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Electrochemistry - Corrosion

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Corrosion is the process of slow conversion of metals into their undesirable compounds (like oxides, carbonates, or sulfides) by interaction with the environment (air and moisture).

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Mechanism of Rusting: It is considered an electrochemical phenomenon. At a particular spot on the iron object, oxidation takes place, and that spot behaves as an anode: Fe(s)→Fe2+(aq)+2e−Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-.

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At the cathodic spot, electrons released at the anode move through the metal and reduce atmospheric oxygen in the presence of H+H^+ ions: O2(g)+4H+(aq)+4e−→2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l).

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The H+H^+ ions involved in the cathodic reaction are usually available from H2CO3H_2CO_3, formed due to the dissolution of CO2CO_2 from the air into water: H2O(l)+CO2(g)→H2CO3(aq)H_2O(l) + CO_2(g) \rightarrow H_2CO_3(aq).

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The ferrous ions (Fe2+Fe^{2+}) are further oxidized by atmospheric oxygen to ferric ions (Fe3+Fe^{3+}), which come out as hydrated ferric oxide (Fe2O3⋅xH2OFe_2O_3 \cdot xH_2O), known as rust.

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Factors promoting corrosion: Presence of impurities in the metal, presence of moisture and air, presence of electrolytes (like NaClNaCl in seawater), and the reactivity of the metal.

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Prevention Methods: Barrier protection (painting, oiling), Sacrificial protection (coating iron with a more active metal like ZnZn, known as Galvanization), and Cathodic protection (connecting the metal to a more reactive metal like MgMg or ZnZn acting as a sacrificial anode).

📐Formulae

Anodic half−reaction:Fe(s)→Fe2+(aq)+2e−,E(Fe2+/Fe)∘=−0.44 VAnodic \, half-reaction: Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-, \quad E^\circ_{(Fe^{2+}/Fe)} = -0.44 \, V

Cathodic half−reaction:O2(g)+4H+(aq)+4e−→2H2O(l),E∘=1.23 VCathodic \, half-reaction: O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l), \quad E^\circ = 1.23 \, V

Overall cell reaction:2Fe(s)+O2(g)+4H+(aq)→2Fe2+(aq)+2H2O(l),Ecell∘=1.67 VOverall \, cell \, reaction: 2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l), \quad E^\circ_{cell} = 1.67 \, V

Chemical formula of Rust:Fe2O3⋅xH2OChemical \, formula \, of \, Rust: Fe_2O_3 \cdot xH_2O

ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}

💡Examples

Problem 1:

Explain why the rusting of iron is faster in saline water than in ordinary water.

Solution:

Saline water contains a high concentration of dissolved electrolytes like Na+Na^+ and Cl−Cl^- ions.

Explanation:

These ions increase the electrical conductivity of the water film on the iron surface. Since corrosion is an electrochemical process, the presence of these electrolytes facilitates the movement of ions and the flow of current between the anodic and cathodic sites, thereby accelerating the rate of corrosion.

Problem 2:

Calculate the standard Gibbs free energy change (ΔG∘\Delta G^\circ) for the initial step of iron corrosion given Ecell∘=1.67 VE^\circ_{cell} = 1.67 \, V and n=4n = 4 for the balanced overall reaction.

Solution:

Using the formula ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}: ΔG∘=−(4)×(96500 C mol−1)×(1.67 V)≈−644,620 J mol−1=−644.62 kJ mol−1\Delta G^\circ = -(4) \times (96500 \, C \, mol^{-1}) \times (1.67 \, V) \approx -644,620 \, J \, mol^{-1} = -644.62 \, kJ \, mol^{-1}.

Explanation:

The large negative value of ΔG∘\Delta G^\circ indicates that the corrosion (rusting) process is highly spontaneous under standard conditions.

Corrosion - Revision Notes & Key Formulas | ICSE Class 12 Chemistry