How do you calculate the theoretical cell voltage?

1) Zn|Zn^+2 | Cu|Cu^+2
2) Cu|Cu^+2 | Zn|Zn^+2
3) Fe|Fe^+2 | Cu|Cu^+2
4) Zn|Zn^+2 | Cu|Cu^+2
5) Cu|Cu^+2 | Cu|Cu^+2

http://www.ausetute.com.au/calcelemf.html

but for example for question 4 Zn is 1 M and Cu is 0.1 M will that make a difference? and can you just do one example I want to see if I'm on the right track. Thanks.

To calculate the theoretical cell voltage, you need to know the standard reduction potentials of the half-reactions involved in the cell. The cell voltage can be calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode. Here are the steps to calculate the theoretical cell voltage for each of the given cell configurations:

1) Zn|Zn^+2 | Cu|Cu^+2:
- Look up the reduction potentials for the half-reactions: Zn^2+ + 2e^- → Zn and Cu^2+ + 2e^- → Cu.
- Subtract the reduction potential of the anode (Zn) from the reduction potential of the cathode (Cu): (E° Cu) - (E° Zn).

2) Cu|Cu^+2 | Zn|Zn^+2:
- Look up the reduction potentials for the half-reactions: Cu^2+ + 2e^- → Cu and Zn^2+ + 2e^- → Zn.
- Subtract the reduction potential of the anode (Zn) from the reduction potential of the cathode (Cu): (E° Cu) - (E° Zn).

3) Fe|Fe^+2 | Cu|Cu^+2:
- Look up the reduction potentials for the half-reactions: Fe^2+ + 2e^- → Fe and Cu^2+ + 2e^- → Cu.
- Subtract the reduction potential of the anode (Fe) from the reduction potential of the cathode (Cu): (E° Cu) - (E° Fe).

4) Zn|Zn^+2 | Cu|Cu^+2:
- Same as case 1).

5) Cu|Cu^+2 | Cu|Cu^+2:
- Since the half-reactions involve the same species, Cu^2+ + 2e^- → Cu, the reduction potentials are the same for both electrodes.
- In this case, the cell voltage will be zero, as there is no difference between the reduction potentials.

By following these steps, you can calculate the theoretical cell voltage for each of the given cell configurations.