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oxidation value of each N atom in hydrazine (H2NNH2) is -2 and in NO3- it is +5

(standard in compounds: H is +1, O is -2, but these stay unchanged)

The difference in oxidation value is (-2) -(+5) = -7 = 7 electrons (e-) per N atom (or 14 e- per 2N's in one N2H4).

So 7 electrons are to be donated to - (and taken or accepted by) one NO3- ion.

oxidator: 2NO3- + 16 H+ + 14 e- --> N2H4 + 6 H2O

reductor (donator): XXX --> YYY + 14 e-

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