Equivalent weights of acting as an oxidant in acidic, neutral, and strong basic media respectively a — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

The n-factors of $KMnO_4$ in acidic ( $Mn^{2+}$ ), neutral ( $MnO_2$ ), and basic ( $MnO_4^{2-}$ ) media are 5, 3, and 1 respectively. Equivalent weight is M/n. Thus, the ratio is $M/5 : M/3 : M/1$ . Multiplying by 15 gives the ratio $3 : 5 : 15$ .