How much potassium permanganate is needed to completely oxidize ferrous oxalate?

Question

Amount of substance of $\ce{MnO4-}$ required to oxidize one mole of ferrous oxalate completely in acid medium will be:

(a) 7.5 moles

(b) 0.2 moles

(c) 0.6 moles

(d) 0.4 moles

In acidic medium, 1 mole of $\ce{MnO_4^-}$ accepts 5 moles of electrons. Oxalate is dibasic so will give 2 moles of electrons when oxidized to $\ce{CO2}$.

$\ce{Fe^2+ -> Fe^3+ + e-} $ so total 3 moles of electrons.

$\text{Amount of substance of } \ce{MnO_4^-}=\frac{3}{5}=0.6$

However, the given answer is 0.4 moles indicating that only oxalate is considered to be oxidized.

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Does $\ce{Fe^{2+} -> Fe^{3+} + e^-} $ actually occur in this reaction?

What would be the net reaction when $\ce{KMnO_4}$ is used(not just the ions)?

Answer 1

See here.

The "green salt" is $\ce{K3Fe(C2O4)3.3 H2O}$. \begin{align} \ce{MnO4- + 8H+ +5e- &-> Mn^2+ + 4 H2O}\\ \ce{C2O4^2- &-> 2 CO2 + 2e-}\\ \ce{2 MnO4^- + 16 H+ + 5 C2O4^2- &-> 2 Mn^2+ + 8 H2O + 10 CO2 (g)}\\ \end{align}

One expects all the oxalate is oxidized to $\ce{CO2}$ and $\ce{Fe(II)}$ is oxidized to $\ce{Fe(III)}$.