Does the EMF of a battery change with time?

OK, so... I know that batteries provide less and less energy with time. My question is- is that a result of a decrease in the EMF of a cell or an increase in the internal resistance of the cell or both?!

I know the terminal voltage (V) is whatever is left after some potential drop within the cell So, V = E - Ir Now, is the decrease in V with time a result of decrease in E or an increase in r or both?

Extending this question, when a battery is charged, does the EMF increase or is it the internal resistance that decreases?

Furthermore, consider a 10V battery with internal resistance r connected to An external Voltage Supply of 100V for charging. We know I = (100-10)/r What is the terminal voltage of the battery? If the answer to the previous question, i.e. whether charging increases EMF is false. Then it's Obvious... the Potential Drop or work done is 10V + Ir (Actually 10V - Ir but I is negative) Ir is again, like in the case of discharge the energy lost to the internal resistance and 10V is what is being overcome to push the electrons against the EMF of 10V battery.

But if in fact, charging does increase the EMF, shouldn't there be an extra amount of work done on the 10V battery, to increase its EMF? Hence, the actual Potential Drop Across the battery, i.e. the Terminal Voltage of the battery would in fact be greater than 10V + Ir.

What's true? What am I missing?

is that a result of a decrease in the EMF of a cell or an increase in the internal resistance of the cell of both?

It's both.

Batteries are surprisingly simple things, at least in principle. A battery is powered by a type of reaction called a redox reaction and it is the energy of the reaction that provides the electrical energy to the electrons. When an electron passes through a potential difference of $$E$$ its energy changes by $$eE$$ and this is the same as the energy per electron of the chemical reaction. Indeed equating the electron and reaction energy gives us the Nernst equation, which is the fundamental equation describing electrochemical cells:

$$\Delta G = -zFE$$

Here $$\Delta G$$ is the Gibbs free energy per mole of the reaction and $$E$$ is the battery EMF.

The reason that the battery EMF falls as the battery runs down is that the reaction energy $$\Delta G$$ depends on the concentrations of the chemicals in the battery. As the battery runs down these concentrations change and this reduces the EMF. The equation that describes this is:

$$E = E° - \frac{RT}{zF}\ln Q_r$$

where $$E°$$ is the standard cell emf and $$Q_r$$ is the reaction quotient. The reaction quotient depends on the concentrations of the reacting chemicals, and as the reaction goes and the chemicals are used up the concentrations change and this lowers the EMF.

As you say in the question, when the battery is connected and supplying a current the terminal voltage is lower than the EMF due to the internal resistance:

$$V = E - Ir$$

As the battery runs down the internal resistance generally increases, but this is a much more complicated property than the EMF and there isn't any simple equation to describe it. Different types of batteries have different internal resistances and those internal resistances change at different rates.

You asked about charging: when you charge a battery it drives the reaction inside the battery in the reverse direction. This increases the EMF because it's changing the concentrations of the chemicals inside the battery back to their original state. It will also affect the internal resistance, but as I said above describing the changes in the internal resistance is complicated.

Charging also increases the terminal voltage and it is as you suggested in your question:

$$V_\text{charge} = E + I_\text{charge}r$$

where $$I_\text{charge}$$ is the charging current.

• Yes. I am aware of all that. But I mean to ask V = E + Ir doesn't involve the fact that E is increasing during charging. Or does it? Perhaps the increase in Work done I was wondering about was the very change in E. Hence V is indeed greater than E_initial + Ir but is in fact given absolutely by E_final + Ir? I think that makes sense. Anyways, thank you for your answer! Commented Jul 4, 2023 at 18:53
• @Maddy During charging $E$ will increase. The terminal voltage $V = E + Ir$ could increase or decrease because both the charging current and the internal resistance will decrease as the battery charges (unless you're using a constant current charger). Which way $V$ changes would be hard to predict. Commented Jul 4, 2023 at 19:04
• During charging , E may decrease due to heating. Batteries are simple in principle, but they are complicated, tricky devices in practice. Commented Jul 4, 2023 at 19:52
• E may as well not depend on the history of charging or discharging, depending on the particular battery chemistry. There are cells made with exactly this intent (e.g. the Weston cell). Then again, charge/discharge currents create concentration gradients in the electrolyte that may affect E short-term (e.g. minutes or even hours). Complex, the batteries are. Commented Jul 5, 2023 at 0:45
• With nickel batteries, both NiCd and NiMH, during fast charge the terminal voltage initially increases due to an increase in E, but then decreases again due to an increase in temperature that reduces the internal resistance and so reduces IR. It's this drop (NiCd) or flattening (NiMH) of the terminal voltage curve that signals 'full charge' to the charger. Commented Jul 5, 2023 at 7:10

Usually, it's both EMF and (non-ohmic) resistance that change. Temperature and aging also change these. Cold cells generally have higher EMF and resistance. Old cells generally have lower EMF and higher resistance. And yes, charging current raises the cell's terminal voltage above its open-circuit EMF. However, sometimes you see voltage decline during charging as the temperature rises. Real batteries are complicated things, and even technical books about them aren't entirely reliable on the details. Textbook idealizations are, well, idealizations, not real world physics. If you want to understand batteries, get some and connect them to other things.