posted by amy .
summarize part a ( the data i showed you) in a data table containing heading: anode, cathode, overall cell reaction, and cell potential . for each cell combination, write the anode half reaction, the cathode half reaction, the overall reaction in the cell notation ( the concentration of all solutions is 0.1M) and the potential in volts
CuSO4(cathode)and Al2(SO4)3 (anode)--[0.050V]
cuSO4(anode) and ZnSO4(cathode)-- [0.048 V]
MgSO4(anode)and ZnSO4(cathode)--[0.284 V]
ZnSO4(anode) and AlSO4(cathode)-- [-0.056V]
Al2(SO4)3(cathode) and MgSO4(anode)--[0.966 V]
I get the anode and cathode part but i dunno what the rest really means and what im supposed to show
I DON KNOW
I am assuming that the second one you have is actually a copper electrode dipping into CuSO4 soln and an Al electrode dipping into a ZnSO4 soln. If not, then I don't know what the problem is talking about either. For the Cu/CuSO4 and Zn/ZnSO4, this is the cell.
anode is Zn, cathode is Cu
cell reaction is
Zn + Cu^+2 ==> Cu + Zn^+2 but written is cell notation is as above
Anode half reaction is
Zn ==> Zn^+2 + 2e E = +0.762
cathode half reaction is
Cu^+2 + 2e ==> Cu E = 0.337
Ecell = 0.762 + 0.337 = 1.099 (This is the cell potential that is calculated; what you measured as 0.048 v I don't know.)
It is interesting to me that the second and the last equation are the same but reversed. Yet the voltages are not reversed. I suspect you made a typo in one or the other of them.
I can't look at that table anymore i finally finished with it. Thank you. How do i create a reduction potential table with cells containing copper for part a. My question says assume the reduction potential for Cu^2+ + 2e- -->Cu is 0.00 V and calculate the remaining reduction potentials relative to Cu^2+/ Cu. List in order of decreasing reduction potential.
Is there an equation i could use to plug and chug?
You have only two Cu cells. One is Zn/Cu and the other is Al/Cu. Are you supposed to use calculated values from the Nernst equation or are you to use the measured potentials that you have? I will assume you are to use the measured values
The easiest way to see it it to write the half cells and the cell reaction. For example, the Zn/Cu cell would be as follows:
Zn ==> Zn^^+2 + 2e Eo = ??
Cu^+2 + 2e ==> Cu Eo = 0.00
cell reaction is the sum of the two
Zn + Cu^+2 ==> Zn^+2 + Cu Eocell = 0.048
If the cell voltage is 0.048 and Cu is 0.00, then Eo for Zn ==>Zn^+2 + 2e must be 0.048. Al/Cu is done the same way.
Writing as oxidations,
Zn ==> Zn^+2 + 2e Eo = 0.048
Al ==> Al^+3 + 3e Eo = 0.050
The more positive number has the greater tendency to occur; therefore, Al would be on top and Zn last ranked in decreasing order of oxidations. The reverse would be true if you wrote the reactions in order of reductions; i.e.,
Al+3 + 3e ==> Al -0.050
Zn+2 + 2e ==> Zn -0.048