Question

The colour of $KMn{O}_4$ disappears when oxalic acid is added to its solution in an acidic medium. This can be explained as :

• A. the pH of the solution changes on adding oxalic acid, hence $KMn{O}_4$ is decolourised
• B. $KMn{O}_4$ oxidizes oxalic acid to $C{O}_2$ and itself changes to $M{n}^{2+}$ ions which are colourless
• C. $KMn{O}_4$ is oxidised to potassium sulphate which is colourless
• D. on exposure to air the acidic solution of $KMn{O}_4$ becomes colourless

Answer 1

The correct option is B $KMn{O}_4$ oxidizes oxalic acid to $C{O}_2$ and itself changes to $M{n}^{2+}$ ions which are colourless

In acidic condition

$5e^{-}+8H^{+}+{Mn{O}_4}^{-}⟶M{n}^{2+}+4H_{2}O\times 2$

colourless

$H_2{C}_2{O}_4⟶2C{O}_2+2H^{+}+2e^{-}\times 5$

$reduced$

$\downarrow$

${2Mn{O}_4}^{-}+\underset{oxalicacid}{5H_2{C}_2{O}_4}+6H^{+}⟶2M{n}^{2+}+10C{O}_2+8H_{2}O$

Here $KMn{O}_4$ oxidizes oxalic acid to $C{O}_2$