I don't know what you are using for Eo. Texts vary so the voltage calculated will vary. Here is how to do the first one using my values. You should substitute values from your text.
Zn ==> Zn^+2 + 2e Eo = 0.762 as written.
Cu^+2 + 2e ==> Cu Eo = 0.345 as written.
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Zn + Cu^2+ ==> Cu + Zn^2+ Eocell = 0.762 + 0.345 = 1.107 v calculated. Since the concns are 1 M we know Ecell is the same as Eocell; therefore, Ecell = 1.107
Note that I added an oxidn half rxn to a redn half rxn to arrive at the cell rxn; therefore, I added the oxidn half potential to the redn half potential to arrive at the cell potential.
#2 is -1.107v.
#3.
Fe^2+ ==> Fe^3+ + e Eo = -.771
Cu^2+ + 2e ==> Cu Eo = 0.345
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2Fe^2+ + Cu^2+ ==> 2Fe^3+ + Cu Eocell = -0.771 + 0.345 = -1.533 v and this cell is not spontaneous. As in #1 and #2, this is also at 1M concn for both; therefore, Ecell = Eocell.
By the way, I wonder why you have written the cell notation as you have. For example, I would have written #1 as
Zn|Zn^+(1M)||Cu^2+|Cu.
#4 is not 1M so I will do it as I described before.
Zn ==> Zn^2+ + 2e Eo = 0.762 as written
Cu^2+ + 2e ==> Cu Eo= -0.345 as written
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Zn + Cu^2+ ==> Zn ^2+ + Cu Eocell = 1.107v. Then we correct for the fact that Cu^2+ is 0.1M and not 1 M.
Ecell = Eocell -(0.0592/2)log Q
log Q = (Zn^2+)(Cu)/(Zn)(Cu^2+)
Substitute Zn^2+ = 1M
(Cu)(s) = 1--that's the standard state for SOLID copper and it is 1 by definition.
(Cu^2+) = 0.1M
(Zn)(s) = 1 by definition.
log Q then is log (1*1/1*0.1 = log10 = 1
Ecell = 1.107 - 0.0296(1) = 1.077 v.
the measured cell voltages are supposed to be smaller than the theoretical cell voltages, but when I'm calculating the theoretical cell voltages they are smaller. And I don't know why I'm getting these. Can you please check if I'm doing it right. Thank you.
Measured Cell voltages
1) Zn|Zn^+2 (1M)| Cu|Cu^+2 (1M): 0.947
2) Cu|Cu^+2 (1M)| Zn|Zn^+2 (1M):-0.947
3) Fe|Fe^+2 (1M)| Cu|Cu^+2 (1M):0.703
4) Zn|Zn^+2 (1M)| Cu|Cu^+2 (0.1M):0.906
5) Cu|Cu^+2 (1M)| Cu|Cu^+2 (0.1 M):-0.0143
Theoretical Cell Voltage( my calculataions)
1) 0.42
2) -0.42
3)-0.11
4)-0.450
5)0.31
#5 you did it but the rest I'm not sure.
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