# Electrolyte and internal resistance

If you have a Daniell cell(zinc / copper), excess charges will build up on the electrodes making the open circuit voltage until the reactions halt.

Now in normal situation when you connect their terminals with a light bulb on the negative terminal there will be this happening:

• two electrons leave

• the reaction can continue

• they get replaced by two other

• the reactions halts again

• two electrons leave and so forth

Maybe not be two and maybe all those steps happen at the same time .

Where is the internal resistance in all of this, and in a short circuit if more electrons leave then they enter back in the electrode this will eventually make the plate neutral and have 0 excess charges , thus 0 electric field but that's not the case ,

TL;DR: What causes the internal resistance in a copper /zinc Daniell cell in both short circuit and in series with a light bulb?

In both the normal situation and the "short circuit" situation the same thing will happen. What will be different is the total resistance available in the circuit. By that I mean both the internal and external resistance. In the idealized case we consider $$R_\text{internal}=0$$ and hence (as far as net resistance is concerned) our results only depend on the $$R_\text{external}$$ and when you directly connect the terminals of the battery, an infinite current runs through your circuit as suggested by the famous Ohm's law,

$$\lim\limits_{R \to 0}I(R)=\infty$$

where

$$I(R)=\dfrac V {R_\text{total}}$$

But actually that's not what happens. In reality a short circuit happens whenever the current gets undesirably high which causes damage and is very dangerous.

-two electrons leave

-the reaction can continue

-they get replaced by two other

-the reactions halts again

-two electrons leave and so forth

I personally think that the 2 electrons stuff isn't a very good way to think. Rather what you may think like is that there are many electrons involved in the reaction at a particular instant(when the cell is in use) but for every 2 electrons lost by the $$Zn$$ electrode the $$Cu$$ electrode gains 2 little electrons.

Now coming to the internal resistance part, each part of the Daniell cell has a resistance(most of it coming from the nature of the electrolyte used and the salt bridge). If the electrodes are left in the open for so long, part of their surfaces will suffer oxidation which will further increase the resistance. As all these components are considered internal to the cell system we say all of these factors contribute to the internal resistance of the circuit.

and in a short circuit if more electrons leave then they enter back in the electrode this will eventually make the plate neutral and have 0 excess charges

That doesn't happen because the charges on the opposite plates are always equal and opposite. The charges on the plates will decrease as the cell voltage decreases with use and eventually reach $$0$$.

TLDR : what causes the internal resistance in a copper /zinc daniell cell in both short circuit and in series with a light bulb

In either of the cases, the internal resistance is offered by the contents inside the cell though it will slowly increase with time as the electrolyte will lose ions.

• The question is really about what is the internal resistance at the microscopic level part if you could offer more details about that Commented Jul 4, 2019 at 17:29
• "What causes the internal resistance in a copper /zinc Daniell cell in both short circuit and in series with a light bulb?" i will edit my question for it to become more clear Commented Jul 4, 2019 at 18:43
• You've entirely modified your question which makes my answer obsolete. At first you were asking about some short- circuit related stuff. Commented Jul 5, 2019 at 4:21
• I think i will roll back the edit and ask an other question Commented Jul 5, 2019 at 4:28

I don't find the question totally clear because it doesn't give us much of an idea of what you do or don't already know and what specifically you want clarified. But I'll attempt to answer it anyway. There are "simple" answers and not so simple ones. I'll give the most basic answer I can, though it won't be terribly short. I don't think there is a satisfactory short answer.

We don't have to specify a Daniell cell. The answer is more or less the same for any battery, though the details would be different. I'll try to give a pretty general answer which would apply to any battery. Really, the answer is not much different for the internal resistance in a battery than it is for the resistance in anything else, like a resistor. So I'll start there.

First let's establish what resistance is, in terms of basic measurements. I'll start by talking about resistance in something like a wire or resistor because it is the simplest case. Recall that electric potential is potential energy per unit charge, so voltage (which is potential difference) is the difference in potential energy per unit charge between two locations. Take Ohm's Law:

$$\Delta V = I R$$.

Since $$\Delta V$$ is just a change in potential energy per unit charge and $$I$$ is just a quantity of charge passing by per unit time ($$I = \Delta Q / \Delta t$$), then $$I \Delta V$$ is just (energy/charge) X (charge/time) = energy/time. That's power. So this gives the equation which you've certainly seen, $$P = I^2 R$$, which is valid for a resistor.

