Balance the following redox reaction:

$$\ce{FeS2 + O2 -> Fe2O3 + SO2}$$

My Efforts:

  1. I tried balancing with oxidation number method.

First of all i determined the oxidation state as follows: $$\ce{Fe^{(2)}S2^{(-1)} + O2^{(0)} -> Fe2^{(3)}O3^{(-2)} + S^{(4)}O2^{(-2)} }$$

Here, all the element are either oxidized or reduced, so how to move ahead?

  1. I tried balancing with Half reaction method.

$$\ce{Fe^{(2)} -> Fe2^{(3)}} \text Oxidation$$ $$\ce{S2^{(-1)} -> 2S^{(4)} } \text Oxidation$$
$$\ce{O2^{(0)} -> 2O^{(-2)}} \text Reduction$$

Same here, how to move ahead?

P.S. Oxidation state are mentioned in the brackets. Correct me if i am wrong anywhere.

  • $\begingroup$ A couple things... Are you sure it's FeS2 and not FeS? Also, in the oxidation state method, the charges don't balance in FeS2. $\endgroup$
    – jerepierre
    Commented Sep 22, 2014 at 12:34
  • $\begingroup$ @jerepierre Yes i am sure it is FeS2. If you can show it by half reaction method then also it will work, because i have test tomorrow. $\endgroup$
    – Freddy
    Commented Sep 22, 2014 at 13:40

2 Answers 2


FeS2 has an odd structure; the iron atom has a +2 oxidation number and each of the sulfurs has a -1 oxidation number.

This can be balanced by inspection.

4 FeS2 + 11 O2 --> 2 Fe2O3 + 8 SO2

Just to check, using oxidation numbers we get:

Sulfur goes from -1 to +4. (Total change from sulfur = +40)

Iron goes from +2 to +3. (Total change from iron = +4)

(I'm not sure where you got 4 for the oxidation number of iron in FeS2.)

Oxygen goes from 0 to -2. (Total change from oxygen = -44)

ETA: Half reactions are messy here, since both iron and sulfur are oxidized. Similarly, using oxidation numbers is problematic (except to check the solution) because there are three substances changing oxidation states. You could write a system of equations to describe it, but that's a lot more trouble than it's worth.

You know from the structure of iron(II) disulfide that there are twice as many sulfur atoms as iron atoms. That means that the number of sulfur dioxide molecules must be four times the number of iron(III) oxide molecules. With that relationship in mind, the smallest ratio of molecules that fits the pattern is the one I wrote above.

If you really want to use oxidation numbers, here's what I've come up with:

Let a equal the number of iron atoms, b equal the number of sulfur atoms, and c equal the number of oxygen atoms.

Iron increases its oxidation state by 1, sulfur by 5, and oxygen decreases by 2. So:

1a + 5b - 2c = 0

We also know that iron and sulfur are in a 1:2 ratio because they come from pyrite.

2 a = b

Substitution yields:

1a + 5(2a) - 2c = 0

11a - 2c = 0 Iron and oxygen are in a 2:11 ratio.

Oxygen must be even since it comes as O2, so its smallest possible value is 22 atoms. Using the ratios listed above, we get 22 O: 4 Fe: 8 S.

ETA #2: Half Reactions

Fe+2 --> Fe+3 + e-

S-1 --> S+4 + 5 e-

O20 + 4 e- --> 2 O-2

We know from FeS2 that there must be twice as many sulfur atoms as iron, so the second equation has to be multiplied by two.

2 S-1 --> 2 S+4 + 10 e-

If we add together all of the species that are being oxidized (Fe and S) then we get an oxidation half reaction of:

Fe+2 + 2 S-1 --> Fe+3 + 2 S+4 + 11 e-

This must be multiplied by 4 and the reduction equation by 11 to balance the number of electrons.

4 Fe+2 + 8 S-1 --> 4 Fe+3 + 8 S+4 + 44 e-

11 O20 + 44 e- --> 22 O-2

Adding this mess together gives:

4 Fe+2 + 8 S-1 + 11 O20 --> 4 Fe+3 + 8 S+4 + 22 O-2

Since we know the structure of the molecules, putting it all back together is fairly straightforward, and it gives the same result as listed above.

  • $\begingroup$ I would be thankful if you can saw me by using oxidation number method or by half reaction method. Thank you for correcting my mistake. $\endgroup$
    – Freddy
    Commented Sep 22, 2014 at 14:06
  • 1
    $\begingroup$ Welcome to Chemistry.SE! Please have a look at this and the documentation for mhchem $\endgroup$ Commented Sep 22, 2014 at 14:07
  • $\begingroup$ Think of $\ce{FeS2}$ as the $\ce{Fe(II)}$ salt of hydrogen disulfide. In nature, you'll find it as Pyrite. Looks nice, unless you're in for the real gold. $\endgroup$ Commented Sep 22, 2014 at 14:39

Balancing With Half Reaction method

Step 1: Determine oxidation number of each element.

$$\ce{Fe^{(2)}S2^{(-1)} + O2^{(0)} -> Fe2^{(3)}O3^{(-2)} + S^{(4)}O2^{(-2)} }$$

Step 2: Determine total increase and decrease in oxidation number. Also maintain $\ce{Fe}$ and $\ce{S}$ ratio(1:2) on both side.

$$\ce{Fe^{(2)} -> Fe2^{(3)}} \text Oxidation -----------(A)$$ $$\ce{S2^{(-1)} -> 2S^{(4)} } \text Oxidation -----------(A)$$

$$\ce{O2^{(0)} -> 2O^{(-2)}} \text Reduction -----------(B)$$

Step 3: Balance Oxidation number of of reaction (A) and (B)

There are total 11 electron transfer take place in (A) and 4 electron in (B) $$4\ce{[Fe, S2] ->4[Fe, 2S] }$$

$$\ce{11[O2] -> 11[2O]}$$

Step 5: Finally combine above two reactions

$$\ce{4FeS2 + 11O2 -> 2Fe2O3 + 8SO2}$$


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