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I am trying to understand the chemistry that occurs in an iron carbon battery during charging.

The negative electrode is iron, the positive electrode is carbon. The electrolyte is iron(II) chloride. I understand that ferrous chloride dissociates in water to $\ce{Fe^2+}$ and $\ce{Cl-}$ ions. During charge the $\ce{Fe^2+}$ ions are attracted to to the negative iron electrode and the $\ce{Cl-}$ ions are attracted to the positive carbon electrode. The $\ce{Fe^2+}$ ions combine with electrons from the power supply to form metallic iron that plates on to the iron electrode.

My problem is understanding what happens at the positive carbon electrode. I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride. Fine, but how does this happen if all the $\ce{Fe^2+}$ ions are at the negative electrode? How does this happen if the iron(II) chloride has dissociated in to ions, there is no iron(II) chloride anymore?

If the power supply drives all the $\ce{Fe^2+}$ ions to the negative electrode and $\ce{Cl-}$ ions to the positive electrode how can any $\ce{Fe^2+}$ ions even be at the positive electrode to somehow recombine with $\ce{Cl-}$ ions back into iron(II) chloride that then loses electrons to the positive electrode and becomes iron(III) chloride?

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Ions in solution

[OP] I understand that ferrous chloride dissociates in water to $\ce{Fe^2+}$ and $\ce{Cl-}$ ions.

Yes. The same goes for the iron(III) species.

[OP] I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

No, iron(II) loses electrons and becomes iron(III). The chloride is less relevant, and could be exchanged for a different ion.

[OP] How does this happen if the iron(II) chloride has dissociated in to ions, there is no iron(II) chloride anymore?

The reactant is $\ce{Fe^2+(aq)}$, not $\ce{FeCl2(s)}$.

Ion movement

[OP] During charge the $\ce{Fe^2+}$ ions are attracted to to the negative iron electrode and the $\ce{Cl-}$ ions are attracted to the positive carbon electrode.

Yes, but a small fraction, only. Otherwise the repulsion between like ions would be too great.

[OP] My problem is understanding what happens at the positive carbon electrode. I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

Just $\ce{Fe^3+(aq)}$. There might be a double layer, but no solid iron(III) chloride.

[OP] Fine, but how does this happen if all the $\ce{Fe^2+}$ ions are at the negative electrode?

There are plenty of cations in the bulk of the solution, and plenty near either electrode.

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    $\begingroup$ I apologize for the delay in responding. I would like to thank you very much for the answer to my question. It has been very helpful to me. I now have a much better understanding of the chemistry. My main stumbling block was the assumption that the ions would segregate to each electrode far more than actually happens. Many thanks. I will now be attempting to understand how to determine the specific energy output by the iron carbon battery by measuring the mass of the iron active material and in some way taking into account the role played by the electrolyte as an active material. Thanks again. $\endgroup$ – Mike Nash May 21 '20 at 16:15
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The fully charged cell has an iron anode in a solution of ferric chloride. When it discharges, iron dissolves to produce ferrous ion while ferric ion (the oxidizer) is reduced to ferrous ion. Chloride ion is essentially a spectator ion which happens to be especially good at assisting corrosion of iron. So as the cell discharges, the solution takes 2 ferric ions to 3 ferrous ions by dissolution of one iron atom. The iron electrode is the anode and is the source of electrons to the external circuit. When all the ferric ions have been reduced to ferrous, the cell is discharged and there is no voltage. This is the condition stated in the second sentence: “The electrolyte is iron(II) chloride.”

Charging reverses this: the iron electrode is made negative by an outside source, that is, it becomes the cathode during recharge. Ferrous ions are attracted to this iron cathode and plate out on the iron, tending to restore the original condition of the electrode.

At the same time, electrons are being withdrawn from the solution thru the carbon electrode. This is where it gets a little tricky: the standard potential to produce Cl2 from Cl- is 1.36 V; but it only takes 0.77 V to produce Fe+++ from Fe++. So you can convert essentially all the ferrous ions to ferric ions before liberating any Cl2 gas. It seems a bit counterintuitive to get electrons from a positive ion rather than from a chloride ion, but those chloride ions are very reluctant to give up an electron, and this is what makes the battery so interesting. Other anions have been used also, like sulfate. The reactions are not all so simple, but the battery is cheap, and economy is attractive.

