I'm having a hard time trying to figure out n factor of $\ce{SO2}$ in the reaction

$$\ce{FeS2 + O2 → Fe2O3 + SO2}\qquad \text{(unbalanced)}$$

To balance it, I saw that in 1 mole of $\ce{FeS2}$, it looses a net total of 11 moles of electrons, while $\ce{O2}$ gains 4 moles of electrons per mole of it. That means they will have to react in a $4:11$ ratio, from this, the equation balances out to

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

The question was (wrt to this equation), what will be the equivalent weight of $\ce{SO2}$, if the molecular weight of $\ce{SO2}$ is $M$?

This means I will need the $n$ factor of it to divide by $M$,

I tried to reverse this reaction

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

But the problem is that now it's hard to tell from which $\ce{O}$ atoms (of the now-products), the oxidation state is getting changed to $\pm0$ (zero). It's now confusing me.

The answer is given as $M/5$. I don't know why, according to the solution they only concidered the oxidation number change of $\ce S$ while ignoring the others, I don't think it's right.

  • $\begingroup$ All oxygen atoms undergo the same change in oxidation state. $\endgroup$
    – Jan
    Nov 25, 2017 at 14:19
  • $\begingroup$ I was thinking of the definition of n as the moles of electrons supplied or used up by a mole of substance in a reaction. For SO2, i saw the change in ON of S but for O, I got confused, even that will give electrons when the ON changes from -2 to 0. Why are we not considering that change in ON? Also which ones, as even Fe203 was formed from the same O2? $\endgroup$
    – Rick
    Nov 25, 2017 at 14:27
  • $\begingroup$ All oxygen atoms on the $\ce{O2}$ side of the reaction have an oxidation state of $\pm0$. All oxygen atoms on the other side of the reaction have an oxidation state of $\mathrm{-II}$. I am confused where you are confused. (Incidentally, is that iron(IV) sulphide or iron(II) disulphide? That is confusing me.) $\endgroup$
    – Jan
    Nov 25, 2017 at 14:31
  • $\begingroup$ It's pyrite Fe is in the +2 state and S_2 has 2- $\endgroup$
    – Rick
    Nov 25, 2017 at 14:38
  • $\begingroup$ So iron(II) disulphide (or should I call it bissulphide). Thanks! $\endgroup$
    – Jan
    Nov 25, 2017 at 14:38

1 Answer 1


As this is clearly a homework question, I will only point you in the right direction.

Balance the oxidation reaction:

$$\ce{S2^2- -> 2 SO2 + $n$e^-}$$

Balancing the charges will give you $n$. (I'll leave the balancing to the OP, as this is homework).

The equivalent mass of $\ce{SO2}$ in this reaction is then:


The factor $2$ is needed because the electronically balanced reaction has two molecules of sulphur dioxide in the equation.

  • $\begingroup$ Thanks , but even Fe^2+ is getting oxidized to Fe^3+, right? Why aren't we considering that? In your equation, balancing charges should give n = 2, but that's not matching. $\endgroup$
    – Rick
    Nov 24, 2017 at 18:15
  • $\begingroup$ You're looking at the oxidation of $S_2^{2-}$ to $SO_2$, not the ferro to ferric. The half reaction requires 6 electrons but you need to half that because there are 2 $SO_2$. So you get $n=3$ and $N=\frac{M}{3}$. You might want to look into what $n$ and the equivalent weight really stand for. $\endgroup$
    – Gert
    Nov 24, 2017 at 18:31
  • $\begingroup$ The $5$ comes from the fact that $S$ in $S_2^{2-}$ has $OS=-1$ and in $SO_2$ has $OS=4$, so the difference is $5$. But $S_2^{2-}$ contains 2 $S$, so you also need 2 $SO_2$. That then give a difference of $2\times 4-2\times(-1)=6$. So $5$ is wrong. $\endgroup$
    – Gert
    Nov 24, 2017 at 18:58
  • 1
    $\begingroup$ Well, actually you made a slight calculation mistake.. that last difference will be 10 so 5 per molecule. Thanks alot, I understood it now. I was confused about why we aren't involving the changes in OS of the other atom O, I guess it's because.. it's kinda like a spectator.. but it isn't right? , it's the one taking electrons in the reaction to oxidize FeS2. Why are we not considered it? (Sorry if this sounds like a stupid question... I mean, at one time I feel it's obvious since we are writing an oxidation half reaction , but I'm not able to think of an exact reason.) $\endgroup$
    – Rick
    Nov 24, 2017 at 20:10
  • 2
    $\begingroup$ For the half reaction of $\ce{S_2^{2-}}$ you still need to add the oxygen. The atoms must balance in a valid half reaction. So it should be $$\dfrac{1}{2}\ce{S_2^{2-} + O_2 -> SO2 + ne^-}$$ $\endgroup$
    – MaxW
    Nov 24, 2017 at 22:45

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