# At an atomic level, what happens when you connect two batteries in series so that their voltages are added?

I can't for the life of me figure this out. I feel like i'm missing some crucial detail about how batteries work.

Imagine two batteries connected in series, like this:

Circuit <= -(Battery A)+ <= -(Battery B)+ <= Circuit


As far as I've studied, this is what happens:

1. The negative side of Battery B has a surplus of electrons, while the positive side of Battery A has a "normal" concentration of electrons. Since electrons repel each other, they kinda push themselves to Battery A. (I am imagining a tube with one side filled with "elastic" balls tightly packed and the other side with elastic balls but not pressed together. The balls get pushed until they are all at an equal "pressure").
2. A number of electrons constantly reach the positive end of Battery A with a "force" of x Volts. In practical terms, the electrons are colliding with the battery with a higher intensity. For some reason, this causes the electrons coming out of the negative end of Battery A to have a "force" of 2x Volts. I don't understand why.

Is it because there is now two times more free electrons at the negative end of Battery A? And thus the difference between this end and Battery B is two times the normal?

1. The electrons get pushed again to the positive end of Battery B. From what I understand, the material in the battery "absorbs" them and releases a free electron on the negative end of Battery B.

Please avoid using water analogies. I specifically want to know what happens at the atomic, per-electron level.

I don't care much about the differences between each battery. If possible, pick a battery type you prefer or just talk about batteries in general (characteristics shared by all batteries).

Chemistry and Physics of Batteries in Series

What happens at the atomic level inside a cell when putting two batteries in series?

Characteristics of a single cell vs. two cells in series: (i.e., compare the voltage, current, $$E$$ field, stored energy, power, charge, and operating time in a circuit with the same resistive load.):

• Twice the voltage, due to the summation of voltages with two cells in series (i.e., twice the $$E$$ field acting on a resistive load through the conductor).
• Twice the current with the same load, due to the doubled voltage.
• Twice the reaction rate at both the anode and cathode in all cells in series.
• Twice the rate of $$\mathrm{H^+}$$ ion migration from the anode to cathode to maintain charge neutrality at both anode and cathode.
• Twice the rate of battery charge depletion. When the current doubles and the charge capacity of each cell is unchanged, the duration of flow is reduced to half.
• Within the circuit with series batteries: the same current flows through all the current elements. (Electron current flows through the circuit elements: originating in the outer anode, travels through the conductor, load, outer cathode, inner anode, and finally neutralized at by the inner cathode reaction).
• Within the electrolytes of the cells in series, the magnitude of ion flow rate is the same as the current flow rate through the electrolytes of both cells, ($$\text{cell}_1$$ and $$\text{cell}_2$$).

What force drives current through the electrolyte and conductor between the cells in series?

• Ultimately, the sum of the battery cell voltages drives current through the cells, but its action is indirect. The terminal battery voltages drive current through the load, but ion migration and reaction potentials are responsible for driving current within the batteries.
• The doubled voltage of two batteries in series drive twice the current through the load.
• This increased current draws more electrons off the outer anode and conducts it through the load to the outer cathode. (Note: the outer anode and outer cathode refer to the external electrodes of the two cells in series. And obviously, the terms "inner anode" and "inner cathode" refer to the terminals connected between batteries in series.)
• The increased rate of removal of electrons from the outer anode, and delivery to the outer cathode, disturbs the equilibrium state of the reactions in both the outer anode and outer cathode, resulting in the production of more electrons by the anode reaction, and the consumption of more electrons by the cathode reaction.
• The disturbance of equilibrium state of the outer anode and outer cathode is communicated to the inner anode and inner cathode of the series cells by charge accumulation around the terminals. The charge attraction between ions in the electrolyte results in an ion migration between terminals. The increased current and ion flow results in the adjustment of the rate of reaction. Eventually, the current transient dampens, and the same current flows through the whole circuit (conductors, load, and cells in series), thereby meeting and matching the current flow through the load.

