Chapter 21, Problem 21.19P

(a)

Interpretation Introduction

Interpretation:

The given skeleton reaction has to be balanced.  The reducing and the oxidizing agents has to be identified.

$S{O}_3^{2-}\left(aq\right)+C{l}_2\left(g\right)\to S{O}_4^{2-}\left(aq\right)+C{l}^{-}\left(aq\right)\left[Basic\right]$

Concept Introduction:

Oxidation half-reaction: An oxidation half-reaction is a part of redox reaction that shows only for the oxidized species with electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: A reduction half-reaction is a part of redox reaction (counterpart of oxidation half-reaction) that shows only for the reduced species with electrons and its oxidation state decreases with the gain of electrons.

Anode: The electrode where the oxidation occurs is called as an anode.  It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode.  It is a positively charged electrode.

Steps in balancing organic redox reactions:

1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction
2. 2) Balance atoms other than $\text{O}$ and $\text{H}$.
3. 3) Balance $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ as needed
4. 4) Balance $\text{H}$ by adding ${\text{H}}^{\text{+}}$ ion at the required side.
5. 5) Balance charges by adding, as needed number of electrons,
6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
9. 9) For the reaction in acidic medium, the presence of $H^{+}$ ion allowed.
10. 10) For the reaction in basic medium, the $H^{+}$ ion should be neutralized by the addition of required amount of $O{H}^{-}$ ions on both the sides.
11. 11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Reducing agent: The substance that reduces another substance and gets oxidized is called as a reducing agent.

Oxidizing agent: The substance that oxidizes another substance and gets reduced is called as an oxidizing agent.

Explanation of Solution

The given cell reaction is balanced as follows,

The atoms other than oxygen and hydrogen are balanced first.  Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using $H^{+}$.  Finally, the charges are balanced using electrons.

$S{O}_3^{2-}\left(aq\right)+C{l}_2\left(g\right)\to S{O}_4^{2-}\left(aq\right)+C{l}^{-}\left(aq\right)\left[Basic\right]$

Examining the above equation it is clear that reduction of $Cl$ and oxidation of S are taking place.

The two half- reaction:

Reduction half-reaction:

$C{l}_2\left(g\right)\to C{l}^{-}\left(aq\right)$

Oxidation half-reaction:

$S{O}_3^{2-}\left(aq\right)\to S{O}_4^{2-}\left(aq\right)$

$\begin{array}{cc}Balancereductionhalf-reaction& Balanceoxidationhalf-reaction\\ \begin{array}{l}C{l}_2\left(g\right)\to C{l}^{-}\left(aq\right)\\ \\ C{l}_2\left(g\right)\to 2C{l}^{-}\left(aq\right)\\ \\ C{l}_2\left(g\right)+2e^{-}\to 2C{l}^{-}\left(aq\right)\end{array}& \begin{array}{l}S{O}_3^{2-}\left(aq\right)\to S{O}_4^{2-}\left(aq\right)\\ \\ S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)\to S{O}_4^{2-}\left(aq\right)\\ \\ S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)\to S{O}_4^{2-}\left(aq\right)+2H^{+}\left(aq\right)\\ \\ S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)\to S{O}_4^{2-}\left(aq\right)+2H^{+}\left(aq\right)+2e^{-}\end{array}\end{array}$

Then add both the reactions and cancel electrons.

$\begin{array}{l}\underset{\_}{\begin{array}{l}S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)\to S{O}_4^{2-}\left(aq\right)+2H^{+}\left(aq\right)+\cancel{2e^{-}}\\ \\ C{l}_2\left(g\right)+\cancel{2e^{-}}\to 2C{l}^{-}\left(aq\right)\end{array}}\\ S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)+C{l}_2\left(g\right)\to S{O}_4^{2-}\left(aq\right)+2H^{+}\left(aq\right)+2C{l}^{-}\left(aq\right)\end{array}$

Add $2O{H}^{-}$ to both sides to neutralize $H^{+}$ and form $H_{2}O$:

$S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)+C{l}_2\left(g\right)+2O{H}^{-}\left(aq\right)\to S{O}_4^{2-}\left(aq\right)+2H^{+}\left(aq\right)+2O{H}^{-}\left(aq\right)+2C{l}^{-}\left(aq\right)$

Or

$S{O}_3^{2-}\left(aq\right)+H_{2}O\left(l\right)+C{l}_2\left(g\right)+2O{H}^{-}\left(aq\right)\to S{O}_4^{2-}\left(aq\right)+2H_{2}O\left(l\right)+2C{l}^{-}\left(aq\right)$

The balanced chemical reaction in basic medium is

$S{O}_3^{2-}\left(aq\right)+C{l}_2\left(g\right)+2O{H}^{-}\left(aq\right)\to S{O}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)+2C{l}^{-}\left(aq\right)$

Hence,

Oxidizing agent: $C{l}_2$

Reducing agent: $S{O}_3^{2-}$

(b)

Interpretation Introduction

Interpretation:

The given skeleton reaction has to be balanced.  The reducing and the oxidizing agents has to be identified.

