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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?

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

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