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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Byju's Answer

Standard XII

Chemistry

Chemical Properties of Phenols

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:

8

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4

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16

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24

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Solution

The correct option is D $24$
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. Calculate

Δ H

for the following reaction:

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

Given

Δ H_{f}

for

A l_{2} O_{3} (s)

and

F e_{3} O_{4} (s)

are -1669.8 kJ and -1120.9 kJ respectively.

Q. Calculate

Δ H

for the following reaction:

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

Given

Δ H_{f}

for

A l_{2} O_{3} (s)

and

F e_{3} O_{4} (s)

are -1669.8 kJ and -1120.9 kJ respectively.

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

The correct option is D 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 D $24$
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.