Answer:
Cu
Explanation:
We can use oxidation numbers to figure out what is happening with the electrons.
[tex]\rm \stackrel{\hbox{0}}{\hbox{Cu}} + 4H\stackrel{\hbox{+5}}{\hbox{N}}O_{3} \longrightarrow \, \stackrel{\hbox{2+}}{\hbox{Cu}}(NO_{3})_{2} + 2\stackrel{\hbox{+4}}{\hbox{N }}O_{2} + 2H_{2}O[/tex]
The oxidation number of Cu increases from 0 to +2 in Cu(NO₃)₂. The copper has gained two positive charges, so it has lost two electrons. It has been oxidized, so it is the reducing agent.
NO₃⁻ is wrong. The oxidation number of N decreases from +5 in NO₃⁻ to +4 in NO₂, so it has been reduced. It is the oxidizing agent.
Cu(NO₃)₂ and NO₂ are wrong. They are the products of the reaction. They cannot be either oxidizing or reducing agents in this reaction
