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.

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.

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 E_cell for each stage of reduction of vanadium species to determine its extent of reduction.

E_cell 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⁺.

E_cell 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.

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.

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.

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.

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

Therefore Cu²⁺ will be reduced at the cathode.

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

Hence the answer to this question is option C.

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.

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.

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.

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

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

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.

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.

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.

The E_cell is given by E_reduction - E_oxidation.

To have a higher cell potential we can either have:

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

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

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 E_oxidation will cause the E_cell to be smaller.

Hence option A is wrong.

B. increasing the hydrogen ion concentration of the hydrogen electrode

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

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 E_cell.

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⁺.

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 E_oxidation will cause the E_cell to be larger.

Hence option D is our answer to this question.

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

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