(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). \((\mathbf{b})\) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to $\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)$.

: During iron corrosion, iron in the solid form undergoes oxidation to form iron(II) ions in the aqueous phase. To write the anode reaction, we will write the oxidation process of iron: \[ \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2e^-\]

: In the corrosion process, it is usually oxygen and water that react with iron. Oxygen is reduced as it accepts electrons in the cathode reaction. To write the cathode reaction, we will follow the reduction process of oxygen in the presence of water: \[ \mathrm{O}_2(g) + 2 \mathrm{H}_2 \mathrm{O}(l) + 4e^- \rightarrow 4 \mathrm{OH}^-(aq) \] Now let's proceed to part (b).

: The oxidation of Fe2+(aq) to form Fe2O3 involves the loss of electrons by iron(II) ions, as the oxidation number of each iron atom changes from +2 to +3. The half-reaction for this process is: \[2 \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}_2 \mathrm{O}_3(s) + 2e^-\]

: Since oxygen, which is involved in this reaction, is typically found in the environment as O2 in the air, the reduction half-reaction involves oxygen gaining electrons. As Fe2O3.3H2O is formed, the molecular oxygen gains electrons and combines with the ions already present in the oxidation equation: \[ \mathrm{O}_2(g) + 2 \mathrm{H}_2 \mathrm{O}(l) + 2e^- \rightarrow 2 \mathrm{OH}^-(aq) \] Since both half-reactions must balance out the number of electrons, we need to multiply the reduction half-reaction by 2 to have the same number of electrons as the oxidation half-reaction. The balanced reduction half-reaction is: \[ 2 \mathrm{O}_2(g) + 4 \mathrm{H}_2 \mathrm{O}(l) + 4e^- \rightarrow 4 \mathrm{OH}^-(aq) \]