Question:

Write the ionic equations for the oxidizing action of $\mathrm{MnO_4^-}$ in acidic medium for :
(I) $\mathrm{Fe^{2+}}$ ion
(II) $\mathrm{C_2O_4^{2-}}$ ion

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MnO₄− → Mn²⁺ (gains 5e−) in acid, oxidising Fe²⁺ and C₂O₄²−.
Updated On: Jun 16, 2026
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Solution and Explanation

(I) answer: In acidic medium permanganate $\mathrm{MnO_4^-}$ is a strong oxidiser. It takes 5 electrons and is itself reduced from Mn(+7) to $\mathrm{Mn^{2+}}$, using $\mathrm{H^+}$ ions to mop up its oxygen as water. It oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$. Balancing electrons, one $\mathrm{MnO_4^-}$ oxidises five $\mathrm{Fe^{2+}}$ ions:\[ MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \]

(II) answer: In the same way $\mathrm{MnO_4^-}$ oxidises oxalate $\mathrm{C_2O_4^{2-}}$ to carbon dioxide, while it is reduced to $\mathrm{Mn^{2+}}$. Balancing the electrons gives two permanganate ions reacting with five oxalate ions:\[ 2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O \]
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