Voltage drop of galvanic cell

[chiddler]

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How is this proven? I would have thunk that it would be an exponential drop (like a y = 1/x graph) since a galvanic cell uses most of its energy at the beginning.

thank you.

 

[Buttafuoco]

It's a graph of the Nernst equation basically, right? Like a graph of y = (1-X) / (X). Does seem like it should be different, but maybe there's something more subtle.

 

[SaintJude]

Does this represent a galvanic cell without a salt bridge?

Edit: Lol, sorry chiddler--you posted a question expecting an answer and all you get is question back. My bad.

 

[chiddler]

Does this represent a galvanic cell without a salt bridge?

Edit: Lol, sorry chiddler--you posted a question expecting an answer and all you get is question back. My bad.

if i understand correctly, a salt bridge is necessary to exchange ions and keep the reaction from accumulating charge in one of the cells. yes?

in other words, a galvanic cell cannot proceed for very long without a salt bridge.

so it should be with a bridge.

 

[Buttafuoco]

Okay on second thought, I'm a dummy. I totally forgot that the Nernst Equation is a log function. And since it has the negative part, you should totally expect it to look like a negative log equation.

 

[chiddler]

A negative log equation is similar to a 1/x graph, yes?

click

that still doesn't really answer my question, anyway.

Why is it that the graph looks like what I drew given that most energy in the cell is consumed initially?

 

[SaintJude]

This is what a negative log function graph.

This is what a natural log function looks like:

edit: This indeed I think represent the negative log sign of a nernst equation

 

[SaintJude]

Actually chiddler, when I look at a y= 1/x graph, it strikes me as a less appropriate trend b/c it suggest that the voltage would drop really quickly in a really short amount of time. Then the voltage would supposedly hover at a low point over a really long duration. Doesn't fit with the profile of a galvanic cell. What do you think?

edit: so it look like that the picture you drew does represent the negative log portion of the Nernst equation.

 

[chiddler]

i'm sure you're correct, but i'm more concerned with my question than the details of the graph lol

after all, the graphs given on "which graph?" questions are usually radically different.

i don't mean to be unappreciative though. thank you.

but if you're asking me generally, then according to nernst equation, it should be the log graph. so you're right.

 

[milski]

What you have drawn is the correct graph. Keep in mind that you are plotting something like y=A-ln(x/(1-x)), not y=A-ln(x)

1/x and -ln(x) look somewhat similar but have one very distinct and different characteristic. 1/x is asymptotic towards 0 and will never cross below it. Anything 1/x like will have a horizontal asymptote and will stay above/below that line. -ln(x) is not limited in any way in the negative direction.

ln(x/(1-x)) is undefined for x>1 (or whatever constant we use to scale).

 

[chiddler]

What you have drawn is the correct graph. Keep in mind that you are plotting something like y=A-ln(x/(1-x)), not y=A-ln(x)

1/x and -ln(x) look somewhat similar but have one very distinct and different characteristic. 1/x is asymptotic towards 0 and will never cross below it. Anything 1/x like will have a horizontal asymptote and will stay above/below that line. -ln(x) is not limited in any way in the negative direction.

ln(x/(1-x)) is undefined for x>1 (or whatever constant we use to scale).

yes, my question is why?

specifically

Why is it that the graph looks like what I drew given that most energy in the cell is consumed initially?

 

[Left2right]

Op, I'm not sure what you mean by galvanic cells using up most of its energy in the beginning.

As far as I understand it, galvanic cells maintain its voltage for the majority of its reaction until the relative end. So I believe your original graph is accurate. It shouldn't look like a 1/x graph because at the potential we are dealing with in galvanic cells, the standard potential (E naught) is relatively large compared to the value in the portion of the equation after the negative sign. So even if the concentration of reactants decrease a lot while the products increase in a galvanic cell, the potential only lowers by a little bit compared to the original standard potential. The negative term only makes a significant difference when the ratio of products to reactants are dramatically different.

If that was confusing, think of batteries. They power the electronics for the majority of its life until the very end, when it dies very quickly. Notice how flashlights are powered for so long until it starts dimming towards the end and is dead soon after.

Does this help at all?

 

[chiddler]

Op, I'm not sure what you mean by galvanic cells using up most of its energy in the beginning.

As far as I understand it, galvanic cells maintain its voltage for the majority of its reaction until the relative end. So I believe your original graph is accurate. It shouldn't look like a 1/x graph because at the potential we are dealing with in galvanic cells, the standard potential (E naught) is relatively large compared to the value in the portion of the equation after the negative sign. So even if the concentration of reactants decrease a lot while the products increase in a galvanic cell, the potential only lowers by a little bit compared to the original standard potential. The negative term only makes a significant difference when the ratio of products to reactants are dramatically different.

