I'm trying to balance the following redox equation. I think this is happening in acidic solution, the textbook doesn't specify anything more. Can you please help me understand if I got the half reactions correct? $$\ce{KMnO4 + FeSO4 + H2SO4 -> K2SO4 + MnSO4 + Fe2(SO4)3 + H2O}$$

I found out that manganese is reduced as the oxidation number goes from $+7$ to $+2$ and iron gets oxidized as the oxidation number goes from $+2$ to $+3$. I wrote the two half reactions, \eqref{Q:red} for the reduction and \eqref{Q:ox} for the oxidation. Did I get them right?

\begin{align} \tag{A}\label{Q:red}\ce{KMnO4 + H2SO4 &-> K2SO4 + MnSO4}\\ \tag{B}\label{Q:ox}\ce{FeSO4 &-> Fe2(SO4)3}\\ \end{align}

I started working on the reduction half-equation and this is what I came up with after balancing atoms and charge:

$$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 8 H2O}$$

Now comes the trouble. How do I balance iron and sulfur in the oxidation half-reaction \eqref{Q:ox}?

  • $\begingroup$ You may have marked this question as a duplicate but I still didn't understand what is it that I got wrong with the oxidation half-reaction, or why do I need to write the sulfuric acid twice in order to balance both half-reactions and obtain the correct answer. Can anyone care to explain? Please? $\endgroup$ – WobblyWindows Sep 11 '15 at 16:57
  • $\begingroup$ Well you can ignore all the spectator ions: $\ce{K+, SO4^2-}$ Your half equations are wrong because they don't balance. I would get rid of all the clutter and then balance the half equations using $\ce{H2O, H+}$ and $\ce{e-}$. $\endgroup$ – bon Sep 11 '15 at 17:46
  • $\begingroup$ You mean I should re-write the given reaction this way? $$\ce{(MnO4)^- +Fe^{2+} -> Mn^{2+} \,+Fe^{3+}} $$ …and write the reduction and oxidation half-reactions from there on? How would I then get the solution (see below, in a comment to @oryza's advice)? $\endgroup$ – WobblyWindows Sep 11 '15 at 22:21

Your first half reaction, for the reduction, is correct: $$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\label{red}$$

For the second half reaction, the oxidation, start by balancing iron: \begin{align} \ce{FeSO4 &-> Fe2(SO4)3 }\tag{2a}\\ \ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2b} \end{align}

Add $\ce{H2SO4}$ on the left so you can balance sulfur: \begin{align} \ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2c}\\ \ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d} \end{align}

Now balance the protons and electrons: \begin{align} \ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d}\\ \ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 + 2 H+ + 2 e-}\tag{2e}\label{ox}\\ \end{align}

Now add \eqref{red} and five times \eqref{ox} so that the electrons are equal on every side and coincidentally the protons also balance: \begin{align} \ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 &-> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\\ \ce{5 H2SO4 + 10 FeSO4 &-> 5 Fe2(SO4)3 + 10 H+ + 10 e-}\tag{$5\times$2e} \end{align}

And the final result is: \begin{align} \ce{2 KMnO4 + 8 H2SO4 + 10 FeSO4 &-> K2SO4 + 2 MnSO4 + 8 H2O + 5 Fe2(SO4)3}\tag3 \end{align}


Yes you are correct saying that manganese gets reduced from $+7$ to $+2$ and iron gets oxidised from $+2$ to $+3$.

However, if I were you, since they already gave you the complete redox equation, – it is just unbalanced –, I wouldn't bother with writing the half equations. I would just focus on balancing the equation that they have already given you.

Here is a guide for balancing most redox equations:

  1. Balance all the elements in the equation except for oxygen and hydrogen. For polyatomic ions that are spectator ions, just think as them as one big element.
  2. Check if oxygen is balanced. If it isnt, add water to the side which is missing oxygen.
  3. Check if hydrogen is balanced. If it isn't, add protons $(\ce{H+})$ to the side with doesn't have enough hydrogen.
  4. Now that all the elements have been balanced, make sure that the charges are balanced, too. If they are not, add electrons to the side which is more positive.
  5. Check your equation now to make sure that everything is balanced correctly.
  6. When you have balanced both half-reactions, add them so that electrons cancel on each side.

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