- Electrochemistry questions
- Redox reaction from dissolving zinc in copper sulfate
- Introduction to galvanic/voltaic cells
- Electrodes and voltage of Galvanic cell
- Shorthand notation for galvanic/voltaic cells
- Free energy and cell potential
- Standard reduction potentials
- Voltage as an intensive property
- Using reduction potentials
- Spontaneity and redox reactions
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential edited
- Nernst equation
- Using the Nernst equation
- Concentration cell
- Introduction to electrolysis
- Quantitative electrolysis
- Electrolysis of molten sodium chloride edited
- Lead storage battery
- Nickel-cadmium battery
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- Is this concept entropy driven?(4 votes)
- 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. :)(3 votes)
- I don't get it that why do we need to make concentrations equal to produce voltage?And at8:15as 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.(3 votes)
- 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.(2 votes)
- 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?(1 vote)
- 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)
- 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?(1 vote)
- A concentration cell is a cell that has the same electrodes on both sides. So here we have zinc electrode on the left, and zinc electrode on the right. The only difference is the concentration. On the left side there's a .10 molar solution of zinc sulfate. On the right side, there's a 1.0 molar solution of zinc sulfate. So the left side is the less concentrated side, and the right side is the more concentrated side. There's a tendency for the concentrations to be equalized, and that's enough to create a small voltage. So how can we make these concentrations more equal? Let's focus on the less concentrated side first. The less concentrated side needs to get more concentrated. So it can do that by increasing the concentration of zinc to plus ions in solution. So how can we increase the concentration of zinc two plus ions? Well, if solid zinc turned into zinc two plus ions, that increases the concentration. Solid zinc turning into zinc two plus is oxidation, so oxidation occurs on the less concentrated side. So let's write that down here. So we have solid zinc turning into zinc two plus. I'm going to write .10 molar to distinguish this from the other side, plus 2 electrons. So we lose two electrons. Solid zinc loses two electrons to turn into zinc two plus. Those two electrons move in our wire here, and we create a current. Now let's think about the more concentrated side. The more concentrated side needs to decrease its concentration, so it needs to decrease the concentration of zinc two plus ions in solution. It can do that if zinc two plus ions come out of solution, so if they gain electrons to form solid zinc. So that's a reduction. ^So reduction occurs on the more concentrated sides. ^Let's write that, reduction right here. ^So this would be zinc two plus ions. ^I'll write 1.0 molar concentration once again to ^distinguish it from the other one, so this would be ^gaining two electrons to form solid zinc. ^So overall, what is happening overall here? ^So let's draw a line, so we have solid zinc ^on both sides. We can cancel that out. ^We have two electrons on both sides. ^So on the left side we would have zinc two plus, ^zinc two plus at an initial concentration of ^1.0 molar, and this is going to zinc two plus at .10 molar. ^So this is zinc two plus at .10 molar. ^How do we find the voltage of our concentration cell? Remember from the last few videos that the nernst equation allows us to calculate the potential of the cell. So let's get some more room down here. Let's write down the nernst equation. Cell potential, which is what we're trying to find, E, is equal to the standard cell potential E zero minus .0592 over the number of moles of electrons transferred, which is N, times the log of Q. So this is one form of the nernst equation from the last few videos. Let's think about Q. So what would Q be for our concentration cell? So Q would be equal to the concentration of zinc two plus, This would be the conentration of zinc two plus, on the less concentrated side. So this is the concentration on the less concentrated side, over the concentration of zinc two plus on the more concentrated side, so over the concentration on the more concentrated side. So right now, that would be .10, right now that's point .10, over 1.0. So .10 over 1.0, so this is what Q is equal to. Next, let's think about the standard cell potential. So the standard cell potential, E zero. What's the standard cell potential here? Well, remember the standard cell potential is the potential under standard conditions, so one molar of concentration of zinc two plus. So let's write down the reduction half reaction. So zinc two plus, this would be at one molar, so this is a reduction half reaction, so gaining two electrons to give us solid zinc. If you look at a table of standard reduction potentials, the standard reduction potential for this half reaction is negative .76 volts. For the oxidation half reaction, we need to show solid zinc, turning into zinc two plus ions, and this would need to be a one molar concentration of zinc two plus ions, because we're talking about standard cell potentials, standard conditions. This is oxidation, so losing two electrons. The standard oxidation potential would be just the negative of the standard reduction potential. So the standard oxidation potential is positive .76. So we've done this several times in earlier videos. Therefore, the standard cell potential, the standard cell potential would be equal to negative .76 ^plus positive .76 which is equal to zero. ^So the standard cell potential is equal to zero. That makes sense, because under standard conditions, you're starting with the same concentrations, so you shouldn't get a voltage difference. So the standard cell potential is equal to zero. We're going to plug that into here in the nernst equation. Alright, let's go ahead and plug everything in. ^So the cell potential of E is equal to the standard ^cell potential which is equal to zero, ^minus .0592 over N. What is N? We go back up here to remind ourselves that N is equal to two. We're talking about two electrons transferred. ^So we write N is equal to two. ^So let me make sure we keep that nernst equation ^up there, so N is equal to two, times the log of Q, ^and Q is equal to .10 over 1.0, so .10 over 1.0, ^so the cell potential is equal to .0296 volts. ^So that's positive. It's positive, indicating this ^is spontaneous. So this is our instantaneous cell voltage. So when we're talking about these concentrations, when we're talking about these concentrations right here, this is our instantaneous cell potential. So we get a positive, we get a positive voltage. It's small, but it is there. And it's due to the difference in the concentrations. It's due to the difference in the concentrations. What happens when the concentrations approach each other? So as time goes on, Q is going to change. Q is going to change. What happens as the concentrations approach each other? Q should increase. So, Q increases as the concentrations approach each other, and therefore the instantaneous cell potential decreases. So the cell potential decreases here. So, again, we talked about this in the video on using the nernst equation. What happens when the concentrations are equal? Alright, let's go back up here to remind ourselves about Q. So what happens when the concentrations are equal? If the same number up here and here, then Q would be equal to one. so when the concentrations are equal, ^let me go ahead and write that, ^when the concentrations are equal, Q is equal to one. and what happens when Q is equal to 1? W'ed be taking the log of one, and the log of one is equal to zero. ^So let me go ahead and write that down. ^So E would be equal to zero minus .0592 over 2, ^times the log of one. ^And the log of one is zero, so all of this goes to zero, ^and therefore your cell potential, your cell potential ^would now be zero. ^And that makes sense, because the concentrations are equal. ^So there's no longer any tendency for the concentrations ^to be equalized, and so you're no longer producing a voltage.