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2021 Paper 1 Suggested Solution

2021 P1 Q30 - Compare Properties between s-block and Transition Metals

Let's take a look at our final question for 2021 A Levels H2 Chemistry Paper 1 Question 30.

2021 p1 q30 question

We are required to deduce which properties of rhenium are typical of a transition element.

Let's compare the melting point and density between s-block metal and transition metal.

2021 p1 q30 compare properties

1. Melting Point

Transition metals have higher melting points as they can delocalise electrons from s-subshell and d-subshell.

They will form more positively charged cations attracting a bigger sea of delocalised electrons, metallic bonds are stronger hence higher melting point than s-block metal like calcium.

Rhenium with high melting point of 3180oC is typical of a transition element.

2. Density

Transition metals have higher density since they have bigger mass number and smaller radii.

This means the transition metal atoms are more closely packed per unit volume which contributes to their higher density as compared to s-block metals.

Rhenium with high density of 21 020 kg m-3 is typical of a transition element.

Therefore the answer to this question is option D (both properties 1 and 2).

2021 p1 q30 answer

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

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2021 P1 Q29 - Determine Redox Reaction between Zn and VO2+

Let's take a look at 2021 A Levels H2 Chemistry Paper 1 Question 29.

2021 p1 q29 question

We need to determine the redox reaction when excess zinc is added to acidified VO2+ ions.

In the options vanadium can be reduced to different oxidation states.

Let us find all the relevant half equations in the data booklet.

2021 p1 q29 E values for Zn and V

Notice zinc metal can be oxidised to Zn2+.

VO2+ (+5 oxidation state) can be reduced to VO2+ (+4 oxidation state).

VO2+ can then be further reduced to V3+, V3+ can be reduced to V2+, and finally V2+ reduced to vanadium metal.

Therefore we need to calculate the Ecell for each stage of reduction of vanadium species to determine its extent of reduction.

2021 p1 q29 calculate Ecell

Ecell for redox reactions between Zn and VO2+ (+5 oxidation state), VO2+ (+4 oxidation state) and V3+ are all positive. This means all these reactions are feasible and VO2+ can be reduced all the way to V2+.

Ecell for redox reaction between Zn and V2+ is negative. This means Zn cannot reduce V2+ to vanadium metal.

Therefore the final oxidation state of vanadium should be V2+ and answer to this question is option C.

2021 p1 q29 answer

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

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2021 P1 Q28 - Calculate Increase in Mass of Cathode during Electrolysis

Let's take a look at 2021 A Levels H2 Chemistry Paper 1 Question 28.

2021 p1 q28 question

We are required to determine the increase in mass of cathode.

First let us calculate the number of moles of electrons involved during this electrolysis using Faraday's equations.

2021 p1 q28 calculate mole of electrons

Next we have to deduce the reaction at the cathode.

Since electrolyte is 1.0 moldm-3 of Cu2+ and Ni2+, we have to compare their standard electrode potentials and determine which is more likely reduced at the cathode.

2021 p1 q28 compare E value

Since Cu2+ has a more positive E value, it is more likely reduced than Ni2+.

Therefore Cu2+ will be reduced at the cathode.

2021 p1 q28 reduction of Cuii

Now we can compare mole ratio between copper and electrons to determine moles and mass of copper deposited at the cathode.

2021 p1 q28 calculate mass of Cu deposited

Hence the answer to this question is option C.

2021 p1 q28 answer

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

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2021 P1 Q27 - Calculate Avogadro Constant from Electrolysis Data

Let's take a look at 2021 A Levels H2 Chemistry Paper 1 Question 27.

2021 p1 q27 question

We are required to calculate Avogadro's constant from the electrolysis of molten aluminium oxide.

Let's recap the 3 Faraday's equations that we use for calculation questions in electrolysis.

2021 p1 q27 faradays equations

Q = I.t
where Q = charge (C)
I = current (A)
t = time (s)

Q = n.F
where Q = charge (C)
n = number of moles of electrons (mol)
F = Faraday's constant (96500 C mol-1)

F = e.L
where F = Faraday's constant (C mol-1)
e = charge of 1 electron (1.60 x 10-19 C)
L = Avogadro's constant (6.02 x 1023 mol-1)

The values of F, e and L can all be found in the Data booklet.

First we need to determine the moles of electrons involved in this electrolysis.

From the mass of Al we can calculate moles of Al.

Using reduction half equation of Al3+ to Al, we can compare mole ratio between electron to Al to determine moles of electrons.

2021 p1 q27 calculate mole of electrons

Next we can use equation Q = n.F to determine Faraday's constant, since charge is given.

2021 p1 q27 calculate faraday constant

Finally we can use equation F = e.L to solve for Avogadro's constant L.

2021 p1 q27 calculate avogadro constant

The calculated value (6.05 x 1023) is quite close to the actual value of Avogadro's constant which is 6.02 x 1023.

Therefore the answer to this question is option B.

2021 p1 q27 answer

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

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2021 P1 Q26 - How to increase Cell Potential

Let's take a look at 2021 A Levels H2 Chemistry Paper 1 Question 26.

2021 p1 q26 question

We want to determine which of the changes will lead to a higher cell potential.

First let us decide the anode and cathode for this electrochemical cell.

The E value for standard copper electrode is +0.34V and that for standard hydrogen electrode is 0.00V.

Since the copper electrode has a more positive E value, it is more likely reduced hence the cathode.

The standard hydrogen electrode has a more negative E value, more likely oxidised hence the anode.

2021 p1 q26 determine anode and cathode

The Ecell is given by Ereduction - Eoxidation.

To have a higher cell potential we can either have:

1. Higher E value for copper electrode which is the cathode (higher Ereduction)
2. Lower E value for hydrogen electrode which is the anode (lower Eoxidation)

2021 p1 q26 how to have greater Ecell

Let's look at the options and see which will increase cell potential.

A. decreasing the pressure of the hydrogen gas of the hydrogen electrode

2021 p1 q26 option A

Decreasing pressure of hydrogen gas will shift the position of equilibrium to the right.

Reduction is favoured hence E value for hydrogen electrode increases.

Since hydrogen electrode is the anode, a larger Eoxidation will cause the Ecell to be smaller.

Hence option A is wrong.

B. increasing the hydrogen ion concentration of the hydrogen electrode

2021 p1 q26 option B

Increasing hydrogen ion concentration will shift the position of equilibrium to the right.

The result will be exactly the same as option A hence option B is also wrong.

C. using 1.0 moldm-3 copper(II) nitrate instead of copper(II) sulfate solution

2021 p1 q26 option C

Both 1.0 moldm-3 copper (II) nitrate and copper (II) sulfate will give the same concentration of 1.0 moldm-3 Cu2+(aq).

Since there is no change to Cu2+ concentration, there is no change to E value for copper electrode and overall Ecell.

Hence option C is wrong.

D. using 1.0 moldm-3 ethanoic acid instead of hydrochloric acid in the hydrogen electrode

1.0 moldm-3 strong acid HCl is fully dissociated hence will give 1.0 moldm-3 hydrogen ions.

1.0 moldm-3 weak acid ethanoic acid is partially dissociated hence will give a lower concentration of H+.

2021 p1 q26 option D

Decreasing hydrogen ion concentration will shift the POE to the left.

Reduction is disfavoured hence E value for hydrogen electrode decreases.

Since hydrogen electrode is the anode, a smaller Eoxidation will cause the Ecell to be larger.

Hence option D is our answer to this question.

2021 p1 q26 answer

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

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