Question

Calculate the number of electrons exchanged in the reaction given below.
$Al + F{e_3}{O_4} \to A{l_2}{O_3} + Fe$.

Answer 1

Hint: It is a redox reaction. Balance the given chemical equation. Then calculate the oxidation states of Fe and Al both in the product as well as on the reactant side followed by calculate the number of electrons gained or lost.

Complete step by step answer:
Balance the chemical reaction
$8Al + 3F{e_3}{O_4} \to 4A{l_2}{O_3} + 9Fe$
This above reaction is a Redox reaction. A redox reaction is a chemical reaction in which electrons are transferred between two reactants participating in it. This transfer of electrons can be identified by observing the changes in the oxidation states of the reacting species.

Now we calculate the Oxidation states of Al and Fe

Fe is reduced by reductant Al with oxidation state from +8/3 to 0. Al acts as reducing agent because it reduces Fe and gets oxidised

In the case of Al, on the left side of the equation it has 0 oxidation state and on the right side has +3 . Here Fe acts as an oxidising agent because it oxidises Al and gets reduced.
On product side,
$2x - 2( - 3) = 0$
$2x = - 6$
$x = + 3$
$A{l^0} \to A{l^{ + 3}} + 3{e^ - }$
As there are 4 $A{l_2}{O_3}$on product side, so there are 8 Al
Therefore, The number of electrons lost by $Al = 8 \times 3 = 24{e^ - }_{}$

Therefore, Al loses 24 electrons and Fe also gains 24 electrons. So, there are $24{e^ - }$ exchanged

Note: Another method can be used where in the stoichiometric coefficients are added to calculate the number of electrons exchanged. So in this equation 8, 3, 4, and 9 are stoichiometric coefficients of Al, $F{e_2}{O_3}$, $A{l_2}{O_3}$and Fe respectively.

$8Al + 3F{e_3}{O_4} \to 4A{l_2}{O_3} + 9Fe$
Here, $8 + 3 + 4 + 9 = 24{e^ - }$
Hence any of these methods can be used.