Battery Chemistry Example: (single cell)

• The voltage of a single cell is $$2.05\ \mathrm V=1.60\ \mathrm V+0.36\mathrm V$$
• The anode half-cell oxidation reaction is: $$\mathrm{Pb+SO_4^{2-}\to PbSO_4+2e^-}$$
• The cathode half-cell reduction reaction is: $$\mathrm{PbO_2+SO_4^{2-}+2e^-+4H^+\to PbSO_4+H_2O}$$
• Anode electrolyte Reaction: $$\mathrm{H^+}$$ and $$\mathrm{SO_4^{2-}}$$ ions migrate as needed to maintain charge neutrality. $$\mathrm{SO_4^{2-}}$$ ions are removed from the electrolyte by the anode reaction with $$\mathrm{Pb}$$. Two electrons are left on the anode, resulting in a net positive charge in the electrolyte around the anode. The excess $$\mathrm{H^+}$$ ions migrate toward the cathode, where $$\mathrm{O_2^{2-}}$$ ions react with $$\mathrm{H^+}$$ ions to produce $$\mathrm{H_2O}$$, which neutralizes the net charge in the electrolyte.
• Why does the anode reaction stop in an open circuit? The anode reaction spontaneously proceeds and liberates electrons, which accumulate on the surface of the anode. $$\mathrm{H^+}$$ ions in solution are attracted to the electrons on the anode, effectively creating a net positively charged ion layer covering the entire anode. This positive layer of $$\mathrm{H^+}$$ ions attracts $$\mathrm{SO_4^{2-}}$$ ions from the electrolyte, but they cannot penetrate the $$\mathrm{H^+}$$ ion layer to react with the $$\mathrm{Pb}$$. When the battery circuit is closed, current flows and the excess electrons on the plate move to the cathode, the $$\mathrm{H^+}$$ ion layer disperses, and the $$\mathrm{SO_4^{2-}}$$ ions migrate to the $$\mathrm{Pb}$$ plate and react.
• Cathode electrolyte reaction: the overall reaction is: $$\mathrm{4H^+O^{2-}\to2H_2O}$$. This reaction neutralizes, a) $$\mathrm{2H^+}$$ from the anode, b) $$\mathrm{2H^+}$$ from the cathode, with c) $$\mathrm{O^{2-}}$$ from the cathode, d) resulting in $$\mathrm{H2O}$$.
• Electrolyte current: Electrons flow from anode to cathode through a conductor, but there is also charge movement in the electrolyte by ion migration where $$\mathrm{H^+}$$ ions migrate from anode to cathode. A big picture view of battery charge movement is 1) electrons move from anode to cathode through a conductor, and 2) those electrons combine with $$\mathrm{H^+}$$ ions at the cathode. The battery maintains overall net electric charge neutrality, but internally the anode and cathode are reservoirs of separated positive and negative charge, both on the terminals and in the electrolyte. The charge differential between terminals produces the associated $$E$$ field potential energy gradient, which is utilized to perform work. The battery converts the electrostatic-bond energy of one atomic/molecular species into another lower energy species. The energy is converted into an $$E$$ field by creating a charge separation scenario, which produces electric charge neutrality in the electrolyte. However, the reaction sequence is more complex than that. The cathode releases $$\mathrm{O^{2-}}$$ from the $$\mathrm{PbO_2+SO_4^{2-}+2e^-}$$ reaction, the $$\mathrm{H^+}$$ ions react with the $$\mathrm{O^{2-}}$$ and convert it to water. If the $$\mathrm{H^+}$$ had reacted with a free electron at the cathode, it would have formed $$\mathrm{H_2}$$ gas.
• Series Cell Electrostatics and Electrochemistry: The same chemical reactions occur in the individual cells whether in a circuit as a single cell or in series. The difference is the amount of current drawn by the series batteries vs. the single cell through the same resistive load. An electrical potential is generated by the half-cell reactions composing a battery.
The anode and cathode reactions generate an open circuit potential, which results in a current flow proportional to the voltage when the circuit is closed. When cells are placed in series, the voltage of the two cells adds, resulting in a doubled $$E$$ field in the space surrounding the batteries. As a result, the higher voltage across the load draws twice the current. This current/rate of electron removal doubles the rate of reaction in the cells. Thus, because the voltage is (essentially) constant from a single battery the controlling factor in current flow is the load (lower resistance, higher current). But, if batteries were placed in series and the load was kept constant, the current would double and the reaction rate would double because the voltage had doubled.

