 # 2019 A Level H2 Chemistry Paper 1 Question 30 - Calculate Time Taken for Copper Electrolysis

We are finally at our last question (Q30) for 2019 A Level H2 Chemistry Paper 1!

Let's take a look at the question. We need to calculate the time taken to cover both sides of the copper with new copper to a total depth of 1000 atoms.

First let's determine the number of copper atoms per layer.

Given the dimensions of the square copper cathode is 0.1m x 0.1m and of each copper atom is 3.0 x 10-12m x 3.0 x 10-12m. Hence the number of copper atoms per layer will be 1.111 x 1021

Next we have to interpret the statement "cover both sides of the piece of copper with new copper to a total depth of 1000 atoms".

Many students see this as a depth of 1000 atoms to the lefthand side plus a depth of another 1000 atoms to the righthand side to give a total depth of 2000 atoms.

However since the question states a "total depth of 1000 atoms", we should interpret it as a depth of 500 atoms to the LHS plus a depth of another 500 atoms to the RHS to give a total depth of 1000 atoms. We can now find the total number of copper atoms and then the moles of copper discharged will be 1.845 mol. Next we can work out the moles of electron involved in the electrolysis to be 3.69 mol using the half equation for discharge of Cu2+ to Cu metal. Finally we can use both Faraday's equations (Q = n.F and Q = I.t) to find time taken for the electrolysis to be 24.7 hours. Hence the answer to this question will be option B.

This was a pretty controversial question as many students who interpreted the total depth to be 2000 atoms (as discussed earlier) will get option C as their answer.

In my opinion this question was not very well crafted as challenging questions should be set based on chemistry concepts and not due to ambiguous statements.

Topic: Electrochemistry, Physical Chemistry, A Level Chemistry, Singapore

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## 2019 A Level H2 Chemistry Paper 1 Question 29 - Measure Standard Electrode Potential

Let's take a look at 2019 A Level H2 Chemistry Paper 1 Question 29. We are asked to determine the set of concentrations that would give the correct value of standard electrode potential for an electrode system.

First we will have to determine which species are taking part in reduction/oxidation in this reversible electrode. Notice the oxidation state of Cl is +1 in HOCl and zero in Cl2.

This means the species that we should focus on is HOCl and Cl2 since there is a change on oxidation state for Cl.

In the forward direction HOCl(+1) is reduced to Cl2(0), while in the reverse direction Cl2(0) is oxidised to HOCl(+1).

The standard conditions required for standard electrode potential should apply to both HOCl and Cl2.

Standard conditions: Temperature at 298K, Pressure at 1 atm, Concentration of solutions at 1.0 moldm-3

Hence concentration of HOCl and Cl2 has to be both at 1.0 moldm-3.

Notice the concentration of H+ is not that important since it just provides the acidic medium for the system. By comparing the options we can conclude the correct answer to this question should be option B.

Topic: Electrochemistry, Physical Chemistry, A Level Chemistry, Singapore

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## 2019 A Level H2 Chemistry Paper 1 Question 27 - Deduce Formula of Chromium Complex in Excess Hydroxide

Let's take a look at 2019 A Level H2 Chemistry Paper 1 Question 27. We are required to deduce the formula of the chromium containing species in the deep green solution.

The starting chromium complex in chromium (III) chloride should be [Cr(H2O)6]3+ where chromium (III) ion is surrounded by water ligands. When hydroxide is added, the precipitate formed should be [Cr(H2O)3(OH)3] due to ligand exchange between H2O and OH-.

When more OH- is added, the remaining H2O ligands will be substituted with OH- to form [Cr(OH)6]3- complex which should give us the deep green solution.

Now we can go through the options.

A. [CrCl4]2-

This complex is not formed here but we can get this complex if concentrated HCl is added to provide a high concentration of Cl- ligands.

B. [Cr(H2O)6]3+

This is our starting complex so is not the answer.

C. [Cr(OH)6]3-

This is the complex formed on adding excess OH- ligands so this will be the answer to this question.

D. [Cr(OH)3(H2O)3]

This is the initial precipitate formed hence is not the answer.

Topic: Transition Elements, Inorganic Chemistry, A Level Chemistry, Singapore

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## 2019 A Level H2 Chemistry Paper 1 Question 28 - Deduce Products of Hydrolysis by Chymotrypsin

Here's question 28 for 2019 A Level H2 Chemistry Paper 1. Given the activity of chymotrypsin, we need to deduce which option will form 2 dipeptides when hydrolysed by the enzyme.

First let's interprete the struture of pentapeptide W.

Recall the primary structure of proteins has a very predictable repeat unit: Position 1 - amine group
Position 2 - carbon with R group
Position 3 - acid group

Therefore we can find out the positions 1, 2 and 3 for W. Next, we know that chymotrypsin will hydrolyse the peptide bond on the acid or carboxyl side of a residue with an aromatic ring.

So we can look at W, find the R groups with aromatic ring, break the peptide bond on the acid side, and figure out the products X (dipeptide), Y (dipeptide) and Z (amino acid) formed. Once we understand how chymotrypsin functions, we can now apply this to the options and see which one gives us 2 dipeptides when hydrolysed with chymotrypsin.

A will give 1 tripeptide and 1 amino acid, hence not the answer. B will give 1 dipeptide and 2 amino acids, hence not the answer. C will give 2 dipeptides hence this is the answer that we want. D will give 2 amino acids and 1 dipeptide, hence not the answer. Topic: Nitrogen Compounds and Proteins, Organic Chemistry, A Level Chemistry, Singapore

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## 2019 A Level H2 Chemistry Paper 1 Question 26 - Deduce Electronic Configuration of Mo(IV) Ion

Here's 2019 A Level H2 Chemistry Paper 1 Question 26 on transition elements. We need to deduce the electronic configuration of molybdenum (IV) ion, given its electronic configuration is similar to those of first row of transition elements and ions.

So the first thing to do is to look up the Periodic Table and find the first row transition element that is directly above molybdenum, which is chromium. We can figure out the electronic configuration of Cr4+ which is [Ar] 3d2 4s0. Students might also be interested in knowing how to write out electronic configuration for the first 30 elements.

Therefore the electronic configuration for Mo4+ will be the same as Cr4+ but the principal quantum number increases by 1 since it is in the next Period, ie [Kr] 4d2 5s0. Hence the answer to this question will be option B.

Topic: Transition Elements, Inorganic Chemistry, A Level Chemistry, Singapore

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