posted by Anna .
Let me try! Let me try! :)
nickel(II)nitrate and zinc nitrate
-->Ni|Ni^+2 (??M) ||Zn^+2 (??M)|Zn
Iron(II) nitrate and zinc nitrate
-->Fe|Fe^+2 (??M) ||Zn^+2 (??M)|Zn
Silver nitrate and zinc nitrate
-->Ag|Ag^+2 (??M) ||Zn^+2 (??M)|Zn
Is this correct?
Also, how would I find the calulated cell potential?
1. The ??M must be completed when you know what the concentrations of the solutions are. The concentrations may well be in the instructions for performing the lab.
2. I am somewhat confused by your notation. You can't have a cell with nickel(II) nitrate and zinc nitrate. That confused me when you first posted. Now if you place a Ni electrode in the Nickel nitrate solution and a zinc electrode in the zinc nitrate solution, as you show in your cell notation, THEN you have can have a cell. With that understanding, the ones you have done are correct, EXCEPT that you don't always have the anode on the left. You really won't know which way to write them until you perform the experiment and know which is the anode.
3. How do you find the calculated cell potential? When you perform the experiment (or you can look these up if you want to do them before you perform the experiment). As an example, take the Zn|Zn^+2||Cu^+2|Cu cell.
Look up the Zn cell in your set of tables. I'm sure there is one in the back of your text. All are written as standard reduction potentials. The one you will find in your table will be Zn^+2 + 2e = Zn and the voltage will be given as approximately -0.762. That isn't the one you want. You want the Zn ==> Zn^+2 + 2e so you just turn the half equation around (reverse it) and change the sign to +0.762. Easy!!. Almost the same for Cu. You will find Cu^+2 + 2e ==> Cu and the half cell voltage will be given as +0.337. That is the direction you want it to go so you leave it alone. Then just add the two half equations together and add the two voltages together to get the complete cell and the cell potential. "But I don't know which is the anode yet so what do I do," you say. Not to worry. Look up the half cells in your table. You will find them listed this way.
Zn^+2 + 2e ==> Zn E = -0.762 v
Cu^+2 + 2e ==> Cu E = +0.337 v
NOW, and this is important, determine which half cell is the more negative, reverse that one half cell, then add the other one in as is and you will have it.(By the way the texts I have seen give a different set of instructions for this but I like this better. Follow the instructions of your teacher.) The one with the higher + voltage (in this case the Zn electrode) WILL BE THE ANODE, WILL BE THE NEGATIVE ELECTRODE, AND WILL BE THE ELECTRODE WHERE OXIDATION OCCURS and will be the cell set up for it to occur spontaneously.
4. Finally, what if you think you've done everything right and you get this.
Zn^+2 + 2e ==> Zn E = -0.762 volts
Cu ==> Cu^+2 + 2e E = -0.337 volts
Add them and you get
Zn^+2 + Cu ==> Zn + Cu^+2 E = -1.009 v.
If you get this with a negative cell voltage you know it will not go (it must be a positive cell voltage to occur spontaneously) BUT you know it will occur spontaneously if you reverse it. So you just take the eraser out, turn each equation around (reverse each), change the sign of each and you will have it.
I hope this helps