Chapter 17, Problem 6QAP

Write balanced equations for the following reactions in acid solution.
(a) ${\text{P}}_4\left(s\right)+{\text{Cl}}^{-}\left(aq\right)\to {\text{PH}}_3\left(g\right)+{\text{Cl}}_2\left(g\right)$
(b) ${\text{MnO}}_4{}^{-}\left(aq\right)+{\text{NO}}_2{}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{NO}}_3{}^{-}\left(aq\right)$
(c) ${\text{HBrO}}_3\left(aq\right)+\text{Bi}\left(s\right)\to {\text{HBrO}}_2\left(aq\right)+{\text{Bi}}_2{\text{O}}_3\left(s\right)$
(d) ${\text{CrO}}_4{}^{2-}\left(aq\right)+{\text{SO}}_3{}^{2-}\left(aq\right)\to {\text{Cr}}^3{}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)$

Interpretation Introduction

(a)

Interpretation:

Write balanced equations for the following reaction in acid solution.

${\text{P}}_{\text{4}}\left(\text{s}\right){\text{ + Cl}}^{\text{-}}\left(\text{aq}\right)\to {\text{PH}}_{\text{3}}\left(\text{g}\right){\text{ + Cl}}_{\text{2}}\left(\text{g}\right)$

Concept introduction:

Here are the rules to balance redox reactions in acid solution.

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

Answer to Problem 6QAP

${\text{P}}_{\text{4}}\left(\text{s}\right)+{\text{12Cl}}^{\text{-}}\left(\text{aq}\right)+12{\text{H}}^{+}\to 4{\text{PH}}_{\text{3}}\left(\text{g}\right){\text{+ 6Cl}}_{\text{2}}\left(\text{g}\right)$

Explanation of Solution

${\text{P}}_{\text{4}}\left(\text{s}\right){\text{ + Cl}}^{\text{-}}\left(\text{aq}\right)\to {\text{PH}}_{\text{3}}\left(\text{g}\right){\text{ + Cl}}_{\text{2}}\left(\text{g}\right)$

Determining the oxidation numbers:

The oxidation of Cl in Cl-1 is -1, whereas the oxidation number of Cl in Cl₂ (free elemental form) is zero. As the oxidation number of Cl is increasing from -1 to 0 it is the oxidation half of the reaction.

Oxidation half reaction:

${\text{2 Cl}}^{\text{-}}\left(\text{aq}\right)\to {\text{ Cl}}_{\text{2}}\left(\text{g}\right)+2{\text{e}}^{-}$

First balance the number of Cl atoms on both side of the reaction. Then balance the charge by adding two electrons on the product side of the half reaction.

Reduction half reaction:

${\text{P}}_{\text{4}}\left(\text{s}\right)+12{\text{H}}^{+}+12{\text{e}}^{-}\to 4{\text{PH}}_{\text{3}}\left(\text{g}\right)$

(1)

First balance the number of P atoms on both side of the reaction. Then we balance the excess H atom on the product side by adding twelve H+ on the reactant side. Finally we balance the charge by adding twelve electrons to the reactant side of the half reaction.

Net reaction:

We multiply the oxidation half reaction by six, in order to cancel out the electrons in the net reaction. Finally add both the half reaction to get the net reaction.

${\text{12Cl}}^{\text{-}}\left(\text{aq}\right)\to {\text{ 6Cl}}_{\text{2}}\left(\text{g}\right)+12{\text{e}}^{-}$

-(2)

Adding equation (1) and (2):

${\text{P}}_{\text{4}}\left(\text{s}\right)+{\text{12Cl}}^{\text{-}}\left(\text{aq}\right)+12{\text{H}}^{+}\to 4{\text{PH}}_{\text{3}}\left(\text{g}\right){\text{+ 6Cl}}_{\text{2}}\left(\text{g}\right)$

Interpretation Introduction

(b)

Interpretation:

Write balanced equations for the following reaction in acid solution.

${\text{MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{ + NO}}_{\text{2}}{}^{\text{-}}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right){\text{ + NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)$

Concept introduction:

Here are the rules to balance redox reactions in acid solution.

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

Answer to Problem 6QAP

${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right)+{\text{5NO}}_{\text{2}}{}^{\text{-}}\left(\text{aq}\right)+6{\text{H}}^{\text{+}}\to 2{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)\text{ +}5{\text{NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right){\text{+ 3H}}_{\text{2}}\text{O}$

Explanation of Solution

${\text{MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{ + NO}}_{\text{2}}{}^{\text{-}}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right){\text{ + NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)$

Determining the oxidation numbers:

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

Oxidation half reaction:

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

First balance the number of O atoms by adding one H₂ O on the reactant side. Then we balance the excess H atom on the reactant side 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:

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

First balance the number of O atoms by adding four H₂ O on the product side. 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. Finally add both the half reaction to get the net reaction.

