Consider the reaction that occurs when copper is added to nitric acid. Cu(s)+4HNO3(aq) > Cu(NO3)2(aq)+2NO2(g)+2H2O(I). What is the reducing agent in this reaction? Cu NO3– Cu(NO3)2 NO2

Cu

Explanation:

We can use oxidation numbers to figure out what is happening with the electrons.

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

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its going to be mutualism cuz they both benefit from the relationship.

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