Question

(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 \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)\).

Short answer

The anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II) are: Anode reaction: \[ \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2e^-\] Cathode reaction: \[ \mathrm{O}_2(g) + 2 \mathrm{H}_2 \mathrm{O}(l) + 4e^- \rightarrow 4 \mathrm{OH}^-(aq) \] The balanced half-reactions involved in the air oxidation of $\mathrm{Fe}^{2+}(a q)$ to $\mathrm{Fe}_{2}\mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)$ are: Oxidation half-reaction: \[2 \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}_2 \mathrm{O}_3(s) + 2e^-\] Reduction half-reaction: \[ 2 \mathrm{O}_2(g) + 4 \mathrm{H}_2 \mathrm{O}(l) + 4e^- \rightarrow 4 \mathrm{OH}^-(aq) \]

Step-by-step solution

(a) Anode reaction of iron corrosion

: 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: \[ \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2e^-\]

(a) Cathode reaction of iron corrosion

: 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).

(b) Oxidation half-reaction of Fe2+ to Fe2O3.3H2O air oxidation

: 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: \[2 \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}_2 \mathrm{O}_3(s) + 2e^-\]

(b) Reduction half-reaction of Fe2+ to Fe2O3.3H2O air oxidation

: 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: \[ 2 \mathrm{O}_2(g) + 4 \mathrm{H}_2 \mathrm{O}(l) + 4e^- \rightarrow 4 \mathrm{OH}^-(aq) \]

Key concepts

Anode Reaction

In the process of iron corrosion, an important step is the anode reaction. This is where oxidation takes place. When iron (Fe) in its solid form interacts with the environment, it loses electrons. This is what we call the oxidation of iron. During this process, iron solid (Fe) is converted into aqueous iron(II) ions (\(\mathrm{Fe}^{2+}(aq)\)), releasing electrons in the process.
This transformation is crucial in corrosion because it leads to the formation of rust. The generalized equation for this reaction is:

• \[\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2e^-\]

Oxidation is a loss of electrons, often encapsulated in the mnemonic "OIL" (Oxidation Is Loss). Understanding this part of the process is essential in grasping why metal structures weaken over time in corrosive environments.

Cathode Reaction

Parallelly to the anode reaction in iron corrosion, the cathode reaction occurs. Here, reduction takes place. While the anode reaction leads to the liberation of electrons, the cathode phase involves a different player—oxygen. In this reduction process, molecular oxygen present in the environment gains electrons and interacts with water to form hydroxide ions.
This is represented by the equation:

• \[\mathrm{O}_2(g) + 2 \mathrm{H}_2 \mathrm{O}(l) + 4e^- \rightarrow 4 \mathrm{OH}^-(aq)\]

Reduction is a gain of electrons, encapsulated in the mnemonic "RIG" (Reduction Is Gain). The cathode reaction is essential as it complements the anodic process, completing the electrochemical cycle of corrosion.

Iron Oxidation

Iron oxidation is at the heart of the rusting process. It's during this stage that iron transforms from a metallic state to various oxidized states, like iron(II) oxide or iron(III) oxide. When iron becomes \(\mathrm{Fe}^{2+}\) in aqueous form, it sets the stage for further reactions with oxygen and water, leading to well-known compounds like iron(III) oxide (rust).
This process is responsible for the characteristic reddish-brown color of rust and significantly affects the structural integrity of iron or steel objects.
Ensuring adequate protection against this oxidative process is vital in industries to prevent material deterioration and ensure the longevity of structures and tools.

Half-Reactions

The concept of half-reactions is a cornerstone in understanding electrochemical processes like corrosion. In corrosion studies, the reaction is typically split into two halves: one focusing on oxidation and the other on reduction. Each half-reaction clearly outlines the transfer of electrons involved.
For instance, in the corrosion of iron:

• The oxidation half-reaction shows iron losing electrons.
• The reduction half-reaction shows oxygen gaining electrons to form hydroxide ions.\[\text{Oxidation: } \quad \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2e^-\]
• \[\text{Reduction: } \quad \mathrm{O}_2(g) + 2 \mathrm{H}_2 \mathrm{O}(l) + 4e^- \rightarrow 4 \mathrm{OH}^-(aq)\]

These reactions are balanced to ensure that the number of electrons lost in oxidation equals the number gained in reduction. Understanding how to split a full reaction into these components can greatly simplify the study of electrochemistry and corrosion.