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The neutralization reaction, such as $\ce{HCl + NaOH → NaCl + H2O}$, generates heat, which is around $\pu{57.30 kJ/mol}$. Can we use this energy to design a battery? This idea occurs in my mind for a long time.


---------- Selection for Acid and Base -------------

For acid, $\ce{HCl}$ is a good choice, $\pu{36.46g}$ for $\pu{1 mol}$ of $\ce{H^+}$.

For base, $\ce{NaOH}$ is Okay, but not the best. Problems for both $\ce{NaOH}$ and $\ce{LiOH}$ is that they both absorb water when forming crystal from water.

$\ce{LiOH·H2O → LiOH + H2O}$, ($\pu{100-110 °C}$)

In my idea, I consider hydrogen chloride ($\ce{HCl}$) as the acid and ammonia ($\ce{NH3}$) as the base. Surprisingly, it turns out they have a very high energy density. Also, they both can be stored in gas phase, which is hard to solve for sulfuric acid ($\ce{H2SO4}$) with high concentration.

$\ce{NH3(g) + HCl(g) -> NH4Cl(s)}, \quad ΔH = \pu{-176.0 kJ/mol}$

Energy density $= \pu{176.0 kJ}/\pu{53.49g} = \pu{3.29 MJ/kg} = \pu{913.9 W·h/kg}$, which is better than lithium-ion battery ($\pu{265 W·h/‎kg}$‎).

As a comparison, for Hydrogen Combustion, $\ce{H2(g) + 1/2 O2(g) → H2O(l)} + \pu{572 kJ}$ ($\pu{286 kJ/mol}$), it gives energy density of $\pu{286kJ}/\pu{18g} = \pu{15.89MJ/kg}$.

To be noted, $\pu{1eV} = \pu{96.485 kJ/mol}$, which means the battery should give $176.0/96.485 = \pu{1.82 V}$.

In water, higher than $\pu{1V}$ might break $\ce{H2O}$ into $\ce{H2}$ and $\ce{O2}$. In my design, the energy in each proton is released in two steps, where each gives $1.82/2 = \pu{0.91V}$.


---------- Properties for Acid and Base -------------

$\{1\} \; \ce{HCl(g) + H2O(l) -> H3O+(aq) + Cl-(aq)}, \quad ΔH = -92.31 - (-167.16) = \pu{-74.85 kJ/mol}$

$\{2\} \; \ce{NH3(g) + H2O(l) -> NH_4+(aq) + OH-(aq)}, \quad ΔH = -46.11 - (-80.29) = \pu{-34.18kJ/mol}$

$\{3\} \; \ce{H3O+(aq) + OH-(aq) -> 2H2O(l)}, \quad ΔH = \pu{−57.30 kJ/mol}$

$\{4\} \; \ce{NH4+(aq) + Cl-(aq) -> NH4Cl(s)}, \quad ΔH = \pu{–20 kJ/mol}$

$\{4.1 ~Lattice ~energy\}$:

$\ce{NH4+(g) + Cl-(g) -> NH4Cl(s)}, \quad ΔH = \pu{–705 kJ/mol}$

$\{4.2 ~Hydration ~enthalpy\}$:

$\ce{NH4+(aq) -> NH4+(g)}, \quad ΔH = \pu{307 kJ/mol}$

$\ce{Cl-(aq) -> Cl-(g)}, \quad ΔH = \pu{378 kJ/mol}$


---------- Battery Design -------------

With the aid of a small electrolytic capacitor, we can release the energy in neutralization reaction into electricity. The reactions highly rely on proton transmission.

Solvent: all reactions are in aqueous solution (water).

Capacitor: anode with $\ce{HCl(aq)}$, cathode with $\ce{NH3(aq)}$.

enter image description here

$\ce{HCl}$ gas is connected to anode, and $\ce{NH3}$ gas to cathode, both connections are controlled by valve (switch). Moreover, anode and cathode are connected by two channels (with switch): PEM membrane channel, pipe channel.

--------- energy release -----------

The battery energy is released by repeating the following two steps.