Single Cell Example: Lead-Acid battery, $$10\ \mathrm{Ah}$$ charge

• Load $$R=4.1\ \mathrm\Omega$$
• Voltage $$V=2.05\ \mathrm V$$
• Total stored charge $$q_\text{total} = 10\ \mathrm{Ah} =10\ \mathrm{C/s}\times3\,600\ \mathrm{s/h}\times1\ \mathrm h=36\,000\ \mathrm C$$
• Current flow $$I=V/R=2.05\ \mathrm V/4.1\ \mathrm\Omega=0.5\ \mathrm A$$
• Operating time $$q_\text{total}/I=36\,000\ \mathrm C/0.5\ \mathrm A=72\,000\mathrm s=72\,000/3\,600=20\ \mathrm h$$
• Power consumption: $$P=IV=0.5\ \mathrm A\times2.05\ \mathrm V=1.025\ \mathrm W$$
• Total energy storage: $$\Delta E=P\times\Delta t=1.025\ \mathrm W\times20\ \mathrm h=20.5\ \mathrm{Wh}$$
• Total energy storage: $$\Delta E=IV\times\Delta t=V\times(I\cdot\Delta t)=2.05\ \mathrm V\times10\ \mathrm{Ah}=20.5\ \mathrm{Wh}$$

Series Cell Example: Lead-Acid battery, $$10\ \mathrm{Ah}$$ per cell, same load

• Load $$R=4.1\ \mathrm\Omega$$ (i.e. same circuit, but with two cells in series)
• Voltage $$V=2.05\ \mathrm V+2.05\ \mathrm V=4.1\ \mathrm V$$
• Total charge storage = $$q_\text{total}=10\ \mathrm{Ah}=36\,000\ \mathrm{As}=36\,000\ \mathrm C$$ (Note: Full discharge after delivering $$36\,000\ \mathrm C$$ of charge. Both batteries have stored $$36\,000\ \mathrm C$$ each, but the discharge of one cell passes its electrons from the anode of one the the cathode of the second. Thus, the discharge of 36K coulombs depletes both batteries at the same time.)
• Current flow: $$I=V/R=4.1\ \mathrm V/4.1\ \mathrm\Omega=1\ \mathrm A$$
• Operating time = $$q_\text{total}/I=36\,000\ \mathrm C/1\ \mathrm A=36\,000\ \mathrm s=10\ \mathrm h$$
• Power delivery/consumption: $$P=IV=1\ \mathrm A\times4.1\ \mathrm V=4.1\ \mathrm W$$
• Total energy storage: $$\Delta E=P\times\Delta t=4.1\ \mathrm V\times10\ \mathrm{Ah}=41.0\ \mathrm{Wh}$$
• Total energy storage: $$\Delta E=IV\times\Delta t=V\times(I\cdot\Delta t)=4.1\ \mathrm V\times10\ \mathrm{Ah}=41.0\ \mathrm{Wh}$$

Battery Principles:

Conservation of Charge: The above single battery contains $$10\ \mathrm{Ah}$$ of charge, which equals $$36\,000\ \mathrm C$$. This is the total amount of charge that the battery can deliver as current. Batteries in series, both deplete their electron stores at the same time because all current flows through both batteries. In a closed circuit, at the anode, the reactants react and generate excess electrons. At the cathode, the reactants react with electrons to create products. This proceeds spontaneously and generates a state of electron deficiency at the cathode. The reaction cannot proceed without an outside source of electrons. When electrons conduct from the anode to the cathode, they satisfy the cathode's need for electrons, allowing the cathode reaction to proceed as fast as the electrons are supplied (with limits at upper limits that compete with the maximum reaction rate). When all the reactants at the anode or cathode have reacted to become products is spent, as the battery can no longer supply electrons to, or accept electrons from, the load. The increased current from a series battery source depletes the battery's charge-store twice as fast. The benefit of a higher voltage is a hotter resistor, brighter incandescent lamp, faster DC motor...

