Number of moles of KMnO_4 required to oxidize | vedclass

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(A) In an acidic medium,the oxidation of $Fe(C_2O_4)$ by $KMnO_4$ involves the following changes in oxidation states:$Fe^{2+} \rightarrow Fe^{3+} + e^

Vedclass · Accepted Answer

$0.6$

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(A) In an acidic medium,the oxidation of $Fe(C_2O_4)$ by $KMnO_4$ involves the following changes in oxidation states:$Fe^{2+} ightarrow Fe^{3+} + e^-$$C_2O_4^{2-} ightarrow 2CO_2 + 2e^-$Total electrons lost per mole of $Fe(C_2O_4) = 1 + 2 = 3$.Thus,the $n$-factor for $Fe(C_2O_4)$ is $3$.For $KMnO_4$ in acidic medium,$Mn^{+7} + 5e^- ightarrow Mn^{2+}$,so the $n$-factor is $5$.Using the law of equivalence: $	ext{Equivalents of } KMnO_4 = 	ext{Equivalents of } Fe(C_2O_4)$.Let $x$ be the number of moles of $KMnO_4$.$x 	imes 5 = 1 	imes 3$$x = \frac{3}{5} = 0.6$.

Number of moles of $KMnO_4$ required to oxidize one mole of $Fe(C_2O_4)$ in acidic medium is

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