KMnO_4 oxidises oxalic acid to CO_2 in acidic | vedclass

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(B) In acidic medium,$KMnO_4$ acts as an oxidizing agent and undergoes reduction as follows:$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$The cha

Vedclass · Accepted Answer

$31.6$

Vedclass · Answer

(B) In acidic medium,$KMnO_4$ acts as an oxidizing agent and undergoes reduction as follows:$MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$The change in oxidation state of $Mn$ is from $+7$ to $+2$,which is a change of $5$ units.Molecular mass of $KMnO_4 = 39 + 55 + (4 	imes 16) = 158 \ g/mol$.Equivalent weight = $\frac{	ext{Molecular mass}}{	ext{n-factor}} = \frac{158}{5} = 31.6$.

$KMnO_4$ oxidises oxalic acid to $CO_2$ in acidic medium. The equivalent weight of $KMnO_4$ is $(Mn = 55, K = 39, O = 16)$:

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