Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: TheVanquished on November 04, 2009, 12:13:16 PM
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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
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
Eo values if needed:
Br2 + 2e ::equil:: 2Br- Eo= +1.07
Na+ + e ::equil:: Na Eo= -2.71
O2 +4H+ + 4e ::equil:: 2H2O Eo= +1.23
O2 + 2H2O + 4e ::equil:: 4OH- Eo= +0.40
2H2O + 2e ::equil:: H2 + 2OH- Eo= -0.83
The ones in red are the equations in question. Which one should i use to compare Eo 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 ::equil:: 2H2O equation to compare instead of O2 + 2H2O + 4e ::equil:: 4OH- why is this so?
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*Ignore me, I am impatient*. Sorry had to *Ignore me, I am impatient* this thread as this question is quite urgent
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O2 +4H+ + 4e ::equil:: 2H2O Eo= +1.23
O2 + 2H2O + 4e ::equil:: 4OH- Eo= +0.40
Have you checked if they don't give the same potential for pH=7?
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these are the standard data we use for electrolysis questions, regardless of ph i think
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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.
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sorry but i dont get what you mean
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Write Nernst equation for both reactions, calculate formal potential for neutral solution.
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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 Eo value and for reduction (cathode) , the most positive
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O2 +4H+ + 4e ::equil:: 2H2O Eo= +1.23
O2 + 2H2O + 4e ::equil:: 4OH- Eo= +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 ::equil:: H3O+ + OH-
or
in an electrolytic cell
Hydrogen ions are discharged to hydrogen gas
2H+ + 2e- :rarrow: H2
and
4 moles of hydroxide are discharged to oxygen gas and water
4OH- :rarrow: O2 + 2H2O + 4e-
Thanks
Clive
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O2 + 2H2O + 4e ::equil:: 4OH- Eo= +0.40
4OH- :rarrow: O2 + 2H2O + 4e-
Strangely these both look identical to me, yet one is absurd and the other is not ;)
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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 Eo 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.
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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?
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O2 +4H+ + 4e ::equil:: 2H2O Eo= +1.23
O2 + 2H2O + 4e ::equil:: 4OH- Eo= +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 ::equil:: H3O+ + OH-
or
in an electrolytic cell
Hydrogen ions are discharged to hydrogen gas
2H+ + 2e- :rarrow: H2
and
4 moles of hydroxide are discharged to oxygen gas and water
4OH- :rarrow: 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.
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O2 + 2H2O + 4e ::equil:: 4OH- Eo= +0.40
4OH- :rarrow: 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? :P
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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- ::equil:: 2H2O Eo= +1.23V
relates to any cell that could be measured against the hydrogen electrode.
Clive
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probably an acidic solution where u bubble oxygen into it?
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probably an acidic solution where u bubble oxygen into it?
Very good answer :) Have a snack ;)
Inserting platinum electrode into solution will help measure the potential, but it won't change redox reaction.
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?
IMHO it is impossible to answer this question correctly at this level. That's why different sources give conflicting answers.
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Firstly thanks for the snack :)
Well is there an answer which is "more correct"?
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Yes - but this the one that requires use of Nernst equation. You should write Nernst equation for the oxygen production, use known concentration of H+ of OH- to calculate formal potential of the water/oxygen half reaction, and use this potential to predict what will be the product.
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oh my. i have no idea how to do that
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Do you know what reaction quotient is?
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sorry but i have no idea
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No need to be sorry if you have not learnt about it yet.
Take a look at these pages:
http://en.wikipedia.org/wiki/Reaction_quotient
http://www.ph-meter.info/pH-Nernst-equation
(you may finish reading after Nernst equation for permanganate reaction).
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probably an acidic solution where u bubble oxygen into it?
Very good answer :) Have a snack ;)
Inserting platinum electrode into solution will help measure the potential, but it won't change redox reaction.
OK, I get it !
Thanks to both of you.
Just to finish that off, what happens at the platinum surface that enables the voltage to be measured?
Clive
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Just to finish that off, what happens at the platinum surface that enables the voltage to be measured?
Last time I asked answer was "we are not sure". But it was about 20 years ago.
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Just to finish that off, what happens at the platinum surface that enables the voltage to be measured?
Last time I asked answer was "we are not sure". But it was about 20 years ago.
For some reason what you wrote brought a smile to my face (strange person that I am)
Evidence >> :)
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Last time I asked answer was "we are not sure". But it was about 20 years ago.
I always thought the platinum was acting as a catalytic surface to facilitate the conversion of H2 to 2H+ back and forth ???
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I always thought the platinum was acting as a catalytic surface to facilitate the conversion of H2 to 2H+ back and forth ???
Yes, especially if covered with platinum blak it lover overpotantial of such reaction. But as I understand the question ikt asks about why pltinum electrode has the same potential as the redox couple present in the solution - in the other words, why it can work as ORP electrode - and here answer is not that obvious.