Chapter 17, Problem 9QAP

Write balanced net ionic equations for the following reactions in acid solution.
(a) Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed.
(b) Solid phosphorus (P₄) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate (H₂PO₄^-) ions.
(c) Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese(II) ions are formed.

Interpretation Introduction

(a)

Interpretation:

The net ionic equation needs to be determined and balanced in the acidic medium.

Concept introduction:

The rules to balance redox reactions in basic medium are as follows:

1. Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
2. Balance the atoms of elements other than H and O atoms. Use H₂ O to balance the number of O atoms, H+ to balance the number of H atoms and the electrons for charge balance. Equalize the number of electrons on both the half reactions and then add the half reactions.

Answer to Problem 9QAP

${\text{6N}}_{\text{2}}{\text{H}}_{\text{4}}\left(l\right)\text{+}{\text{4BrO}}_{\text{3}}{}^{\text{-}}\left(aq\right)\to {\text{ 6N}}_{\text{2}}\left(g\right)+{\text{4Br}}^{\text{-}}\left(aq\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The reaction of liquid hydrazine with an aqueous solution of sodium bromate results in the formation of nitrogen gas and bromide ions as follows:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(l\right){\text{ + BrO}}_{\text{3}}{}^{\text{-}}\left(aq\right)\to {\text{ N}}_{\text{2}}\left(g\right){\text{+ Br}}^{\text{-}}\left(aq\right)$

Determining the oxidation numbers:

The oxidation number of Br in BrO₃ - and Br- is +5 and -1 respectively. As the oxidation number is decreasing this is the reduction half of the reaction.

Oxidation half reaction:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(l\right)\to {\text{ N}}_{\text{2}}\left(g\right)\text{+}{\text{4H}}^{\text{+}}\left(aq\right)\text{+ 4e}$

To balance the excess H atom on the reactantside by adding four H+ on the product side. Finally we balance the charge by adding four electrons to the product side of the half reaction.

Reduction half reaction:

${\text{BrO}}_{\text{3}}{}^{\text{-}}\left(aq\right)+{\text{6H}}^{\text{+}}\left(aq\right)\text{+}\text{6e}\to {\text{ Br}}^{\text{-}}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

To balance the excess O atomson the reactant side we add three H₂ O on the product side, and then we balance the excess H atom on the product side by adding six H+ on the reactant side. Finally we balance the charge by adding six electrons to the reactant side of the half reaction.

Net reaction:

We multiply the reduction half reaction by four and the oxidation half reaction by sic, in order to cancel out the electrons in the net reaction. Add the half reactions to get the net reaction.

${\text{4BrO}}_{\text{3}}{}^{\text{-}}\left(aq\right)+24{\text{H}}^{\text{+}}\left(aq\right)\text{+}24\text{e}\to {\text{ 4Br}}^{\text{-}}\left(aq\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)$

${\text{6N}}_{\text{2}}{\text{H}}_{\text{4}}\left(l\right)\to {\text{ 6N}}_{\text{2}}\left(g\right)\text{ +}2{\text{4H}}^{\text{+}}\left(aq\right)\text{+ 24e}$

${\text{6N}}_{\text{2}}{\text{H}}_{\text{4}}\left(l\right)\text{+}{\text{4BrO}}_{\text{3}}{}^{\text{-}}\left(aq\right)\to {\text{ 6N}}_{\text{2}}\left(g\right)+{\text{4Br}}^{\text{-}}\left(aq\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(b)

Interpretation:

The net ionic equation needs to be determined and balanced in the acidic medium.

Concept introduction:

The rules to balance redox reactions in basic medium are as follows:

1. Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
2. Balance the atoms of elements other than H and O atoms. Use H₂ O to balance the number of O atoms, H+ to balance the number of H atoms and the electrons for charge balance. Equalize the number of electrons on both the half reactions and then add the half reactions.

Answer to Problem 9QAP

${\text{3P}}_{\text{4}}\left(s\right)+{\text{20NO}}_{\text{3}}{}^{\text{-}}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+8{\text{H}}^{\text{+}}\left(aq\right)\to 12{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right)+\text{20NO}\left(g\right)$

Explanation of Solution

The reaction of solid phosphorus (P₄ ) with an aqueous solution of nitrate results in the formation of nitrogen oxide and dihydrogen phosphate (H₂ PO₄ ) ions as follows:

${\text{P}}_{\text{4}}\left(s\right){\text{+ NO}}_{\text{3}}{}^{\text{-}}\left(aq\right)\to \text{ NO}\left(g\right){\text{ +H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right)$

Determining the oxidation numbers:

The oxidation number of N in NO and NO₃ - is +2 and +2 respectively. As the oxidation number is increasing this is the oxidation half of the reaction.

