Why iron reacts differently with concentrated and dilute sulfuric acid?

When $\ce{H2SO4}$ is dilute,



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> $ \ce{Fe +H2SO4 -> FeSO4 + H2} $
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and when $\ce{H2SO4}$ is concentrated,



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




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.




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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.




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More quantitatively, the redox reaction above can be investigated using standard reduction potentials and the Nernst equation. The following data is taken from [Wikipedia](https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)) (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.


 15

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.


 3

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


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