Electrolysis of mgbr2

Have you compared standard potentials?

I have.
O2+2h20+2e➡️4oh- +0.4

Br2 +2e⬅️2br- +1.07

I thought that the at the anode oxidation occurs thus the least oxidising will be liberated at the anode.
In this case, isn't it oxygen?

OK. Now we come to the hard part.



Standard potential assumes [OH^-] = 1 M, in a typical solution that's not the case. MgBr₂ solution is more or less neutral, so [OH^-] = 10^-7 M. That means we have to turn to formal potentials, calculated from the Nernst equation:

$E = E_0 + \frac {RT}{nF} ln \frac {p_{O_2}}{[OH^-]^4}$

it gets ugly, doesn't it? But if you plug all numbers in, you will get something like

$E = 0.4 + 0.03 log \frac {0.21}{(10^{-7})^4} = 1.21 V$

Actually formal potential for the bromine is not 1.07V either, its exact value will depend on the concentration of the bromide and on the concentration of the free bromine.

There is also another factor that gets into the way. Electrode reactions involving production of oxygen are pretty slow. That means to observe production of substantial amounts of oxygen we need to speed them up by applying additional potential. That means other reactions - despite occurring at nominally lower potentials - will be fast enough to be observed first.