2 states of aggregation should not be subscripted
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I have the redox reaction $\ce{N_2H_4_{(g)} + N_2O_4_{(g)} \rightarrow N_2{(g)} + H_2O_{(g)}}$$\ce{N_2H_4 {(g)} + N_2O_4 {(g)} -> N_2 {(g)} + H_2O {(g)}}$.

In $\ce{N_2O_4_{(g)}}$$\ce{N_2O_4 {(g)}}$, the oxidation state of nitrogen is $+4$. In $\ce{N_2_{(g)}}$$\ce{N_2 {(g)}}$, the oxidation state of $\ce{N}$ is $0$. Thus, $\ce{N_2O_4_{(g)}}$$\ce{N_2O_4 {(g)}}$ is reduced.

In $\ce{N_2H_4_{(g)}}$$\ce{N_2H_4 {(g)}}$, the oxidation state of nitrogen is $+2$, because the oxidation state of hydrogen when bonded to nonmetals is $-1$. In $\ce{N_2_{(g)}}$$\ce{N_2 {(g)}}$, the oxidation state of $\ce{N}$ is $0$. Thus, $\ce{N_2H_4_{(g)}}$$\ce{N_2H_4 {(g)}}$ is also reduced.

Clearly, both $\ce{N_2O_4_{(g)}}$$\ce{N_2O_4 {(g)}}$ and $\ce{N_2H_4_{(g)}}$$\ce{N_2H_4 {(g)}}$ cannot both be reduced. WhyWhat is wrong with this?

I have the redox reaction $\ce{N_2H_4_{(g)} + N_2O_4_{(g)} \rightarrow N_2{(g)} + H_2O_{(g)}}$.

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. Why is wrong with this?

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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Why is this redox reaction possible?

I have the redox reaction $\ce{N_2H_4_{(g)} + N_2O_4_{(g)} \rightarrow N_2{(g)} + H_2O_{(g)}}$.

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. Why is wrong with this?