1. Predicting the Best Metal for Coating Iron:
Given Data:
- \( E^\circ_{X^{2+}/X} = -2.36 \, \text{V} \)
- \( E^\circ_{Y^{2+}/Y} = -0.14 \, \text{V} \)
- \( E^\circ_{Fe^{2+}/Fe} = -0.44 \, \text{V} \)
Explanation:
The \( E^\circ \) values represent the standard electrode potentials of the half-reactions. The more negative the \( E^\circ \) value, the more easily the substance will undergo oxidation. In the case of corrosion protection, a metal that is more easily oxidized (has a more negative \( E^\circ \)) can protect iron by sacrificing itself and forming a protective coating.
Step 1: Analyzing the Electrode Potentials:
- The more negative the \( E^\circ \) value, the more readily the substance will lose electrons and undergo oxidation. Therefore, the metal with the most negative \( E^\circ \) will be more easily oxidized and form a sacrificial layer to protect iron from corrosion.
- Iron (\( Fe^{2+}/Fe \)) has a standard electrode potential of \( -0.44 \, \text{V} \), which means iron can be oxidized in the presence of a more easily oxidized metal.
Step 2: Comparing X and Y for Coating:
- \( E^\circ_{X^{2+}/X} = -2.36 \, \text{V} \): Metal X has a very negative electrode potential, meaning it is highly prone to oxidation. This makes it a good candidate for sacrificial protection.
- \( E^\circ_{Y^{2+}/Y} = -0.14 \, \text{V} \): Metal Y has a less negative \( E^\circ \), meaning it is less prone to oxidation compared to metal X.
Conclusion:
Since metal X has a more negative \( E^\circ \) value, it is more likely to undergo oxidation and form a protective sacrificial coating on iron. Thus, metal X is a better choice for coating the surface of iron to prevent corrosion.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.