Question

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

Answer 1

Answer:

Cu

Explanation:

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Answer 2

Answer:

Cu

Explanation:

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

$\rm \stackrel{\hbox{0}}{\hbox{Cu}} + 4H\stackrel{\hbox{+5}}{\hbox{N}}O_{3} \longrightarrow \, \stackrel{\hbox{2+}}{\hbox{Cu}}(NO_{3})_{2} + 2\stackrel{\hbox{+4}}{\hbox{N }}O_{2} + 2H_{2}O$

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