To determine the correct statement(s) regarding the electrochemical oxidation of hydrazine ($$\mathrm{N}_2 \mathrm{H}_4$$) by $$\mathrm{O}_2$$, let's analyze each option in detail.

Option A: $$\mathrm{OH}^{-}$$ ions react with $$\mathrm{N}_2 \mathrm{H}_4$$ at the anode to form $$\mathrm{N}_2$$$$(\mathrm{~g})$$ and water, releasing 4 electrons to the anode. This statement is accurate because in an electrochemical cell, the anode is where oxidation occurs. Hydrazine can be oxidized in the presence of $$\mathrm{OH}^{-}$$ ions to yield nitrogen gas and water, releasing electrons as a part of this redox reaction:

$$ \mathrm{N}_2 \mathrm{H}_4 + 4 \mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4 \mathrm{H}_2 \mathrm{O} + 4e^{-} $$

Option B: At the cathode, $$\mathrm{N}_2 \mathrm{H}_4$$ breaks to $$\mathrm{N}_2$$ $$(\mathrm{~g})$$ and nascent hydrogen released at the electrode reacts with oxygen to form water. This statement is incorrect because, usually, the cathode is where reduction occurs, not a breaking down of hydrazine. Instead, reduction reactions involve the gain of electrons. Additionally, at the cathode, it's more common for oxygen to be reduced rather than hydrazine itself breaking down.

Option C: At the cathode, molecular oxygen gets converted to $$\mathrm{OH}^{-}$$. This statement is correct. At the cathode in an alkaline medium, oxygen undergoes reduction to form hydroxide ions $$\mathrm{OH}^{-}$$, as depicted by the half-reaction:

$$ \mathrm{O}_2 + 2H_2O + 4e^{-} \rightarrow 4 \mathrm{OH}^{-} $$

Option D: Oxides of nitrogen are major by-products of the electrochemical process. This statement is incorrect because the primary product from the electrochemical reaction of hydrazine is nitrogen gas ($$\mathrm{N}_2$$), and not oxides of nitrogen. The formation of oxides of nitrogen would be notable only under different specific conditions not mentioned in this context.

Therefore, the correct statements regarding the electrochemical oxidation of hydrazine by $$\mathrm{O}_2$$ are:

Option A and Option C.