I think that on cathode (minus, Pt) is produced hydrogen. On the anode (plus, Ag) is produced chlorine ( and AgCl).
Cathode:
Na+ on the cathode is more difficult to reduce than H+, thus H2 evolution is occurred and NaOH is produced.
Anode:
Cl- is oxidized and Cl2 gas is produced. Meanwhile also AgCl is produced. If electrolyte is not divided by semipermeable membrane which protect anode area against NaOH, NaOH reacts with Cl2 and NaCl + NaClO + H2O are produced.
To sum up, on cathode hydrogen evolution is occurred and on anode chlorine evolution is occurred. If you seen bubbles on one side only it was hydrogen on cathode (platinum) side. On anode you were producing AgCl and then also NaClO.
I agree with the above two researchers. Although your setup is invisible, based on your description, it is clear that the Ag electrode experienced oxidation reaction, i.e., Ag - e + Cl- = AgCl. The electrons lost by Ag have to be consumed on Pt.
In order to form Cl2 or O2, both Cl- or H2O must be oxidized, i.e., lose electrons. This statement contracts with the above analysis that the reaction on Pt must take electrons, i.e., reduction. Therefore, the most likely reaction on Pt is reduction of water that generates H2.
Cl2 is lightly green, you might distinguish it if it indeed forms and accumulates for a while.