How much potassium permanganate is needed to completely oxidize ferrous oxalate? 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+ -&gt; 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.

Does  $\ce{Fe^{2+} -&gt; Fe^{3+} + e^-} $  actually occur in this reaction?
What would be the net reaction when $\ce{KMnO_4}$ is used(not just the ions)?
See here.

The "green salt" is $\ce{K3Fe(C2O4)3.3 H2O}$.
\begin{align}
\ce{MnO4- + 8H+ +5e- &amp;-&gt; Mn^2+ + 4 H2O}\\
\ce{C2O4^2- &amp;-&gt; 2 CO2 + 2e-}\\
\ce{2 MnO4^- + 16 H+ + 5 C2O4^2- &amp;-&gt; 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 $\ce{Fe(III)}$.