Half reaction question- standard cell potential

Question

For the 1/2 rxn of F2(g) + 2e- --> 2F-, why is it written like that? Since it is reduced going from 0 to -1, I thought that it should be written as F2---> 2F- +1e-. I realized that the cell potential value is only for the 1/2 rxn that I mentioned originally, but I do not know why. Please explain.

Answer 1

There are a few reasons why it is written F₂ + 2 e^- --> 2F^- rather than F₂ --> 2 F^- + 1 e^-.

1) As you correctly identified, fluorine is being reduced (ox # from 0 to -1). That means it gains an electron (per atom). So, we show that as the atom plus the electron(s) (from the reducing agent, whoever that is) giving rise to the ion.

2) The second equation above violates conservation of charge, as there is no net charge on the left, but a negative 3 total charge on the right.

3) There are two fluorine atoms reduced, giving rise to two fluoride ions, and that requires two electrons, where you've only shown one.