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 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)
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 Eoxidation will cause the Ecell 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 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+.
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.
Topic: Electrochemistry, Physical Chemistry, A Level Chemistry, Singapore
Back to list of questions for 2021 A Level H2 Chemistry Paper 1.
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