6 Fe_3O_4 + 2 MnO_4^- + H_2O to 9 Fe_2O_3 + 2 | vedclass

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(A) In $Fe_3O_4$,the average oxidation state of $Fe$ is $+\frac{8}{3}$.In $Fe_2O_3$,the oxidation state of $Fe$ is $+3$.The change in oxidation state

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$M$

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(A) In $Fe_3O_4$,the average oxidation state of $Fe$ is $+\frac{8}{3}$.In $Fe_2O_3$,the oxidation state of $Fe$ is $+3$.The change in oxidation state per $Fe$ atom is $3 - \frac{8}{3} = \frac{1}{3}$.Since there are $3$ atoms of $Fe$ in one molecule of $Fe_3O_4$,the total change in oxidation state ($n$-factor) is $3 	imes \frac{1}{3} = 1$.The equivalent weight $E_w = \frac{	ext{Molar mass } (M)}{n	ext{-factor}} = \frac{M}{1} = M$.

$6 Fe_3O_4 + 2 MnO_4^- + H_2O \to 9 Fe_2O_3 + 2 MnO_2 + 2 OH^-$ Calculate the equivalent weight of $Fe_3O_4$ in the above equation.

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