Chapter 17, Problem 7QAP

Interpretation Introduction

(a)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

${\text{SO}}_{\text{2}}\left(\text{g}\right){\text{ + I}}_{\text{2}}\left(\text{aq}\right)\to {\text{SO}}_{\text{3}}\left(\text{g}\right){\text{ + I}}^{\text{-}}\left(\text{aq}\right)$

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

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.
3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Answer to Problem 7QAP

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

Explanation of Solution

${\text{SO}}_{\text{2}}\left(\text{g}\right){\text{ + I}}_{\text{2}}\left(\text{aq}\right)\to {\text{SO}}_{\text{3}}\left(\text{g}\right){\text{ + I}}^{\text{-}}\left(\text{aq}\right)$

Determining the oxidation numbers:

The oxidation number of I in I₂ and I- is 0 and -1 respectively. As the oxidation number of I is decreasing this is the reduction half reaction.

Reduction half reaction: ${\text{ I}}_{\text{2}}\left(\text{aq}\right)\to {\text{ I}}^{\text{-}}\left(\text{aq}\right)$

Oxidation half reaction: ${\text{SO}}_{\text{2}}\left(\text{g}\right)\to {\text{SO}}_{\text{3}}\left(\text{g}\right)$

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 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.

Thus, the balanced oxidation half reaction:

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

Now, balance the charge by adding two electrons to the reactant side of the half reaction.

Thus, the balanced reduction half reaction:

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

Net reaction:

By adding both the half reaction, the net reaction is obtained as follows:

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

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH^-. 2 OH- ions are added on both side of the equation to even out the charges. H+ and OH- combines to form H₂ O, which is then cancelled out on both side of the net reaction.

Thus, the balanced net reaction in the basic medium will be:

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

Or,

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

Interpretation Introduction

(b)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

$\text{ Zn}\left(\text{s}\right){\text{ + NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)\to {\text{NH}}_{\text{3}}\left(\text{aq}\right){\text{ + Zn}}^{\text{2+}}\left(\text{aq}\right)$

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

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.
3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Answer to Problem 7QAP

$\text{ 4Zn}\left(\text{s}\right)+{\text{NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)+{\text{6H}}_{\text{2}}\text{O}\to {\text{4Zn}}^{\text{2+}}\left(\text{aq}\right)+{\text{NH}}_{\text{3}}\left(\text{aq}\right){\text{+ 9OH}}^{\text{-}}$

Explanation of Solution

$\text{ Zn}\left(\text{s}\right){\text{ + NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)\to {\text{NH}}_{\text{3}}\left(\text{aq}\right){\text{ + Zn}}^{\text{2+}}\left(\text{aq}\right)$

Determining the oxidation numbers:

The oxidation number of Zn in Zn and Zn²+ is 0 and +2 respectively. As the oxidation number of Zn is increasing this is the oxidation half of the reaction.

Oxidation half reaction: $\text{ Zn}\left(\text{s}\right)\to {\text{ Zn}}^{\text{2+}}\left(\text{aq}\right)$

Reduction half reaction: ${\text{ NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)\to {\text{NH}}_{\text{3}}\left(\text{aq}\right)$

Balance the charge by adding two electrons to the product side of the half reaction.

Thus, the balanced oxidation half reaction will be:

$\text{ Zn}\left(\text{s}\right)\to {\text{ Zn}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}$

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

Thus, the balanced reduction half reaction will be:

${\text{ NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)+{\text{9H}}^{\text{+}}{\text{+8e}}^{-}\to {\text{NH}}_{\text{3}}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}$

Net reaction:

Multiply the oxidation half reaction by 4, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.

$\text{ 4Zn}\left(\text{s}\right)\to {\text{ 4Zn}}^{\text{2+}}\left(\text{aq}\right)+8{\text{e}}^{-}$

$\text{ 4Zn}\left(\text{s}\right)+{\text{NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)+{\text{9H}}^{\text{+}}\to {\text{4Zn}}^{\text{2+}}\left(\text{aq}\right)+{\text{NH}}_{\text{3}}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}$

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH^-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H₂ O, which is then cancelled out on both side of the net reaction. Thus, the balanced net reaction will be:

$\text{4Zn}\left(\text{s}\right)+{\text{ NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)+{\text{9H}}^{\text{+}}+9{\text{OH}}^{\text{-}}\to {\text{4Zn}}^{\text{2+}}\left(\text{aq}\right)+{\text{NH}}_{\text{3}}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}{\text{+ 9OH}}^{\text{-}}$

Or,

$\text{ 4Zn}\left(\text{s}\right)+{\text{NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)+{\text{6H}}_{\text{2}}\text{O}\to {\text{4Zn}}^{\text{2+}}\left(\text{aq}\right)+{\text{NH}}_{\text{3}}\left(\text{aq}\right){\text{+ 9OH}}^{\text{-}}$

