JEE Chemistry Club

John Rennie

04:58

@Abcd hi

Abcd

@JohnRennie See he has shown equilibrium sign in the second pic.

I dont understand how these cells can be in equilibirum.

John Rennie

@Abcd if the cell is open circuit then it will be in equilibrium.

Abcd

@JohnRennie open circuit (here) means?

John Rennie

That's because the reaction causes charge to flow, but charge cannot flow if the cell is open circuit.

@Abcd it means the two terminals of the cell, the anode and cathode, aren't connected to an external load.

Abcd

@JohnRennie not getting

Then what is the significance of $E^o$??

John Rennie

05:05

The reaction in a cell transports electrons from the cathode to the anode.

Abcd

No, anode to cathode

John Rennie

So if the cell is open circuit a negative charge builds up on the anode and a positive charge on the anode. That prevents any more electrons being transported so the reaction stops.

Abcd

2 mins ago, by Abcd

Then what is the significance of $E^o$??

John Rennie

I always get mixed up about the anode and cathode in cells. As I recall it's the opposite way round from what we call the anode and cathode in the external circuit - or something like that ...

Abcd

anode is electron supplier everytime - nice way to remember

John Rennie

05:08

@Abcd context?

Abcd

@JohnRennie in the above picture why has he shown iverall reaction to be in equilbrium?? What does he mean?

John Rennie

I think you might be taking it too literally. The author may just mean it's a reversible reaction.

By $E^0$ do you mean the standard EMF?

Abcd

yes

@JohnRennie See that is measured by connecting our half cell to hydrogen half cell.

But

Will that cell never die?

Then what exactly is $E^o$

Also, whats the use of electrochemical cells when their current produced is going to vary so much

John Rennie

Aha, OK, that's the EMF when the cell is open circuit and the reagents are all at unit activity.

The cell is in equilibrium when it's open circuit.

Abcd

but how can there be emf

when cell is in equilbrium

John Rennie

05:15

Let's take an analogy.

Suppose we have a gas AB that dissociates into A and B. If we put some AB in a sealed vessel it dissociates until the pressure increase stops the dissociation going any further. Then we get equilibrium with some pressure P inside the vessel.

In a cell we put some chemicals in an open circuit cell. As they react a potential difference develops across the terminals and the PD builds up until it stops the reaction going any further. Then we get equilibrium with some potential difference E.

Abcd

@JohnRennie Why is equilbirum attained in cell?

Why cant pd keep on increasing?

John Rennie

The cell reaction transports electrons. The reaction in cells is a redox reaction and that is always associated with the transport of electrons.

Abcd

ikr

John Rennie

The cell EMF is just the free energy of the reaction divided by some constant - Faraday's constant?

Abcd

lets not go in that direction, I'll come to it later.

@JohnRennie Is there any reason why equilibirum is being attained?

John Rennie

05:22

But the PD developed as the electrons are transported from the anode to the cathode oposes the motion of electrons. The overall free energy change is then $EF + eE$.

$E$ is the cell EMF and $F$ is Farady's constant.

Hmm, something like that anyway ...

Anyhow the point is that the magnitude of the eE term increases until the total free energy change becomes zero. At that point the cell is in equlibrium.

Abcd

@JohnRennie what does it mean for the cell to be in equilbrium?

like what is happening there?

John Rennie

It means the energy per electron due to the reaction is the same as the energy per electron due to the potential difference developed between the electrodes.

Abcd

:/

John Rennie

The cell reaction is trying to push an electron from the anode to the cathode, but the PD is trying to push the electron back the other way.

We get equilibrium when the two effects balance out.

Abcd

@JohnRennie which pd are you talking about?

28 mins ago, by Abcd

John Rennie

05:29

The PD measured by the voltmeter

Abcd

how is that pd developed?

Even charge is balanced in both half cells

Then how can there be pd?

John Rennie

@Abcd can you expand on Even charge is balanced in both half cells?

If the cell is open ciruit there is a charge imbalance

We get a build up of electrons on the cathode and a deficit on the anode.

Abcd

@JohnRennie See both half cells are neutral. That neutrality is maintained by salt bridge.

So charge is balanced on both sides

John Rennie

Neutrality just means no charge flows.

