Chemistry - Electrochemistry Question with Solution | TestHub

The reduction reaction of ${\text{Al}}^{+3}$ to $\text{Al}$ is given by:

${\text{Al}}^{+3}+3e^{-}\to \text{Al}$

From the stoichiometry of the reaction, $1$ mole of ${\text{Al}}^{+3}$ requires $3$ moles of electrons for complete reduction.

Therefore, the number of moles of electrons required for the reduction of $2$ moles of ${\text{Al}}^{+3}$ is $2\times 3=6$ moles.

Answer: $6$