$Fe{\left(CN\right)}_6^{3-}\left(aq\right)+\mathrm{Re}\left(s\right)\to Fe{\left(CN\right)}_6^{4-}\left(aq\right)+\mathrm{Re}{O}_4^{-}\left(aq\right)\left[Basic\right]$

Concept Introduction:

Oxidation half-reaction: An oxidation half-reaction is a part of redox reaction that shows only for the oxidized species with electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: A reduction half-reaction is a part of redox reaction (counterpart of oxidation half-reaction) that shows only for the reduced species with electrons and its oxidation state decreases with the gain of electrons.

Anode: The electrode where the oxidation occurs is called as an anode.  It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode.  It is a positively charged electrode.

Steps in balancing organic redox reactions:

1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction
2. 2) Balance atoms other than $\text{O}$ and $\text{H}$.
3. 3) Balance $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ as needed
4. 4) Balance $\text{H}$ by adding ${\text{H}}^{\text{+}}$ ion at the required side.
5. 5) Balance charges by adding, as needed number of electrons,
6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
9. 9) For the reaction in acidic medium, the presence of $H^{+}$ ion allowed.
10. 10) For the reaction in basic medium, the $H^{+}$ ion should be neutralized by the addition of required amount of $O{H}^{-}$ ions on both the sides.
11. 11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Reducing agent: The substance that reduces another substance and gets oxidized is called as a reducing agent.

Oxidizing agent: The substance that oxidizes another substance and gets reduced is called as an oxidizing agent.

Explanation of Solution

The given cell reaction is balanced as follows,

The atoms other than oxygen and hydrogen are balanced first.  Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using $H^{+}$.  Finally, the charges are balanced using electrons.

$Fe{\left(CN\right)}_6^{3-}\left(aq\right)+\mathrm{Re}\left(s\right)\to Fe{\left(CN\right)}_6^{4-}\left(aq\right)+\mathrm{Re}{O}_4^{-}\left(aq\right)\left[Basic\right]$

Since both products contain phosphorus, divide the half-reactions so each includes $P_4$ as the reactant.

The two half- reaction:

Reduction half-reaction:

$Fe{\left(CN\right)}_6^{3-}\left(aq\right)\to Fe{\left(CN\right)}_6^{4-}\left(aq\right)$

Oxidation half-reaction:

$\mathrm{Re}\left(s\right)\to \mathrm{Re}{O}_4^{-}\left(aq\right)$

$\begin{array}{cc}Balancereductionhalf-reaction& Balanceoxidationhalf-reaction\\ \begin{array}{l}Fe{\left(CN\right)}_6^{3-}\left(aq\right)\to Fe{\left(CN\right)}_6^{4-}\left(aq\right)\\ \\ Fe{\left(CN\right)}_6^{3-}\left(aq\right)+1e^{-}\to Fe{\left(CN\right)}_6^{4-}\left(aq\right)\end{array}& \begin{array}{l}\mathrm{Re}\left(s\right)\to \mathrm{Re}{O}_4^{-}\left(aq\right)\\ \\ \mathrm{Re}\left(s\right)+4H_{2}O\left(l\right)\to \mathrm{Re}{O}_4^{-}\left(aq\right)\\ \\ \mathrm{Re}\left(s\right)+4H_{2}O\left(l\right)\to \mathrm{Re}{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)\\ \\ \mathrm{Re}\left(s\right)+4H_{2}O\left(l\right)\to \mathrm{Re}{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+7e^{-}\end{array}\end{array}$

Multiply the reduction half-reaction by 7 to balance the no. of electrons involved in both of the reaction. Then add the reactions and cancel the electrons.

$\begin{array}{l}\underset{\_}{\begin{array}{l}7Fe{\left(CN\right)}_6^{3-}\left(aq\right)+\cancel{7e^{-}}\to 7Fe{\left(CN\right)}_6^{4-}\left(aq\right)\\ \\ \mathrm{Re}\left(s\right)+4H_{2}O\left(l\right)\to \mathrm{Re}{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+\cancel{7e^{-}}\end{array}}\\ 7Fe{\left(CN\right)}_6^{3-}\left(aq\right)+\mathrm{Re}\left(s\right)+4H_{2}O\left(l\right)\to 7Fe{\left(CN\right)}_6^{4-}\left(aq\right)+\mathrm{Re}{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)\end{array}$

Add $\text{8 }O{H}^{-}$ to both sides to neutralize $H^{+}$ and form $H_{2}O$:

$7Fe{\left(CN\right)}_6^{3-}\left(aq\right)+\mathrm{Re}\left(s\right)+4H_{2}O\left(l\right)+8O{H}^{-}\left(aq\right)\to 7Fe{\left(CN\right)}_6^{4-}\left(aq\right)+\mathrm{Re}{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+8O{H}^{-}\left(aq\right)$

