What is $F e_{3} O_{4}$ $Fe_3O_4$ , and how do we represent its formation from iron, and water?

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What is $F e_{3} O_{4}$ $Fe_3O_4$ , and how do we represent its formation from iron, and water?

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Answer 1 · 2017-08-28T13:00:23

This is a mixed valence oxide of $F e O$ $FeO$ and $F e_{2} O_{3}$ $Fe_2O_3$ ........

Explanation:

In the original redox reaction, elemental iron is OXIDIZED to $F e O + F e_{2} O_{3}$ $FeO + Fe_2O_3$ , i.e. $F e O \cdot F e_{2} O_{3} \equiv F e_{3} O_{4}$ $FeO*Fe_2O_3-=Fe_3O_4$ .

And so we write 2 separate oxidation rxns....

$F e + H_{2} O \to F e O + 2 e^{-} + 2 H^{+}$ $Fe +H_2O rarr FeO+2e^(-) +2H^+$ , and....

$2 F e + 3 H_{2} O \to F e_{2} O_{3} + 6 H^{+} + 6 e^{-}$ $2Fe +3H_2O rarr Fe_2O_3+6H^(+) + 6e^(-)$

We adds these together and get.....

$3Fe +4H_2O rarr underbrace(Fe_2O_3+FeO)_(Fe_3O_4) +8H^(+) + 8e^(-)$ $(i)$

And for every oxidation, electron loss, there is a corresponding reduction, electron gain, and the most likely oxidant is dioxygen gas.....

$1/2O_2 + 2H^+ + 2e^(-)rarrH_2O$ $(ii)$

And so we take $4xx(ii)+(i)$ to eliminate the electrons from our final redox equation.......

$3Fe +2O_2 + cancel(4H_2O +8H^(+) + 8e^(-)) rarr underbrace(Fe_2O_3+FeO)_(Fe_3O_4) +cancel(8H^(+) + 8e^(-)+4H_2O)$

To give finally.......

$3Fe +2O_2 rarr underbrace(Fe_2O_3+FeO)_(Fe_3O_4)$

$Fe_3O_4$ is a mixed valence iron oxide..... $Fe_2O_3*FeO$

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What is $F e_{3} O_{4}$ $Fe_3O_4$ , and how do we represent its formation from iron, and water?

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What is $F e_{3} O_{4}$ $Fe_3O_4$ , and how do we represent its formation from iron, and water?