What is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid? [duplicate]








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[How to balance more complex redox reactions?](/questions/36970/how-to-balance-more-complex-redox-reactions)

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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}?




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}


 4

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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