Conservation of Energy The total energy available to deliver by cells in series is the sum of the stored energy of the individual cells. As per the above examples: a single cell with $$10\ \mathrm{Ah}$$ of charge has $$2.05\ \mathrm V\times10\ \mathrm{Ah}=20.5\ \mathrm{Wh}$$ of stored electrical potential energy. Two identical cells in series have $$41\ \mathrm{Wh}$$ of stored electrical potential energy. Comparing the two circuits: a series battery doubles the voltage and current, resulting in 4 times the power consumption. Thus, even though a circuit with two cells in series holds two times the stored energy, the power consumption increases by four times, resulting in half the battery life. Conversion into any other form of energy satisfies the principle of energy conservation.

Energy Conversion: As a battery discharges, it converts electrical potential energy into various types of energy in the load. Common conversions of electrical energy include heat, kinetic energy, or gravitational, electrical, and magnetic potential energy. Examples of the types of circuit loads include a resistor (thermal energy), motor (kinetic energy, gravitational potential, electrical potential, magnetic potential), an inductor/electromagnet ($$B$$ field potential energy), or a capacitor ($$E$$ field potential energy).

Battery terminal reactions: In the Lead-Acid battery, before reaction, energy is stored as potential energy by the charge attraction between the chemical reactants at each of the terminals.

• Anode $$\mathrm{e^-}$$ liberation: $$\mathrm{Pb_{(s)}}$$ and $$\mathrm{SO_4{^{2-}}_{(aq)}}$$ are in metal-electrolyte contact, and will spontaneously react to form $$\mathrm{PbS{O_4}_{(s)}}$$ and $$2$$ extra electrons. The positive oxidation potential of this half-cell reaction is $$+1.69\ \mathrm V$$. The positive potential indicates the reaction proceeds spontaneously at room temperature. The reactants attract and react because of electrostatic forces, such as energetically favorable orbital bonding.
• Anode $$\mathrm{H^+}$$ Shielding: After $$\mathrm{Pb}$$ and $$\mathrm{SO_4^{2-}}$$ react, the liberated electrons accumulate on the anode, creating a negative $$E$$ field. The surface electrons attract a layer of $$\mathrm{H^+}$$ ions. This stops the anode $$\mathrm{Pb+SO_4^{2-}}$$ reaction very quickly. The $$\mathrm{H^+}$$ ion layer bonds with the $$\mathrm{SO_4^{2-}}$$ ions preventing them from migrating close enough to the $$\mathrm{Pb}$$ to react. For the reaction to proceed on the anode, the electrons must be removed. When a conductor connects anode and cathode, electrons flow from a place of excess to a place of deficiency. As the electrons leave the anode, the anode reaction resumes.
• Anode Energy Transfer Sequence: All energy begins as reactant bond-energy and is converted into another form of energy by the load:
1) Batteries store energy as the bond energy of reactants. The reactants have more bonding energy than the products, so breaking the stronger bonds and remaking weaker bonds mobilizes energy. That energy can be used to convert into other types of energy. An electrostatic force between reactants draws them to react.
2) In an open circuit battery, a small number of the anode half-cell reactions proceed spontaneously to completion at any moment, resulting in the production of excess electrons, which give the battery its characteristic voltage (in combination with its other half-reaction at the cathode.
3) At the anode, the energy differential between the old bonds within the reactants and new bonds within the products is converted to a concentration of free electrons and an associated $$E$$ field.
4) The anode's $$E$$ field adds to the cathode's $$E$$ field (with its electron deficiency), creating a net total $$E$$ field which permeates the space around the battery. The $$E$$ field drives electrons as a current through the high permittivity conductor to a load.
5) Passage of current through a load converts current into another form of energy.
• Cathode Reaction: the equivalent but opposite process of reaction proceeds at the cathode (i.e., reduction of reactants). The free energy is positive for the bonding of $$\mathrm{PbO_2+SO_4^{2-}+H^++2e^-}$$ to produce $$\mathrm{PbSO_4+H_2O}$$, which means this reaction will proceed spontaneously at STP. The reactants require electrons to proceed, and it does so to a small extent, creating a positive $$E$$ field since the cathode is now electron deficient after the reactant conversion to products consumed electrons. Note, this reaction requires electrons to complete. Even when the circuit is open, a small number of reactants scavenge loosely bound electrons from other atoms or ions. Reaction sequesters those electrons in neutral end products, which results in an electron deficit, a net positive charge on the cathode, and an associated positive $$E$$ field. The electron deficiency at the cathode is the other half of the battery, which creates a net positive charge and adds to the $$E$$ field created by the negative charge on the anode.

