moles of a solution containing an ion require moles of for the oxidation of to in acidic medium. Wha — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1:** In acidic medium, $MnO_4^-$ is reduced to $Mn^{2+}$, so $n$-factor = 5. <br>**Step 2:** Equivalents of $MnO_4^-$ = $1.61 \times 10^{-3} \times 5 = 8.05 \times 10^{-3}$. <br>**Step 3:** For $A^{n+} \rightarrow AO_3^-$, change in O.S. = $(5 - n)$. <br>**Step 4:** Equivalents of $A^{n+}$ = $2.68 \times 10^{-3} \times (5 - n)$. <br>**Step 5:** Equating equivalents: $2.68 \times (5 - n) = 8.05 \Rightarrow 5 - n \approx 3 \Rightarrow n = 2$.