How is francium metal oxidized by oxygen gas or by chlorine gas?

And so....

#Fr(s) rarr Fr^(+) + e^(-)# #(i)#

Francium as a metal loses an electron, and its oxidation number goes from #0# as the metal to #+I# in the metal ion..... This increase in oxidation number betokens an #"oxidation"#.

Now we could go further than this, in that for every electron loss, there is a corresponding electron gain. Why? Because charge is CONSERVED in a chemical reaction, just as mass is... And something, maybe oxygen gas, maybe chlorine gas, undergoes a corresponding reduction....

#1/2Cl_2(g) +e^(-) rarr Cl^(-)# #(ii)#...else...

#1/2O_2(g)+2e^(-)rarrO^(2-)# #(iii)#

In each instance, the non-metal gains electrons, and undergoes a FORMAL REDUCTION, a decrease in oxidation number. And as is typical, we add the half-equations so that the electrons, virtual particles of convenience are eliminated....

For these process we would take #(i)+(ii)#

#Fr(s) +1/2Cl_2(g) +cancel(e^(-))rarr underbrace(Fr^(+) + Cl^(-))_"francium chloride"+ cancel(e^(-))#

i.e. #Fr(s) +1/2Cl_2(g) rarr FrCl(s)#

...or #2xx(i)+(iii)#

#2Fr(s) +1/2O_2(g)rarr Fr_2O(s)#

The key to doing these equations is to realize that mass and charge are ALWAYS conserved in a chemical process. We do the same thing when we balance a cheque book, or a credit card; for every debit, there must be a corresponding credit. Capisce?

Well, good luck with that... francium is radioactive.

Suppose we only looked at the moment before it decays. Then it is an alkali metal, and typically forms a #+1# cation.

#"Fr"(s) -> "Fr"^(+)(aq) + e^(-)#

But again, this only exists for a short amount of time.

Oxidation occurs upon loss of an electron, so the electron is a "product". But we must pair this with some reduction half-reaction and then add them together later.