Answer:
2 Al(s) + 3 Fe²⁺(aq) → 2 Al³⁺(aq) + 3 Fe(s)
6 electrons are transferred.
Step-by-step explanation:
Let's consider the following standard reduction potentials.
E°(Al³⁺/Al) = -1.66 V
E°(Fe²⁺/Fe) = -0.44 V
The one with the highest standard reduction potential will occurs as a reduction and the other will occur as an oxidation. The corresponding half-reactions are:
Oxidation: Al(s) → Al³⁺(aq) + 3 e⁻
Reduction: Fe²⁺(aq) + 2 e⁻ → Fe(s)
To get the global equation, we have to multiply both half-reactions by numbers that assure that the number of electrons gained and lost are the same, and then add them.
2 × (Al(s) → Al³⁺(aq) + 3 e⁻)
3 × (Fe²⁺(aq) + 2 e⁻ → Fe(s))
----------------------------------------------------
2 Al(s) + 3 Fe²⁺(aq) + 6 e⁻ → 2 Al³⁺(aq) + 6 e⁻ + 3 Fe(s)
2 Al(s) + 3 Fe²⁺(aq) → 2 Al³⁺(aq) + 3 Fe(s)