Current Flow through the Series Batteries: Batteries in series increase the voltage and current through the load. The current passing through the load also passes through every cell in series. Each cell contributes an $$E$$ field to the total $$E$$ field of the batteries in series. And, the sum of the individual $$E$$ fields drives the current through the load. The outer anode and cathode initially supply the electron source and sink. But, after a transient period, the rate of reaction at every anode and cathode in series equilibrates, and the same current flows through all the cells in series.

Ion migration in the electrolyte of a Lead-Acid battery: In the lead-acid battery, both anode and cathode reactions produce the same neutral compound, $$\mathrm{PbSO_4}$$. In the process, both anode and cathode consume $$\mathrm{SO_4^{2-}}$$ ions from solution. At the anode, the reaction leaves a net positive charge in the electrolyte, which is carried by the $$\mathrm{H^+}$$ ions.
- At the cathode, the consumption/bonding of $$\mathrm{SO_4^{2-}}$$ with $$\mathrm{PbO_2}$$ results in the accumulation of a negative charge in the electrolyte. This is because the reaction of $$\mathrm{PbO_2}$$ with $$\mathrm{SO_4^{2-}}$$ and $$\mathrm{2e^-}$$ results in the liberation of $$\mathrm{O_2^{2-}}$$. The negative charge in the electrolyte around the cathode attracts the $$\mathrm{H^+}$$ ions, from the anode, resulting in the reaction of $$\mathrm{4H^+}$$ ions with one $$\mathrm{O_2^{2-}}$$ ion to produce $$\mathrm{2H_2O}$$. Thus, in lead-acid batteries, charge neutrality in the electrolyte solution surrounding the solution is maintained by the migration of unpaired $$\mathrm{H^+}$$ ions from the anode to cathode.

Battery as Capacitor: The battery acts as a capacitor when in an open circuit. Opposite charges populate the anode and cathode, and this charge differential is separated by a thin dielectric layer of $$\mathrm{H^+}$$ and $$\mathrm{SO_4^{2-}}$$ ions surrounding the anode. This dielectric layer disperses when current flows through a conductor from the anode to the cathode. Conduction removes the excess electrons from the anode, thereby breaking the ionic bond between the $$\mathrm{H^+}$$ and $$\mathrm{SO_4^{2-}}$$ ions. Dispersal of the $$\mathrm{H^+}$$ layer around the anode enables $$\mathrm{SO_4^{2-}}$$ ions to flow past the previously impermeable $$\mathrm{H^+}$$ ion layer.

