Chapter 19, Problem 89A

Interpretation Introduction

(a)

Interpretation:

Chemical equation is to be written. Oxidation states are to be determined. Half cell reactions are to be written. Then balancing is to be done.

Concept introduction:

Oxidation involves loss of electron which is represented by oxidation half reaction and reduction involves gain of electron which is represented by reduction half reaction.

Answer to Problem 89A

$HgO\to Hg+O_2$

Oxidation states:

$\stackrel{+2}{Hg}\stackrel{-2}{O}\to \stackrel{0}{Hg}+\stackrel{0}{O_2}$

Oxidation half equation:

$O^{2-}\to O_2+2e^{-}$

Reduction half reaction:

$H{g}^{2+}+2e^{-}\to Hg$

Balanced equation:

$2HgO\to 2Hg+O_2$

Explanation of Solution

$HgO\to Hg+O_2$

Oxidation states:

$\stackrel{+2}{Hg}\stackrel{-2}{O}\to \stackrel{0}{Hg}+\stackrel{0}{O_2}$

Oxidation half equation:

$O^{2-}\to O_2+2e^{-}$

Reduction half reaction:

$H{g}^{2+}+2e^{-}\to Hg$

Balanced equation:

$2HgO\to 2Hg+O_2$

Interpretation Introduction

(b)

Interpretation:

Chemical equation is to be written. Oxidation states are to be determined. Half cell reactions are to be written. Then balancing is to be done.

Concept introduction:

Oxidation involves loss of electron which is represented by oxidation half reaction and reduction involves gain of electron which is represented by reduction half reaction.

Answer to Problem 89A

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

Oxidation states:

$\stackrel{0}{Cu}+\stackrel{+1}{Ag}\stackrel{+5}{N}{\stackrel{-2}{O}}_3\to \stackrel{0}{Ag}+\stackrel{+2}{Cu}\stackrel{+5}{(N}{\stackrel{-2}{O}}_3{)}_2$

Oxidation half equation:

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

Reduction half reaction:

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

Balanced equation:

$Cu+2A{g}^{+}\to 2Ag+C{u}^{2+}$

Or

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

Explanation of Solution

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

Oxidation states:

$\stackrel{0}{Cu}+\stackrel{+1}{Ag}\stackrel{+5}{N}{\stackrel{-2}{O}}_3\to \stackrel{0}{Ag}+\stackrel{+2}{Cu}\stackrel{+5}{(N}{\stackrel{-2}{O}}_3{)}_2$

Oxidation half equation:

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

Reduction half reaction:

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

Adding both equations after multiplying reduction half cell with 2

Balanced equation:

$Cu+2A{g}^{+}\to 2Ag+C{u}^{2+}$

Or

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