Step 1: Capacitor charge with reactions {1}, {2}, and {3}:

$\ce{anode + Cl- - e- -> anode^+(Cl-)}$

$\ce{cathode + NH4+ + e^- -> cathode^-(NH4+)}$

Switch on PEM membrane channel, and switch off pipe channel. $\ce{H3O+}$ flow through PEM membrane and react with $\ce{OH-}$ {3}, then $\ce{Cl-}$ flow to anode, and $\ce{NH4+}$ to cathode, causing potential ($<\pu{1V}$) in capacitor, and electrons flow from the anode to the cathode.

Step 2: Capacitor discharge with reaction {4}:

$\ce{anode^+(Cl-) + e^- -> anode + Cl-}$

$\ce{cathode^-(NH4+) - e^- -> cathode + NH4+}$

Switch off PEM membrane channel, and switch on pipe channel. $\ce{Cl-}$ and $\ce{NH4+}$ flow through pipe channel and produce $\ce{NH4Cl(s)}$, causing potential ($<\pu{1V}$) in capacitor, then electrons flow from the cathode to the anode.

------------battery recharge---------------------

To charge the battery (input energy), reverse the above two process.


Possible Issues

i) Not environmental friendly, $\ce{HCl}$ and $\ce{NH3}$ gas are toxic.

ii) Too much pressure in $\ce{HCl}$ gas and $\ce{NH3}$ gas.

iii) Electron exchange rate limits the battery power density.

iv) Hard to design capacitor for $\ce{NH4Cl}$.

v) switches of the membrane or pipe are hard to design.

vi) energy efficiency might not be as good as expected.

Welcome to discuss the above issues and other practical problems.

----------------------------update on 14.08.2019--------------

This is respond to the comments by @Oscar Lanzi, about the issue of low potential.

Actually, the concept of storing energy in acid and base for battery is already studied in hybrid battery. Check this research paper "Three electrolyte high voltage acid–alkaline hybrid rechargeable battery", they use electrolyte configuration of 1 M H2SO4|0.2 M K2SO4|2 M KOH to increase the voltage. (https://www.sciencedirect.com/science/article/pii/S001346861101228X)

Go back to the design of HCl,NH3 battery, the low potential is indeed a problem as mentioned by @Oscar Lanzi, and it also brings many practice issues such as the membrane. But I still want to discuss about the possibility or physical/chemical principles behind those problems, regardless of the limitation of materials.

For the voltage potential, 0.8V is the best for HCl + NH3 reaction in water solution, the reason is that water is so easily to break down into H+ ions, self-ionization of water is only $10^{-14}$ at room temperature. However pure NH3 itself does not produce H+ easily,

$2NH_3 ⇌ NH_2^− + NH_4^+, k = 1.9 \times 10^{-33} ~( - 50 °C)$

This effect is called Leveling Effect.

Therefore, can we just select another approriate solvent and achieve $1.82V$ potential?

One possible solution might be to mix water with other solvent so that the ionic product of water (or equivalently pKa) is reduced from $10^{-14}$ to $10^{-32}$, one example is to mix water with Dimethyl Sulfoxide (DMSO). As is shown here, pKa value for $H_2O$ in liquid water and DMSO are 15.7 and 32 respectively, that means $[H3O+][OH-]/[H2O] = 10^{-32}$ in the mixture solution of water and DMSO (if I understand correctly). (http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf)

Well, I am not sure about whether nernst equation still works in this situation since the solvent inside and outside the membrane is not the same, i.e. (water) / (water DMSO mixture).

The following calculation is based on the assumption that nernst equation still works.

Illustration of new design

~==========================================~

~   $<A>$   ||     $<B>$     ||     $<C>$

~        ||        ||

~  HCl+H2O  ||   Cl-   NH4+  ||   NH3 + H2O

~        || ↗      ↖ ||

~- - - - - - - - - - - - - ||xxxxxxxxxxxxxxxx|| - - - - - - - - - - - - -~

~    ↓↓   ||xxxxxxxxxxxxxxxx||    ↓↓

~   [H+]   ||xxxxxxxxxxxxxxxx||   [OH-]    ~

~    ↓↓   ||xxxxxxxxxxxxxxxx||    ↓↓

~- - - - - - - - - - - - - ||xxxxxxxxxxxxxxxx|| - - - - - - - - - - - - -~

   $<D>$   H2O   +   DMSO

~==========================================~

Then, assume we have the following ideal battery design where

1) $H^+$ penetrates from HCl(aq) to DMSO(aq),

2) $OH^-$ penetrates from NH3(aq) to DMSO(aq),

3) $Cl^-$$<B>$

4) $NH_4^+$$<B>$

Here is the calculation of potential induced by the ion transmission, based on Nernst Equation.

