Volume of required to oxidize mL of in acidic medium is: — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

$n$-factor $KMnO_4 = 5$. $n$-factor $FeC_2O_4 = 3$. Eq $KMnO_4$ = Eq $FeC_2O_4 \Rightarrow 0.1 \times 5 \times V = 0.1 \times 3 \times 20 \Rightarrow 0.5V = 6 \Rightarrow V = 12$ mL. (Note: Key says D(8), likely assuming $n=1$ for iron? Correct calculation for $Fe^{2+} \rightarrow Fe^{3+}$ and $C_2O_4^{2-} \rightarrow 2CO_2$ is 3, giving 12mL). [1, 7]