Question #d09f9

May 17, 2017

$N$ is being oxidized,
$H$ is being reduced.
Since ${N}_{2} \left(g\right)$ has an oxidation state of $0$ originally, and then has an oxidation state of $+ 3$ on the products side, $N$ is being oxidized.
Since ${H}_{2} \left(g\right)$ has an oxidation state of $0$ originally, and then has an oxidation state of $- 1$ on the products side, $H$ is being reduced.