2021 P1 Q26 - How to increase Cell Potential

Let Chemistry Guru, Singapore's top choice for JC Chemistry tuition, guide you through 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

Back to list of questions for 2021 A Level H2 Chemistry Paper 1.

Found this A Level Chemistry video useful?

This free chemistry video lesson is brought to you by Chemistry Guru, Singapore's top JC Chemistry tuition centre since 2010.

Please like this video and share it with your friends!

Join my 18,000 subscribers on my YouTube Channel for new A Level Chemistry video lessons every week.

Check out other A Level Chemistry Video Lessons here!

Need an experienced tutor to make Chemistry simpler for you?

Do consider signing up for my JC Chemistry Tuition classes at Bishan or on-demand video lessons!