• Difference between a battery and capacitor: The comparison between a capacitor and battery is incomplete. When in an open circuit, the two behave as charge-storing devices, with both storing charge on either side of a dielectric. The battery has two modes: 1) open circuit: the capacitor and battery configuration are comparable, and 2) closed circuit, the dielectric disperses, and the battery becomes a charge differential generator, using bond-energy liberation from reactants to create products with a differential concentration of charge.
• Capacitive, Field, and Power Flow effects: in the air around a battery and conductor. The open circuit voltage across a battery's terminals acts as an air dielectric capacitor. The charge between terminals polarizes the dielectric of the air/space, producing a small displacement current until charged. However, a conductor is also a capacitor (in the most general sense of the word) since a conductor has a dielectric constant. Placing a conductor between a battery' terminals produces a large displacement current. The Electric Displacement Field is a capacitive effect, causing polarization of the metallic dielectric. The current flowing through the conductor is an attempt by the battery voltage to charge the dielectric of the metal. Since the dielectric constant is so large, the electric displacement field is essentially unlimited. Therefore, the current in the circuit is limited by the resistance of the load, rather than the electric permittivity of the conductor.
• Electric Permittivity and Conductivity: The conductor provides a high electric permittivity $$\epsilon_\text{conductor}$$ path for the $$E$$ field to act. The equivalent circuit to the conductor is a huge capacitor in series with a tiny resistor. The massive electric permittivity $$\epsilon_\text{conductor}$$ of the metallic conductor provides insignificant capacitive impedance. The current flow of a conductor is not similar to the high-velocity trajectory of particle accelerator. Rather, current flow through a conductor is like an atomic executive pendulum, with the microscopic motion of individual electrons at the anode transmitting the repulsive force of their incremental change in proximity from electron to electron from the anode to the cathode. The conductor is a capacitor with a dielectric constant so high that the size of the capacitor is infinite (for practical purposes), as the capacitive effects of a conductor are significant only at high frequencies. At low frequencies/DC, the equivalent size of the conductor-capacitor is so large that it will never fully charge.
• Electric Displacement Field $$D$$: A conductor provides a high electrical permittivity path in which the electric field $$E$$ acts (units: volt/meter = newton/coulomb) act on charge to exert a force on electrical charge. The $$E$$ field of a battery polarizes the dielectric between the plates of a capacitor. The electric permittivity of the vacuum of space is a measure of the capacitance of space and is denoted as $$\epsilon_0$$ and is $$8.85\times10^{-12}\ \mathrm{F/m}$$). Vacuum is the smallest possible electric permittivity and all other materials are expressed as a ratio, a multiple of $$\epsilon_0$$. An $$E$$ field acting in a space with $$\epsilon_0)$$ will have a electric displacement field $$D=\epsilon_0E$$, which has units of $$\mathrm{C/m^2}$$. The $$\epsilon_\text{air}$$ is approximately equal to vacuum permittivity, and that permittivity is very low, so minimal charge is stored on the terminals. However, in a conductor, the $$\epsilon_\text{conductor}$$ is large, which means that for any given $$E$$ field the electric displacement field (in $$\mathrm{C/m^2}$$) will be large, and the closed circuit current through a conductor alone is very high. In a series connection of two cells, the $$E_\text{series}$$ field is double the electric potential of a single cell, hence drives and double the current.

Forces acting inside series cells: Placing two lead-acid cells in series doubles the voltage, and hence doubles the current across the load. That same current will pass through the conductor between the two cells in series, and the ions in the electrolyte of both cells will likewise conduct charge at the same rate. All of the cathodes accept electrons to complete the $$\mathrm{PbO_2+SO_4^{2-}+2e^-+4H^+\toPbSO_4+2H_2O}$$ reaction. And, all the anodes generate electrons in the $$\mathrm{Pb+SO_4^{2-}\toPbSO_4+2e^-}$$ reaction.

• Two batteries in series will draw twice the current through the same load. The higher current increases the demand for electrons from the anode and increases the supply of electrons to the cathode.
• The increased demand for electrons at the anode removes electrons from the products side of the reaction, accelerating the anode reaction, causing it to liberate more electrons at a rate commensurate with demand. The anode supplies current at the rate demanded by the energy consumption rate at the anode for the reaction voltage.
• The increased supply of electrons at the cathode supplies more electrons to the reactant side of the reaction, allowing it to increase its reaction rate by producing more products. The increased current causes the cathode reaction to increase its rate of consumption.
• Consider the chemical reactions acting at the inner anode and inner cathode of the batteries in series. (Note: the terms "inner anode" and "inner cathode" refer to the terminals which make contact between the cells.)
• Placing a second cell in series, the voltage across the load increases, which doubles the current through the load. The increased current draws away the excess electrons, which disperses the $$\mathrm{H^+}$$ ion layer, which allows $$\mathrm{SO_4^{2-}}$$ to react with the $$\mathrm{Pb}$$ at a greater rate, and increases the concentration of $$H^+$$ ions. The positive charge around the anode draws $$\mathrm{O_2^{2-}}$$ ions to migrate from the inner cathode to the outer anode. By increasing the concentration of reactants, by Le Chatlier's principle, the cathode reaction is driven toward products, making the inner cathode more positive, and drawing electrons from the inner anode, which increases its reaction rate, and draws $$\mathrm{O_2^{2-}}$$ ions from the outer cathode in the same way as the outer anode.
• And, at the outer cathode, electrons arrive from the outer anode through the load, resulting in a current of electrons which supplies the outer cathode with electrons to for its reaction.
• In sequence, the increased voltage increases the current flow through the load, which draws off electrons from the outer anode and supplies them to the cathode, which increases the rate of reaction at both the outer anode and outer cathode. In turn, the outer anode affects the inner cathode reaction, and the outer cathode affects the inner anode, both via the modification of the migration rate of ions, which in turn changes the reaction rate at the inner anode and cathode.

