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# Concentration cell

Concentration cells are relevant when varying zinc sulfate concentrations generate voltage. Oxidation and reduction play key roles in balancing these concentrations. The Nernst equation steps in to calculate cell potential, providing a complete understanding of electrochemistry basics.
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Created by Jay.

## Want to join the conversation?

• Is this concept entropy driven?
• In a way, that is indeed the case! Stoichiometrically, the two half reactions are identical and opposite, however the concentrations are unequal and therefore it is entropically favorable for them to equalize. This is analogous to allowing two systems of different temperatures to interact; due to the tendency for the entropy of the universe to increase, there would a flow of thermal energy from one system to another until their temperatures are equal. Here, electrons and ions are the medium of exchange, rather than heat. :)
• Why is the 1.0M solution of zinc on the right side of the equation, and the 0.10M solution on the left side? Obviously Q would be different if these were switched, so how do we know which side they should go on?
• I don't get it that why do we need to make concentrations equal to produce voltage?And at as the reaction proceeds Zn(0.10) concentration should decrease as it is continuously loosing electrons i.e Q should decrease, then why is zinc's(0.10) concentration increasing?
(1 vote)
• Keep in mind that the Standard redox reaction at equal concentrations of Zn gives 0V. So the fact that we have an imbalance of concentrations of Zn in the solutions means that the solutions want to come to equilibrium, which drives them to produce an electric potential.

That being said:Zn(0.1M) is increasing because there is less of a concentration of Zn in the solution. So Zn comes off the plate and into the solution, which increases the molarity of the solution. In the Zn (1M), since electrons are being transferred via the wire due to the oxidation happening in the other cell, the Zn(1M) wants to precipitate out of the aqueous solution to reduce the Molarity of it until both sides come to the same concentration.
• So is this actually an electromotive force or is it a diffusion force? This concept brings to mind plasma membranes and how diffusion gradients and electrical gradients each have an influence on how an ion travels into/out of a cell.
(1 vote)
• You bring up a good point. This is an example of an electromotive force. Note that electromotive force is worded in a confusing manner because it is actually potential energy at play here.
• He draws a salt bridge in this setup, but a salt bridge would quench the charge differential between the two cells, which is what drives the reaction
(1 vote)
• Is the more concentrated side always reduced for concentration cells?
(1 vote)
• I'm assuming the cathode and anode are the same as a voltaic cell?
(1 vote)
• Why is Q increasing as the concentrations approach each other?
Also, why would E decrease if Q increases?

Sorry for the basic questions, but I would really appreciate any help! Thanks!
(1 vote)
• Q is the reaction quotient, it is the ratio of products over reactants (raised to the power of their coefficients). You start of with a concentration of reactants and you have something happening making a certain amt of products. The reaction proceeds and you get more products and less reactants. more products mean a bigger numerator means a higher Q. It is a way of keeping track of where you are in relation to your equilibrium.
(1 vote)
• How do you determine n? For instance under what conditions would n be more than 2 or less than 2?
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• I have a funny question. If one half cell consists of Zn and 0.1M ZnSO4, the other Cu and 1M CuSO4, what would happen? The zinc copper galvanic cell will make a typical Galvanic cell and generate 1.10V. How about 0.1M ZnSO4 and 1M CuSO4? I understand Zn2+ might not want to equalize concentration with Cu2+, but would 0.1M SO4 2- want to equalize with 1M SO4 2-? In that case, we can make a Galvanic and concentration cell right?
(1 vote)