Number of moles of MnO_4^{-} required to oxid | vedclass

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(C) In acidic medium,$MnO_4^{-}$ acts as an oxidizing agent and gets reduced to $Mn^{2+}$.Ferrous oxalate $(FeC_2O_4)$ contains $Fe^{2+}$ and $C_2O_4^

Vedclass · Accepted Answer

$0.6$

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(C) In acidic medium,$MnO_4^{-}$ acts as an oxidizing agent and gets reduced to $Mn^{2+}$.Ferrous oxalate $(FeC_2O_4)$ contains $Fe^{2+}$ and $C_2O_4^{2-}$.Both $Fe^{2+}$ and $C_2O_4^{2-}$ are oxidized by $MnO_4^{-}$:$Fe^{2+} ightarrow Fe^{3+} + e^-$$C_2O_4^{2-} ightarrow 2CO_2 + 2e^-$Total electrons lost per mole of $FeC_2O_4 = 1 + 2 = 3 \ e^-$.Reduction half-reaction of $MnO_4^{-}$:$MnO_4^{-} + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$.For complete oxidation,the number of electrons lost must equal the number of electrons gained.$n_{MnO_4^{-}} 	imes 5 = n_{FeC_2O_4} 	imes 3$.Given $n_{FeC_2O_4} = 1 \ mole$,then $n_{MnO_4^{-}} = \frac{3}{5} = 0.6 \ mole$.

Number of moles of $MnO_4^{-}$ required to oxidise one mole of ferrous oxalate completely in acidic medium will be $:-$ (in $mole$)

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