Summary: With the higher voltage delivered by cells in series, the load draws an increased current. The reaction rate at the anode increases the current supplied in response. As current arrives at the cathode, the cathode reaction rate increases. The inner anode and inner cathode follow the lead of the outer anode and cathode, increasing their reaction rates - influenced by ion migration in the electrolytes of the two cells. After a short transient, the systems reach a steady state where the reaction rates at all anodes and cathodes match the current demand at the load.

One way to think of this is as follows: for your two-battery system with the batteries in series, the positive terminal of battery "A" doesn't know or care what voltage the negative terminal of battery "B" is presenting to it. All battery "A" is going to do is pull electrons out of its positive terminal, lift them to a potential that is N volts greater than this (where N is the battery's rated voltage) and present them to its negative terminal. In practice, this means that the voltages of batteries in series are linearly additive.

• I was under the impression that the potential from a negative pole of a battery was solely due to the surplus of electrons in that region. Is this correct? Could you expand on how/why exactly the battery is increasing the potential? – Lucas Cioffi Aug 2 '18 at 1:29
• the electrons there are being actively pushed out of the battery by the release of chemical energy inside the battery. connecting the battery to a load allows the electrons to flow out of the terminal and unleashes the chemical reaction inside, causing it to pump out more electrons. the "water analogy" is actually pretty useful in this context. – niels nielsen Aug 2 '18 at 4:36
• Then the question seems to boil down to "what causes voltage?" – user253751 Aug 2 '18 at 5:48
• @immibis, see esl's response below. – niels nielsen Aug 2 '18 at 5:52

I think that your analogy of "surplus" of electrons is wrong in one sense, and incomplete in another sense. First, what do you consider a "normal" and a "surplus" of electrons? Second, if you have a "surplus" in one side, in the other you don't have a normal, you have a deficit.

Think in a classic Lead-Acid battery. When fully charged you have lead in the negative plate, and lead dioxide in the positive one. In both plates at start you have the usual amount of electrons (82 e per Pb atom in the negative, and 98 per PbO2 molecule).

But they are submerged in a aqueous solution of sulfuric acid, given the possibility to the following redox reaction:

• At anode (negative): Pb (s) + HSO4 (aq) → PbSO4 (s) + H+ (aq) + 2 e
• At cathode (possitive): PbO2 (s) + HSO4 (aq) + 3 H+ (aq) + 2 e → PbSO4 (s) + 2 H2O (l)

As you can see, for this hemireacctions occur, you have to do take the electrons in the anode and put an equivalent amount in the cathode. It does not work if you do just one, because you need equilibrate charge and matter: An H2SO40 transforms into H+ and HSO4-, you can't have one without the the other (nor have positive without negative, neither transform a complete molecule 98 g of H2SO4 into 1 g of H+.

The voltage is measure of potential job diference. If you have 2 volts between cell electrodes means that the cell is capable of do a 2 J job moving a 1 C negative charge from the anode to the cathode. If you put two cells in series, the job, each one with 2 v, among both are able to do a 4 J job moving a 1 C negative charge from the free anode of one to the free cathode of the other.

Edit 3/8: Someone asks "So if you connect two, do you get a neutral state in the middle?". Initially, no.

Let's start disconnected from the circuit. Electrons can't flow if the hemireaction does not happens. The hemireaction can't happen on its own, without its counterpart. And the counterpart can't happen without its own electron flow.

So, when connected, the carges are going to start neutralising as the electrons flow through the circuit and both hemireactions happen. And that neutralisation is going to happen in all four electrodes, discharging the batteries; because some hemireaction happens in all four electrodes. When the neutralisation is complete, the batteries are completely discharged.

