When KMnO_4 is reduced with oxalic acid in ac | vedclass

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Question

(C) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in an acidic medium is:$5(COOH)_2 + 2KMnO_4 + 3H_2SO_4 	o K_2SO_

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$7$ to $2$

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(C) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in an acidic medium is:$5(COOH)_2 + 2KMnO_4 + 3H_2SO_4 	o K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1 + 4(-2) = 0$,which gives $x = +7$.In $MnSO_4$,the oxidation state of $Mn$ is $+2$ because the sulfate ion $(SO_4^{2-})$ has a charge of $-2$.Therefore,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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