Do you set up testing conditions properly?

Giang Nguyen, yes I do it properly. OCV is checked for 24 hrs(2.92V).

Most of the times it happens for the first cycles. Sometimes curve becomes fine with 2nd or 3rd loop.

If not voltage keeps decreasing until it becomes less than zero and ultimately cell is useless.

Ritesh Yadava Can you reveal your testing conditions and your materials?

Since the basic makeup of the cell hasn't been disclosed, forgive my pointing out the obvious: in a half cell using lithium metal as cathode, the voltage will naturally decrease with charging, since the potential between the anode and the reference (lithium metal, also the cathode in this case) will decrease as the anode becomes lithiated.

Assuming your cells are not anode half-cells, I would suggest contacting Metrohm sales support, since your graph is clearly made by Nova. They tend to be helpful with advice on using their equipment/software.

Sure, It is two electrode cell.With Lithium reference electrode and graphite powder as working/active material.

I struggle to understand how to do formation process. I think that's the part of problem here

One advice I got is to discharge the cell first and then charge it for initial cycles.

Currently I am doing charging(I: +ve) first followed by discharging (I:-ve)(pic attached with question)

All the

It would be helpful if you share some references /paper which explains the problem I face.

Thanks & regards

Ritesh Yadava You need to do the discharge step first. In the Li||Graphite half cell, graphite will act as a cathode, but without lithium inside. So, if you charge it first, what kind of matter can be released???

Therefore, it needs to be lithiated (discharge) to receive Li from the Li anode at the initial step.

I must respectfully disagree with the previous post, since graphite is widely used as an anode for Li-ion cells. Since it is not a source of lithium, it is not used as a cathode. Even the terms "anode" and "cathode" are used cheekily by the battery research community, since the electrodes switch roles when charging, but I will use these terms conventionally to explain.

Your cells are made with a graphite electrode (unlithiated) and a lithium electrode, therefore the lithium is serving as the source of lithium, and as the cathode, and also as the reference electrode. Upon construction of your half-cell, the OCP (2.92V) indicates the combined potential of each electrode/electrolyte interface, so I can assume your electrolyte is a conventional LiPF6-in-carbonates liquid. However, this half-cell is not ready to provide energy to a load, since the graphite must first be lithiated. Although some galvanostatic equipment (like my Maccor cycler) considers any procedure decreasing the OCP as a "discharge", this term is counter-intuitive with the graphite-lithium half-cell. Before this cell can release any stored energy, the graphite must be lithiated, which means energy must be added to the system to overcome the energy barrier of intercalating lithium ions into the graphite. Some might call this "discharge" because the OCP is decreasing as it would if a full cell were being discharged, but since one must add energy to perform this procedure, I avoid that term with anode half-cells.

Here are some well-written papers that explain this better than I have:

https://www.sciencedirect.com/science/article/pii/S2352484721001463

https://iopscience.iop.org/article/10.1149/2.1171902jes

https://www.nature.com/articles/s41467-020-16259-9

And here is some lit on the Metrohm website about using Autolab to peform these tests: https://www.metrohm.com/en_gb/applications/application-notes/autolab-applikationen-anautolab/an-bat-002.html

I know anode and cathode terminologies can be confused and dependent on the specific electrochemical process. However, these terms are widely used in the scientific community.

As I understand, you may think the electrode that stores lithium is the cathode. However, it's not common to refer to lithium metal as a cathode material. The distinction between anode and cathode also takes into account the potential of each electrode in the cell. We need to consider in both electrochemical and thermodynamic aspects. Using positive and negative terminologies may make the distinction clearer.

For the Li||Gr half-cell, it is actually ready for use without requiring any lithiation process. If you do lithiation, the cell potential would reach approximately 0V or exhaust. The question that arises is: how can it generate energy in this state? Note that this is only for demonstration purposes, as the energy output for Li||Gr cell is too low. Similarly, for an actual full-cell configuration like Gr||LFP, we need first de-lithiated LFP before using.

Thanks for adding to my knowledge. I also believe the terminologies are confusing sometimes in literature and as a beginner it gets even difficult to grasp the actual point.That's why I like to talk in term of current I=+ve(charge) or I=-ve(discharge) terms.

Image attached in question is for the first cycle(Charge cycle first(I=+ve) )

Ritesh Yadava As aforementioned, you need to apply a negative current for the first step.