Substances that are oxidized and reduced in the following reaction are respectively
N₂H_4(l)
+ 2H₂O_2(l)

→ N_2(g) + 4H₂O_(l)

${\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{l}\right)+2{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

${\mathrm{N}}_2{\mathrm{H}}_4$ is oxidized and ${\mathrm{H}}_2{\mathrm{O}}_{2 }$is reduced.

Hence,  ${\mathrm{H}}_2{\mathrm{O}}_{2 }$is an oxidizing agent and ​${\mathrm{N}}_2{\mathrm{H}}_4$ is a reducing agent.