Question

Corrosion of pure iron takes place in an acidic electrolyte by forming \(Fe^{2+}\) ions at ambient condition. The corrosion current density is measured to be \(2 \times 10^{-4} \, A \, cm^{-2}\). The corrosion rate (in mm per year) of iron is (rounded off to one decimal place). Given: \[ \text{Atomic weight of iron} = 55.85, \rho = 7.86 \, g/cm^3 \] \[ \text{1 year} = 365 \times 24 \times 3600 \, s, 1F = 96500 \, C/mol \]

Hint:

Convert corrosion current density to charge per year → use Faraday’s law to get moles of Fe lost → convert to thickness loss using density.

Answer 1

Correct Answer: 2

Step 1: Current density and charge per year.
Corrosion current density: \[ i = 2 \times 10^{-4} \, A/cm^2 = 2 \times 10^{-4} \, C/s \cdot cm^2 \] Time in one year: \[ t = 365 \times 24 \times 3600 = 3.1536 \times 10^7 \, s \] Charge passed per cm\(^2\) in one year: \[ Q = i \times t = (2 \times 10^{-4})(3.1536 \times 10^7) = 6.307 \times 10^3 \, C/cm^2 \]

Step 2: Moles of electrons passed.
\[ \text{Moles of } e^- = \frac{Q}{F} = \frac{6.307 \times 10^3}{96500} \approx 0.0654 \, mol/cm^2 \]

Step 3: Moles of iron oxidized.
Reaction: \[ Fe \to Fe^{2+} + 2e^- \] So, 2 moles of electrons are needed for 1 mole of Fe. \[ \text{Moles of Fe} = \frac{0.0654}{2} = 0.0327 \, mol/cm^2 \]

Step 4: Mass of iron lost.
\[ m = (0.0327)(55.85) = 1.83 \, g/cm^2 \, \text{ per year} \]



Step 5: Thickness loss.
Density of Fe: \[ \rho = 7.86 \, g/cm^3 \] Volume loss per cm\(^2\): \[ V = \frac{m}{\rho} = \frac{1.83}{7.86} = 0.233 \, cm^3/cm^2 \] Since area = 1 cm\(^2\), thickness = volume/area: \[ \Delta h = 0.233 \, cm = 2.33 \, mm \]

Step 6: Correction check.
Wait — recalculation: - Mass loss = 1.83 g/cm\(^2\) - Divide by density = 1.83/7.86 = 0.233 cm = 2.33 mm/year Thus, the correct thickness loss is ~ \(\boxed{2.3 \, mm/year}\).