How do you calculate the theoretical cell voltage?
1) Zn|Zn^+2 (1M)| Cu|Cu^+2 (1M)
2) Cu|Cu^+2 (1M)| Zn|Zn^+2 (1M)
3) Fe|Fe^+2 (1M)| Cu|Cu^+2 (1M)
4) Zn|Zn^+2 (1M)| Cu|Cu^+2 (0.1M)
5) Cu|Cu^+2 (1M)| Cu|Cu^+2 (0.1 M)
Now 4 and 5 have different moles how do you calculate those ones? And please can you do one I want to see if I'm on the right track, thank you so much.
Rather than us typing up one or more of these to let you check your work why don't you show how you've done one or two and let us critique your work?
I think the easy way to solve those with different concns is to solve for Eo cell, then use Ecell = Eocell -(0.059/n)*log Q. Log Q is where the different concns go and Ecell comes from that.
To calculate the theoretical cell voltage for a given cell, you need to use the standard electrode potentials (also known as reduction potentials) for the half-reactions involved in the cell. These potentials can be found in tables or reference books.
The cell voltage (Ecell) is determined by subtracting the reduction potential of the anode (Eanode) from the reduction potential of the cathode (Ecathode):
Ecell = Ecathode - Eanode
Here's how you can calculate the theoretical cell voltage for the given cells:
1) Zn|Zn^+2 (1M)| Cu|Cu^+2 (1M):
The reduction potential for Zn^+2 and Cu^+2 can be found in the table. Let's assume the values are +0.76 V and +0.34 V, respectively.
Ecell = (+0.34 V) - (+0.76 V) = -0.42 V
2) Cu|Cu^+2 (1M)| Zn|Zn^+2 (1M):
The reduction potentials for Cu^+2 and Zn^+2 remain the same as in the previous case, but the anode and cathode are reversed, so we interchange the values.
Ecell = (+0.76 V) - (+0.34 V) = +0.42 V
3) Fe|Fe^+2 (1M)| Cu|Cu^+2 (1M):
Similarly, suppose the reduction potentials for Fe^+2 and Cu^+2 are +0.44 V and +0.34 V.
Ecell = (+0.34 V) - (+0.44 V) = -0.10 V
4) Zn|Zn^+2 (1M)| Cu|Cu^+2 (0.1M):
When the concentrations of the species involved in the cell reaction are different, the Nernst equation needs to be used to account for the concentration effect. This equation is given by:
Ecell = Ecathode - Eanode + (0.0592V/n) * log([Cu^+2]/[Zn^+2])
Where [Cu^+2] and [Zn^+2] are the concentrations of Cu^+2 and Zn^+2, respectively, and n is the number of electrons transferred in the balanced cell reaction.
5) Cu|Cu^+2 (1M)| Cu|Cu^+2 (0.1 M):
In this case, both the anode and cathode have the same species (Cu|Cu^+2), which means there is no net electron transfer occurring. Therefore, the cell voltage is essentially zero.
Now, let's do an example to check if you're on the right track:
Example: Ag|Ag^+1 (0.1 M)| Cu|Cu^+2 (1M)
Assuming the reduction potentials for Ag^+1 and Cu^+2 are +0.80 V and +0.34 V, respectively.
Ecell = (+0.34 V) - (+0.80 V) = -0.46 V
I hope this explanation helps!