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  • $\begingroup$ I would like to thank you very much for the answer to my question. I have a much better appreciation for the cell reactions. Drawing my attention to the standard potentials of chloride to chlorine versus fe3+ to fe2+ were the key for me to understand why no chlorine gas was forming. I should have realized this myself. My next task is to understand how to determine the cell's specific energy output. I would usually measure the mass of the active materials but here it seems the mass of the electrolyte, as an active material itself, must be considered. Any suggestions? Thank you, again. $\endgroup$ – Mike Nash May 21 '20 at 16:50
  • $\begingroup$ There would be at least 3 ways to determine the energy output of this cell. 1) Fe + 2 FeCl3 (aq) --> 3 FeCl2 (aq). Look up heats of formation (CRC Handbook) for both sides of the equation. 2) Electrochemically (theoretical). Standard potential, corrected for concentration, times current (1 Faraday = 96500 amp sec). 3) My favorite (experimental): start with a discharged cell (FeCl2 solution), charge it and record the actual voltage vs current vs time. Then discharge it, recording the same. You can get the efficiency this way, and how many cycles of utility. Compare with theoretical. $\endgroup$ – James Gaidis May 22 '20 at 13:52
  • $\begingroup$ Yes, I see the energy stored and output from the cell reactions could be calculated from 1) and 2). I have spent a lot of time on 3). I have built and cycled (charge/discharge) several iron carbon cells. I have experimentally determined cell out power versus time (mW/h) and efficiency (Ein vs Eout). My problem is how to determine the watt/hours per kg of the cell. For simpler batteries I would measure the mass of active materials. In this case the electrolyte acts as active material. Should I simply take the mass of the electrolyte add the mass of iron and calculate the watt/hours per kg? $\endgroup$ – Mike Nash May 24 '20 at 16:53
  • $\begingroup$ Is there a way, other than this comment bar, to discuss this cell more fully? $\endgroup$ – Mike Nash May 24 '20 at 17:04
  • $\begingroup$ My "expertise" has run out, I'm afraid. I'm running now on wisdom acquired by age. For your question about Whr/kg, your final cell would consist of electrodes, electrolyte, any containment and separators. All those should be included for a realistic Whr/kg. $\endgroup$ – James Gaidis May 25 '20 at 16:09
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I have tried to understand the chemical reactions occurring in this Iron-Carbon cell. It seems to me that this cell will have difficulty to work at all. Here is my interpretation.

The initial solution contains $\ce{Fe^{2+}}$ and $\ce{Cl^-}$ ions. During charging, the $\ce{Fe^{2+}}$ ions are moving to the negative electrode made of Iron, where they are discharged and transformed into metallic Fe. Simultaneously, the $\ce{Cl^-}$ ions are moving to the positive electrode made of Carbon. When touching this electrode, they are discharged, converted into Chlorine molecules $\ce{Cl_2}$, but this molecule is a powerful oxidant which reacts easily with $\ce{Fe^{2+}}$ ions and transform them into $\ce{Fe^{3+}}$ according to :$$\ce{2 Cl^- -> Cl_2 + 2e^-}$$ $$\ce{Cl_2 + 2 Fe^{2+} -> 2 Fe^{3+} + 2 Cl^-}$$ Another explanation may be stated at this Carbon electrode, which avoids the presence of $\ce{Cl_2}$. If the existence of $\ce{Cl_2}$ is denied, the two previous equations may be simply replaced by the oxidation of $\ce{Fe^{2+}}$ into $\ce{Fe^{3+}}$ $$\ce{Fe^{2+} -> Fe^{3+} + e- }$$ Whatever the correct phenomena happening at this Carbon electrode, the result is the same : the ions $\ce{Fe^{2+}}$ are gradually consumed and disappear on both electrodes, and new $\ce{Fe^{3+}}$ ions will appear at the Carbon electrode, and then in the whole solution. In the long range, the ions $\ce{Fe^{2+}}$ will be slowly replaced by $\ce{Fe^{3+}}$ in the solution. Unfortunately $\ce{Fe^{3+}}$ cannot exist in solution if some metallic Iron is present. $\ce{Fe^{3+}}$ ions react spontaneously with $\ce{Fe}$ according to : $$\ce{2 Fe^{3+} + Fe -> 3 Fe^{2+}}$$

As a consequence, the "charging process" will have no chemical effect, as the composition of the solution remains more or less constant during the "charging" process. The only way for this cell to work is preventing $\ce{Fe^{3+}}$ ions to be dissolved at the Carbon electrode. Maybe this ion is simply hydrolyzed and transformed into a precipitate like $\ce{Fe(OH)_3}$ by reaction with water, as for example : $$\ce{Fe^{3+} + 3 H_2O -> Fe(OH)_3 + 3 H^+}$$ But this reaction will only happen at relative high pH values, which is improbable as this reaction produced $\ce{H+}$ ions. So my reasoning must be flawed. Where ?

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  • $\begingroup$ I apologize for the delay in responding. I would also like to thank you so very much for the answer to my question. My main problem was thinking that the ions involved would be isolated to the immediate areas around the electrodes. Having built several of these cells, I have observed the formation over time of a light brown precipitate in the electrolyte. This may indeed be iron hydroxide via reaction with water. With a charged battery, the Fe3+ reaction with the iron electrode that forms Fe2+ may form a barrier to further interaction with remaining Fe3+ ions? Many thanks. $\endgroup$ – Mike Nash May 21 '20 at 16:32

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