Question

Using \( E^\circ \) values of X and Y given below, predict which is better for coating the surface of iron to prevent corrosion and why?

Given:
\( E^\circ_{X^{2+}/X} = -2.36 \, \text{V} \)
\( E^\circ_{Y^{2+}/Y} = -0.14 \, \text{V} \)
\( E^\circ_{Fe^{2+}/Fe} = -0.44 \, \text{V} \)

Hint:

To prevent corrosion, the coating material should have a reduction potential that is less negative than that of the metal being protected, so it will be more stable and less likely to corrode.

Answer 1

To prevent iron corrosion, the electrode potentials of the coating metal are critical. A metal with a higher (less negative) reduction potential is more readily reduced and less prone to corrosion.

A metal with a more negative reduction potential will function as a sacrificial anode, corroding instead of the iron.

1. Metal X:
The reduction potential of \(X^{2+}/X\) is \(-2.36 \, \text{V}\). This highly negative value indicates that metal X is easily oxidized and will corrode rapidly, making it unsuitable for coating iron.

2. Metal Y:
The reduction potential of \(Y^{2+}/Y\) is \(-0.14 \, \text{V}\). This is less negative than metal X, signifying that metal Y is more stable and less likely to corrode.

3. Iron (Fe):
The reduction potential of \(Fe^{2+}/Fe\) is \(-0.44 \, \text{V}\). This value is less negative than X but more negative than Y. Iron will corrode over time if unprotected.

Conclusion:
Metal Y is the superior choice for coating iron to prevent corrosion due to its less negative reduction potential compared to iron and metal X. Consequently, metal Y offers greater stability and reduced corrosion, providing enhanced protection to the iron surface.