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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...