Abcd

its so hard I dont understand

John Rennie

05:33

A cell reaction is basically a pump. Imagine the electrons as water flowing ina pipe, then the cell reaction is analogous to a water pump.

If we connect the pipe in a circuit then the pump transports water round the circuit just like a cell makes electrons flow round the circuit.

But now suppose we close a valve to stop the water flowing. The pump will try to pump water, but since there's no flow it will just compress the water and the pressure of the water will build up until it stops the pump from transporting any more water.

That's exactly what happens in an open circuit cell.

Abcd

@JohnRennie i dont understand which electric field exactly is responsible for flow of electrons in outer circuit

John Rennie

A battery is just a device that creates a potential energy difference between its terminals. The voltage of a battery is just the potential energy change per coulomb of charge that moves from the anode to the cathode round the circuit.

Although we all it a voltage that potential is just a free energy change.

So as the electron flows through the external circuit its free energy changes by some value, which I think is equal to $\Delta G = VF$, where $F$ is the Faraday constant.

Inside the battery there is a chemical reaction, and that chemical reaction also has a free energy.

And basically that reaction increases the free energy of an electron, then as the electron flows round the external circuit its free energy decreases again.

The point is that a reaction free energy per electron and a voltage are basically the same thing.

Abcd

31 mins ago, by Abcd

Also, whats the use of electrochemical cells when their current produced is going to vary so much

John Rennie

06:06

@Abcd I don't understand what you mean when you say the current will vary

One mole of reaction produces $z$ moles of electrons . The current just depends on how fast those electrons flow through the external circuit, and that depends on what the external load is.

But as far as the cell is concerned it has produced a mole of electrons with an associated free energy change of $\Delta G = zFE$.

Abcd

06:24

@JohnRennie Are you there?

@JohnRennie Can you please explain the working without bringing $\Delta G$ into the picture.?

I have tried hard but not getting.

John Rennie

You have to include the free energy. The point is that the free energy is effectively a potential energy just like the cell voltage. The two are basically the same thing. Understanding that the cell voltage is the free energy per mole of electrons is fundamental to understanding how cells work.

Abcd

@JohnRennie Okay consider a simple Daniel cell

what will happen at the beginning?

See Zn will oxidise to Zn2+

and 2 electrons will move towards Cu

Then Cu will be deposited there.

Now till what time will this continue?

@JohnRennie ??

John Rennie

06:42

I need to work now for about 20 minutes ...

Abcd

07:12

@JohnRennie Its been 30 minutes already??

John Rennie

@Abcd it's taking me a bit longer than I expected this morning but I will be finished soon

Abcd

...OK

John Rennie

07:36

@Abcd I'm back!

Abcd

@JohnRennie Please explain then....

59 mins ago, by Abcd

Now till what time will this continue?

John Rennie

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

07:44

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

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

07:49

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

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

07:55

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

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

08:01

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

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

08:10

@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

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

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

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 ...

Abcd

Okay.

@JohnRennie I don't understand why Physics textbook says some entirely different thing.

Are the two things same? If yes, then how?

John Rennie

That's a rather abstract view of a battery. In real life there is a rather messy chemical reaction going on, and that reaction is the internal mechanism referred to in the furst paragraph.

Abcd

09:07

he says that in battery $F_b = F_e$ I dont understand how that is possible for our electrochemical cell.

John Rennie

I'm not sure how useful it is to talk about the force $F_b$. You can sort of define a force because if the free energy change is $\Delta G$ and the distance between the electrodes is $x$ you can say $F_b = \Delta G/x$ i.e. just using work = force times distance. But there isn't really a force acting on the electrons.

Abcd

@JohnRennie OK, even if we consider it as Fb then what is Fe ?

John Rennie

In the example we started with the cell reaction in effect pumps electrons through the cell. Those electrons then flow through the lamp and back into the cell where they get pumped through the cell again. OK so far?

Abcd

yes'

John Rennie

Now suppose we disconnect the lamp. The reaction is still trying to pump electrons, but now those electrons have nowhere to go. The result is that we end up with a (very small) negative charge on the zinc electrode and an equal and opposite (very small) positive charge on the copper electrode.

Abcd

09:15

hmm

@JohnRennie ??