${\text{5NO}}_{\text{2}}{}^{\text{-}}\left(\text{aq}\right)+5{\text{H}}_{\text{2}}\text{O}\to 5{\text{NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)+10{\text{H}}^{\text{+}}+10{\text{e}}^{-}$

${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right)+16{\text{H}}^{\text{+}}+10{\text{e}}^{-}\to 2{\text{Mn}}^{\text{2+}}\left(\text{aq}\right){\text{ + 8H}}_{\text{2}}\text{O}$

${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right)+{\text{5NO}}_{\text{2}}{}^{\text{-}}\left(\text{aq}\right)+6{\text{H}}^{\text{+}}\to 2{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)\text{ +}5{\text{NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right){\text{+ 3H}}_{\text{2}}\text{O}$

Interpretation Introduction

(c)

Interpretation:

Write balanced net ionic equations for the following reaction in acidic medium.

${\text{HBrO}}_{\text{3}}\left(\text{aq}\right)\text{ + Bi}\left(\text{s}\right)\to {\text{HBrO}}_{\text{2}}\left(\text{aq}\right){\text{ + Bi}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Concept introduction:

Here are the rules to balance redox reactions in acid solution.

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

Answer to Problem 6QAP

${\text{3HBrO}}_{\text{3}}\left(\text{aq}\right)\text{+}\text{2Bi}\left(\text{s}\right)\to 3{\text{HBrO}}_{\text{2}}\left(\text{aq}\right){\text{ + Bi}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Explanation of Solution

${\text{HBrO}}_{\text{3}}\left(\text{aq}\right)\text{ + Bi}\left(\text{s}\right)\to {\text{HBrO}}_{\text{2}}\left(\text{aq}\right){\text{ + Bi}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Determining the oxidation numbers:

The oxidation number of Bi in Bi is 0 and +3 in Bi₂ O₃ . As the oxidation number of Bi is increasing this is the oxidation half of the reaction.

Oxidation half reaction:

$\text{2Bi}\left(\text{s}\right)+3{\text{H}}_{\text{2}}\text{O}\to {\text{Bi}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+6{\text{H}}^{\text{+}}\text{+}{\text{6e}}^{-}$

-(1)

First balance the number of Bi atoms on both side of the reaction. Then balance the excess O atoms by adding three H₂ O on the reactant side. Then we balance the excess H atom on the reactant side by adding six H+ on the product side. Finally we balance the charge by adding six electrons to the product side of the half reaction.

Reduction half reaction:

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

First balance the number of O atoms by adding one H₂ O on the product side. Then we balance the excess H atom on the product side by adding two H+ on the reactant side. Finally we balance the charge by adding two electrons to the reactant side of the half reaction.

Net reaction:

We multiply the reduction half reaction by three, to cancel out the electrons in the net reaction. Finally add both the half reaction to get the net reaction.

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

-(2)

Adding equation (1) and (2):

${\text{3HBrO}}_{\text{3}}\left(\text{aq}\right)\text{+}\text{2Bi}\left(\text{s}\right)\to 3{\text{HBrO}}_{\text{2}}\left(\text{aq}\right){\text{ + Bi}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Interpretation Introduction

(d)

Interpretation Write balanced equations for the following reaction in acid solution.

${\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right){\text{ + SO}}_{\text{3}}{}^{\text{2-}}\left(\text{aq}\right)\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right){\text{ + SO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)$

Concept introduction:

Here are the rules to balance redox reactions in acid solution.

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

Answer to Problem 6QAP

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

Explanation of Solution

${\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right){\text{ + SO}}_{\text{3}}{}^{\text{2-}}\left(\text{aq}\right)\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right){\text{ + SO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)$

Determining the oxidation numbers:

The oxidation number of Cr inCr+3 is +3 and +6 in CrO₄ ^2-. As the oxidation number of Cr is decreasing this is the reduction half of the reaction.

Oxidation half reaction:

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

First balance the number of O atoms by adding one H₂ O on the reactant side. Then we balance the excess H atom on the reactant side 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:

${\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right){\text{+8H}}^{\text{+}}\text{+}3{\text{e}}^{-}\to {\text{Cr}}^{\text{3+}}\left(\text{aq}\right)+4{\text{H}}_{\text{2}}\text{O}$

First balance the number of O atoms by adding four H₂ O on the product side. 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 three electrons to the reactant side of the half reaction.

Net reaction:

We multiply the oxidation half reaction by three and reduction half reaction by two, to cancel out the electrons in the net reaction. Finally add both the half reaction to get the net reaction.

${\text{2CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right){\text{+16H}}^{\text{+}}\text{+}6{\text{e}}^{-}\to 2{\text{Cr}}^{\text{3+}}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}$

${\text{ 3SO}}_{\text{3}}{}^{\text{2-}}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}\to {\text{ 3SO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)+6{\text{H}}^{\text{+}}\text{+}6{\text{e}}^{-}$

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