Chapter 18, Problem 18.1QP

Balance the following redox equations by the half-reaction method.

(a) H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O (in acidic solution)

(b) Cu + HNO₃ → Cu²⁺ + NO + H₂O (in acidic solution)

(c) CN⁻ + MnO₄⁻ →     CNO⁻ + MnO₂ (in basic solution)

(d) Br₂ → BrO₃⁻ + Br⁻ (in basic solution)

(e) S₂O₃²⁻ + I₂ →I⁻ + S₄O₆²⁻ (in acidic solution)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by using half-reaction.

Concept Introduction:

Redox equation: balancing redox equation in basic solution is identified to that of acidic solution except for the last few steps as shown below

To balance the equation, use the following steps:

Firstly, divide the equation into two halves an oxidation half-reaction and reduction half reaction by grouping appropriate species.

Lastly Balance the oxygen atom using water molecule.

Answer to Problem 18.1QP

${\text{2Fe}}^{\text{2+}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+2H}}^{\text{+}}\stackrel{}{\to }{\text{2Fe}}^{\text{3+}}{\text{+2H}}_{\text{2}}\text{O}$

Explanation of Solution

Draw the balanced half –reaction and type of charge.

Balancing redox equation

Balancing each half reaction and type of atoms and charges.

$\begin{array}{l}{\text{2Fe}}^{\text{2+}}\stackrel{}{\to }{\text{2Fe}}^{\text{2+}}\text{+2e}\\ \underset{\_}{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{2H}}_{\text{2}}\text{O}}\\ {\text{2Fe}}^{\text{2+}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+2H}}^{\text{+}}\stackrel{}{\to }{\text{2Fe}}^{\text{3+}}{\text{+2H}}_{\text{2}}\text{O}\end{array}$

Interpretation Introduction

Interpretation: The given redox equation should be balanced by using half-reaction.

Concept Introduction:

Redox equation: balancing redox equation in basic solution is identified to that of acidic solution except for the last few steps as shown below

To balance the equation, use the following steps:

Firstly, divide the equation into two halves an oxidation half-reaction and reduction half reaction by grouping appropriate species.

Lastly Balance the oxygen atom using water molecule.

Answer to Problem 18.1QP

${\text{3Cu+2HNO}}_{\text{3}}{\text{+6H}}^{\text{+}}\to {\text{3Cu}}^{\text{2+}}{\text{+2NO+4H}}_{\text{2}}\text{O}$

Explanation of Solution

Draw the balanced half –reaction and type of charge.

Balancing redox equation

Balancing each half reaction and type of atoms and charges

$\begin{array}{l}\text{3Cu}\stackrel{}{\to }{\text{3Cu}}^{\text{2+}}{\text{+6e}}^{\text{-}}\\ \underset{\_}{{\text{2HNO}}_{\text{3}}{\text{+6H}}^{\text{+}}\stackrel{}{\to }{\text{2NO+4H}}_{\text{2}}\text{O}}\\ {\text{3Cu+2HNO}}_{\text{3}}{\text{+6H}}^{\text{+}}\stackrel{}{\to }{\text{3Cu}}^{\text{2+}}{\text{+2NO+4H}}_{\text{2}}\text{O}\end{array}$

Interpretation Introduction

Interpretation: The given redox equation should be balanced by using half-reaction.

Concept Introduction:

Redox equation: balancing redox equation in basic solution is identified to that of acidic solution except for the last few steps as shown below

To balance the equation, use the following steps:

Firstly, divide the equation into two halves an oxidation half-reaction and reduction half reaction by grouping appropriate species.

Lastly Balance the oxygen atom using water molecule.

Answer to Problem 18.1QP

${\text{3CN}}^{\text{-}}{\text{+2MnO}}^{\text{-}}{}_{\text{4}}{\text{+2H}}^{\text{+}}\to {\text{2MnO}}_{\text{2}}{\text{+3CNO}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}$

Explanation of Solution

Draw the balanced half –reaction and type of charge.

Balancing redox equation

Balancing each half reaction and type of atoms and charges

$\begin{array}{l}{\text{3CN}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{3CNO}}^{\text{-}}{\text{6H}}^{\text{+}}{\text{+6e}}^{\text{-}}\\ \underset{\_}{{\text{2MnO}}_{\text{4}}{}^{\text{-}}{\text{+8H}}^{\text{+}}{\text{+6e}}^{\text{-}}\stackrel{}{\to }{\text{2Mno}}_{\text{2}}{\text{+4H}}_{\text{2}}\text{O}}\\ {\text{3CN}}^{\text{-}}{\text{+2MnO}}^{\text{-}}{}_{\text{4}}{\text{+2H}}^{\text{+}}\stackrel{}{\to }{\text{2MnO}}_{\text{2}}{\text{+3CNO}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\end{array}$

Interpretation Introduction

Interpretation: The given redox equation should be balanced by using half-reaction.

Concept Introduction:

Redox equation: balancing redox equation in basic solution is identified to that of acidic solution except for the last few steps as shown below

To balance the equation, use the following steps:

Firstly, divide the equation into two halves an oxidation half-reaction and reduction half reaction by grouping appropriate species.

Lastly Balance the oxygen atom using water molecule.

Answer to Problem 18.1QP

${\text{6Br}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\to {\text{2BrO}}_{\text{3}}{}^{\text{-}}{\text{+12H}}^{\text{+}}{\text{+10Br}}^{\text{-}}$

Explanation of Solution

Draw the balanced half –reaction and type of charge.

Balancing redox equation

Balancing each half reaction and type of atoms and charges

$\begin{array}{l}{\text{Br}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{2BrO}}_{\text{3}}{}^{\text{-}}{\text{+12H}}^{\text{+}}{\text{+10e}}^{\text{-}}\\ \underset{\_}{{\text{5Br}}_{\text{2}}{\text{+10e}}^{\text{-}}\stackrel{}{\to }{\text{10Br}}^{\text{-}}}\\ {\text{6Br}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{2BrO}}_{\text{3}}{}^{\text{-}}{\text{+12H}}^{\text{+}}{\text{+10Br}}^{\text{-}}\end{array}$

Interpretation Introduction

Interpretation: The given redox equation should be balanced by using half-reaction.

Concept Introduction:

Redox equation: balancing redox equation in basic solution is identified to that of acidic solution except for the last few steps as shown below

To balance the equation, use the following steps:

Firstly, divide the equation into two halves an oxidation half-reaction and reduction half reaction by grouping appropriate species.

Lastly Balance the oxygen atom using water molecule.

Answer to Problem 18.1QP

${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}{\text{+I}}_2\to {\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}{\text{+2I}}^{\text{-}}$

Explanation of Solution

Draw the balanced half –reaction and type of charge.

Balancing redox equation

Balancing each half reaction and type of atoms and charges

$\begin{array}{l}{\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}{\text{+2e}}^{\text{-}}\\ \underset{\_}{{\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{2I}}^{\text{-}}}\\ {\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}{\text{+I}}_{\text{2}}\stackrel{}{\to }{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}{\text{+2I}}^{\text{-}}\end{array}$