What happens when a few drops of alkaline \[{\text{KMn}}{{\text{O}}_4}\] are added to ethanol?
Answer
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Hint: Alkaline \[{\text{KMn}}{{\text{O}}_4}\] is known to be a strong oxidising agent and hence will first oxidise ethanol. This might cause the reaction to occur in a different way than expected.
Complete step by step solution:
When \[{\text{KMn}}{{\text{O}}_4}\] is added to a solution of ethanol, and heated, the colour of the solution disappears. This is because, \[{\text{KMn}}{{\text{O}}_4}\] causes oxidation of ethanol to ethanoic acid by donating nascent oxygen.
Following reaction occurs -
\[{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH + 2[O] }}\xrightarrow[{{\text{Heat}}}]{{{\text{Alkaline KMn}}{{\text{O}}_4}}}{\text{ C}}{{\text{H}}_3}{\text{COOH + }}{{\text{H}}_2}{\text{O}}\]
Due to the acidic environment of ethanoic acid, the dark purple colour of \[{\text{KMn}}{{\text{O}}_4}\]disappears.
Hence, when alkaline\[{\text{KMn}}{{\text{O}}_4}\]is added to a solution of ethanol and warmed a little, the colour of potassium permanganate disappears.
Additional information: \[{\text{KMn}}{{\text{O}}_4}\]is dark purple in colour originally. It changes its colour to brown when used in a redox titration. This is because, in \[{\text{KMn}}{{\text{O}}_4}\] the oxidation state of \[{\text{Mn}}\]is \[{\text{M}}{{\text{n}}^{ + 7}}\]which changes to \[{\text{M}}{{\text{n}}^{ + 2}}\]when it is reduced. Hence it is known as a self-indicator. It is also called ‘Bayer’s reagent’.
In acidic solutions, \[{\text{KMn}}{{\text{O}}_4}\]with is reduced to pale pink with \[{\text{M}}{{\text{n}}^{ + 2}}\]oxidation state of manganese ion. The reaction occurs is –
\[8{{\text{H}}^ + }{\text{ + Mn}}{{\text{O}}^ - }_4{\text{ + 5}}{{\text{e}}^ - }{\text{ }} \to {\text{ M}}{{\text{n}}^{ + 2}}{\text{ + 4}}{{\text{H}}_2}{\text{O}}\]
In basic solutions, \[{\text{KMn}}{{\text{O}}_4}\]with \[{\text{M}}{{\text{n}}^{ + 7}}\]is reduced to green with \[{\text{M}}{{\text{n}}^{ + 6}}\] oxidation state of manganese. The reaction that occurs is –
\[{\text{Mn}}{{\text{O}}^ - }_4{\text{ + }}{{\text{e}}^ - }{\text{ }} \to {\text{ Mn}}{{\text{O}}^{ - 2}}_4\]
In neutral solutions, it gets reduced to brown colour with \[{\text{M}}{{\text{n}}^{ + 4}}\]oxidation state of manganese ion. The reaction that occurs is –
\[{\text{2}}{{\text{H}}_2}{\text{O + Mn}}{{\text{O}}_4}{\text{ + 3}}{{\text{e}}^ - }{\text{ }} \to {\text{ Mn}}{{\text{O}}_2}{\text{ + 4O}}{{\text{H}}^ - }\]
Note: We must keep in mind that if the environment becomes extremely acidic, the colour of potassium permanganate disappears completely.
Complete step by step solution:
When \[{\text{KMn}}{{\text{O}}_4}\] is added to a solution of ethanol, and heated, the colour of the solution disappears. This is because, \[{\text{KMn}}{{\text{O}}_4}\] causes oxidation of ethanol to ethanoic acid by donating nascent oxygen.
Following reaction occurs -
\[{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH + 2[O] }}\xrightarrow[{{\text{Heat}}}]{{{\text{Alkaline KMn}}{{\text{O}}_4}}}{\text{ C}}{{\text{H}}_3}{\text{COOH + }}{{\text{H}}_2}{\text{O}}\]
Due to the acidic environment of ethanoic acid, the dark purple colour of \[{\text{KMn}}{{\text{O}}_4}\]disappears.
Hence, when alkaline\[{\text{KMn}}{{\text{O}}_4}\]is added to a solution of ethanol and warmed a little, the colour of potassium permanganate disappears.
Additional information: \[{\text{KMn}}{{\text{O}}_4}\]is dark purple in colour originally. It changes its colour to brown when used in a redox titration. This is because, in \[{\text{KMn}}{{\text{O}}_4}\] the oxidation state of \[{\text{Mn}}\]is \[{\text{M}}{{\text{n}}^{ + 7}}\]which changes to \[{\text{M}}{{\text{n}}^{ + 2}}\]when it is reduced. Hence it is known as a self-indicator. It is also called ‘Bayer’s reagent’.
In acidic solutions, \[{\text{KMn}}{{\text{O}}_4}\]with is reduced to pale pink with \[{\text{M}}{{\text{n}}^{ + 2}}\]oxidation state of manganese ion. The reaction occurs is –
\[8{{\text{H}}^ + }{\text{ + Mn}}{{\text{O}}^ - }_4{\text{ + 5}}{{\text{e}}^ - }{\text{ }} \to {\text{ M}}{{\text{n}}^{ + 2}}{\text{ + 4}}{{\text{H}}_2}{\text{O}}\]
In basic solutions, \[{\text{KMn}}{{\text{O}}_4}\]with \[{\text{M}}{{\text{n}}^{ + 7}}\]is reduced to green with \[{\text{M}}{{\text{n}}^{ + 6}}\] oxidation state of manganese. The reaction that occurs is –
\[{\text{Mn}}{{\text{O}}^ - }_4{\text{ + }}{{\text{e}}^ - }{\text{ }} \to {\text{ Mn}}{{\text{O}}^{ - 2}}_4\]
In neutral solutions, it gets reduced to brown colour with \[{\text{M}}{{\text{n}}^{ + 4}}\]oxidation state of manganese ion. The reaction that occurs is –
\[{\text{2}}{{\text{H}}_2}{\text{O + Mn}}{{\text{O}}_4}{\text{ + 3}}{{\text{e}}^ - }{\text{ }} \to {\text{ Mn}}{{\text{O}}_2}{\text{ + 4O}}{{\text{H}}^ - }\]
Note: We must keep in mind that if the environment becomes extremely acidic, the colour of potassium permanganate disappears completely.
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