I have been desperately trying to balance the following equation, and finally (ultima ratio) used an online program to get it done (posted the same question there as well). $$\ce{KMnO4 + CaC2O4 + H2SO4 -> MnSO4 + K2SO4 + CaSO4 + CO2 + H2O}\tag{I}$$

No worries, balancing the K, Mn, Ca, C and S - but by then the H and O got out of my control. So this is how far I got myself: $$\begin{multline}\ce{2KMnO4 + CaC2O4 + 4H2SO4 ->\\ 2MnSO4 + K2SO4 + CaSO4 + 2CO2 + 4H2O}\tag{II}\end{multline}$$

So the only thing left to be balanced, was the O, having 28 on the reagents' side and 24 on the product side. I started off, getting up to 36:33 O (reagents:products), and I felt that I was getting nowhere.

Anyone who would guide me through the steps following the balancing until equation II above?

  • 2
    $\begingroup$ You should balance this like any other redox: identify the oxidant, look how many electrons it requires, identify the reducing agent, look... $\endgroup$ Apr 11, 2016 at 10:21

1 Answer 1


This is a redox reaction. While it can be balanced by the method you describe, it is much more reliable to balance it by the half-reaction method.

Step 1 - Identify your spectator ions (so we do not spend time trying to balance them)

Three ions appear unchanged on both sides of the reaction: $\ce{K+},\ \ce{Ca^2+},\ \& \ \ce{SO4^2-}$

Step 2 - Identify your redox pairs.

You have one manganese species on each side. They must constitute one half-reaction:

$$\ce{MnO4- -> Mn^2+}$$

Likewise, you have on carbon species on each side. They must constitute the other half reaction:

$$\ce{C2O4^2- -> CO2}$$

For each half reaction:

Step 3 - Balance O by adding $\ce{H2O}$

Step 4 - Balance H by adding $\ce{H+}$

Step 5 - Balance charge by adding $\ce{e-}$

Step 6 - Combine the half reactions

Can you take it from here?


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