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? Eo values if needed: Br2 + 2e ::equil:: 2Br- Eo= +1.07V Na+ + e ::equil:: Na Eo= -2.71V O2 + 4H+ + 4e ::equil:: 2H2O Eo= +1.23V O2 + 2H2O + 4e ::equil:: 4OH- Eo= +0.40V 2H2O + 2e ::equil:: H2 + 2OH- Eo= -0.83V The ones in red are the equations in question. Which one should i use to compare Eo 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?