2025 A Level H2 Chemistry Paper 1 Solutions - Questions 21 to 30

Question 21

Answer: B

Topic: Carboxylic Acid and Derivatives

Explanation:

For each of the compounds 1 to 3, we hydrolyse the ester with sulfuric acid to form carboxylic acid and alcohol, followed by oxidation of alcohol with dichromate(VI) to form carboxylic acid. If there is only 1 organic compound product then it could be ester J.

Ester 1 will form only ethanoic acid on hydrolysis followed by oxidation.

Ester 2 will form propanoic acid and propanone as final products.

Ester 3 will form butanoic acid and ethanoic acid as final products.

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Question 22

Answer: B

Topic: Nitrogen Compounds and Polymers

Explanation:

The building blocks of proteins are α-amino acids where the amine group is bonded to the carbon directly bonded to acid group.

Only 2,6-diaminohexanoic acid is α-amino acid. The other two are β-amino acids hence not found in proteins.

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Question 23

Answer: A

Topic: Carboxylic Acid and Derivatives

Explanation:

A is correct. Product formed is ethyl ethanamide.

B is wrong. Product formed should be phenyl ethanoate with 2 oxygen atoms instead of 1.

C is wrong. Product formed should be 1-methylethyl ethanoate instead of 1-methylethyl propanoate.

D is wrong. Product formed should be ethanoic acid instead of ethanal.

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Question 24

Answer: B

Topic: Nitrogen Compounds and Polymers

Explanation:

To compare base strengths between the 3 compounds, we focus on lone pair availability on nitrogen for donation.

Phenylamine Y is the least basic since lone pair of N is delocalised into benzene ring and less available for donation.

Z is a primary amine with 1 electron donating alkyl group which increase nitrogen's electron density and lone pair availability for donation, hence more basic than Y.

X is a secondary amine with 2 electron donating groups therefore should be more basic than Z.

Finally the base strength in increasing order should be Y, Z, X.

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Question 25

Answer: C

Topic: Organic Chemistry Synthesis

Explanation:

The synthesis routes with reagents and conditions are given so we just need to verify if the correct final product is formed for each route.

Route 2 is wrong since we start off with chloromethylbenzene to form benzoic acid but we want to form phenylethanoic acid (with an additional carbon) instead.

Route 3 is wrong since in step 3 there is no reaction between 2 alcohols to form ester.

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Question 26

Answer: C

Topic: Electrochemistry

Explanation:

The chlorine half cell should undergo reduction and silver half cell oxidation and E_cell can be calculated to be +0.56V.

On adding sodium chloride to the silver half cell, AgCl ppt is formed which lowers the Ag⁺ concentration. This shifts the position of equilibrium for silver half cell to the left, and E value should decrease (less than +0.8V).

Since silver half cell is the anode, a less positive E for oxidation means E_cell will increase (greater than +0.56V).

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Question 27

Answer: A

Topic: Electrochemistry

Explanation:

From the balanced equation we can deduce 6 moles of electrons are transferred between the reactants.

Using formula ΔG = -nFE_cell we can calculate ΔG to be -625 kJ mol^-1.

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Question 28

Answer: C

Topic: Electrochemistry

Explanation:

To be reduced by Ni, the E value for M²⁺/M half cell must be greater than -0.25V. Therefore answer must be Sn²⁺.

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Question 29

Answer: B

Topic: Electrochemistry

Explanation:

To find charge of copper ion, we need to determine mole ratio of electrons to Cu.

Mole of electrons can be determined via Faraday's equations Q = I.t = n.F

Mole of copper is its mass over atomic mass.

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Question 30

Answer: B

Topic: Transition Elements

Explanation:

Element M must be a transition metal with larger atomic mass and smaller radius than Ca. Hence M must have greater density than Ca.

MCl₂ will have more exothermic lattice energy compared to CaCl₂ since M²⁺ is smaller than Ca²⁺ and magnitude of lattice energy for MCl₂ should be greater.

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