The dark purple colour of $\mathrm{KMnO}_{4}$ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is :
Solution
$2 \overset{+7}{\mathrm{KMnO}_{4}}+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow$
$$
\mathrm{K}_{2} \mathrm{SO}_{4}+2 \overset{+2}{\mathrm{MnSO}_{4}}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}
$$
<br/><br/>
Change is oxidation state $M n$ is 5.
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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