Number of moles of MnO_4^- required to oxidiz | vedclass

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(C) The balanced redox reaction involves the oxidation of $Fe^{2+}$ to $Fe^{3+}$ and $C_2O_4^{2-}$ to $CO_2$,while $MnO_4^-$ is reduced to $Mn^{2+}$.T

Vedclass · Accepted Answer

$0.6$

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(C) The balanced redox reaction involves the oxidation of $Fe^{2+}$ to $Fe^{3+}$ and $C_2O_4^{2-}$ to $CO_2$,while $MnO_4^-$ is reduced to $Mn^{2+}$.The valence factor $(v.f.)$ for $MnO_4^-$ is $5$ (change in oxidation state from $+7$ to $+2$).The valence factor $(v.f.)$ for ferrous oxalate $(FeC_2O_4)$ is $3$ ($Fe^{2+} 	o Fe^{3+}$ is $1$ electron,and $C_2O_4^{2-} 	o 2CO_2$ is $2$ electrons; total $1 + 2 = 3$).Using the principle of equivalence: $n_{MnO_4^-} 	imes (v.f.)_{MnO_4^-} = n_{FeC_2O_4} 	imes (v.f.)_{FeC_2O_4}$.$n 	imes 5 = 1 	imes 3$.$n = \frac{3}{5} = 0.6 	ext{ moles}$.

Number of moles of $MnO_4^-$ required to oxidize one mole of ferrous oxalate completely in acidic medium will be .............. $moles$.

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