Remember that no oxidation can occur without a reduction happening at the same time, and vice versa. If you apply a positive potential to your working electrode, and something is oxidised on it, a reduction reaction must occur simultaneously at your counter electrode. Obviously, the current at the working electrode and that at the counter electrode must be identical (and of opposite sign). If the surface area of your counter electrode is small compared to the surface area of your working electrode, this will translate into a high current density at your counter electrode, that can only be attained if its potential is made very negative (very positive if your working electrode is negative). On platinum, this can easily lead to hydrogen evolution. Incidentally, if the current density is very high, the potential of the counter electrode should be so negative (positive, if your working electrode is negative), that the voltage between the working and the counter electrode would exceed the maximum voltage that the potentiostat can apply (in most potentiostats, this is around 10 V). That is why it is usually recommended to use counter electrodes with a surface area larger than that of the working electrode.

Usually this happen when your counter electrode is not compensation the current flow well enough, probably because your working electrode have a large surface area, and a massive current, you should try to use a larger counter electrode, another possibility is that you connect the electrodes in a wrong way, but I think the first option is the correct.

Hydrogen evolution when Pt used as cathode.

PL SEE A GOOD BOOK ON ELECTROCHEMISTRY

It seems you had gas evolution reaction. A positive potential is your working electrode. The counter electrode will have potential different from the working electrode. The potential reading (vs. Hg/HgO) is for the working electrode. I would go with HER.

Remember that no oxidation can occur without a reduction happening at the same time, and vice versa. If you apply a positive potential to your working electrode, and something is oxidised on it, a reduction reaction must occur simultaneously at your counter electrode. Obviously, the current at the working electrode and that at the counter electrode must be identical (and of opposite sign). If the surface area of your counter electrode is small compared to the surface area of your working electrode, this will translate into a high current density at your counter electrode, that can only be attained if its potential is made very negative (very positive if your working electrode is negative). On platinum, this can easily lead to hydrogen evolution. Incidentally, if the current density is very high, the potential of the counter electrode should be so negative (positive, if your working electrode is negative), that the voltage between the working and the counter electrode would exceed the maximum voltage that the potentiostat can apply (in most potentiostats, this is around 10 V). That is why it is usually recommended to use counter electrodes with a surface area larger than that of the working electrode.

Thank you for good answers, professors and doctors. I will change Pt electrode with more large surface area.

Hydrogen evolution. If the potential is to high also hydrogen evolution may occur. Has negative impact during measurements. The pH of solution also is critical. If the buffer is not suitable, some protons can interfere in measurement. Good Job.

please cheek also your connection, if there is problem with your reference electrode connection

Hi.

The most common cause is that you have connected the electrodes in a bad position, make sure the working and reference are at the right plug.