Chemistry - What is formed when you leave iron(II) sulfate in plain air?

Solution 1:

Usually, the original iron(II) sulfate is present as green $\ce{FeSO4.7H2O}$.

Oxygen can oxidize $\ce{Fe(II)}$ salts to $\ce{Fe(III)}$ salts; for example, at $\mathrm{pH}=0$:

$$\begin{alignat}{2} \ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\ \ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V} \end{alignat}$$

$\ce{Fe(II)}$ is even easier oxidized under alkaline conditions; for example, at $\mathrm{pH}=14$:

$$\begin{alignat}{2} \ce{FeO(OH) + H2O + e- \;&<=> Fe(OH)2 + OH-}\quad &&E^\circ = -0.69\ \mathrm{V}\\ \ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad &&E^\circ = +0.401\ \mathrm{V} \end{alignat}$$

However, you cannot simply oxidize iron(II) sulfate $\left(\ce{FeSO4}\right)$ to iron(III) sulfate $\left(\ce{Fe2(SO4)3}\right)$ in dry air since you would need additional sulfate to balance the equation:

$$\ce{2FeSO4 + SO4^2- -> Fe2(SO4)3 + 2e-}$$

Nevertheless, $\ce{Fe(II)}$ can be oxidized to $\ce{Fe(III)}$.

The resulting $\ce{Fe(III)}$ is subject to hydrolysis. By way of comparison, the ion $\ce{[Fe(H2O)6]^3+}$ is only stable under strong acidic conditions. Already at $\mathrm{pH}=0{-}2$, it turns into yellow $\ce{[Fe(OH)(H2O)5]^2+}$ and begins to form $\ce{[Fe(OH)2(H2O)4]+}$:

$$\begin{align} \ce{[Fe(H2O)6]^3+ \;&<=> [Fe(OH)(H2O)5]^2+ + H+}\\ \ce{[Fe(OH)(H2O)5]^2+ \;&<=> [Fe(OH)2(H2O)5]+ + H+}\\ \end{align}$$

Further addition of base causes precipitation of amorphous iron(III) hydroxide.

Accordingly, the likely product when iron(II) sulfate is oxidized is basic iron(III) sulfate, i.e. approximately:

$$\ce{4FeSO4 + O2 + 2H2O -> 4Fe(OH)SO4}$$

However, the real weathering and aging of iron(II) sulfate in dry air actually yields a mixture of various compounds, including iron(III) sulfate and iron(III) oxide-hydroxide.

You may observe this reaction in some iron fertilizers for lawns. The fresh product typically contains green iron(II) sulfate, which gradually becomes yellow.

Solution 2:

From wikipedia:

Upon exposure to air, it oxidizes to form a corrosive brown-yellow coating of basic ferric sulfate, which is an adduct of ferric oxide and ferric sulfate: $$\ce{12FeSO4 + 3O2 -> 4Fe2(SO4)3 + 2Fe2O3}$$

The Pubchem page for ferrous sulphate says that this reaction occurs rapidly in moist air and that the rate of oxidation is increased by increasing the pH, temperature, or by exposure to light. However, I don't have access to the original sources.

As @Greg pointed out, oxygen appears to be acting as the oxidiser here and it forms $\ce{Fe2O3}$. There is not enough sulphate present in the starting materials to form only $\ce{Fe2(SO4)3}$ and hence the oxide is formed as well.