okay, so here's another question: the reaction for G-3-P oxidizing to 1,3-BPG is:
G-3-P + NAD+ + Pi --> 1,3-BPG + NADH + H+.
I understand that the hydride ion leaves G-3-P and donates two electrons and one proton, which neutralizes the charge on NAD+ and gives NADH. But it is only giving one...