[TheVanquished]

Here is the question:

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

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 said H₂(cathode) , H₂O and O₂ (anode) are formed

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

The ones in red are the equations in question. Which one should i use to compare E^o values with Br? 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  2H₂O equation to compare instead of O₂ + 2H₂O + 4e  4OH^- why is this so?

[TheVanquished]

*Ignore me, I am impatient*. Sorry had to *Ignore me, I am impatient* this thread as this question is quite urgent

[Borek]

O₂ +4H⁺ + 4e  2H₂O   E^o= +1.23
O₂ + 2H₂O + 4e  4OH^-   E^o= +0.40

Have you checked if they don't give the same potential for pH=7?

[TheVanquished]

these are the standard data we use for electrolysis questions, regardless of ph i think

[Borek]

Formal potential in both cases must be pH dependent - standard is given for activities 1, at pH 7 activities of H⁺ and OH^- are 10^-7.

IMHO this is basically the same reaction.

[TheVanquished]

sorry but i dont get what you mean

[Borek]

Write Nernst equation for both reactions, calculate formal potential for neutral solution.

[TheVanquished]

Write Nernst equation for both reactions, calculate formal potential for neutral solution.

whats a nernst equation? dont think i learnt it before

the stuff given in the question above are basically the only things you need to solve it. To find which product is oxidised (anode) what i learnt is to compare which is the most negative E^o value and for reduction (cathode) , the most positive

[cliverlong]

O₂ +4H⁺ + 4e  2H₂O   E^o= +1.23
O₂ + 2H₂O + 4e  4OH^-   E^o= +0.40

Can someone explain what those two equations represent?
What I'm asking is how would you set up a cell or reaction that would actually cause the following changes or equilibria to be established?

One mole of oxygen gas dissolves in an acidic solution to form two moles of water

One mole of oxygen gas dissolves in two moles of water to form 4 moles of hydroxide ions

Those two reactions seem absurd and unachievable to me as I have only come across
 the disassociation of water

2H₂O  H₃O⁺  +  OH^-

or

in an electrolytic cell

Hydrogen ions are discharged to hydrogen gas
2H⁺ + 2e^-    H₂
and
4 moles of hydroxide are discharged to oxygen gas and water
4OH^-  O₂ + 2H₂O + 4e^-

Thanks

Clive

[Borek]

O₂ + 2H₂O + 4e  4OH^-   E^o= +0.40

4OH^-  O₂ + 2H₂O + 4e^-

Strangely these both look identical to me, yet one is absurd and the other is not

[Borek]

whats a nernst equation? dont think i learnt it before

http://en.wikipedia.org/wiki/Nernst_equation

Basically potential of the reaction depends on the concentration of ions involved. Standard potentials - as given in tables - are for solution in which concentrations of all ions equal 1 (in reality is not about concentrations, it is about their activities, but as you don't know what Nernst equation is, activities are most likely way above your head; as first approximation concentration and activity are identical).

the stuff given in the question above are basically the only things you need to solve it. To find which product is oxidised (anode) what i learnt is to compare which is the most negative E^o value and for reduction (cathode) , the most positive

Trick is, when you put NaBr into solution, even if concentrations of Na⁺ and Br^- are 1 (which means their reduction/oxidation will occur at potential listed in tables), concentrations of H⁺ and OH^- are not 1M, thus potential at which H₂ and O₂ evolve will differ from given in tables. Hence you can't simply compare numbers given in table of standard potentials, as it will give incorrect results.

[TheVanquished]

well probably i may be inaccurate at a higher level but this will do for my level. honestly, this is the only information i have for the question. So say if you only have these equations to compare with what do you think would be the appropriate answer?

[TheVanquished]

O₂ +4H⁺ + 4e  2H₂O   E^o= +1.23
O₂ + 2H₂O + 4e  4OH^-   E^o= +0.40

Can someone explain what those two equations represent?
What I'm asking is how would you set up a cell or reaction that would actually cause the following changes or equilibria to be established?

One mole of oxygen gas dissolves in an acidic solution to form two moles of water

One mole of oxygen gas dissolves in two moles of water to form 4 moles of hydroxide ions

Those two reactions seem absurd and unachievable to me as I have only come across
 the disassociation of water

2H₂O  H₃O⁺  +  OH^-

or

in an electrolytic cell

Hydrogen ions are discharged to hydrogen gas
2H⁺ + 2e^-    H₂
and
4 moles of hydroxide are discharged to oxygen gas and water
4OH^-  O₂ + 2H₂O + 4e^-

Thanks

Clive

well those 2 equations are merely data which shows the voltage obtained when the particular half cell is compared with a standard hydrogen electrode. Basically the data shows how good a reducing/oxidizing agent the substance in question is so it is used to compare which products are produced at the anode/cathode.

[cliverlong]

O₂ + 2H₂O + 4e  4OH^-   E^o= +0.40

4OH^-  O₂ + 2H₂O + 4e^-

Strangely these both look identical to me, yet one is absurd and the other is not

Of course! 

How about the other one? 

[cliverlong]

well those 2 equations are merely data which shows the voltage obtained when the particular half cell is compared with a standard hydrogen electrode. Basically the data shows how good a reducing/oxidizing agent the substance in question is so it is used to compare which products are produced at the anode/cathode.

I understand that the voltage obtained when the particular half cell is compared with a standard hydrogen electrode without

My question still is I can't see how the half-cell

O₂ +4H⁺ + 4e^-    2H₂O   E^o= +1.23V

relates to any cell that could be measured against the hydrogen electrode.

Clive