Question

In the chemical reaction between stoichiometric quantities of KMnO_4 and KI in weakly basic solution, what is the number of moles of I_2 released for 4 moles of KMnO_4 consumed?

Answer 1

In weak basic solution, permanganate ion converts into MnO₂ (reduction of potassium permanganate) and iodine ion converts into iodide ( oxidation of KI)

i.e., chemical reaction is written as ..

MnO₄¯ + I¯ ⇔ MnO₂ + I₂

so equivalents of potassium permanganate = equivalents of I₂

given no of moles of potassium permanganate = 4

let n moles of I₂ released for 4 moles of potassium permanganate consumed.

we know, equivalents = no of moles × n - factor

so no of moles of KMnO₄ × n - factor of it = no of moles of I₂ × n - factor of it

⇒4 × 3 = n × 2

⇒n = 6

Therefore 6 moles of I₂ released for 4 moles of KMnO₄ consumed.