Or

$7Fe{\left(CN\right)}_6^{3-}\left(aq\right)+\mathrm{Re}\left(s\right)+4H_{2}O\left(l\right)+8O{H}^{-}\left(aq\right)\to 7Fe{\left(CN\right)}_6^{4-}\left(aq\right)+\mathrm{Re}{O}_4^{-}\left(aq\right)+8H_{2}O\left(l\right)$

The balanced chemical reaction in basic medium is

$7Fe{\left(CN\right)}_6^{3-}\left(aq\right)+\mathrm{Re}\left(s\right)+8O{H}^{-}\left(aq\right)\to 7Fe{\left(CN\right)}_6^{4-}\left(aq\right)+\mathrm{Re}{O}_4^{-}\left(aq\right)+4H_{2}O\left(l\right)$

Hence,

Oxidizing agent: $Fe{\left(CN\right)}_6^{3-}$

Reducing agent: $\mathrm{Re}$

(c)

Interpretation Introduction

Interpretation:

The given skeleton reaction has to be balanced. The reducing and the oxidizing agents has to be identified.

$Mn{O}_4^{-}\left(aq\right)+HCOOH\left(aq\right)\to M{n}^{2+}\left(aq\right)+C{O}_2\left(g\right)\left[Acidic\right]$

Concept Introduction:

Oxidation half-reaction: An oxidation half-reaction is a part of redox reaction that shows only for the oxidized species with electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: A reduction half-reaction is a part of redox reaction (counterpart of oxidation half-reaction) that shows only for the reduced species with electrons and its oxidation state decreases with the gain of electrons.

Anode: The electrode where the oxidation occurs is called as an anode.  It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode.  It is a positively charged electrode.

Steps in balancing organic redox reactions:

1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction
2. 2) Balance atoms other than $\text{O}$ and $\text{H}$.
3. 3) Balance $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ as needed
4. 4) Balance $\text{H}$ by adding ${\text{H}}^{\text{+}}$ ion at the required side.
5. 5) Balance charges by adding, as needed number of electrons,
6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
9. 9) For the reaction in acidic medium, the presence of $H^{+}$ ion allowed.
10. 10) For the reaction in basic medium, the $H^{+}$ ion should be neutralized by the addition of required amount of $O{H}^{-}$ ions on both the sides.
11. 11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Reducing agent: The substance that reduces another substance and gets oxidized is called as a reducing agent.

Oxidizing agent: The substance that oxidizes another substance and gets reduced is called as an oxidizing agent.

Explanation of Solution

The given cell reaction is balanced as follows,

The atoms other than oxygen and hydrogen are balanced first.  Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using $H^{+}$.  Finally, the charges are balanced using electrons.

$Mn{O}_4^{-}\left(aq\right)+HCOOH\left(aq\right)\to M{n}^{2+}\left(aq\right)+C{O}_2\left(g\right)\left[Acidic\right]$

The two half- reaction:

Reduction half-reaction:

$Mn{O}_4^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)$

Oxidation half-reaction:

$HCOOH\left(aq\right)\to C{O}_2\left(g\right)$

$\begin{array}{cc}Balancereductionhalf-reaction& Balanceoxidationhalf-reaction\\ \begin{array}{l}Mn{O}_4^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)\\ \\ Mn{O}_4^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)\\ \\ Mn{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)\\ \\ Mn{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+5e^{-}\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)\end{array}& \begin{array}{l}HCOOH\left(aq\right)\to C{O}_2\left(g\right)\\ \\ HCOOH\left(aq\right)\to C{O}_2\left(g\right)+2H^{+}\left(aq\right)\\ \\ HCOOH\left(aq\right)\to C{O}_2\left(g\right)+2H^{+}\left(aq\right)+2e^{-}\end{array}\end{array}$

Multiply reduction half-reaction by $2$ and oxidation half-reaction by $5$ to balance the no. of electrons involved in both the reaction.  Then add both the reactions and cancel $H_{2}Oand{H}^{+}$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}2Mn{O}_4^{-}\left(aq\right)+\cancel{16H^{+}\left(aq\right)}+\cancel{10e^{-}}\to 2M{n}^{2+}\left(aq\right)+8H_{2}O\left(l\right)\\ \\ 5HCOOH\left(aq\right)\to 5C{O}_2\left(g\right)+\cancel{10H^{+}\left(aq\right)}+\cancel{10e^{-}}\end{array}}\\ 2Mn{O}_4^{-}\left(aq\right)+5HCOOH\left(aq\right)+6H^{+}\left(aq\right)\to 2M{n}^{2+}\left(aq\right)+8H_{2}O\left(l\right)+5C{O}_2\left(g\right)\end{array}$

Hence,

Oxidizing agent: $Mn{O}_4^{-}$

Reducing agent: $HCOOH$