Number of moles of $\mathrm{MnO}_4^{-}$required to oxidize one mole of ferrous oxalate completely in acidic…

Number of moles of $\mathrm{MnO}_4^{-}$required to oxidize one mole of ferrous oxalate completely in acidic medium will be
  1. 0.2 moles
  2. 0.6 moles
  3. 0.4 moles
  4. $7.5 \mathrm{moles}$

Solution

$\begin{aligned} & \left.\mathrm{FeC}_2 \mathrm{O}_4 \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{CO}_2+3 \mathrm{e}^{-}\right] \times 5 \\ & \left.\mathrm{MnO}_4{ }^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 3 \\ & 5 \mathrm{FeC}_2 \mathrm{O}_4+3 \mathrm{MnO}_4^{-}+24 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+10 \mathrm{CO}_2+3 \mathrm{Mn}^{2+}+12 \mathrm{H}_2 \mathrm{O} \\ & 5 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv 3 \mathrm{~mol} \mathrm{MnO}_4^{-} \\ & \therefore 1 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv \frac{3}{5} \mathrm{~mol} \mathrm{MnO}_4^{-} \\ & =0.6 \mathrm{~mol} \mathrm{MnO}_4^{-} \\ & \end{aligned}$

Asked in: NEET 2008 (Mains)

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