Oxidation half reaction:

${\text{NO}}_{\text{3}}{}^{\text{-}}\left(aq\right)\text{+}{\text{4H}}^{\text{+}}\left(aq\right)+3{\text{e}}^{-}\to \text{ NO}\left(g\right){\text{+ 2H}}_{\text{2}}\text{O}\left(l\right)$

To balance the excess O atoms on the reactant side we add two H₂ O on the product side, and then we balance the excess H atom on the product side by adding four H+ on the reactant side. Finally we balance the charge by adding three electrons to the reactant side of the half reaction.

Reduction half reaction:

${\text{P}}_{\text{4}}\left(s\right)+16{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 4{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right)+24{\text{H}}^{\text{+}}\left(aq\right)+20{\text{e}}^{-}$

First we balance the P atom on both side of the reaction. To balance the excess O atoms on the product side we add sixteen H₂ O on the reactant side, and then we balance the excess H atom on the reactant side by adding twenty four H+ on the product side. Finally we balance the charge by adding twenty electrons to the product side of the half reaction.

Net reaction:

We multiply the oxidation half reaction by twenty and the reduction half reaction by 3 in order to cancel out the electrons in the net reaction. Add both the half reaction to get the net reaction.

${\text{20NO}}_{\text{3}}{}^{\text{-}}\left(aq\right)\text{+}80{\text{H}}^{\text{+}}\left(aq\right)+60{\text{e}}^{-}\to \text{ 20NO}\left(g\right){\text{+ 40H}}_{\text{2}}\text{O}\left(l\right)$

${\text{3P}}_{\text{4}}\left(s\right)+48{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 12{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right)+72{\text{H}}^{\text{+}}\left(aq\right)+60{\text{e}}^{-}$

${\text{3P}}_{\text{4}}\left(s\right)+{\text{20NO}}_{\text{3}}{}^{\text{-}}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+8{\text{H}}^{\text{+}}\left(aq\right)\to 12{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{-}\left(aq\right)+\text{20NO}\left(g\right)$

Interpretation Introduction

(c)

Interpretation:

The net ionic equation needs to be determined and balanced in the acidic medium.

Concept introduction:

The rules to balance redox reactions in basic medium are as follows:

1. Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
2. Balance the atoms of elements other than H and O atoms. Use H₂ O to balance the number of O atoms, H+ to balance the number of H atoms and the electrons for charge balance. Equalize the number of electrons on both the half reactions and then add the half reactions.

Answer to Problem 9QAP

${\text{2MnO}}_4{}^{\text{-}}+{\text{5SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+6{\text{H}}^{\text{+}}\left(aq\right)\to 2{\text{Mn}}^{\text{2+}}\left(aq\right)\text{+}{\text{5SO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

Aqueous solution of potassium sulfite and potassium permanganate react. Sulfate and manganese II ions are formed.

${\text{SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right){\text{ + MnO}}_4{}^{\text{-}}\left(aq\right)\to {\text{ SO}}_{\text{4}}{}^{\text{2-}}\left(aq\right){\text{ + Mn}}^{\text{2+}}\left(aq\right)$

Determining the oxidation numbers:

The oxidation number of Mn in MnO₄ - and Mn²+ is +7 and +2 respectively. As the oxidation number is decreasing this is the reduction half of the reaction.

Oxidation half reaction:

${\text{SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ SO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)\text{ + }{\text{2H}}^{\text{+}}\left(aq\right)+2{\text{e}}^{-}$

To balance the excess O atoms on the product side we add one H₂ O on the reactant side, and then we balance the excess H atom on the reactantside by adding two H+ on the product side. Finally we balance the charge by adding two electrons to the product side of the half reaction.

Reduction half reaction:

${\text{MnO}}_4{}^{\text{-}}\left(aq\right)+8{\text{H}}^{\text{+}}\left(aq\right)+5{\text{e}}^{-}\to {\text{Mn}}^{\text{2+}}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

To balance the excess O atoms on the reactant side we add four H₂ O on the product side, and then we balance the excess H atom on the product side by adding eight H+ on the reactant side. Finally we balance the charge by adding five electrons to the reactant side of the half reaction.

Net reaction:

We multiply the oxidation half reaction by five and the reduction half reaction by two, in order to cancel out the electrons in the net reaction.

${\text{5SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+5{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{ 5SO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)\text{ + }10{\text{H}}^{\text{+}}\left(aq\right)+10{\text{e}}^{-}$

${\text{2MnO}}_4{}^{\text{-}}\left(aq\right)+16{\text{H}}^{\text{+}}\left(aq\right)+10{\text{e}}^{-}\to 2{\text{Mn}}^{\text{2+}}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Add both the half reaction to get the net reaction.

${\text{2MnO}}_4{}^{\text{-}}+{\text{5SO}}_{\text{3}}{}^{\text{2-}}\left(aq\right)+6{\text{H}}^{\text{+}}\left(aq\right)\to 2{\text{Mn}}^{\text{2+}}\left(aq\right)\text{+}{\text{5SO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$