What is the n-factor of ferrous oxalate ( ) when it undergoes complete oxidation in an acidic medium — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

In ferrous oxalate, both the cation and the anion undergo oxidation. $Fe^{2+}$ oxidizes to $Fe^{3+}$ , losing 1 mole of electrons. The oxalate ion ( $C_2O_4^{2-}$ ) oxidizes to $2CO_2$ , losing 2 moles of electrons. Total moles of electrons lost per mole of $FeC_2O_4 = 1 + 2 = 3$ . Therefore, the n-factor is 3.