Final state of concentration cell?

In summary, it is possible to measure the voltages between two voltmeters if each voltmeter is connected to one compartment of a concentration cell. If one of the voltmeters is not connected to a compartment of the cell, then the voltages between the voltmeters must be lower than the voltages between the two vats if the cell is in equilibrium.
  • #1
somasimple
Gold Member
766
5
Hi,
here is an example of a concentration cell
http://www.jce.divched.org/jcedlib/qbank/collection/conceptests/echem.html [Broken]
1/ The final state is assumed when concentrations are equal in each compartment, right?
2/ Graphic evolution of the voltage is a decreasing curve, right?
 
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  • #2
Yes2.
 
  • #3
1) technically, it's when the activities are the same. If the ionic atmospheres in each half cell are different, activity of the ion in question is modified and the final state may have somewhat different concentrations in each half cell (assuming copper electrodes in both for example)
2) applying ohm's law, current will be linearly related to voltage in the cell assuming constant load. From the nernst equation the voltage is proportional to the log of one concentration divided by the other. For initial half cell concentrations C1 and C2, the electric current will add the stoichiometric amount of ions to one cell solution and subtract an equal amount from the other. cell voltage ~ log((C1-integral(current, time 0 to t))/(C2+integral(current, time 0 to t)))

I wrote a short loop to see what the functional dependence of current is (setting resistance to 1 ohm and neglecting faraday constant etc) and it turns out the functional dependence is an exponential decay, which is not surprising.

For[t = 0, t < 200, t++,
v[[t + 1]] =
Log[(1 - Total[Take[v,t]*.005])/(2 + Total[Take[v,t]*.005])]]

1 and 2 were taken to be the initial half cell concentrations and the Total[] is just an integral over time of how many ions have been removed or added to solution. Mathematica code btw.
 
  • #4
Thanks you²

Another question:
Let's take 3 vats:
1/ The first (A) contains a copper sulfate solute at 0.1 M and a copper electrode.
2/ The second(B) contains a copper sulfate solute at 0.1 M and a copper electrode.
3/ The third (C) contains a copper sulfate solute at 1 M and a copper electrode.

They are connected by salt bridges A to C and B to C.
And with voltmeters by A to C and a second voltmeter B to C.

Is the voltage of each voltmeter a fraction (1/2) of the equilibrium voltage of my first example (the voltage is divided by the number of vats that contain the lower concentrations)?
How is it possible to rearrange the Nersnt equation?
 
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  • #5
No chemistry here, this is a simple circuit question.

attachment.php?attachmentid=60085&stc=1&d=1373019733.png


Note: could be I got signs wrong (battery symbols may need to be reversed), but it doesn't matter much.
 

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  • #6
I understand the diagram but I seem to fail why it is not chemistry:
Compartment C is the one where is the chemical reduction, right?
Compartment A and B have chemical oxidation, right?

Note: B is not connected to A (as in your picture)
 

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  • #7
You're right it doesn't matter at all.
It may matter if the experiment is modified like the picture?
Electroneutrality must be balanced between C //A and B (because electrons flow to C from A and B)?
 

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  • #8
Sorry, I mislabeled the points.

attachment.php?attachmentid=60088&stc=1&d=1373028373.png


Compartment C is the one where is the chemical reduction, right?
Compartment A and B have chemical oxidation, right?

No much other choice - you need to reduce concentration of 1M solution and increase concentrations of 0.1M solutions to reach the equilibrium.
 

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  • #9
So, if we remove the salt bridges, there is an electroneutrality violation that stops very quickly the chemical reactions, and, equalization of concentrations is made impossible?
Since 2 vats (A, B) give 1 electron to a single vat (C) that must react to this violation but can't compensate it, the voltage measured between AC or BC must be lower than a quite same situation where B is removed (like the picture below) ?
 

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1. What is a final state of concentration cell?

A final state of concentration cell refers to the end state of a concentration cell where the concentrations of the two electrodes are equal. This results in a zero potential difference between the two electrodes and the cell reaches equilibrium.

2. How does a concentration cell work?

A concentration cell works by utilizing the difference in concentration between two electrodes to generate an electric potential. This potential difference is then used to power a reaction or an electrical device.

3. What factors affect the final state of concentration cell?

The final state of concentration cell can be affected by factors such as temperature, concentration gradient, and the nature of the electrolyte solution. These factors can alter the rate of diffusion and ultimately impact the final state of the cell.

4. What is the purpose of a concentration cell?

The purpose of a concentration cell is to convert chemical energy into electrical energy. They can be used as batteries or fuel cells to power devices and can also be used in electrochemical experiments to study the effects of concentration gradients on potential difference.

5. How can the final state of concentration cell be predicted?

The final state of concentration cell can be predicted by using the Nernst equation, which takes into account the concentration gradient, temperature, and the nature of the electrolyte solution. This equation can be used to calculate the potential difference between the two electrodes at equilibrium.

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