I am not an experienced electrical engineer, so don't take anything I say as necessarily true, but instead check everything. If any reader is an experienced electrical engineer, please check what I wrote. I'll be happy to accept corrections.
First of all, your amplifier circuit looks really weird. The inputs to the op amp are simply labeled "lead", but it's not clear if they are the same signal on two wires; a differential signal, where one is the negative of the other; or perhaps one is a signal and the other is cable (signal) ground. In any case, this circuit will not work, as several others have already pointed out:
- In a (non-differential) amplifier, there is only one input signal, not two. In your case, it looks like that ought to be the left lead. Connecting both leads (or one lead twice) to the inputs of an op amp is weird (outside of differential amplifiers).
- I'm going on about differential amplifiers because it seems to me that an ECG lead might well pick up common-mode noise, i.e., noise that affects both the signal lead and the signal ground equally and in the same direction, and differential amplifiers are very good at getting rid of such noise. But the picture you show is not a differential amplifier.
- The feedback voltage seems to be added to the right lead without a coupling capacitor. That means that if the feedback voltage contains a nontrivial DC component (as it seems it should in your case, see below), current could flow into your lead. If there aren't any protections in place in the ECG probe to prevent that from happening, the result could be unpleasant.
Then you say that your signal has a "potential" of 2 mV, but the picture on your oscilloscope shows a hefty DC bias of nearly 25 mV (if I'm reading this correctly) with a 2 mV signal riding on top of it. Do you want to amplify that bias as well, or just the raw AC signal?
What you also don't show is how you have connected the power. Have you connected the positive rail to the positive terminal of the battery (+9 V) and the negative to the negative terminal (0 V)? Or have you created a balanced supply of +/-4.5 V with a divider?
If you want to amplify the DC bias along with the signal in a non-differential amplifier, you can use the circuit depicted here. Don't worry if the op amp shown is not what you have. It's a bog-standard noninverting amplifier with gain of 101.
However, if I'm right about that +25 mV DC bias, that will be amplified as well. With a gain of 101, your output signal will have a bias of about +2.5 V, with a 0.2 V signal riding on top of that. If you want to remove that, put a small capacitor on the output. You can use an electrolytic cap, just make sure that the polarity is right.
I've put a bypass capacitor on the power connection in the circuit, which might or might not be needed, depending on what else that battery is powering.