Q21.142 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution:

${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{OH + Cr}}}_{\mathbf{\text{2}}}{{\mathbf{\text{O}}}_{\mathbf{\text{7}}}}^{\mathbf{\text{2 - }}}\mathbf{\to }{\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{COOH + Cr}}}^{\mathbf{\text{3 + }}}$

Step-by-Step Solution

Answer

The balanced equation for the conversion of ethanol to acetic acid in acid solution is:

${\text{3CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH + 2Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}{\text{ + 16H}}^{\text{ + }}\to {\text{3CH}}_{\text{3}}{\text{COOH + 4Cr}}^{\text{3 + }}{\text{ + 11H}}_{\text{2}}\text{O}$

Step 1: Define chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions

Step 2: Balancing the equation for the conversion of ethanol to acetic acid

The reaction is:

 ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH + Cr}}_{\text{2}}{{\text{O}}_{\text{7}}}^{\text{2 - }}\to {\text{CH}}_{\text{3}}{\text{COOH + Cr}}^{\text{3 + }}$

To balance the equation, we first will separate the half-reactions

 $$\begin{gathered} {\text{(i) Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}\text{(aq)}\to {\text{Cr}}^{\text{3 + }}\text{(aq)} \\ {\text{(ii) CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\to {\text{CH}}_{\text{3}}\text{COOH} \end{gathered}$$

The reduction and oxidation half reaction then can be written as:

 $$\begin{gathered} {\text{6e}}^{\text{ - }}{\text{ + 14H + Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}\to {\text{2Cr}}^{\text{3 + }}{\text{ + 7H}}_{\text{2}}\text{O(reduction)} \\ {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH + H}}_{\text{2}}\text{O}\to {\text{CH}}_{\text{3}}{\text{COOH + 4e}}^{\text{ - }}{\text{ + 4H}}^{\text{ + }}\text{(oxidation)} \end{gathered}$$

To balance the number of charges, we should multiply the reduction reaction by two and three for the oxidation reaction:

$$\begin{gathered} {\text{12e}}^{\text{ - }}{\text{ + 28H + 2Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}\to {\text{4Cr}}^{\text{3 + }}{\text{ + 14H}}_{\text{2}}\text{O} \\ {\text{3CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH + 3H}}_{\text{2}}\text{O}\to {\text{3CH}}_{\text{3}}{\text{COOH + 12e}}^{\text{ - }}{\text{ + 12H}}^{\text{ + }} \end{gathered}$$ 

The balanced equation then will be:

 ${\text{3CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH + 2Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}{\text{ + 16H}}^{\text{ + }}\to {\text{3CH}}_{\text{3}}{\text{COOH + 4Cr}}^{\text{3 + }}{\text{ + 11H}}_{\text{2}}\text{O}$

Therefore, the balanced reaction is:  .

${\text{3CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{OH + 2Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 - }}{\text{ + 16H}}^{\text{ + }}\to {\text{3CH}}_{\text{3}}{\text{COOH + 4Cr}}^{\text{3 + }}{\text{ + 11H}}_{\text{2}}\text{O}$