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Offline PFScience

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Electrochemistry query...
« on: July 22, 2016, 01:29:08 PM »
Hi all

This sounds like an easy query but its deceptive and I've seen this question come
up on several other forums and caused quite a debate among people of all science knowledge levels. To the extent that I have yet to see an answer that truly made sense....

It is commonly known that when cells are in series, the voltages are added to make the combined total. But why exactly is this?

Lets say you have 3 x 1.5 V dry cells stacked anode to cathode as expected. This as we know will provide a combined potential difference of 4.5 V.

When dealing with a single cell, the situation is relatively simple - electrons flow from the anode (eg zinc) to the cathode (via wire) where they pass through the metal (copper/brass) cap and into the carbon electrode inside the cell. The de-energised electrons in the electrode then react with the surrounding electrolytic paste inside the cell in order to avoid the build up of negative charge (which would prevent further flow of current) and to avoid the increase of pressure that would occur due to the formation of gaseous species (typically ammonia and hydrogen).

How does this work when dealing with multiple cells?

For there to be a combined 4.5 V this must mean that each coulomb of charge leaving the final anode into the circuit now has 3x the energy as it did before (4.5 V = 4.5 J / 1 C).

But how does this voltage build up from the top cell's anode to the bottom cell's anode which is connected to the conducting wire? As the cells are in series, each anode is in direct contact with a cathode (apart from the last one), but how do the energised electrons pass through each cell in order to build up the voltage? They cant pass through the carbon electrode because they would react with positively charged species such as NH4+ in the surrounding electrolytic paste and also the carbon electrode isn't in direct contact with both the cathode and anode.


I've thought about this and how it could work, and this is the best I could come up with! Feedback/thoughts/opinions as always appreciated....  :)

Because each coulomb is now entering the circuit with 4.5 V instead of 1.5 V this must mean it is not the actual coulombs of charge that have been transferred and built up in each cell, but the energy stored within them.

Starting from the 'top' cell....

The anode at the bottom is in direct contact with a cathode below - due to the difference in reduction potentials, electrons are transferred from the zinc anode to the cathode.
 
Unlike a single cell, where the returning electrons are 'de-energised' here the electrons have done no work at all. When de-energised electrons pass into the carbon electrode and react with the surrounding electrolyte paste (typically with NH4+), there is little if any energy released.
 
However, when energised electrons react there is a significant release of energy. It is this energy that is transferred to the electrons leaving the anode into the cathode of the final cell below - these electrons now have 2x their normal energy (ie 3 V). In the final cell, these 'super-energised' electrons again pass into the carbon electrode and react with the surrounding electrolyte paste. This time even more energy is released which is transferred finally to the electrons at the anode connected to the conducting wire - these electrons now have 3x their normal energy (ie 4.5 V).

Each coulomb going into the circuit now has 4.5 V, and as expected has now increased the amps by x 3 also due to the greater repulsive effect of the higher energy coulombs.
« Last Edit: July 22, 2016, 02:02:07 PM by PFScience »

Offline mjc123

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Re: Electrochemistry query...
« Reply #1 on: July 25, 2016, 08:55:09 AM »
Quote
It is commonly known that when cells are in series, the voltages are added to make the combined total. But why exactly is this?
In the first cell, the cathode is at +1.5V relative to the anode. This cathode is connected by a conductive path to the anode of the next cell. Therefore they are at the same potential. The second cathode is at +1.5V relative to the second anode, and therefore at +3V relative to the first anode. And so on.
Consider the following analogy: Two buildings; one has 4 floors (ground, 1st, 2nd,  3rd), and 3 staircases, one between each pair of adjacent floors. Each time you climb a staircase, you get 1 floor higher. If you climb all 3, you get 3 floors higher. That's analogous to the voltage being additive for cells in series.
The second has 2 floors (ground & 1st), with 3 separate staircases linking them. Here you can only get 1 floor higher, but the 3 staircases allow more people to climb up at the same time. That's analogous to the capacity being additive for cells in parallel.

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The de-energised electrons in the electrode
What do you mean by this?
Quote
to avoid the increase of pressure that would occur due to the formation of gaseous species (typically ammonia and hydrogen).
How do the electrons know there would be an increase in pressure, and why do they want to avoid it?
Quote
how do the energised electrons pass through each cell in order to build up the voltage?
Electrons don't pass through cells. Current is carried through the electrolyte by ions, not electrons. Electrochemical reactions consume electrons at the cathode and release electrons at the anode. The same electrons don't go in at the cathode and emerge at the anode.
Quote
Unlike a single cell, where the returning electrons are 'de-energised' here the electrons have done no work at all.
I have no idea what this means.
Quote
This time even more energy is released which is transferred finally to the electrons at the anode connected to the conducting wire - these electrons now have 3x their normal energy (ie 4.5 V).
It is true that the electrons are moving faster - the current is 3x what it would be with a single cell - but the current is the same throughout the stack, and throughout the circuit - it doesn't build up from the top cell to the bottom.
Quote
due to the greater repulsive effect of the higher energy coulombs.
This is nonsense. First, don't talk of coulombs as if they were things. Electrons are things (the quantum physicists might want to dispute that, but for present purposes we'll ignore them). A coulomb is a measure of the quantity of charge. Electrons move through circuits and repel each other, coulombs don't. A six-foot tall man is a person; six feet aren't.
Second, current is not caused by the repulsion of electrons.
Third (irrelevant, perhaps, in the light of second), repulsion of electrons depends on their charge and distance - not on their energy.

Offline PFScience

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Re: Electrochemistry query...
« Reply #2 on: July 29, 2016, 04:27:39 PM »
Thanks for the reply.... much appreciated. I'll go through things in number format:

1) Surely it would be the anode that is at +1.5 V relative to the cathode?

2) The conductive path connecting the cathode to the anode is just simply the metallic casing of the cell?

3) "De-energised electrons" - well these are electrons returning to the cell after they have done work (eg powering bulb, heating the copper wiring... etc)

4) The electrons absolutely wouldn't know that there would be an increase in pressure but the designers do and would want to avoid it surely?!

The following are the reactions in a typical electrolytic paste:

2NH4+ + 2e = 2NH3 + H2
ZnCl2 + 2NH3 = Zn(NH3)2Cl2
2MnO2 + H2 = Mn203 + H20

5) I get that the electrons themselves dont pass through the cells, but by the maths you do have a p.d of 4.5 V between the first cathode and the last anode and this means that each coulomb entering the circuit has 4.5 J (as opposed to 1.5 J for a single cell). so each coulomb HAS surely tripled its energy?   

6) But in essence a coulomb is a "thing" - the charge on 6.242×1018 protons/electrons. We don't talk about a coulomb of bread or trips to the shops...its basically only used in reference to charge... ipso facto it is a "thing". 

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