Urgent a level help- electrochemistry! (cie)

Hi,
My A2 exams begin in a few days so help would be greatly appreciated!

This is related to solving problems using half-equations for cells.
In my textbook it says that if the half-equation shows oxidation, we need to reverse the sign of the electrode potential given in the data booklet because in the data booklet, values are given for the reduction reaction. All seemed fine until this question came along in the past papers- May/June 2010 paper 41, Question 5.b)

I've attached the question, mark scheme and data booklet values below.

So basically my question is that, for the first reaction, shouldn't 0.13 be made positive from negative as oxidation is taking place? But then again the reaction is at the cathode...so how can oxidation take place at the cathode?
Also, can somebody please explain how you get the overall cell reaction in the presence of sulphuric acid??

Please help!

Your textbook is wrong. You can't reverse the sign on a reduction potential.

You can see that the first equation is oxidation since electrons are being lost, likewise the second equation is reduction since electrons are gained. $E_{cell}=E_{red}-E_{ox}$ which gives E=1.47-(-0.13)

Working from the overall cell reaction there are 2Pb2+ produced so if we add 2SO4 to the left hand side then all of the ions will be precipitated as PbSO4

Thank you so much! That was really helpful

So can oxidation occur at the cathode? Because I thought only the positive and negative electrodes are interchanged between electrolysis and electrochemical cells...

No, oxidation only happens at the anode.

Posted from TSR Mobile

Exactly....but in the question they've written cathode...

There are two phases in a rechargeable battery, the charging and discharging phase.

Technically the anode changes as the reverse reactions occur, but I suspect that they are keeping the terms incorrectly the same.

It should be negative and cathode for the charging phase and negative and anode for the discharging phase.

During the charging phase there is oxidation at the electrode connected to the positive terminal of the external potential difference. This is the anode.
Pb²⁺ --> Pb⁴⁺ + 2e
During the discharging phase there is reduction at this electrode (the reverse reaction) so it becomes the cathode and is positive as it is attracting electrons.
Pb⁴⁺ + 2e --> Pb²⁺