Why is this redox reaction possible?

I have the redox reaction $\ce{N_2H_4 {(g)} + N_2O_4 {(g)} -> N_2 {(g)} + H_2O {(g)}}$.

In $\ce{N_2O_4 {(g)}}$, the oxidation state of nitrogen is $+4$. In $\ce{N_2 {(g)}}$, the oxidation state of $\ce{N}$ is $0$. Thus, $\ce{N_2O_4 {(g)}}$ is reduced.

In $\ce{N_2H_4 {(g)}}$, the oxidation state of nitrogen is $+2$, because the oxidation state of hydrogen when bonded to nonmetals is $-1$. In $\ce{N_2 {(g)}}$, the oxidation state of $\ce{N}$ is $0$. Thus, $\ce{N_2H_4 {(g)}}$ is also reduced.

Clearly, both $\ce{N_2O_4 {(g)}}$ and $\ce{N_2H_4 {(g)}}$ cannot both be reduced. What is wrong with this?

• Well first you need to balance the equation. $\ce{2N2H4_{(g)} + N2O4_{(g)} -> 3N2{(g)} + 4H2O{(g)}}$. Then what are the half cell reactions? – MaxW Jun 29 '18 at 2:02
• I'm not sure, because in both reactants, the nitrogen atom has a positive oxidation state. They can't both be reduced. – coder Jun 29 '18 at 2:15
• Again stop thinking about "oxidation states" and think half cells. Add both $\ce{H+ \text{and} e-}$ to one side to create a balanced equation. Start with hydrazine. $\ce{N2H4 -> \text{?}}$ – MaxW Jun 29 '18 at 2:20
• On an unrelated note, what oxidation state would you assign to F in HF? – Ivan Neretin Jun 29 '18 at 5:07
• The oxidation state of hydrogen is $+1$ when bonded to non-metals and not $-1$. – PolarBear Jul 14 '18 at 4:14