Chapter 19, Problem 67A

Interpretation Introduction

Interpretation:

The redox reaction in the sapphire, oxidizing and reducing agents are to be written

Concept introduction:

Redox reaction is the reaction in which both oxidation and reduction takes place.

Oxidation is loss of electrons and reduction is gain of electrons.

Answer to Problem 67A

The reaction is:

$Cu+AgN{O}_3\to Cu{\left(N{O}_3\right)}_2+Ag$

Oxidation number of each element is:

$\stackrel{0}{Cu}+\stackrel{+1}{Ag}N{O}_3\to \stackrel{+2}{Cu}{\left(N{O}_3\right)}_2+\stackrel{0}{Ag}$

Oxidation half reaction:

$Cu\to C{u}^{2+}+2e^{-}$

Reduction half reaction:

$A{g}^{+}+e^{-}\to Ag$

Balanced equation:

$Cu+2AgN{O}_3\to Cu{\left(N{O}_3\right)}_2+2Ag$

Explanation of Solution

The reaction is:

$Cu+AgN{O}_3\to Cu{\left(N{O}_3\right)}_2+Ag$

Oxidation number of each element is:

$\stackrel{0}{Cu}+\stackrel{+1}{Ag}N{O}_3\to \stackrel{+2}{Cu}{\left(N{O}_3\right)}_2+\stackrel{0}{Ag}$

Oxidation number of copper is increasing by 2 and silver is decreasing by 1

Balanced equation is:

$Cu+2AgN{O}_3\to Cu{\left(N{O}_3\right)}_2+2Ag$

Conclusion

Balanced equation is:

$Cu+2AgN{O}_3\to Cu{\left(N{O}_3\right)}_2+2Ag$