3Fe + 2O₂ ---------- Fe₃O₄
How does this reaction occur, because on cris-crossing the valencies of Fe and O we don't get Fe₃O₄?

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Solution

Oxide ions tend to have a 2- charge. Fe3O4 is a neutral molecule, so that means that the overall charge balance is 3(Fe) + 4(-2) = 0, which gives the iron in Fe3O4 an overall valency of +8/3. But since you cannot have fractional valencies (you can only lose a whole number of electrons, but not fractions of an electron), this is only an average valency for the iron atoms in Fe3O4. It also means that each iron atom has a different valency. Currently, it has been assumed that Fe3O4 is made up of FeO and Fe2O3 in equal parts, thus giving the empirical formula of Fe3O4. This means that you have 2 irons with a valency of +3, and one iron with a valency of +2.

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E_{F e^{2 +} / F e}^{o} = - 0.441 V

E_{F e^{3 +} / F e^{2 +}}^{o} = 0.771 V

E^{o}

F e + 2 F e^{3 +} \to 3 F e^{2 +}

Balance the following reaction

$3 F e$ + $4 H_{2} O$ $\to$ $F e_{3} O_{4}$ + $_H_{2}$

3 F e^{2 +} (a q) + 2 C r (s) = 2 C r^{3 +} (a q) + 3 F e (s)

E_{F e^{2 +} / F e}^{o} = - 0.44

E_{C r^{3 +} / C r}^{o} = - 0.74

(F = 96500 C)

E_{F e^{2 +} / F e}^{0}

-

E_{F e^{3 +} / F e^{2 +}}^{0}

E M F

F E + 2 F E^{3 +} \to 3 F E^{2 +}

E_{F e^{2 +} / F e}^{0} = - 0.441 V

E_{F e^{3 +} / F e^{2 +}}^{0} = 0.771 V

F e + 2 F e^{3 +} \to 3 F e^{2 +}

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