Solving Electrolysis Questions: NaBr(aq) & Comparing Eo Values

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SUMMARY

The discussion focuses on the electrolysis of NaBr(aq) using inert electrodes, specifically addressing the products formed at the anode and cathode. The two main answers presented are Br2 gas and H2 gas at the anode, versus H2 gas and O2 gas at the anode. The standard electrode potentials (Eo values) for the relevant half-reactions are provided, with Br2 having an Eo of +1.07V, while O2 has an Eo of +1.23V. The consensus leans towards Br2 being the correct product at the anode due to its higher reduction potential compared to the formation of O2.

PREREQUISITES
  • Understanding of electrolysis and its principles
  • Familiarity with standard electrode potentials (Eo values)
  • Knowledge of half-reaction equations in electrochemistry
  • Basic concepts of aqueous solutions and ionization
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Thevanquished
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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[/color]
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?[/color]
 
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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)
 
Aren't OH- ions produced in the cathode?
 

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