The number of moles of reduced by one mole of in alkaline medium is: — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

In alkaline medium, $MnO_4^-$ is reduced to $MnO_2$ ($+7 \rightarrow +4, n=3$) and $I^-$ is oxidized to $IO_3^-$ ($-1 \rightarrow +5, n=6$). <br>Equating equivalents: $3 \times \text{moles}(KMnO_4) = 6 \times \text{moles}(KI)$. <br>For 1 mole $KI$, moles of $KMnO_4 = 6/3 = 2$.