Is there an alternative to the half-reaction method to solving red-ox reactions?

Question

The equation for the reaction between a permanganate and iron(II) ions in acidic solution is:

$\text{MnO}_4^- + \text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + \text{Mn}^{2+}$

My book, as well as a certain equation balancer, shows that the answer is:

$5\text{Fe}^{2+} + \text{MnO}_4^- 8\text H^+ \longrightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text H_2\text O$

They check the charges to ensure that it is indeed balanced. My book says that:

$5(2+) + (1-) + 8(1+) = 17+$

$5(3+) + (2+) + 0 = 17+$

And voila, they must be therefore balanced. However, there must be more than one way to balance such a reaction, like many problems. Is there a different, perhaps a more interesting, method than the half-reaction method for balancing this reaction and others?

Answer 1

There is one other way that I have used, using a linear system to solve this reaction. Although it can be used for other more casual reactions too, such as the combustion of ethanol for example.

As you can see, in this special case (because the reaction is relatively easy to solve already) we see that there must be $H_2O$ molecules on the product side of the equation, because there is no oxygen on the products already, but there is oxygen in the reactants. So start by assuming the products include $H_2O$ with a coefficient of at least 4 (because there is 4 oxygen in permanganate ion).

Also, there is no hydrogen anywhere in the equation, so we can assume that to balance the water (which has at least 4 moles) there must be some hydrogen ions added as reactants to balance the reactants side (it is acidic conditions in this case). As it turns out, since there must be at least 4 moles of water, then there must be at least 8 moles of hydrogen ions to satisfy the convention of having all mole coefficients in the lowest terms.

Now we can start building the linear system:

Since we obviously do not know the coefficients yet, we will assign variables to them. Let a b c d e and f be the molar coefficients of each respective molecule in the following reaction we have established so far:

a$H^+$ + b$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + e$Mn^{2+}$ + f$H_2O$

Now how do we build the linear system for this reaction? We might do it as follows:

Starting with the hydrogen atom, simply because it comes first when reading it from left to right. We write this equation to represent the hydrogen atoms in our reaction:

a = 2f

It is as simple as it looks, there is one mole of hydrogen atoms per mole of hydrogen ions, and there is two moles of hydrogen atoms per mole of water molecules.

We can keep doing this for the rest of the elements in our reaction, until we have three more equations along with the one for hydrogen, because there are three more different elements to account for.

b = e This one is pretty straightforward... This accounts for Manganese atoms.

4b = f Oxygen atoms are now accounted for.

c = d Also straightforward, now iron atoms will be accounted for as well. Notice that c and d are lone variables, they do not show up with the other equations, but at least they must be equivalent

Remember that we established some mathematical axioms when we said that there must be at least four moles of water molecules to satisfy the one mole of permanganate ion. An axiom is just something that we assume to be true, such as f = 4 in this case.

Now look at the equation we established for hydrogen atoms, the first equation a = 2f. Since we are assuming that f = 4 we can therefore safely assume that a = 8 without going against our original axiom. Also if we assume that f = 4 we can also go ahead and solve the equation to account for oxygen atoms 4b = f and solving it gives us b = 1 and the rest is pretty straight forward. If b = 1 and b = e then e = 1.

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Given what we have found out so far, lets take a look at our equations again:

a = 8

b = 1

c = d

e = 1

f = 4

So,

8$H^+$ + 1$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + 1$Mn^{2+}$ + 4$H_2O$

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So now we can look at the charges to figure out the correct coefficients for iron(II) and iron(III) ions, which should be the same of course (c = d). As it stands currently (ignoring the iron ions), there is a total charge of 7+ on the reactants side, and 2+ on the products side. These should be the same for the equation to be balanced, so we can start by creating another axiom and saying that c > 1. If there are 2 moles iron then our charges would be 11 on the reactants and 8 on the products. Incrementing it to 3 would increment the charge on the reactants side by 2+ while the product's total charge would increment 3 due to the charge of the ions themselves. Eventually things are balanced when the moles for iron reach 5 making both sides have a total charge of 17 and everything is nicely balanced.

Answer 2

There is oxidation numbers method. Look it up. Shortly, you give any hydrogen atom (in a compound which also has one different element, aside from oxygen) +1 charge, any oxygen atom (in a compound which also has one different element, aside from hydrogen) -- -2 charge. Using these and molecule charge you determine unknown atom's charge. Difference between charges of the same element on the left and on the right should equal difference between charges of another element on the left and on the right, multiplied by -1. Then you balance a reaction so that charge is conserved. Of course, before that you need to use the information given to complete your reaction with compounds, that are initially absent. For example, in this case you see, that there should be oxygen on the right. Also it's said that reaction proceeds in the acidic solution, which means protons are used (add them on the left side), which means they will react with oxygen, which means on the right should be not oxygen, but water. Walkthrough: So we have $MnO^−_4+Fe^{2+}⟶Fe^{3+}+Mn^{2+}$. Obviously, we need to add $4O^{-2}$ on the right side and $H^+$ on the left side (and, consequently, on the right side too): $MnO^−_4+Fe^{2+}+H^+⟶Fe^{3+}+Mn^{2+}+4O^{-2}+H^+$. As we establish by the oxidation numbers, $Mn$'s charge (in $MnO_4^-$) equals $-1 - -2(4) = 7$. While on the right side it's charge is $+2$. $2 - 7 = -5$. Iron also changes it's charge, so we see, that we need to find a coefficient to make a difference between charges of Fe on the right and Fe on the left equal $+5$. That would be $5$: $MnO^−_4+5Fe^{2+}+H^+⟶5Fe^{3+}+Mn^{2+}+4O^{-2}+H^+$ Now only thing, that's left, is to add a coefficient to the hydrogen, so water can be formed: $MnO^−_4+5Fe^{2+}+8H^+⟶5Fe^{3+}+Mn^{2+}+4H_2O$.