bobdring - 15-12-2017 at 01:05
I'm new to this forum trying to learn about electrochemistry.
I've read enough text books to give me a headache but they jump straight into the maths and leave me mystified to the extent of questioning whether
i'm reading the appropriate chapter!
My questions pertain to home made electrochemical cells
1) I often read research papers on batteries where they say something like:-
"showing a capacity which is 50% of the theoretical mAh of the active ingredients of the electrode" How do you calculate this theoretical capacity?
2) say you have a two electrode cell and you try to charge it well beyond maximum how can you measure which electrodes chemistry has limited the
charge?
3) I think water has an overpotential of say 1.3V yet many types of aqueous cell exceed this (e.g. lead acid) so how do you measure the actual
overpotential?
I hope some one will either explain these questions or point me in the direction
where I can find out for myself.
Bob
JJay - 15-12-2017 at 01:30
I guess my question would be where you are as far as knowledge goes.
In general, one coulomb of electrons is the amount transferred by one amp over one second. Coulombs are related to moles by a factor called Faraday's
constant (which says that there are something like 96485 coulombs / mole). You can calculate the number of coulombs available in a cell by multiplying
the number of moles of the limiting reactant times the number of electrons transferred per atom/molecule of the limiting reactant times Faraday's
constant. You can convert coulombs to amp hours by dividing by 3600 seconds.
bobdring - 15-12-2017 at 04:02
Thanks JJAY,
So I need
1) mass of the chemical (g/mol)
2) the number of electrons transferred per molecule reduced or oxidised.
lets say I am using aluminium:
molar mass is 27g/mol
aluminium transfers 3 electrons during redox
so we have aluminium redox needs 96485 x 3 /27 = 10720 coulombs per gram
divide this by 3600 gives 3 amp hours per gram (providing that this electrode is the limiting reaction.)
Did I calculate that correctly?
Using this theoretical value allows question 2, about which is the limiting electrode, to be partially answered since the coulombs have to be the same
for both electrodes. This sets the theoretical mass ratio of the electrodes.
So, if I set up a cell with a 1g aluminium electrode and another unspecified electrode large enough to guarantee the aluminium is the limiting factor.
If I fully charge this cell and measure the mA/h recovered during discharge that would give me the real capacity of the aluminium in my cell? Could I
then do the same for the other electrode at 1g but with an excess for the aluminium electrode this time?
JJay - 15-12-2017 at 08:51
That sounds about right. The capacities would probably be different with different limiting reactants.
One thing I should point out: I made a slight simplification in my explanation given above. It is actually possible though unusual to construct a cell
where the number of electrons transferred by the limiting reactant are not equal to the number of electrons transferred by the cell. In a scenario
like this, you would look at the number of electrons transferred by the total redox equation when the reaction runs to completion with the quantities
present and use the number of electrons transferred by the cell as a whole as the theoretical capacity.
bobdring - 15-12-2017 at 12:19
Hi JJAY,
Thank you for the reply.
If you have time could you please expand on:-
"where the number of electrons transferred by the limiting reactant are not equal to the number of electrons transferred by the cell"? I don't really
understand it. Taking your words "not equal" would "less than" and "greater than" be both applicable?
It's time for me to do some basic experiments.
I will try nickel hydroxide and Iron oxide epoxied onto nickel plated steel as my electrodes. KOH as electrolyte. (NIFE)
photo attached
Bob
JJay - 15-12-2017 at 12:27
You could have a cell where the limiting reactant is an electrolyte in a salt bridge, for example.
bobdring - 15-12-2017 at 16:20
Hi JJAY,
Thank you for taking an interest. I haven't come across salt bridges that take part in the redox yet. But I can see that that would really confuse
things.
You asked about my knowledge, electronics good, chemistry poor.
Could I run a thought by you and check its validity?
If I have one active and one relatively inert electrode (Ti) in a cell and the applied voltage is large enough. The inert electrode will generate gas
and the active electrode will experience redox. If the applied voltage is too low then there will be no gassing and no redox (i.e. little to no
current). Is this correct?
What I'm trying to to do is to set up experiments to evaluate the efficiency of an active electrode by controlling as many other factors as possible.
bob
JJay - 15-12-2017 at 17:54
I think it really depends on what is in the cell. If you experience redox of your electrolyte, you will likely see bubbles. Most electrolytes I think
are pretty good conductors and would support some current flow even if there is no reaction, but if your cell is trying to run in reverse when you
apply a weak current, it seems reasonable to suppose that might stop current flow... I don't know for sure, though. If you post in the Beginnings
section, more people will read your questions, and some of them surely have more experience and expertise on electrochemistry than I do.
bobdring - 15-12-2017 at 21:53
I will follow your suggestion and post in "beginnings"
So, I went there and started looking to see if anyone had asked a similar question...It's going to take a while...This is such an interesting
website...
Thanks JJAY, especially on your help with the theoretical capacity calculation.
Bob