John Rennie

When the cell is open circuit we have a potential difference between the two metal plates equal to the cell emf $E$.

But the cell will have some capacitance $C$. After all it is two metal plates just like the metal plates in a capacitor. That means there will be a charge separation given by the usual equation for a capacitor $Q = CE$.

Abcd

hmm

@JohnRennie You dont seem interested in this. So leave it.

@JohnRennie Could you just tell me about the equilibrium concept? Like why does a state of equilibrium arrive in the cell?

John Rennie

@Abcd it isn't that I'm not interested, but I don't see what the confusion is.

Imagine replacing the lamp by a capacitor, then we don't get a steady current but we get some charge on the plates of the capacitor.

Abcd

@JohnRennie I wanted to know about the $F_e$ and $F_b$ that the author talks about.

John Rennie

That's what I'm getting at.

When it's open circuit the cell behaves like a capacitor so we get a charge separation just like on a capacitor. When there's a charge separation there is always an associated electric field generated by that charge separation. That field creates a force on the electrons in the cell, and that field creates the force $F_e$.

Abcd

09:33

ok

John Rennie

What your physics book is saying is that the cell reaction creates a force (sort of) $F_b$ and the charge separation creates an opposing force $F_e$. The electrons flow in the cell until these two forces balance out. At that point the net force on any electrons in the cell is zero.

Remember that this applies when the cell is open circuit i.e. not connected to a load.

Abcd

ok

John Rennie

Your physics book is taking a high level view i.e. it's saying let's not worry about the details of the cell reaction and how capacitance arises from the cell geometry. Just assume they exist and the forces created by the two effects balance out.

Abcd

ok

John Rennie

Actually all this is also relevant to what it means for equilibrium in the cell.

I don't know if you're ready to move on to discussing equilibrium ...

Abcd

09:44

yes

@JohnRennie Just one thing: Why has he taken Mn^2+ , MnO4, H2O's concentrations as 1 M in second step??

John Rennie

Because the question says: assume other species have no change in concentration

Abcd

Okay its E^o thats why initally all of them were 1 M

@JohnRennie Equilibrium now.

John Rennie

@Abcd if the cell is in equilibrium then that means it doesn't change with time. That means no charge is flowing.

And the only way no charge can flow is if there is no load connected to the cell.

So the only way a cell can be in equilibrium is if it is open circuit.

Abcd

@JohnRennie huh? no

My book says $\Delta G = 0$ in equilibrium condition of cell.

John Rennie

@Abcd really?

Abcd

09:52

@JohnRennie I also heard that cell dies when it is in equilibrium

John Rennie

@Abcd well, yes, a fully discharged cell is also at equilibrium, but that's not a very interesting case.

Abcd

@JohnRennie Okay, fine. Thanks for today's help. Most of my concepts are clear now!

John Rennie

Cool :-)

Abcd

10:28

@JohnRennie Are you there??

John Rennie

@Abcd hi

Abcd

@JohnRennie qr.ae/TUG4o5

Please explain.

Two opposing forces?

John Rennie

It's a somewhat incoherent post. I can't tell if he is talking about a cell that is exhausted of a cell that is open circuit.

Abcd

@JohnRennie exhausted

John Rennie

Well an exhausted cell is certainly at equilibrium ...

Why that is interesting escapes me

Abcd

10:32

@JohnRennie How? (Other than the reason that Delta G is 0)

@JohnRennie You mean you are not sure why exhausted cell is in equi. ??

John Rennie

... on the phone

@Abcd I know why exhausted cells are in equilibrium. We talked about that earlier ... let me gind the post.

2 hours ago, by John Rennie

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

2 hours ago, by John Rennie

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

Abcd

@JohnRennie Waittt

John Rennie

The cell is exhausted when $\Delta G = 0$

Abcd

4 mins ago, by Abcd

@JohnRennie How? (Other than the reason that Delta G is 0)

@JohnRennie The brackets

John Rennie

I think we're getting mixed up.

If a cell is open circuit then by definition nothing in the cell changes with time. This is true no matter whether the cell is fully discharged or not. Yes?

Abcd

10:38

hmm

John Rennie

This is the situation that your physics book was talking about, and that's what I was talking about when I was going on about charge separation and capacitances.

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