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2 states of aggregation should not be subscripted

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?