Question

Write equations for the oxidation of Fe and of Al. Use $\mathbf{∆}{\mathit{G}}^{\mathbf{°}}$ to determine whether either process is spontaneous at${\mathbf{25}}^{\mathbf{°}}\mathit{C}$.

Short answer

At $\text{298K}\left({\text{25}}^{\text{o}}\text{C}\right)$, both Fe and Al oxidation processes occur spontaneously.

Step-by-step solution

Definition of Oxidation

The loss of electrons by a molecule, atom, or ion during a reaction is known as oxidation. When the oxidation state of a molecule, atom, or ion is enhanced, it is called oxidation.

Determining equation for the oxidation of Fe and of Al

(a) Rust formation via Fe oxidation in the presence of water (as catalyzer):

${\text{4Fe(s) + 3O}}_{\text{2}}\text{(g)}\to {\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

The$∆G_{rxn}^{°}$ can be determined using the following formula:

$$\begin{gathered} ∆G_{rxn}^{°}=\stackrel{°}{a}∆G_{products}^{°}-\stackrel{°}{a}∆G_{reac\tan ts}^{°} \\ ∆G_{rxn}^{°}=\left[2mol\times ∆G_{{{F_2}_{e_2}}_{{O_3}_{\left(S\right)}}}^{°}\right]-\left[4mol\times ∆G_{F{e}_{\left(s\right)}}^{°}+3mol\times ∆G_{O_{2_{\left(g\right)}}}^{°}\right] \end{gathered}$$

Since oxygen and iron are in their usual state, use Appendix B.

$$\begin{gathered} ∆G_{rxn}^{°}=\left[2mol\times \left(-743.60kJ/mol\right)\right]-\left[4mol\times \left(0.00kJ/mol\right)+3mol\times \left(0.00kJ/mol\right)\right] \\ ∆G_{rxn}^{°}=-1.487.20kJ \end{gathered}$$

$\text{- 1487.20kJ}$is the$∆G_{rxn}^{°}$ for Fe oxidation. The reaction is spontaneous at standard conditions of $\text{- 1atm}$pressure and $\text{298K}\left({\text{25}}^{\text{o}}\text{C}\right)$temperature because the is$∆G_{rxn}^{°}$ negative.

Determining the process is spontaneous

(b) Oxidation reaction Al:

${\text{4Al(s) + 3O}}_{\text{2}}\text{(g)}\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

The$∆G_{rxn}^{°}$ can be determined using the following formula:

$∆G_{rxn}^{°}=\left[2mol\times ∆G_{A{I_2}_{O_{3_{\left(s\right)}}}}^{°}\right]-\left[4mol\times ∆G_{AI\left(s\right)}^{°}+3mol\times ∆G_{O_{2\left(g\right)}}^{°}\right]$

Since oxygen and aluminum are in their normal form, use Appendix B.

$$\begin{gathered} ∆G_{rxn}^{°}=\left[2mol\times \left(-1582.00kJ/mol\right)\right]-\left[1mol\times \left(0.00kJ/mol\right)\right] \\ ∆G_{rxn}^{°}=-3164.00kJ \end{gathered}$$

$\text{- 3164.00kJ}$is the$∆G_{rxn}^{°}$ for Al oxidation. The reaction is spontaneous at standard conditions of $-1$atm pressure and $\text{298K}\left({\text{25}}^{\text{o}}\text{C}\right)$temperature because $∆G_{rxn}^{°}$the is negative.