When KMnO_4 is reduced with oxalic acid in ac | vedclass

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(C) In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1 + 4(-2) = 0$,so $x = +7$. In acidic medium,$KMnO_4$ acts as an oxidizing agent an

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$7$ to $2$

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(C) In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1 + 4(-2) = 0$,so $x = +7$. In acidic medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $Mn^{2+}$ ions. The balanced chemical equation is: $2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 	o K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2$. Thus,the oxidation number of $Mn$ changes from $+7$ to $+2$.

When $KMnO_4$ is reduced with oxalic acid in acidic solution,the oxidation number of $Mn$ changes from

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