# What is the n-factor of ferric sulfide when it reacts with oxygen to give ferrous sulfate and sulfur dioxide?

What is the n-factor of $$\ce{Fe2S3}$$ in this reaction?

$$\ce{Fe2S3 + O2 -> FeSO4 + SO2}$$

I broke this reaction down into two parts

$$\ce{Fe2S3 -> FeSO4}$$ $$\ce{Fe2S3 -> 3 SO2}$$

And in the second case the change in electrons is 18. But how do we find it in the first reaction?

I think this is a hypothetical reaction. The answer is given to be 20.

• Even if we assume that Fe2S3 is a normal thing and not a typo, I don't believe the reaction will produce FeSO4. Also, 18 looks rather cryptic to me. Why 18? Jun 7 at 6:34
• @IvanNeretin i guess it produces FeO; but anyways, how do we still calculate overall n-factor here for Fe2S3 here? These types of questions do occur in our UG exams like JEE. So I wish this question gets an answer! Jun 7 at 6:53
• I don't believe in FeO either. Jun 7 at 6:54
• @IvanNeretin Well, disregarding what the product should have been and restricting the relevance of the reaction to 'hypothetical purpose only' ,now let us expect an answer! Jun 7 at 7:08
• Then I guess $n$ is the number of electrons transferred. Jun 7 at 7:52

Please note that n-factor is an outdated concept. See: The definition of n-factor, what was it when it was introduced?

We first need to find out which definition of n-factor we need to use here. This is simple, considering that the reaction is a redox reaction. Thus, the definition we use is the number of electrons lost or gained by the compound.

Therefore, we find the need to balance the equation.

Upon balancing, we see that the equation transforms into

$$\ce{Fe2S3 + 5 O2 -> 2 FeSO4 +SO2}$$

This can now be done from two points of views.

### POV 1: From $$\ce{Fe2S3}$$

When you look at the reaction from the point of view of $$\ce{Fe2S3}$$, we see that the main reaction is actually the combination of two reactions

$$\ce{Fe^3+ + S^2- -> 2Fe^2+ + 2S^6+}\tag{1}$$ $$\ce{ S^2- -> S^4+ }\tag{2}$$

Note: I do not care about the balancing of these equations since that has already been done and is not the focus of this question. Here only $$\ce{Fe2S3}$$ is important and so we tunnel vision.

Now, in half-reaction (1), Fe gains two electrons, and sulfur loses $$(8 \times 2)$$. Therefore net change in half-reaction(1) = -16 + 2 = -14.

Now, in half-reaction (2), S loses 6 electrons. Therefore net change in half-reaction (2) = -6

Therefore net change from the POV of $$\ce{Fe2S3}$$ is 20 electrons.

Our second POV is much simpler and also uses the charge neutrality concept - electrons lost by one compound are gained by the other in the reaction.

### POV 2: From $$\ce{O2}$$

From the POV of $$\ce{O2}$$, we only have one reaction.

$$\ce{5O2 -> 10 O^2-}$$

Therefore the net change in electrons from the POV of $$\ce{O2}$$ is 20 electrons gained.

By the principle of charge neutrality, this means that $$\ce{Fe2S3}$$ loses 20 electrons.

Therefore from both POVs, we get the same result. The n-factor of $$\ce{Fe2S3}$$ is 20.