If that was confusing, think of batteries. They power the electronics for the majority of its life until the very end, when it dies very quickly. Notice how flashlights are powered for so long until it starts dimming towards the end and is dead soon after.

Does this help at all?

now that I look at this again, it seems that this is a case of "best answer"-ness. It's right, but not a great description of what is going on. plus the other answers are bad.

the original question was

Q: "When the cell discharges, it will?"
A: "Discharge the most energy initially because it has the highest voltage at the start."

thanks very much.

 

[milski]

The way that the cell will "discharge" is not only a property of the cell - it will depend on what and how you connect to it. In the case where the current is constant, the concentrations will be decreasing linearly and you can use the the formula/graph that I posted (which is pretty much Nernst Equation. Constant current will imply some circuit fancier than just a plane resistor. In the case of a resistor (constant R), you should be able to treat it as a reaction of first order - the area of the electrodes is constant so increasing the concentration increases the likelyhood of an electron transfer. In that case you can replace x with ln(x). The graph in that case looks about the same.

 

[chiddler]

The way that the cell will "discharge" is not only a property of the cell - it will depend on what and how you connect to it. In the case where the current is constant, the concentrations will be decreasing linearly and you can use the the formula/graph that I posted (which is pretty much Nernst Equation. Constant current will imply some circuit fancier than just a plane resistor. In the case of a resistor (constant R), you should be able to treat it as a reaction of first order - the area of the electrodes is constant so increasing the concentration increases the likelyhood of an electron transfer. In that case you can replace x with ln(x). The graph in that case looks about the same.

You meant in the case of a resistor AND constant current, right? And therefore we can treat it as a 1st order reaction because it is decreasing linearly, not as the usual nernst log relation. Thus the replacing x with ln(x).

What's the relevance to my question, though? Are you saying that it depends on the particular situation?

 

[milski]

You meant in the case of a resistor AND constant current, right? And therefore we can treat it as a 1st order reaction because it is decreasing linearly, not as the usual nernst log relation. Thus the replacing x with ln(x).

What's the relevance to my question, though? Are you saying that it depends on the particular situation?

You won't get a constant current for a resistor, since the voltage is continuously dropping.

I am saying if that you can control the resistance between the two electrodes and in that way you can discharge it in any way you find suitable.

For constant current it will actually be 0th order - you'll have a fixed Δe/Δt. That gives you x and 1-x for the corresponding concentrations (with some factors in front of 1 and x - these are just details that won't change the shape of the curve). Plug that in Nernst and you get the first graph I posted.

Constant current is not very realistic since it requires constant change of resistance. If you just put something like a light bulb there, you'll get continuously decreasing voltage and something resembling 1st order reaction. Thus ln(x) instead of x.

It has been a couple of days since we talked about this so I'm not 100% sure what I had in mind at the time. 😀

 

[chiddler]

You won't get a constant current for a resistor, since the voltage is continuously dropping.

I am saying if that you can control the resistance between the two electrodes and in that way you can discharge it in any way you find suitable.

For constant current it will actually be 0th order - you'll have a fixed Δe/Δt. That gives you x and 1-x for the corresponding concentrations (with some factors in front of 1 and x - these are just details that won't change the shape of the curve). Plug that in Nernst and you get the first graph I posted.

Constant current is not very realistic since it requires constant change of resistance. If you just put something like a light bulb there, you'll get continuously decreasing voltage and something resembling 1st order reaction. Thus ln(x) instead of x.

given my error of saying constant current, because voltage should change as cell discharges, I understand what you mean.

It has been a couple of days since we talked about this so I'm not 100% sure what I had in mind at the time. 😀

lol

thank you

 

[Lunasly]

Can someone further explain why the graph looks the way it does?

Right now it is to my understanding that cell potential (or voltage) is measures in Joules per coulomb.

1V = J/C

Current is measured in Amps, where 1 Amp is the number of coulombs that are moving per second.

1 Amp = C/s

So if we rearrange the units, we get:

C = Amp x s

Therefore...

1V = J/(Amp x s).

Now I don't know how J changes over time, but I assume amps stays constant and we have a inverse relationship. As time (seconds) increases, the voltage decreases. This is why I am stuck between the 1/x graph and the graph the OP posted.

Is it, as other have said, because the voltage does not decreases so quickly and it more spread out that the best graph is the one the OP posted? If that's the case, is the formula I proposed above incorrect?

Thanks,
Lunasly.