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Electrochemistry - Corrosion

Grade 12CBSEChemistry

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

🔑Concepts

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Corrosion is an electrochemical phenomenon where the metal surface behaves like an electrochemical cell. For iron, this process is known as rusting.

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At the anodic spot, iron is oxidized: Fe(s)→Fe2+(aq)+2e−Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-. The electrons released move through the metal to another spot on the surface.

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At the cathodic spot, oxygen in the presence of H+H^+ ions (derived from H2CO3H_2CO_3 formed by CO2CO_2 and H2OH_2O) is reduced: O2(g)+4H+(aq)+4e−→2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l).

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The Fe2+Fe^{2+} ions are further oxidized by atmospheric oxygen to form hydrated ferric oxide (Fe2O3⋅xH2OFe_2O_3 \cdot xH_2O), which is the chemical formula for rust.

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Factors that promote corrosion include the presence of impurities in the metal, presence of electrolytes (like NaClNaCl in seawater), and the reactivity of the metal.

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Methods of prevention include barrier protection (painting, oiling), sacrificial protection (galvanization with ZnZn), and cathodic protection (connecting to MgMg or ZnZn blocks).

📐Formulae

Fe(s)→Fe2+(aq)+2e−(EFe2+/Fe∘=−0.44 V)Fe(s) \rightarrow Fe^{2+}(aq) + 2e^- \quad (E^{\circ}_{Fe^{2+}/Fe} = -0.44\text{ V})

O2(g)+4H+(aq)+4e−→2H2O(l)(E∘=1.23 V)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l) \quad (E^{\circ} = 1.23\text{ V})

2Fe(s)+O2(g)+4H+(aq)→2Fe2+(aq)+2H2O(l)(Ecell∘=1.67 V)2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l) \quad (E^{\circ}_{cell} = 1.67\text{ V})

4Fe2+(aq)+O2(g)+4H2O(l)→2Fe2O3(s)+8H+(aq)4Fe^{2+}(aq) + O_2(g) + 4H_2O(l) \rightarrow 2Fe_2O_3(s) + 8H^+(aq) (Formation of rust)

Fe2O3+xH2O→Fe2O3⋅xH2O (Hydrated Ferric Oxide)Fe_2O_3 + xH_2O \rightarrow Fe_2O_3 \cdot xH_2O \text{ (Hydrated Ferric Oxide)}

💡Examples

Problem 1:

Explain why galvanization is preferred over tinning for protecting iron even if the protective coating is scratched.

Solution:

In galvanization, iron is coated with ZnZn (EZn2+/Zn∘=−0.76 VE^{\circ}_{Zn^{2+}/Zn} = -0.76\text{ V}), and in tinning, it is coated with SnSn (ESn2+/Sn∘=−0.14 VE^{\circ}_{Sn^{2+}/Sn} = -0.14\text{ V}). For iron, EFe2+/Fe∘=−0.44 VE^{\circ}_{Fe^{2+}/Fe} = -0.44\text{ V}.

Explanation:

Because ZnZn has a lower reduction potential than iron, it acts as the anode and undergoes oxidation even if the coating is scratched (sacrificial protection). However, SnSn has a higher reduction potential than iron; if the tin coating is scratched, the iron becomes the anode and corrodes even faster.

Problem 2:

Why does the presence of saline water (salt water) accelerate the process of rusting?

Solution:

Rusting is an electrochemical process that requires the movement of ions. Saline water contains dissolved salts like NaClNaCl which increase the conductivity of the water film on the metal surface.

Explanation:

The presence of extra electrolytes (Na+Na^+ and Cl−Cl^- ions) facilitates the flow of current between the anodic and cathodic sites on the iron surface, thereby increasing the rate of the redox reaction that causes corrosion.

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