Urgent help needed for question on electrolysis & comparing Eo values

Thevanquished

Here is the question:

Predict the products formed at the anode and at the cathode when NaBr(aq) is electrolysed using inert electrodes

Answer:

Somehow i got 2 different answers by different answer sheets

One of them stated that Br₂ gas (anode) and H₂ gas (cathode) would be liberated

However, the other one writes H₂(cathode) , H₂O and O₂ (anode) are formed

There is no arguments for the products formed at the cathode as clearly it is H₂ gas which is liberated there. However, the question lies in the products formed at the anode. Which one is correct?

E^o values if needed:
Br₂ + 2e ::equil:: 2Br^- E^o= +1.07V
Na⁺ + e ::equil:: Na E^o= -2.71V
O₂ + 4H⁺ + 4e ::equil:: 2H₂O E^o= +1.23V
O₂ + 2H₂O + 4e ::equil:: 4OH^- E^o= +0.40V
2H₂O + 2e ::equil:: H₂ + 2OH^- E^o= -0.83V

The ones in red are the equations in question. Which one should i use to compare E^o values with Br at the anode (oxidation)? And why?

The answer which states that bromine gas is liberated is most probably the right one (as it is provided by my school) but it used the O₂ + 4H⁺ + 4e ::equil:: 2H₂O equation to compare instead of O₂ + 2H₂O + 4e ::equil:: 4OH^- why is this so?

Fightfish

Where would you get your "4OH-" from in an aqueous solution of NaBr? The equilibrium concentration of it from the self-ionisation of water is going to be very insignificant and negligible (hence E value of the half-cell would be very positive actually, since equilibrium position lies very far to the right)

Thevanquished

Aren't OH- ions produced in the cathode?