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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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Exercise 18.4 Copper metal reacts with nitric acid, HNO3(aq), to give aqueous copper(II) nitrate, water, and nitrogen monoxide gas as products. Write and balance the equation for this reaction.

Interpretation Introduction

Interpretation:

Write and balance the equation for the given reaction.

When copper metal reacts with nitric acid to give aqueous copper(II) nitrate, water, nitrogen monoxide as products.

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

Oxidation is the process in which either loss of electrons, oxidation number increases, or loss of hydrogen atoms takes place. An element is oxidized, when oxidation number increases.

Reduction is the process in which either gain of electrons, oxidation number decreases, or gain hydrogen atoms takes place. An element is reduced, when oxidation number decreases.

The balanced chemical equation is the equation in which number of different atoms of elements in the reactants side is similar to the product side.

Explanation

The unbalanced reaction is given as:

Cu (s)+HNO3(aq)Cu(NO3)2(aq)+H2O(aq)+NO(g)

The oxidation half-reaction is:

Cu+HNO3Cu(NO3)2

Here, oxidation state of copper is 0, hydrogen is + 1, nitrogen is + 5, oxygen is 0 on reactant side and on product side, oxidation state of copper is + 2, nitrogen is + 5 and oxygen is -2.

Oxidation state of copper goes from 0 to + 2. Therefore, copper is oxidized.

Reduction reaction is:

 HNO3NO

Here, oxidation state of hydrogen is + 1, nitrogen is + 5 and oxygen is -2 on reactant side and on product side, oxidation state of nitrogen is + 2 and oxygen is -2.

Oxidation state of nitrogen goes from + 5 to + 2. Therefore, nitrogen is reduced.

First, balance the oxidation-half reaction:

Cu+HNO3Cu(NO3)2

On reactant side, number of nitrogen atoms is one and number of nitrogen atoms on product side is two. Thus, multiply HNO3 by 2. Equation becomes:

Cu+2HNO3Cu(NO3)2

Now, number of oxygen atoms on reactant side is 6 and number of oxygen atoms on product side is 6. Thus, oxygen is balanced.

Balance hydrogen using H+. Equation becomes:

Cu+2HNO3Cu(NO3)2+2H+

Now, balance the charge by using e-. Equation becomes:

Cu+2HNO3Cu(NO3)2+2H++2e

Balance the reduction half reaction:

 HNO3NO

In this reaction, number of nitrogen is on reactant side as well as on product side.

Now, balance the hydrogen atom and oxygen atom.

Add water molecule to balance the reaction:

 HNO3NO+2H2O

Balance the hydrogen using H+. Equation becomes:

 3H++ HNO3NO+2H2O

Balance the charge by using e-

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