When $\ce{H2SO4}$ is dilute,

$ \ce{Fe +H2SO4 -> FeSO4 + H2} $

and when $\ce{H2SO4}$ is concentrated,

$\ce{3Fe +8H2SO4 -> FeSO4 + Fe2(SO4)3 + 4SO2 +8H2O}$

What made the reaction with dilute acid different from the reaction with concentrated acid?

How did the second equation form, and why there are 2 salts - one with Fe[II] and the other with Fe[III]?

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    $\begingroup$ Well, the second equation has the formation of $\ce{SO2}$. How do you think this came about? $\endgroup$ Commented Oct 10, 2015 at 23:27
  • $\begingroup$ related chemistry.stackexchange.com/questions/15612/… chemistry.stackexchange.com/questions/19897/… $\endgroup$
    – Mithoron
    Commented Oct 10, 2015 at 23:29
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    $\begingroup$ @orthocresol i don't know $\endgroup$ Commented Oct 10, 2015 at 23:31
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    $\begingroup$ @bigworld12 Okay.. think about the oxidation states of S in H2SO4 and SO2. What has happened to sulfur? $\endgroup$ Commented Oct 11, 2015 at 10:11
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    $\begingroup$ @bigworld12 Yeah. Actually it's +6 to +4 but close enough. I'll type up an answer. Hold on. $\endgroup$ Commented Oct 11, 2015 at 10:18

3 Answers 3


In the first equation, $\ce{Fe}$ is oxidised to $\ce{Fe^2+}$ by $\ce{H+}$; the $\ce{SO4^2-}$ ions play no role in the reaction. You could replace $\ce{H2SO4}$ with $\ce{HCl}$ and get the same reaction.

In the second equation, the same oxidation of $\ce{Fe}$ to $\ce{Fe^2+}$ by $\ce{H+}$ also occurs. However, on top of that, some of the $\ce{SO4^2-}$ ions are reduced to $\ce{SO2}$, and some of the $\ce{Fe^2+}$ is further oxidised to $\ce{Fe^3+}$. Clearly, in the dilute acid, $\ce{SO4^2-}$ was not strong enough to act as an oxidising agent - but in the concentrated acid it is.

The reason for this can be somewhat explained by Le Chatelier's principle. Consider the redox reaction:

$$\ce{2Fe^2+ + SO4^2- + 4H+ -> 2Fe^3+ + SO2 + 2H2O}$$

In the dilute acid, this equilibrium lies far to the left because the concentrations of $\ce{SO4^2-}$ and $\ce{H+}$ are rather low. However, making the acid more concentrated shifts the equilibrium to the right so that you get some of the products.

As to why you get a mixture of iron(II) and iron(III), the simple answer is that the equilibrium above doesn't lie entirely to the right. $\ce{H2SO4}$ is not that strong an oxidising agent to begin with.

More quantitatively, the redox reaction above can be investigated using standard reduction potentials and the Nernst equation. The following data is taken from Wikipedia (you will get pretty similar values from more "reliable" sources):

  • $E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77 \mathrm{~V}$
  • $E^\circ(\ce{SO4^2-}/\ce{SO2}) = +0.17 \mathrm{~V}$

Under standard conditions, $E^\circ$ for the full reaction is therefore $0.17-0.77 = -0.60 \mathrm{~V}$; the negative value indicates that the forward reaction is non-spontaneous.

The Nernst equation for this reaction can be written:

$$E = E^\circ - \frac{RT}{nF}\ln Q = E^\circ - \frac{RT}{nF}\ln\left[\frac{(a_{\ce{Fe^3+}})^2(a_{\ce{SO2}})(a_{\ce{H2O}})^2}{(a_{\ce{Fe^2+}})^2(a_{\ce{SO4^2-}})(a_{\ce{H+}})^4}\right]$$

I will make several assumptions here:

  • Activity coefficients are taken to be unity, meaning that the activities can be replaced by concentrations.
  • The activities of all the products, as well as $\ce{Fe^2+}$, are unity.
  • The only source of $\ce{H+}$ is from the dissociation of the $\ce{H2SO4}$, and the dissociation is complete. That means that if $c$ is the formal concentration of $\ce{H2SO4}$, we have $[\ce{H+}] = 2c$ and $[\ce{SO4^2-}] = c$.

Substituting in the values of $E^\circ = -0.60 \mathrm{~V}$, $R = 8.314 \mathrm{~J~K^{-1}~mol^{-1}}$, $T = 298 \mathrm{~K}$, $n = 2$, and $F = 96485 \mathrm{~C~mol^{-1}}$, this allows us to simplify the Nernst equation:

$$E / \mathrm{V} = -0.60 + 0.0128\ln (16c^5)$$

Strictly speaking, I should use $c/c^\circ$ instead of $c$ in order to make the logarithm dimensionless, but I'm fed up of technicalities. Now for the actual calculation. Let's assume that you're using $10\mathrm{~M}$ sulfuric acid, such that $c = 10$ (or $c/c^\circ = 10$, if you want to insist). This gives a value of $E = -0.42 \mathrm{~V}$ - as expected, it is less negative than before, which corresponds to a larger value of $K = \exp(nFE/RT)$. The fact that it is still rather negative just means that there is a difference between the real world and my theoretical treatment. Presumably, those assumptions above are not entirely valid.


Just want to be accurate in concepts:

Concentrated sulfuric acid is at least $18~\mathrm{M}$ and contains only 2% water by mass. In this situation, most sulfuric acid molecules do not ionize due to the lack of water. The oxidizing power of concentrated $\ce{H2SO4(l)}$ does not come from $\ce{H+}$ ions or $\ce{SO4^2-}$ ions, but rather the $\ce{H2SO4}$ molecules.


Iron (also steel) is passive in concentrated H2SO4, forming a layer of sulfate on the surface.

  • $\begingroup$ And in dilute? The questions is on the difference between them. $\endgroup$
    – Jon Custer
    Commented May 6, 2016 at 13:04

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