Interpretation Introduction

(c)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

${\text{ClO}}^{\text{-}}\left(\text{aq}\right){\text{ + CrO}}_{\text{2}}{}^{\text{-}}\to {\text{Cl}}^{\text{-}}\left(\text{aq}\right){\text{ + CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)$

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

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.
3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Answer to Problem 7QAP

${\text{ 2CrO}}_{\text{2}}{}^{\text{-}}+{\text{3ClO}}^{\text{-}}\left(\text{aq}\right)+2{\text{OH}}^{\text{-}}\to {\text{ 2CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)+3{\text{Cl}}^{\text{-}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

${\text{ClO}}^{\text{-}}\left(\text{aq}\right){\text{ + CrO}}_{\text{2}}{}^{\text{-}}\to {\text{Cl}}^{\text{-}}\left(\text{aq}\right){\text{ + CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)$

Determining the oxidation numbers:

The oxidation number of Cl in ClO- and Cl- is +1 and -1 respectively. As the oxidation number of Cl is decreasing this is the reduction half of the reaction.

The reduction half reaction: ${\text{ClO}}^{\text{-}}\left(\text{aq}\right)\to {\text{Cl}}^{\text{-}}\left(\text{aq}\right)$

Oxidation half reaction: ${\text{ CrO}}_{\text{2}}{}^{\text{-}}\to {\text{ CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)$

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

Thus, the balanced oxidation half reaction will be:

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

Similarly, to balance the excess O atoms on the reactant side we add one H₂ O on the product side, and 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.

Thus, the balanced reduction half reaction will be:

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

Net reaction:

Multiply the oxidation half reaction by 2 and reduction half reaction by 3, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.

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

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

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

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH^-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H₂ O, which is then cancelled out on both side of the net reaction.

${\text{ 2CrO}}_{\text{2}}{}^{\text{-}}+{\text{3ClO}}^{\text{-}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}+2{\text{OH}}^{\text{-}}\to {\text{ 2CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)+3{\text{Cl}}^{\text{-}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+2{\text{OH}}^{\text{-}}$

Or,

${\text{ 2CrO}}_{\text{2}}{}^{\text{-}}+{\text{3ClO}}^{\text{-}}\left(\text{aq}\right)+2{\text{OH}}^{\text{-}}\to {\text{ 2CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)+3{\text{Cl}}^{\text{-}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

Interpretation Introduction

(d)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

$\text{K}\left(\text{s}\right){\text{ + H}}_{\text{2}}\text{O}\to {\text{K}}^{\text{+}}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\left(\text{g}\right)$

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

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.
3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Answer to Problem 7QAP

$2\text{K}\left(\text{s}\right)+2{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}\left(\text{g}\right)+2{\text{K}}^{\text{+}}\left(\text{aq}\right)+2{\text{OH}}^{\text{-}}$

Explanation of Solution

$\text{K}\left(\text{s}\right){\text{ + H}}_{\text{2}}\text{O}\to {\text{K}}^{\text{+}}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\left(\text{g}\right)$

Determining the oxidation numbers:

The oxidation number of K in K and K+ is 0 and +1 respectively. As the oxidation number of K is increasing this is the oxidation half of the reaction.

Oxidation half reaction: $\text{K}\left(\text{s}\right)\to {\text{K}}^{\text{+}}\left(\text{aq}\right)$

Reduction half reaction: ${\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}\left(\text{g}\right)$

Balance the charge by adding one electron to the product side of the half reaction.

Thus, the balanced oxidation half reaction will be:

$\text{K}\left(\text{s}\right)\to {\text{K}}^{\text{+}}\left(\text{aq}\right){\text{ + e}}^{-}$

To balance the excess O atoms on the reactant side we add one H₂ O on the product side, and 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.

Thus, the balanced reduction half reaction will be:

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

Net reaction:

Multiply the oxidation half reaction by 2, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.

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

$\text{2K}\left(\text{s}\right)\to 2{\text{K}}^{\text{+}}\left(\text{aq}\right){\text{ + 2e}}^{-}$

$2\text{K}\left(\text{s}\right)+2{\text{H}}^{\text{+}}\to {\text{H}}_{\text{2}}\left(\text{g}\right)+2{\text{K}}^{\text{+}}\left(\text{aq}\right)$

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH^-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H₂ O, which is then cancelled out on both side of the net reaction.

$2\text{K}\left(\text{s}\right)+2{\text{H}}^{\text{+}}+2{\text{OH}}^{\text{-}}\to {\text{H}}_{\text{2}}\left(\text{g}\right)+2{\text{K}}^{\text{+}}\left(\text{aq}\right)+2{\text{OH}}^{\text{-}}$

Or,

$2\text{K}\left(\text{s}\right)+2{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}\left(\text{g}\right)+2{\text{K}}^{\text{+}}\left(\text{aq}\right)+2{\text{OH}}^{\text{-}}$