Chapter 6, Problem 6K.1E

(a)

Interpretation Introduction

Interpretation:

From the following redox reaction, which elements undergoing oxidation, reduction and their initial and final oxidation number has to be determined.

  ${\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Concept Introduction:

Net ionic equation:

Net ionic equation is defined as the specific species that only involves to a particular reaction. This type of equations is generally used in acid-base neutralization reactions and redox reactions.

Answer to Problem 6K.1E

The Cr is undergoing reduction reaction and their initial and final oxidation number is +6 and +3, respectively. The C is undergoing oxidation reaction and their initial and final oxidation number is $-2$ and $-1$, respectively.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  ${\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Reduction reaction:

The oxidation number of $\text{Cr}$ is decreased from $+6$ to $+2$, therefore this is reduction reaction.

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}\to {\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}$

Oxidation number of $\text{Cr}$ is calculated from ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}$ is given below;

Oxidation number of $\text{O}$ is $-2$

  $\begin{array}{c}\text{2x+7×(-2)=-2}\\ \text{2x=-2+14}\\ \text{2x=12}\\ \text{x=6}\end{array}$

Therefore, the oxidation number of $\text{Cr}$ present in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}$ is $+6$.

Oxidation number of ${\text{Cr}}^{3+}$ is $+3$.

Oxidation reaction:

The oxidation number of $\text{C}$ is increased from $-2$ to $-1$, therefore this is oxidation reaction

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{C}}_{\text{2}}{\text{H}}_4{\text{O}}_{\text{(aq)}}$

Oxidation number of $\text{C}$ is calculated from ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}$ is given below;

Oxidation number of $\text{O}$ is $-2$

Oxidation number of $\text{H}$ is $+1$

  $\begin{array}{c}\text{2x+5-2+1=0}\\ \text{2x+4=0}\\ \text{2x=-4}\\ \text{x=-2}\end{array}$

Therefore, the oxidation number of $\text{C}$ present in ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}$ is $-2$.

Oxidation number of $\text{C}$ is calculated from ${\text{C}}_{\text{2}}{\text{H}}_4{\text{O}}_{\text{(aq)}}$ is given below;

Oxidation number of $\text{O}$ is $-2$

Oxidation number of $\text{H}$ is $+1$

  $\begin{array}{c}\text{2x+4-2=0}\\ \text{2x+2=0}\\ \text{2x=-2}\\ \text{x=-1}\end{array}$

Therefore, the oxidation number of $\text{C}$ present in ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}$ is $-1$.

Therefore, the $\text{Cr}$ is undergoing reduction reaction and their initial and final oxidation number is $+6$ and $+3$, respectively. The $\text{C}$ is undergoing oxidation reaction and their initial and final oxidation number is $-2$ and $-1$, respectively.

(b)

Interpretation Introduction

Interpretation:

The oxidation half-reaction balanced equation has to be derived.

Concept introduction:

Refer to part (a).

Answer to Problem 6K.1E

The balanced oxidation half-reaction equation is

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}{\text{+2H}}^{\text{+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}$

Explanation of Solution

Oxidation half-reaction:

The oxidation number of $\text{C}$ is increased from $-2$ to $-1$, therefore this is oxidation reaction.

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}$

Balance the $\text{H}$ atom by adding ${\text{H}}^{+}$ ion on the right side

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}{\text{+2H}}^{\text{+}}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $0$ and the right side the net charge is $+2$, therefore it required $2$ electrons on the right side to reduce the charge from $+2$ to $0$.

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}{\text{+2H}}^{\text{+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}$

Therefore, the balanced oxidation half-reaction is,

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}{\text{+2H}}^{\text{+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}$

(c)

Interpretation Introduction

Interpretation:

The reduction half-reaction balanced equation has to be derived.

Concept introduction:

Refer to part (a).

Answer to Problem 6K.1E

The balanced reduction half-reaction equation is

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+14H}}^{\text{+}}{}_{\text{(aq)}}{\text{+6e}}^{\text{-}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+{\text{7H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

Reduction half-reaction:

The oxidation number of $\text{Cr}$ is decreased from $+6$ to $+3$, therefore this is reduction reaction.

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}\to {\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}$

Balance the $\text{O}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ right side

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+{\text{7H}}_{\text{2}}\text{O}$

Balance the $\text{H}$ atom by adding ${\text{H}}^{+}$ ion on the left side

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+14H}}^{\text{+}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+{\text{7H}}_{\text{2}}\text{O}$

Balancing the net charges by adding electrons.

Here, in left side, the net charge is $+12$ and the right side the net charge is $+6$, therefore it required $6$ electrons on the left side to reduce the charge from $+12$ to $+6$.

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+14H}}^{\text{+}}{}_{\text{(aq)}}{\text{+6e}}^{\text{-}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+{\text{7H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Therefore, the balanced reduction half-reaction is

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+14H}}^{\text{+}}{}_{\text{(aq)}}{\text{+6e}}^{\text{-}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+{\text{7H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(d)

Interpretation Introduction

Interpretation:

A balanced redox equation for the combined half reactions has to be derived.

Concept introduction:

Refer to part (a).

Answer to Problem 6K.1E

A balanced redox equation for the combined half reactions is

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+3C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}+8{\text{H}}^{\text{+}}{}_{\text{(aq)}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+3{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}+7{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction $2$ electrons are lost and in reduction half reaction $6$ electron is gained. Therefore, multiply the oxidation half reaction by $6$ and the reduction half reaction by $2$.

  $\begin{array}{c}{\text{2Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+28H}}^{\text{+}}{}_{\text{(aq)}}{\text{+12e}}^{\text{-}}\to 4{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+14{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ {\text{6C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}\to 6{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}{\text{+12H}}^{\text{+}}{}_{\text{(aq)}}{\text{+12e}}^{\text{-}}\end{array}$

Add the equation and cancel the common ions and electrons in each side of the arrow.

  ${\text{2Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+6C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}+16{\text{H}}^{\text{+}}{}_{\text{(aq)}}\to 4{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+6{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}+14{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Divide by 2 each side of the arrow.

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+3C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}+8{\text{H}}^{\text{+}}{}_{\text{(aq)}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+3{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}+7{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Therefore, the balanced net ionic equation the above reaction is,

  ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{+3C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(aq)}}+8{\text{H}}^{\text{+}}{}_{\text{(aq)}}\to 2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}+3{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{(aq)}}+7{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$