inorganic chemistry - Formation of ferrous chloride or ferric chloride

Question
When $\ce{HCl}$ reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?

Answer. Its reaction with iron produces $\ce{H2}$.

$\ce{Fe + 2HCl -> FeCl2 + H2}$

Liberation of hydrogen prevents the formation of ferric chloride.

Now my question is why couldn't it be: $\ce{2Fe + 6HCl -> 2FeCl3 + 3H2}$ ?

Answer

Acids can oxidize iron since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ can oxidize $\ce{Fe}$ to $\ce{Fe^2+}$; but non-oxidizing acids cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:

$$\begin{alignat}{2} \ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V}\\ \ce{[Fe(H2O)6]^2+ + 2e- \;&<=> Fe}\quad &&E^\circ = -0.440\ \mathrm{V}\\ \ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\ \end{alignat}$$

However, oxygen can oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:

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

Therefore, iron is oxidized to $\ce{Fe^3+}$ in dilute hydrochloric acid when in contact with air.