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TERIALS: Solid metals Zn Cu; Fe, Mg, AlSolutions (all M) - KNOs, Zn(NO3)z, CuSO4, FeSO-, MgSO4, Al?Glassware Beakers; U-shaped glass tubesEquipment - PotentiometerV...

Question

TERIALS: Solid metals Zn Cu; Fe, Mg, AlSolutions (all M) - KNOs, Zn(NO3)z, CuSO4, FeSO-, MgSO4, Al?Glassware Beakers; U-shaped glass tubesEquipment - PotentiometerVue passie ateucareretesesletatelreneletettat 3 5 Mlolttel aa ptuen (that requires Iess than 2 Q or less than 2 0 Vato SSaetonicOCEDURE: Design and draw at least three Galvanie cells based on the material provided above:One Galvanic cell you design should have highest possible voltageAnother Galvanic cell you design should have lowest

TERIALS: Solid metals Zn Cu; Fe, Mg, Al Solutions (all M) - KNOs, Zn(NO3)z, CuSO4, FeSO-, MgSO4, Al? Glassware Beakers; U-shaped glass tubes Equipment - Potentiometer Vue passie ateucareretesesletatelreneletettat 3 5 Mlolttel aa ptuen (that requires Iess than 2 Q or less than 2 0 Vato S Saetonic OCEDURE: Design and draw at least three Galvanie cells based on the material provided above: One Galvanic cell you design should have highest possible voltage Another Galvanic cell you design should have lowest possible voltage The last Galvanic cell can be the combination of your choice



Answers

You want to set up a series of voltaic cells with specific cell voltages. The $\mathrm{Ag}^{+}(\mathrm{aq}, 1 \mathrm{M}) | \mathrm{Ag}(\mathrm{s})$ half-cell is one of the compartments. Identify several half-cells that you could use so that the cell voltage will be close to (a) $1.7 \mathrm{V}$ and (b) $0.5 \mathrm{V} .$ Consider cells in which silver can be either the cathode or the anode.

To build an electrochemical cell with the chemicals that are provided, We simply need to construct both a cathode compartment and an an odd compartment. So we would likely have two beakers and then something connecting the two beakers that could hold the potassium nitrate solution serving as a salt bridge. If you can't find in your laboratory, two beakers that could be connected, say, with a membrane, so that the potassium nitrate solution doesn't leak into both sides. Sometimes what people will do is just drape a paper towel Between the two beakers and have the paper towel saturated with potassium nitrate. You'll then add silver nitrate to one side with a silver electrode zinc nitrate to the other side with a zinc electron. You'll then look at the activity series and see that zinc is more active than the silver because it is more active. It will be the one that's oxidized. Whatever is oxidized, serve as as the anodes. The other electrode will then serve as the cathode and reduction will occur there

So next we're looking at electrons, electrolysis, electron, self potential and so on and so forth. So the first thing we're looking at is part A. So this is the 1.1 vault section so that we use the copper as the cathode to oxidize ze and xe and two plus that occurs at the an ode. So the cell potentially is 1.1. So we have the negative. An ode that is the oxidation. So that is the CNN Susie and two plus like we said. And then we have the positive CASS owed, which is C U two plus have two electrons generates. See you. So on that equation here is as follows C n I see you two plus is in equilibrium with C and two plus at C U. So you can see that we've got the electron transfer from sink to cover two plus. So I'm moving on here. R s l is not 0.337 Subtract negative. No 0.763 You sell us 1.11 volts. So in the second part, we're looking for 9.5 boats. So we have C N C N four to minus, which is are an ode, and we use this to reduce ZN two plus two c n at the cathode. So the cell potential is not 20.5. So are honored. We have CNN at four, see and miners that is an equilibrium with CNN C N for two miners on two electrons and then at the cathode we have C and two plus two electrons generates C n so on. That equation is as follows. So we have CNN two plus at four C and minus generates C n CNN for tu minus. Where are S O is not 0.763 volts. Subtract negative 1.26 volts. Eso is not 0.5. Remember? That is in vaults.

Electoral chemistry, where we essentially focused on where electrons are moving from and to in reactions. So we've got a few parts to take a look at hits when our first part we have copper is our cathode in the C N C u voltaic cells. So no cell is possible in which cobalt acts as the an ode. Next part we have silver zinc combination of half cells generates our highest voltage. So we have C n solid state c n a queer state. That's one half of our cell may have a G plus acquiesce state, a g solid state. So the oxidation because our a node that involves all sink on the reduction occurs that are cathode that involves are silver so e no cell value. We have no 0.799 volts. Subtract negative not 0.763 volts. We have a total value off 1.562 Therefore, we have a positive, you know, value. So we will favor our products here. So silver copper combination of Hafsa will generate the lowest voltage. On the other hand, so we have the oxidation at the an ode. So you have See you goes to see you two plus had two electrons. Reduction occurs at the CASS owed to a G plus. I had an electron goes toe a G. You haven't e South off 9.799 Subtract 9.337 volts. We have a value off no 0.462 volts.

Blue. They were going to be looking at this equation, not equation, really. This half reaction for silver and we're gonna be determining what, um, like, what's cell potentials weaken Generate Based off of this half. So? So We know. First of all, electrons being on the left that this is Reducto. All right, so we have two different potential Voltage is we want to reach. That's about zero point. Sorry. About 1.7 worlds. My mistake. 1.7 volts about and about, um, zero planned 50 No. Each of these as two ways we can actually reach that. My lines or not working with me. Good. Good one. There. Okay, so each of these can also be done with, um with Silver as the an ode. All right. In which case, it's an oxidation reaction, so e not equals negative zero points, uh, 994 or silver as the cathode, In which case it is a reduction reaction or half reaction difficulties. And then our e not is gonna be positive. 0.799 So, in going for 1.7, if we have a negative 0.7994 will need about estimating this. A 0.8 or so. We'll need about 2.5 volts from our reduction reaction. If we're going for only 0.5, then we'll need about 1.3 volts from our reduction. React now, um r cathode. This one has a positive sign. So for going for 1.7 here will need about 1.0 volts from our oxidation reaction. This important, right? So we're gonna have to reverse it. Basically, we'll be looking for a negative 1.0 volts reduction. And over here for 0.5 will need about a negative 0.3 volts oxidation, which, for our case, means a positive 0.3 volts reduction. So now, really simple. All we gotta do is dig through, um, the reduction e knock tables and find our values. And I will be back in a second with that. Okay, So, um, while there are multiple different answers you could use thes are the ones that I went with. They might be the ones that you would find in any official record source. So, um, I didn't really find anything for silver as an analog for this one But silver is a cathode. For 1.7 volts, you can use chromium. Your total cell will have a 1.71 voltage for silver. As an analog for a 0.5 wolves, you can use the reduction of chlorine that will give you a 0.56 cell. And for silver as a cathode. For 0.5 holds, you can use the reduction of copper, which will give you a net 0.46 full itself. Thank you, very.


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