The number of electrons lost in the following change is :
$F e + H_{2} O ⟶ {F e}_{3} O_{4} + H_{2}$

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Solution

The correct option is C 8
$3 0 F e + 8 e^{-} ⟶ {({F e}^{8 / 3 +})}_{3}$
$2 H^{\oplus} ⟶_{2} + 2 e^{-} \times 4$
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$3 F e + 4 H_{2} O ⟶ {F e}_{3} O_{4} + 4 H_{2} O$
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Total number of electrons lost or gained is 8.

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Similar questions

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

C u + H_{2} {S O}_{4} - \to C u S O_{4} + {S O}_{2} ↑ + H_{2} O

F e + + H_{2} O - \to {F e}_{3} O_{4} + H_{2} ↑

{C H}_{4} + O_{2} \to C O_{2} + H_{2} O

C u + H_{2} S O_{4} \to C u S O_{4} + {S O}_{2} ↑ + H_{2} O

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

{C H}_{4} + O_{2} \to C O_{2} + H_{2} O

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

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A l + {F e}_{3} O_{4} \to {A l}_{2} O_{3} + F e

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