battery stuff

John Rennie

Oct 5, 2018 07:38

Let's have a think about what is happening physically in this cell ...

Zinc is oxidised to Zn2+ and Cu2+ is reduced to copper. There is a free energy $\Delta G$ associated with this reaction.

So what happens to the (free) energy generated in the reaction? Well, it gets transferred to the two electrons that flow from the anode to the cathode for every Zn/Cu atom that reacts. So the result is that the two electrons gain a free energy $\Delta G/2$ each. Make sense so far?

Abcd

yes

John Rennie

Now what happens is that those two electrons flow through the light bulb, and they transfer that energy to the light bulb and make it glow. So when the electrons get back to the copper electrode they have lost the free energy they got from the reaction and they are back where they started.

Abcd

hmm

John Rennie

So the net result is that for every electron that flows round the circuit a free energy $\Delta G/2$ is transferred from the Zn/Cu to the light bulb.

What confuses people is that free energy $\Delta G$ is normally associated with chemistry and the electrical energy $eV$ is normally associated with electricity.

But they are both just energy changes - different names for the same thing.

Abcd

hmm

John Rennie

Oct 5, 2018 07:47

If you read any textbook on electrochemical cells they will tell you that the fundamental equation is:

$$ \Delta G = -|z|FE $$

Abcd

ikr

nF is charge

So I get that

John Rennie

Where $\Delta G$ is the molar free energy change for the cell reaction, $E$ is the cell potential and $z$ is the number of moles of electrons generated per mole of cell reaction ($z=2$ in this case).

Abcd

ya E i dont understand that

John Rennie

The left side is the energy generated in the cell reaction and the right side is the energy transferred to the circuit as the electrons flow through it.

Abcd

yes

John Rennie

Oct 5, 2018 07:51

The two sides have to be equal because energy is conserved, so that means there must be an EMF $E$ given by that equation.

Abcd

yes

John Rennie

That's where the cell EMF comes from. It's just free energy of electrons.

Abcd

But this means that this should continue for $\infty$ time

Then why does cell die??

14 mins ago, by Abcd

59 mins ago, by Abcd

Now till what time will this continue?

John Rennie

Because for every two electrons one Zn atom becomes a Zn2+ ion and one Cu2+ ion becomes a Cu atom. That means the solution concentrations of [Zn2+] and [Cu2+] change.

Abcd

@JohnRennie Which means changing $\Delta G$ and which ultimately means fluctuating EMF!!

John Rennie

Oct 5, 2018 07:56

Eventually there will be no Cu2+ ions left in solution and the reaction has to stop because it runs out of things to react.

Abcd

Then why do we need electrochemical cells?

Despite fluctuating emfs?

John Rennie

@Abcd I wouldn't say fluctuating because that implies the emf fluctuates up and down. What happens is that the EMF decreases smoothly as the reaction proceeds. And that's exactly what we see. Take any battery, like the one in your laptop, and you'll find that its voltage decreases smoothly to zero as the battery runs down.

Abcd

Ohhh

@JohnRennie So how are they able to measure $E^o$ at 1 M concentration of metal and 1 atm pressure of Hydrogen?

DO they do something like: Lets attach things quickly and quickly check the EMFs

because they wont get the same emf after long time

@JohnRennie What about current? Wont that too decrease? But doesn't the laptop need a definite amount of current every second? How does battery handle that?

John Rennie

Take your circuit and remove the light bulb.

Now no electrons can flow round the circuit, so the reaction cannot proceed.

So now you can attach your voltmeter and measure the emf

Abcd

@JohnRennie but electrons can flow when we attach voltmeter

John Rennie

Oct 5, 2018 08:04

Yes, but voltmeters are designed to have a very high resistance so very few electrons flow.

That means the reaction does proceed when we attach a voltmeter but only very slowly, so the cell emf changes only very slowly.

With a decent voltmeter the resistance is so high that the decrease in the cell emf with time is undetectably small.

Abcd

@JohnRennie Why did we take voltage constant in all physics resistances questions then??

@JohnRennie Please also explain what equilibrium means in case of electrochemical cell.

John Rennie

@Abcd even with a high current flowing the cell emf changes only very slowly.

Again, consider the battery in your laptop. The emf decrease as the battery discharges isn't linear. The battery voltage decreases only very slowly at first then it starts dropping rapidly as the battery nears exhaustion.

In practise we can treat the battery voltage as constant.

Abcd

@JohnRennie ?? Your above statement contradicts that

John Rennie

@Abcd I can't see the contradiction

Abcd

@JohnRennie I think you meant "we can treat battery voltage as constant at the beginning"

6 mins ago, by Abcd

@JohnRennie Please also explain what equilibrium means in case of electrochemical cell.

John Rennie

Oct 5, 2018 08:14

The cell emf looks something like this. The $x$ axis is the total charge that has flowed through the cell.

So for most of its life the battery voltage is effectively constant.

@Abcd you're asking me two different questions, one about cell emf and one about equilibrium. Let's deal with them one at a time. Emf first.

Abcd

@JohnRennie i dont get what voltage has to do with battery charging

23 mins ago, by John Rennie

@Abcd I wouldn't say fluctuating because that implies the emf fluctuates up and down. What happens is that the EMF decreases smoothly as the reaction proceeds. And that's exactly what we see. Take any battery, like the one in your laptop, and you'll find that its voltage decreases smoothly to zero as the battery runs down.

0 charging = 0 voltage? How?

John Rennie

charging?

The $x$ in my graph shows how much charge has flowed through the circuit, and that is proportional to how many moles of the reagents in the battery have reacted.

So in effect the left side of the graph is when all the reagents in the battery are at the initial concentrations and the right end is when all the reagents have reacted as much as they can.

In our Zn/Cu cell the right end would be when all the Cu2+ ions are gone.

Abcd

Oct 5, 2018 08:35

@JohnRennie not necessarily

it might be that delta G of reaction is coming out to be 0?

John Rennie

Possibly. I'd have to sit down and start writing out the equations to investigate that further.

In this case the free energy change would be ...

$$ \Delta G = \Delta G^0 + RT ln \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) $$

(at least I think so)

Ok, so the reaction would stop when the right side became zero and that would be at a non-zero value for the Cu2+ concentration. So yes, you're correct.

Abcd

Oct 5, 2018 09:00

What about current? Wont that decrease too?

@JohnRennie ^^^

John Rennie

@Abcd Assuming some constant load with resistance $R$ the current is simply $E/R$ as usual for an electrical circuits. So as $\Delta G$ changes the current will change as well.

Abcd

@JohnRennie that does no harm?

John Rennie

For most of the life of a battery the cell EMF is roughly constant so the current generated by the battery through a fixed resistance is also roughly constant.

After all, your laptop still works even when the battery is 90% discharged ...

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