This might sound like a stupid question, but I have been unable to wrap my head around it. Recently in a lab I had to titrate H2C2O4 with KMnO4 in acidic medium (H2SO4). However, in the work, it requested for a few drops of a diluted MnSO4 solution. What role does it play in the reaction? It appears as one of the products of the reaction, but doesn't seem to play an active role in the reaction itself.
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Both the oxidating ion $\ce{MnO4-}$ and reducing ion $\ce{C2O4^2-}$ are anions and therefore electrostatically repulsing each other, what makes their reaction slow.
$\ce{Mn^2+}$ ions serve as catalyst, being cycled between $\ce{Mn(II)}$ and $\ce{Mn(III)}$ oxidation states. As manganese ions have the opposite charge that both reagents, it is easy to understand it must speed up the reaction a lot.
Note that the formal standard reductions potential of $\ce{Mn^3+}$ is comparable with the one for $\ce{MnO4-}$.
\begin{align} \ce{4 Mn^2+ + MnO4- + 8 H+ &-> 5 Mn^3+ + 4 H2O}\\ \ce{2 Mn^3+ + C2O4^2- &-> 2 Mn^2+ + 2 CO2} \end{align}
As the parallel mechanism is also possible:
\begin{align} \ce{3 Mn^2+ + 2 MnO4- + 2 H2O &-> 5 MnO2(s) + 4 H+}\\ \ce{MnO2(s) + C2O4^2- + 4 H+ &-> Mn^2+ + 2 CO2 + 2 H2O} \end{align}
See also the Pourbaix diagram for manganese from wikipedia:
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$\begingroup$ Good explanation but in acidic solution the parallel mechanism is doubtful $\endgroup$– jimchmstCommented Dec 6, 2023 at 20:34
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