How is it determined how many electrons are transferred in redox reactions?

Question

For example, iron and tin(II) sulfate will react to produce iron(II) sulfate and tin:

$\ce{Fe + SnSO4 -> FeSO4 + Sn}$

With the following half equations:

$\ce{Fe + SO4^{2-} -> FeSO4 + 2e-}$

and

$\ce{Sn^{2+} + 2e- -> Sn}$

but where does the "2" come from? Why can't it be:

$\ce{Fe + SO4^{5-} -> FeSO4 + 5e-}$

and

$\ce{Sn^{5+} + 5e- -> Sn}$

Answer 1

In short, the species you mention ($\ce{Sn^{5+}}$ and $\ce{SO_4^{5-}}$) do not exist in any chemical system at anywhere near ordinary conditions. The number of electrons shuffled in the reaction is not chosen arbitrarily, but is based on the initial and final oxidation numbers of the elements in the reaction, after the equations are balanced. The only stable compound with formula $\ce{SnSO_4}$ is made of $\ce{Sn^{2+}}$ and $\ce{SO_4^{2-}}$ ions. Hence in your reaction, $\ce{Sn}$ atoms start with oxidation number +2 and end with 0 (therefore gaining two electrons) and $\ce{Fe}$ atoms start with oxidation number 0 and end with +2 (therefore losing two electrons).

As another example of electron counting, here's the reaction between $\ce{Ba}$ and $\ce{Al^3+}$:

Oxidation: $\ce{Ba^0 -> Ba^{2+} + 2e^{-}}$

Reduction: $\ce{Al^{3+} + 3e^{-} -> Al^0}$

To balance the equations, the oxidation half-reaction must be multiplied by 3 and the reduction half-reaction must be multiplied by two. Therefore:

Oxidation: $\ce{3Ba^0 -> 3Ba^{2+} + 6e^{-}}$

Reduction: $\ce{2Al^{3+} + 6e^{-} -> 2Al^0}$

Global: $\ce{3Ba^0 + 2Al^{3+} -> 3Ba^{2+} + 2Al^0}$

In the global reaction, six electrons are involved.

Answer 2

After you calculate everything and balance the $\ce{e-}$ too, then you will know that one side loses a certain number of $\ce{e-}$ and the other gains this certain number of $\ce{e-}$.

There isn't any other way apart from balancing the equation and figuring it out from there.

Check here and here for more info.