Chapter 18, Problem 9STP

(a)

Interpretation Introduction

Interpretation: The balanced oxidation half reaction of the given redox reaction in acidic medium needs to be written.

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

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

Answer to Problem 9STP

  ${\text{Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ Cu}}_{(aq)}^{2+}{\text{ +2e}}^{-}$

Explanation of Solution

Assign oxidation number to each atom:

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

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of Cu increases from 0 to +2. Thus, the oxidation half-reaction can be written as;

  ${\text{Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ Cu}}_{(aq)}^{2+}{\text{ +2e}}^{-}$

(b)

Interpretation Introduction

Interpretation: The balanced reduction half reaction of the given redox reaction in acidic medium needs to be determined.

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

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

Answer to Problem 9STP

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

Explanation of Solution

Assign oxidation number to each atom:

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

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of Ndecreases from +5 to +2. Thus, the reduction half-reaction can be written as;

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

(c)

Interpretation Introduction

Interpretation: The balanced redox reaction of the given redox reaction in acidic medium needs to be determined.

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

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 involved the loss of electrons and oxidized to cation whereas reduction is the process that involved the gain of electrons and reduced to anion.

Answer to Problem 9STP

  ${\text{2NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 12H}}^{\text{+}}+{\text{5Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{2NO}}_{\text{(g)}}{\text{ + 6 H}}_{\text{2}}{\text{O + 5Cu}}_{\text{(aq)}}^{\text{2+}}$

Explanation of Solution

Assign oxidation number to each atom:

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

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of N decreases from +5 to +2. Thus, the reduction half-reaction can be written as;

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

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of Cu increases from 0 to +2. Thus, the oxidation half-reaction can be written as;

  ${\text{Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ Cu}}_{(aq)}^{2+}$

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

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 6H}}^{\text{+}}\stackrel{}{\to }{\text{ NO}}_{\text{(g)}}{\text{ + 3 H}}_{\text{2}}\text{O (Reduction)}\\ {\text{Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ Cu}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation)}\end{array}$

Step-4:

Balance charges:

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 6H}}^{\text{+}}\stackrel{}{\to }{\text{ NO}}_{\text{(g)}}{\text{ + 3 H}}_{\text{2}}\text{O + 5e- (Reduction)}\\ {\text{Cu}}_{\text{(s)}}\text{ + 2 e-}\stackrel{}{\to }{\text{ Cu}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation)}\end{array}$

Step-5:

Add both equation:

  $\begin{array}{l}{\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 6H}}^{\text{+}}\stackrel{}{\to }{\text{ NO}}_{\text{(g)}}{\text{ + 3 H}}_{\text{2}}\text{O + 5e- (Reduction)}\\ {\text{Cu}}_{\text{(s)}}\text{ + 2 e-}\stackrel{}{\to }{\text{ Cu}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 12H}}^{\text{+}}\stackrel{}{\to }{\text{ 2NO}}_{\text{(g)}}{\text{ + 6 H}}_{\text{2}}\text{O + 10e- (Reduction)}\\ {\text{5Cu}}_{\text{(s)}}\text{ + 10 e-}\stackrel{}{\to }{\text{ 5Cu}}_{\text{(aq)}}^{\text{2+}}\text{ (Oxidation)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 12H}}^{\text{+}}+{\text{5Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{2NO}}_{\text{(g)}}{\text{ + 6 H}}_{\text{2}}{\text{O + 5Cu}}_{\text{(aq)}}^{\text{2+}}\end{array}$

Thus, the balance equation is:

  ${\text{2NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 12H}}^{\text{+}}+{\text{5Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{2NO}}_{\text{(g)}}{\text{ + 6 H}}_{\text{2}}{\text{O + 5Cu}}_{\text{(aq)}}^{\text{2+}}$

(d)

Interpretation Introduction

Interpretation: The oxidizing agent and reducing agent in the redox reaction in acidic medium needs to be determined.

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

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved in 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 9STP

• ${\text{Cu}}_{\text{(s)}}$ = Reducing agent
• ${\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{-}$ = Oxidizing agent

Explanation of Solution

Assign oxidation number to each atom:

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

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of N decreases from +5 to +2. Thus, the reduction half-reaction can be written as;

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

Oxidation half-reaction must include oxidation step in which elements show increase in oxidation state. Here the oxidation state of Cu increases from 0 to +2. Thus, the oxidation half-reaction can be written as;

  ${\text{Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{ Cu}}_{(aq)}^{2+}$

The balance equation is:

  ${\text{2NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{\text{-}}{\text{+ 12H}}^{\text{+}}+{\text{5Cu}}_{\text{(s)}}\stackrel{}{\to }{\text{2NO}}_{\text{(g)}}{\text{ + 6 H}}_{\text{2}}{\text{O + 5Cu}}_{\text{(aq)}}^{\text{2+}}$

The substance, which oxidizes is called as reducing agent whereas the substance, which reduces is called as oxidizing agent.

Here ${\text{Cu}}_{\text{(s)}}$ oxidizes so it must be reducing agent whereas ${\text{NO}}_{\text{3}}{}_{{}_{\text{(aq)}}}^{-}$ must be oxidizing agent.