The total number of electrons involved in redox change is :
$3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$

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Solution

The correct option is D 8
In $F e_{3} O_{4}$ , the average oxidation number of Fe is $\frac{8}{3}$ .
For 3 Fe atoms of $F e_{3} O_{4}$ , the total oxidation number is 8.
On the reactants side, the oxidation number of Fe is 0.
Hence, during the reaction, the oxidation number of Fe changes by 8.
Hence, the reaction involves 8 electrons.

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

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

3 F e + 4 H_{2} O ⟶ F e_{3} O_{4} + 4 H_{2}

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

Balance the following reaction

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

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