But what does this equation really mean? Let's boil it down to very basic energy considerations. There is an electric field pointing from locations of high potential to low potential. If that was all that was going on then the electric field would make the electrons in the resistor accelerate and they would lose potential energy and gain kinetic energy as they pass through the resistor. But this is not what happens. Instead the electrons have many (billions per second typically) collisions with the atoms in the resistor. With each collision they "give up" some of their energy to the atoms, making the atoms vibrate. This is thermal energy. So $$P = I^2 R$$ tells us the rate at which electric potential energy is being converted into thermal energy in the resistor, because of collisions between the electrons and the atoms in the resistor. Notice, the electrons never get going very fast between collisions, so there is no buildup of kinetic energy in the circuit. In fact kinetic energy is always negligible in circuits unless one element of the circuit is a particle accelerator. ;-)

Now let's look at a battery. Charge is conserved so there has to be current through the battery as well and this current must equal the current into one terminal and out of the other. But the charge carriers in the battery are not electrons. There will be ions produced and consumed at both terminals. Some of these ions will be the charge carriers in the battery.

Ignoring some complications, in the specific example of a Daniell cell that you ask about the charge carriers are zinc ions (positive) which move from the zinc electrode to the copper one, and sulphate ions (negative) which move from the copper electrode to the zinc one. Note that the zinc electrode is the negative one, so the electric field inside the cell points from the copper electrode to the zinc one. If only electric forces acted then we expect positive charges to move in the direction of the E-field and negative charges to move against the electric field. So both the zinc ions and the sulphate ions are moving in the opposite direction to the way the electric field is pushing them. This is because of diffusion forces.

These diffusion forces are doing work against the electric field to move charges from where they have low potential energy to where they have high potential energy. We interpret the emf, $$\epsilon$$, of the battery as a work per unit charge. This is the work that the diffusion forces do per unit charge moved across the battery. As always a work causes a change in energy. In an ideal battery with zero resistance all of the work done by diffusion forces would be transformed to electric potential energy. But in a real battery there is resistance because once again the charges moving through the battery collide with atoms in the battery and this transfers energy (thermal energy) to the atoms of the battery. But the work per unit charge is fixed (complicated chemistry and thermodynamics which I'll side step...). So if some of the work goes to thermal energy instead of potential energy the voltage (potential energy difference per unit charge) must be smaller than it would be in the ideal case.

So the actual voltage across the battery is

$$\Delta V_{battery} = \epsilon - IR_{battery}$$

where $$IR_{battery}$$ is the thermal energy per unit charge that was produced. That $$R_{battery}$$ is what we call the internal resistance. But it is just talking about a rate at which thermal energy is being produced due to atomic scale collisions in the battery. So in that way it is no different from "external" resistances.

• Correct me if I am wrong: The excess charges on the positive and negative terminals of the cell create the electric field,thus the potential difference, in a short circuit the potential difference accros the two terminals is small,that means one of two possiblities : - either the difference of charges accros the terminals gets lower,Or there is some sort of other electric field counter acting the electric field of the terminals thus reducing it's effects on electron in the wire. which of this two is true? , Thanks Commented Jul 5, 2019 at 4:45
• @mohamedazaiez, the battery maintains a constant emf (approximately). So in the short circuit situation, if we really can approximate the voltage across the wires as zero then the voltage across the battery must also be zero. But this isn't necessarily because of any reduction in charge on the terminals (complicated...). It is that the emf is balanced by the voltage drop due to the internal resistance. The electric field is still there, is still accelerating charges. But those charges lose all of the energy they gain from the E-field due to collisions with atoms, heating up the battery. Commented Jun 24, 2022 at 22:24

When the batteries cell potential is either more(during charging/ requires more energy) or less(during discharging/ deliver less energy) we refer to the phenomenon as overpotential. What happens to this wasted energy? it is lost as heat. There are three main causes of losses in batteries: activation losses (chemical reaction kinetics loss), mass transport loss/concentration loss (slow diffusion of ions) and ohmic losses. These all contribute to the voltage loss or increased resistance with $$P=I^{2}R=I(V_{theoretical}-V_{measured})$$ In addition to this there is another term according to the Bernardi heat equation due to entropy and is the TdS term