Mechanisms Involved When Charging a Battery

[tellmesomething]

Homework Statement

The potential difference between the terminals of a 6V battery is 7.2 V when it is being charged by a current of 2A.

Relevant Equations

None

First let us talk about a case where the battery gets discharged, I have a circuit with a battery and a load.

The emf of the battery is 6V and the potential difference is a little less because the battery has internal resistance.
When a charge moves from the anode and reaches the load it loses all its energy into a useful form of energy for the load. It comes back to the positive potential, and neutralises a charge on the cathode(reduction). Now inside the battery in the electrolyte there are two types of forces on the ions electric field force and non electrostatic force. They were in equilibrium before the circuit was closed. Now since the field becomes a little less the chemical/non electrostatic force forces the positive ions to go to the cathode to sort of "revive this charge which just got neutral" and it forces the negative ions to go to the anode to do the same. However not all of the charge gets revived as theres internal resistance. Hence this chemical force stays always a little more than the electrostatic force.


Now say we want to increase this chemical force more so that even after facing the same amount of resistance it can revive the total charge it loses. I charge the battery. I plug it to a direct current source. The negative terminal of the dc source is connected to the negative electrode of the battery and the positive terminal of the dc source is connected to the positive terminal. This time I have a potential difference of 7.2 volts across the battery and the battery has an emf of 6V. Excess electrons form the negative terminal goes to the negative electrode, reduction occurs. The little charge that was left to revive revives, and it possibly releases ions into the solution?(I am not sure) this increases the chemical force as the solution becomes more ionic which causes the negative ion to move to the anode and the positive ion to move to the cathode despite of the obstruction / internal resistance. The excess electrons in the positive cathode from before discharging goes to the positive terminal of the dc source?



Again I am not sure about ions and how they are neutralised during discharging and separated during charging... Please let me know if the above is any correct.

Also its tempting to think that 1.2 V is the potential drop due to the resistance but I dont concretely know why? If the dc supply was a 6V machine would this charging not have worked?

I understand this might be a very crude way to look at this if not completely wrong. Please let me know if I am thinking even in the right direction

 

[Drakkith]

To understand a battery you must understand the chemistry of the battery and so it helps to discuss a specific type of battery. Let's say we're dealing with the standard lead-acid battery, like a car battery.

When fully discharged, a basic lead-acid battery cell consists of two plates of lead sulfate (PbSO4) with an electrolyte of diluted sulfuric acid (H2SO4). When charging, the negative terminal undergoes the reaction: $PbSO_4 + 2e^- \to Pb + {SO_4}^{2-}$

The positive terminal undergoes: $PbSO_4 + 2H_2O \to PbO_2 + {SO_4}^{2-} + 4H^+ + 2e^-$

The sulfate, water, and hydrogen ions goes into the electrolyte and form more sulfuric acid, increasing the concentration in the electrolyte compared to the discharged state. The electrons are pulled from the positive plate and given to the negative plate. The positive plate ends up as lead-dioxide (PbO2) while the negative plate becomes pure lead.

The driver of this reaction is the voltage provided by the voltage source and the circuit provides both the voltage and the electrons at the negative plate needed to break the lead-sulfate up, releasing the sulfate into the electrolyte. At the same time, the reaction at the positive plate occurs which turns the lead-sulfate into lead-dioxide, releases sulfate and hydrogen ions into the electrolyte, and pulls electrons off of the plate into the circuit where they can replace the electrons that have moved to the negative plate or move to the negative plate themselves.

 

[Drakkith]

Also its tempting to think that 1.2 V is the potential drop due to the resistance but I dont concretely know why?

I hesitate to call it 'resistance' as a charging battery isn't a simple resistive load. Some of the voltage is certainly lost from resistive losses, but most of that voltage drop is going into charging the battery, not being lost as heat.

Again I am not sure about ions and how they are neutralised during discharging and separated during charging... Please let me know if the above is any correct.

It depends on the chemistry of the battery. In a lead-acid battery, the H+ ions are neutralized by reacting with sulfate ions to form sulfuric acid.

 

[Steve4Physics]

I'd like to add this to what @Drakkith has said.

It’s worth remembering what the unit, a volt, means. One volt is ‘shorthand’ for one joule per coulomb. $1V \equiv 1J/C$.

E.g. if points A and B have a potential difference of 5V, this means 5J of electrical energy will be converted to/from 5J of some other form(s) of energy, per coulomb of charge moving between A and B.

If A and B are the two ends of a resistor and 5V is applied, current flows and electrical energy is converted to heat. For each coulomb of charge passing through the resistor, 5J of electrical energy are converted to 5J of heat energy in the resistor.

If you have an ideal cell (say emf = 6.0V, internal resistance =1 $\Omega$) and are charging the cell by applying 7.2V, then for each coulomb passing though the cell:
- 6.0J of electrical energy are converted to chemical energy;
- 1.2J of electrical energy are converted to heat in the internal resistance.

