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: \begin{align*} {Atomic weight of iron} &= 55.85
{Density of iron} &= 7.86\ {g} \cdot {cm}^{-3}
{Number of days in a year} &= 365
1\ {Faraday} &= 96500\ {Coulomb} \cdot {mol}^{-1} \end{align*}

Hint:

To calculate corrosion rate, use the current density, equivalent weight, and other constants. Don't forget to convert units properly when calculating the final rate.

Answer 1

Correct Answer: 2

The corrosion rate can be calculated using the formula:
\[ {Corrosion Rate} = \frac{I \times {Equivalent Weight}}{n \times F \times \rho} \] where:
- \( I \) is the current density, \( 2 \times 10^{-4} \, {A cm}^{-2} \),
- \( {Equivalent Weight} = \frac{{Atomic Weight}}{2} = \frac{55.85}{2} = 27.925 \, {g/mol} \),
- \( n = 2 \) (the number of electrons involved in the reaction for iron),
- \( F = 96500 \, {C/mol} \) is Faraday's constant,
- \( \rho = 7.86 \, {g/cm}^3 = 7860 \, {kg/m}^3 \).
Substitute the values into the formula:
\[ {Corrosion Rate} = \frac{2 \times 10^{-4} \times 27.925}{2 \times 96500 \times 7.86} = \frac{5.585 \times 10^{-3}}{1514.97} \approx 3.7 \times 10^{-6} \, {m/s}. \] Now, convert this to mm per year by multiplying by the number of seconds in a year:
\[ {Corrosion Rate} = 3.7 \times 10^{-6} \times 365 \times 24 \times 3600 \times 1000 = 2.1 \, {mm/year}. \] Thus, the corrosion rate of iron is approximately \( 2.1 \, {mm/year} \).