• I think it would be reasonable to say that a "normal" amount of electrons on a terminal means that terminal has no net electric charge. Doesn't the reaction move electrons from one terminal to another (not the same electrons) until the "back pressure" caused by the inter-electron repulsion causes the reaction to reach equilibrium? – user253751 Aug 2 '18 at 9:59
• That's why I asked, it's not clear to me what he meant, partially because if he meant that, he is wrong. You start with a neutral state, so if then you have a "surplus" in one electrode, you'll have an equivalent deficit in the other. – ESL Aug 2 '18 at 12:21
• Yeah, So if you connect two, do you get a neutral state in the middle? – user253751 Aug 2 '18 at 21:39
• @immibis, I extend the answer thanks to you. Short answer: initially does not, when discharged by electron flow through the circuit, yes. – ESL Aug 3 '18 at 23:35

Consider first two charged capacitors which are connected in series. The potential of plate $A$ relative to plate $B$ is $V_{\rm AB}$ and the potential of plate $C$ relative to plate $D$ is $V_{\rm CD}$ so the potential of plate $A$ relative to plate $D$ is $V_{\rm AB} + V_{\rm CD}$.

Within each capacitor there is an electric field and the work done by an external force in taking unit positive charge from plate $B$ to plate $A$ is $\int_B^A E_{\rm AB} \,d s$ and from plate $D$ to plate $C$ is $\int_D^C E_{\rm CD} \,d s$.
The total work done is $D$ to plate $\int_B^A E_{\rm AB} \,d s + \int_D^CE_{\rm CD} \,d s$.
This is the same result as the addition of potential differences.

What about the conductor between plates $C$ and $D$ as it might appear that the negative charges on plate $B$ should neutralise the positive charges on plate $C$ and due to those charges is there an electric field in the conductor between plates $B$ and $C$.
Well there is no electric field within that conductor and the charges on plates $B$ and $C$ are held in place by the charges on plates $A$ and $D$ respectively.

So what about ideal cells connected in series? If effect cells have two plates which include the terminals.
These terminals/plates get charged up by an electrochemical process which can be thought of a charge pump (sorry about the water analogy).
Initially the plates terminals are uncharged and the charge pump moves charges from one terminal to the other terminal making one terminal having a net positive charge and the other terminal having a net negative charge.
The net charges on the terminals produce an electric field within the cell which opposes the pumping of charges from one terminal to the other by the charge pump (electrochemical process).
Eventually that electric field is enough to stop the migration of charges between the two terminals.
Each cell has a certain potential difference across its terminals and the addition of these potential differences is the same as for the example with capacitors.

So what is different about the cells as compared with the capacitors.
The key difference is that as soon as the terminals of a cell are connected together with a conductor charges will flow from one terminal to the other in the external circuit.
The reduction in charge on the terminals would result in a reduction in the electric field between the terminals but that electric field is no longer sufficiently strong to stop the charge pump pumping charges causing a migration of charges between the two plates within the cell.
The charge pump moves charges within the cell to compensate for those charges which flowed from one terminal to the other terminal via the external conducting circuit.
In effect the charge pump maintains the potential difference across the terminals whereas with a capacitor there is no mechanism for maintaining the amount of charge stored on the plates.

The removal of charges from terminals $D$ and $A$ would result in a migration of charges between terminals $C$ and $B$ ie an electric current flows between the two cells.

Discharge cycle:
Because of electrochemical processes, a battery develops an electric potential difference (voltage) between its electrodes. When you connect two such batteries in series, each battery maintains its individual potential difference (its voltage). When the circuit is completed, by connecting a load between the positive electrode of the first battery and the negative electrode of the second battery, a current, determined by the sum of the voltage of the two batteries (2x) divided by the load resistance added to the internal resistance of the batteries (Rl + 2Ri), flows. This process continues until the batteries are discharged.

If we designate the negative terminal of the second battery as the ground state, then the connection between the two batteries will be at (x) volts (with respect to ground), and the positive electrode of the first battery will be at (2x) volts (with respect to ground).

Charge cycle:
Since the internal resistance of the batteries is the same, the voltage (2x) provided by the charger is divided in half. This is equivalent to each battery receiving x voltage. The first battery will charge to x voltage and so will the second. When both are charged, the current will be 0, since the voltage of the batteries will equal the voltage of the charger.