For ammonia (NH3), 1.0 M aqueous solution (1 mol/L) has a pH of 11.6, which gives $[OH-]_C = [NH4+]_C = 10^{11.6-14} = 10 ^{-2.4} mol/L$.

For hydrogen chloride (HCl), the aqueous solution can achieve pH of -1 (i.e. 10 mol/L), which gives $$[H^+]_A$ = 10 mol/L$.

For ammonium chloride NH4Cl(aq), the saturated aqueous solution gives 39.3g per 100ml at 25°C, which gives $[NH_4^+]_B ≈ [Cl^-]_B = 393/53.49 = 7.35 = 10^{0.866} mol/L$.

For water and DMSO mixture, $[OH^-]_D = [H_3O^+]_D = 10^{-16}$.

Based on Nernst equation, pH sensor will change its output by 59.16mV for every change in one pH unit, i.e. ten fold difference in concentration of $H^+$. (https://www.vernier.com/news/2018/04/06/the-theory-behind-ph-measurements/)

As a result,

For 1) $E_1 = + 59mV * ( (16) - (-1)) = 17 * 59mV :[H^+]_A / [H_3O^+]_D$

For 2) $E_2 = + 59mV * ( (16) - (2.4)) = 13.6 * 59mV :[OH^-]_C / [OH^-]_D$

For 3) $E_3 = + 59mV * ( (-0.866) - (-1)) = 0.134 * 59mV :[Cl^-]_A / [Cl^-]_B$

For 4) $E_4 = + 59mV * ( (-0.866) - (2.4)) = -3.266 * 59mV :[NH_4^+]_C / [NH_4^+]_B$

Note that in 4) the $NH_4^+$ ion needs energy to pump from $<C>$ into $<B>$.

In total, we have $ 27.468 * 59mV = 1.625 V$, still a little far from the best possible potential $1.82V$.

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closed as too broad by Mithoron, Mathew Mahindaratne, Tyberius, Todd Minehardt, Freddy Aug 17 at 13:34

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ My poor eyes, also whole idea isn't particularly practical. $\endgroup$ – Mithoron Aug 9 at 18:44
  • $\begingroup$ @Mithoron Yes, many practical issues. Here, I didn't put much effort in the materials such as PEM membrane and capacitor. $\endgroup$ – Kevin Aug 10 at 23:02
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One problem is there is not enough voltage for this thing to sell. Say you have molar acid and 1 molar base, both strong and both functioning as ideal solutions. Thermodynamically you have one electron transfer associated with the net reaction

$\ce{H^+}_{aq}+\ce{OH^-}_{aq}\to\ce{H2O}$

and the equilibrium constant is $10^{14}$ at room temperature. Put that into the Nernst Equation and you get about 0.8 volt. Not enough to compete with what is commoly used. You need a bigger free energy punch, and generally redox reactions, which involve a greater amount of electron transfer, provide more than acid-base reactions.

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  • $\begingroup$ Thanks for your comments, however, I think it just proves that proton-based battery works as good as electron battery. 0.8 volt indeed comes from the H+, while the other ions Cl- and NH4+ will also contribute to the same amount of voltage. In other words, 0.8 V is the potential voltage for half cell in Nernst Equation. This corresponds to the two steps of energy release in the capacitor where each give 0.8 V. The voltage problem might be solved by increasing the gas pressure since the concentration of HCl and NH3 in water will be increased under high pressure. $\endgroup$ – Kevin Aug 10 at 23:01
  • $\begingroup$ How do you get 0.8 V from the alt ions? Show your reactions and potentials maybe? $\endgroup$ – Oscar Lanzi Aug 11 at 0:13
  • $\begingroup$ Maybe you are right. But this idea is not new, I just found another paper published 3 years ago, based on the same idea (neutralization reaction), "An Acid-Base Electrochemical Flow Battery as energy storage system" (sciencedirect.com/science/article/abs/pii/…), they claim an energy efficiency of 55%, i.e. only achieve half of the ideal potential. $\endgroup$ – Kevin Aug 14 at 18:58

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