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 corrosion of iron, we must consider the electrode potentials of the metal to be used for coating. A metal that has a higher (less negative) reduction potential will be more easily reduced and less likely to corrode.

The metal with the more negative reduction potential will act as a sacrificial anode, corroding in place of the iron.

1. For metal X:
The reduction potential of \(X^{2+}/X\) is \(-2.36 \, \text{V}\), which is very negative. This means that metal X is more easily oxidized and will corrode easily, making it unsuitable for coating the surface of iron.

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

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

Conclusion:
Metal Y is the better choice for coating the surface of iron to prevent corrosion because it has a less negative reduction potential compared to iron and X. As a result, it will be more stable and less likely to corrode, offering better protection to the iron surface.