Answer
Oxidizing agent: $N_{2}$
Reducing agent : $H_{2}$
Work Step by Step
Find oxidation state of each element in the reaction. In $N_{2}$, oxidation state of N = 0. In $H_{2}$, oxidation state of H= 0. In $NH_{3}$, oxidation state of H = +1. Let's say oxidation state of N = $x$. Hence, $x + (+1)\times3 = 0$
By solving we get $x = -3$ So, oxidation state of N = -3.
So, we can see that oxidation state of N is decreased (0 in $N_{2}$ to -3 in $NH_{3}$). Hence, $N_{2}$ is reduced. Therefore, $N_{2}$ is oxidizing agent.
oxidation state of H is increased (0 in $H_{2}$ to +1 in $NH_{3}$). Hence, $H_{2}$ is oxidized. Therefore, $H_{2}$ is the reducing agent.
