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:

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Solution

The correct option is D
24

Elemental metallic Aluminium carries an oxidation number of 0.
From the products, we see that $0 A l$ becomes $3 + A l^{3 +}$ .
Clearly, Aluminium is the reducing agent.

There are 8 Aluminium atoms on the left and they all become $A l^{3 +}$ .
So overall, there are 24 electrons transferred.

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Similar questions

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

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:

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

Δ H

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

Δ H_{f}

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

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

Δ H

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

Δ H_{f}

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

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

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