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

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

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

• Can you clarify whether you're asking about how the number of electrons transferred is determined empirically/experimentally, or merely how that's calculated on paper given skeletal or half equations? Also, how familiar are you with the basics of chemical bonding, electron configurations, etc.? I think most answers would have to presuppose an understanding of those things, and based on the phrasing of your question, I suspect you might be best served by just studying a general chemistry textbook (Oxtoby's is probably the most rigorous I've come across). – Greg E. May 27 '13 at 1:29
• Anybody know why the TeX formatting doesn't seem to be working? The format strings look good, but they're showing verbatim in my Chrome browser. – KeithS May 28 '13 at 17:16
• @Cameron Your first half equation isn't really a half equation... the half equation should have been $\ce{Fe -> Fe^{2+} + 2e^-}$ – Jerry May 28 '13 at 17:37

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.

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.