Chapter 18.2, Problem 7RQ

(a)

Interpretation Introduction

Interpretation: The balancing of oxidation- reduction reaction by using half-reaction method in acidic medium needs to be performed.

  ${\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{-}{\text{+ Br}}_{}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{NO}}_{\text{(g)}}{\text{ + Br}}_{\text{2}}{}_{\text{(l)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved loss or gain electron. Oxidation is the process that involves the loss of electrons and oxidized to cation whereas reduction is the process that involves the gain of electrons and reduced to anion.

Answer to Problem 7RQ

  ${\text{2 NO}}_{\text{3}}{}_{}^{-}+{\text{6 Br}}_{}^{-}+{\text{ 8 H}}^{\text{+}}\stackrel{}{\to }{\text{2 NO + 4 H}}_{\text{2}}{\text{O+ 3 Br}}_{\text{2}}$

Explanation of Solution

To balance the redox reaction by half-reaction or ion-electron method, add both the half-reactions by balancing the charges and atoms.

Step-1:

Assign oxidation number to each atom:

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{-}{\text{+ Br}}_{}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{NO}}_{\text{(g)}}{\text{ + Br}}_{\text{2}}{}_{\text{(l)}}\\ \text{+5 -2 -1 +2-2 0}\end{array}$

Step-2:

Write half reactions:

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{NO}}_{\text{(g)}}\text{ (Reduction)}\\ {\text{Br}}_{}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{Br}}_{\text{2}}{}_{\text{(l)}}\text{ (Oxidation)}\end{array}$

Step-3:

Balance all atoms, O with H₂O and H with H+ ions:

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{}^{\text{-}}{\text{+ 4H}}^{\text{+}}\stackrel{}{\to }{\text{NO + 2 H}}_{\text{2}}\text{O (Reduction)}\\ {\text{2 Br}}_{}^{\text{-}}\stackrel{}{\to }{\text{Br}}_{\text{2}}\text{ (Oxidation)}\end{array}$

Step-4:

Balance charges:

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{}^{-}{\text{+ 4H}}^{\text{+}}\stackrel{}{\to }{\text{NO + 2 H}}_{\text{2}}{\text{O + 3 e}}^{\text{- }}\text{ (Reduction)}\\ {\text{2 Br}}_{}^{-}\stackrel{}{\to }{\text{Br}}_{\text{2}}{\text{ + 2 e}}^{\text{- }}\text{ (Oxidation)}\end{array}$

Step-5:

Add both equation:

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{}^{-}+{\text{4H}}^{\text{+}}\stackrel{}{\to }{\text{NO + 2 H}}_{\text{2}}{\text{O + 3 e}}^{\text{- }}\text{ (Reduction)}\\ {\text{2 Br}}_{}^{-}\stackrel{}{\to }{\text{Br}}_{\text{2}}{\text{ + 2 e}}^{\text{- }}\text{ (Oxidation)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2 NO}}_{\text{3}}{}_{}^{-}+{\text{6 Br}}_{}^{-}+{\text{ 8 H}}^{\text{+}}\stackrel{}{\to }{\text{2 NO + 4 H}}_{\text{2}}{\text{O+ 3 Br}}_{\text{2}}\end{array}$

Thus, the balance equation is:

  ${\text{2 NO}}_{\text{3}}{}_{}^{-}+{\text{6 Br}}_{}^{-}+{\text{ 8 H}}^{\text{+}}\stackrel{}{\to }{\text{2 NO + 4 H}}_{\text{2}}{\text{O+ 3 Br}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation: The balancing of oxidation- reduction reaction by using half-reaction method in acidic medium needs to be performed.

  ${\text{Ni}}_{{}_{\text{(s)}}}^{}{\text{+ NO}}_3{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + NO}}_{\text{2}}{}_{\text{(g)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved loss or gain electron. Oxidation is the process that involves the loss of electrons and oxidized to cation whereas reduction is the process that involves the gain of electrons and reduced to anion.

Answer to Problem 7RQ

  ${\text{2 NO}}_3{}_{{}_{\text{(aq)}}}^{-}+{\text{Ni}}_{{}_{\text{(s)}}}^{}+{\text{ 4 H}}^{\text{+}}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + 2 NO}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2 H}}_{\text{2}}\text{O }$

Explanation of Solution

To balance the redox reaction by half-reaction or ion-electron method, add both the half-reactions by balancing the charges and atoms.

Step-1:

Assign oxidation number to each atom:

  $\begin{array}{l}{\text{Ni}}_{{}_{\text{(s)}}}^{}{\text{+ NO}}_3{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + NO}}_{\text{2}}{}_{\text{(g)}}\\ \text{0 +5 -2 +2 +4-2}\end{array}$

Step-2:

Write half reactions:

  $\begin{array}{l}{\text{ NO}}_3{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{ NO}}_{\text{2}}{}_{\text{(g)}}\text{ (Reduction)}\\ {\text{Ni}}_{{}_{\text{(s)}}}^{}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}\text{ (Oxidation)}\end{array}$

Step-3:

Balance all atoms, O with H₂O and H with H+ ions:

  $\begin{array}{l}{\text{NO}}_3{}_{{}_{\text{(aq)}}}^{-}{\text{+ 2 H}}^{\text{+}}\stackrel{}{\to }{\text{ NO}}_{\text{2}}{}_{\text{(g)}}{\text{+ H}}_{\text{2}}\text{O (Reduction)}\\ {\text{Ni}}_{{}_{\text{(s)}}}^{}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}\text{ (Oxidation)}\end{array}$

Step-4:

Balance charges:

  $\begin{array}{l}{\text{NO}}_3{}_{{}_{\text{(aq)}}}^{-}{\text{+ 2 H}}^{\text{+}}\stackrel{}{\to }{\text{ NO}}_{\text{2}}{}_{\text{(g)}}{\text{+ H}}_{\text{2}}\text{O (Reduction)}\\ {\text{Ni}}_{{}_{\text{(s)}}}^{}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2 e}}^{\text{- }}\text{ (Oxidation)}\end{array}$

Step-5:

Add both equation:

  $\begin{array}{l}{\text{NO}}_3{}_{{}_{\text{(aq)}}}^{-}{\text{+ 2 H}}^{\text{+}}+{\text{e}}^{\text{- }}\stackrel{}{\to }{\text{ NO}}_{\text{2}}{}_{\text{(g)}}{\text{+ H}}_{\text{2}}\text{O (Reduction)}\\ {\text{Ni}}_{{}_{\text{(s)}}}^{}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2 e}}^{\text{- }}\text{ (Oxidation)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2 NO}}_3{}_{{}_{\text{(aq)}}}^{-}+{\text{Ni}}_{{}_{\text{(s)}}}^{}+{\text{ 4 H}}^{\text{+}}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + 2 NO}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2 H}}_{\text{2}}\text{O }\end{array}$

Thus, the balance equation is:

  ${\text{2 NO}}_3{}_{{}_{\text{(aq)}}}^{-}+{\text{Ni}}_{{}_{\text{(s)}}}^{}+{\text{ 4 H}}^{\text{+}}\stackrel{}{\to }{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + 2 NO}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2 H}}_{\text{2}}\text{O }$

(c)

Interpretation Introduction

Interpretation: The balancing of oxidation- reduction reaction by using half-reaction method in acidic medium needs to be performed.

  ${\text{ClO}}_{\text{4}}{}_{{}_{\text{(aq)}}}^{-}{\text{+ Cl}}_{}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{ClO}}_3^{-}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved loss or gain electron. Oxidation is the process that involves the loss of electrons and oxidized to cation whereas reduction is the process that involves the gain of electrons and reduced to anion.

Answer to Problem 7RQ

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

Explanation of Solution

To balance the redox reaction by half-reaction or ion-electron method, add both the half-reactions by balancing the charges and atoms.

Step-1:

Assign oxidation number to each atom:

  $\begin{array}{l}{\text{ClO}}_{\text{4}}{}_{{}_{\text{(aq)}}}^{-}{\text{+ Cl}}_{}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{ClO}}_3^{-}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}\\ \text{+7 -2 -1 +5 -2 0}\end{array}$

Step-2:

Write half reactions:

  $\begin{array}{l}{\text{ClO}}_{\text{4}}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{ClO}}_3^{-}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}\text{ (Reduction)}\\ {\text{Cl}}_{}{}_{{}_{\text{(aq)}}}^{-}\stackrel{}{\to }{\text{ClO}}_3^{-}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}\text{ (Oxidation)}\end{array}$

Step-3:

Balance all atoms, O with H₂O and H with H+ ions:

  $\begin{array}{l}{\text{3 Cl}}_{}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}{\text{+ 6 H}}^{\text{+}}\text{ (Oxidation)}\\ {\text{ClO}}_{\text{4}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 18 H}}^{\text{+}}\stackrel{}{\to }{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}{\text{+9 H}}_{\text{2}}\text{O (Reduction)}\end{array}$

Step-4:

Balance charges:

  $\begin{array}{l}{\text{3 Cl}}_{}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}{\text{+ 6 H}}^{\text{+}}\text{ + 8 e- (Oxidation)}\\ {\text{ClO}}_{\text{4}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 18 H}}^{\text{+}}\text{+ 16 e- }\stackrel{}{\to }{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}{\text{+9 H}}_{\text{2}}\text{O (Reduction)}\end{array}$

Step-5:

Add both equation:

  $\begin{array}{l}{\text{3 Cl}}_{}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 3H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}{\text{+ 6 H}}^{\text{+}}\text{ + 8 e- (Oxidation)}\\ {\text{ClO}}_{\text{4}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 18 H}}^{\text{+}}\text{+ 16 e- }\stackrel{}{\to }{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ + Cl}}_{\text{2}}{}_{\text{(g)}}{\text{+9 H}}_{\text{2}}\text{O (Reduction)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2 Cl}}_{}{}_{{}_{\text{(aq)}}}^{\text{-}}+{\text{ClO}}_{\text{4}}{}_{{}_{\text{(aq)}}}^{\text{-}}+{\text{ 2 H}}^{\text{+}}\stackrel{}{\to }{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ + ClO}}_{\text{2}}{}_{\text{(g)}}{\text{+ H}}_{\text{2}}\text{O }\end{array}$

Thus, the balance equation is:

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