You could charge using a smaller voltage as long as it’s greater than 6.0V, but it would take longer.

Of course real cells behave differently to ideal ones but the principle is the same.

 

[Drakkith]

If you have an ideal cell (say emf = 6.0V, internal resistance =1 Ω) and are charging the cell by applying 7.2V, then for each coulomb passing though the cell:
- 6.0J of electrical energy are converted to chemical energy;
- 1.2J of electrical energy are converted to heat in the internal resistance.

Interesting. Got a reference that goes into more detail about this?

 

[Steve4Physics]

Interesting. Got a reference that goes into more detail about this?

It’s my own explanation. I have no references. I tried a quick search but didn’t find anything useful.

But what I said (in Post #4) follows automatically from the definitions of the volt and emf, along with conservation of energy. It assumes a rechargeable ideal cell in series with its internal resistance. It’s intended to provide the OP with some understanding in terms of basic concepts.

Of course, for a real cell, various electrochemical effects – not present in the ideal case - could muddy the waters, e.g. if the real cell were not rechargeable!

 

[Drakkith]

Ah, okay. I've just never worked through the math for that particular situation, not even in my EE class.

 

[tellmesomething]

To understand a battery you must understand the chemistry of the battery and so it helps to discuss a specific type of battery. Let's say we're dealing with the standard lead-acid battery, like a car battery.

When fully discharged, a basic lead-acid battery cell consists of two plates of lead sulfate (PbSO4) with an electrolyte of diluted sulfuric acid (H2SO4). When charging, the negative terminal undergoes the reaction: $PbSO_4 + 2e^- \to Pb + {SO_4}^{2-}$

The positive terminal undergoes: $PbSO_4 + 2H_2O \to PbO_2 + {SO_4}^{2-} + 4H^+ + 2e^-$

The sulfate, water, and hydrogen ions goes into the electrolyte and form more sulfuric acid, increasing the concentration in the electrolyte compared to the discharged state. The electrons are pulled from the positive plate and given to the negative plate. The positive plate ends up as lead-dioxide (PbO2) while the negative plate becomes pure lead.

The driver of this reaction is the voltage provided by the voltage source and the circuit provides both the voltage and the electrons at the negative plate needed to break the lead-sulfate up, releasing the sulfate into the electrolyte. At the same time, the reaction at the positive plate occurs which turns the lead-sulfate into lead-dioxide, releases sulfate and hydrogen ions into the electrolyte, and pulls electrons off of the plate into the circuit where they can replace the electrons that have moved to the negative plate or move to the negative plate themselves.

Thankyou so much.I am going to try to understand this more by going through electrochemistry

 

[tellmesomething]

I'd like to add this to what @Drakkith has said.

It’s worth remembering what the unit, a volt, means. One volt is ‘shorthand’ for one joule per coulomb. $1V \equiv 1J/C$.

E.g. if points A and B have a potential difference of 5V, this means 5J of electrical energy will be converted to/from 5J of some other form(s) of energy, per coulomb of charge moving between A and B.

If A and B are the two ends of a resistor and 5V is applied, current flows and electrical energy is converted to heat. For each coulomb of charge passing through the resistor, 5J of electrical energy are converted to 5J of heat energy in the resistor.

If you have an ideal cell (say emf = 6.0V, internal resistance =1 $\Omega$) and are charging the cell by applying 7.2V, then for each coulomb passing though the cell:
- 6.0J of electrical energy are converted to chemical energy;
- 1.2J of electrical energy are converted to heat in the internal resistance.

You could charge using a smaller voltage as long as it’s greater than 6.0V, but it would take longer.

Of course real cells behave differently to ideal ones but the principle is the same.

Can you clarify why it would take longer when a smaller voltage is used? The amount like 6J will already be provided and used to convert to chemical energy rest is going to heat anyways isnt that a waste?

 

[Drakkith]

Can you clarify why it would take longer when a smaller voltage is used? The amount like 6J will already be provided and used to convert to chemical energy rest is going to heat anyways isnt that a waste?

6 joules per coulomb. If you increase the voltage, you increase the current, so the battery charges faster. Assuming you need more than 6 volts to even start a flow of current, then just a volt or two over that already provides a drastic increase in charge speed over, say, 6.1 volts.

 

[Steve4Physics]

Can you clarify why it would take longer when a smaller voltage is used?

@tellmesomething, if you haven't 'got it' yet, try the following questions for yourself...

You have an ideal rechargeable cell, emf = 6.0V, which you can treat as being in series with it internal resistance of 1.0 $\Omega$.

Q1. You apply 7.2V.

a) How much charge flows through the cell per second? (Hint if needed: work out the current through the resistor.)

b) By how much does the chemical energy in the cell increase per second? (Hint if needed: re-read Post #4!)

Q2. Repeat Q1 for applied voltage = 7.0V