is good support from the medium to reaction thatswhy it take only 3e from other source remaning electron provided by solution.

In acidic medium

MnO4- + 8H+ + 5e = Mn2+ + 4H2O

from this reaction we can deduce that Mn7+ reduced to Mn2+ so “ n factor” for this reaction is 5

Eq.wt = molecular weight / n factor

So for KMnO4 molecular weight is 158.g

So Eq.wt = 158/5 = 31.6 g per equivalent.