In the reaction- 8Al - 3Fe3O4 - 4Al2O3 - 9Fe- number of electrons transferred from reductant to oxidant is-

On balancing the charges of two half reactions we get 8Al → 8Al^3+ + 24e^ (Oxidation half) 9Fe^83+ + 24e^ → 9Fe (Reduction half) Hence, 24 nbsp;electr ...

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Standard XII

Chemistry

Oxidation and Reduction

In the reacti...

Question

In the reaction, $8 A l + 3 F e_{3} O_{4} \to 4 A l_{2} O_{3} + 9 F e,$ number of electrons transferred from reductant to oxidant is:

4

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24

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16

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8

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Solution

The correct option is B $24$
On balancing the charges of two half-reactions we get
$8 A l \to 8 A l^{3 +} + 24 e^{-} (Oxidation half) 9 F e^{\frac{8}{3} +} + 24 e^{-} \to 9 F e (Reduction half)$
Hence, $24$ electrons are transferred from reductant to oxidant.

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Similar questions

In the reaction, $8 A l + 3 F e_{3} O_{4} \to 4 A l_{2} O_{3} + 9 F e$ , the number of electrons transferred from reductant to oxidant is:

Q. In the reaction

8 A l + {3 F e}_{3} O_{4} \to {4 A l}_{2} O_{3} + 9 F e

, the number of electrons transferred from reductant to oxidant is:

Q. Identify the reaction undergoing oxidation
a) H₂S + Cl₂ ------------> 2HCl + S
b) 3 Fe₃O₄ + 8AL--------------> 9Fe + 4Al₂O3
C) 2Na (S) +H₂ -----------> 2NaH

Q. In the reactions given below, identity the species undergoing oxidation and reduction:

A)

H_{2} S (g) + C l_{2} \to 2 H C l (g) + S (s)

B)

3 F e_{3} O_{4} (s) + 8 A l (s) \to 9 F e (s) + 4 A l_{2} O_{3} (s)

C)

2 N a (s) + H_{2} (g) \to 2 N a H (s)

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In the reaction, 8Al+3Fe3O4→4Al2O3+9Fe, number of electrons transferred from reductant to oxidant is:

The correct option is B 24On balancing the charges of two half-reactions we get 8Al→8Al3++24e− (Oxidation half)9Fe83++24e−→9Fe (Reduction half) Hence, 24 electrons are transferred from reductant to oxidant.

In the reaction, $8 A l + 3 F e_{3} O_{4} \to 4 A l_{2} O_{3} + 9 F e,$ number of electrons transferred from reductant to oxidant is:

The correct option is B $24$
On balancing the charges of two half-reactions we get
$8 A l \to 8 A l^{3 +} + 24 e^{-} (Oxidation half) 9 F e^{\frac{8}{3} +} + 24 e^{-} \to 9 F e (Reduction half)$
Hence, $24$ electrons are transferred from reductant to oxidant.