[quill]

The voltaic cell's setup is similar to this picture:


The overall cell reaction is Zn(s) + 2H⁺(aq) --> Zn²⁺(aq) + H₂(g)


The half reactions are:
Zn²⁺(aq) + 2e^- --> Zn(s)    E° = -0.763 V
2H⁺(aq) + 2e^- --> H₂(g)       E° = 0.000 V

What is the effect on the cell's emf for each of the following changes and why?
a) Pressure of H₂ gas is decreased
b) Area of the anode is doubled
c) Sulfuric acid is added to the cathode compartment, increasing [H⁺]
d) Sodium nitrate is added to the anode compartment.

I am at a complete loss of what to do   Any help would be greatly appreciated!

[Reptilezoo]

The overall cell equation according to the diagram and the title is actually:

Zn (s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)

What you have stated is the zinc half cell connected to the standard hydrogen half cell.
As for the other points in your questions, I am not sure.

As for pressure, I think (am very uncertain), that this would not alter the equilibrium constant (which determines how far a reaction proceeds). The increase in pressure usually pushes the reaction to side with lower number of molecules.

As for the acid and nitrate bits, this would probably affect the concnetrations of the ions in the half cells, by decreasing them, causing equilibrium shifts to the side to best counteract this change.

I'm not entirely sure on this, maybe someone can correct me if I'm wrong.

Luke

[Rabn]

You would do us a huge favor in our endeavors to help you out by giving us the composition of each half-cell in your particular problem (please use standard notation, if you don't know it you should definitely figure it out). This will let us know what your half-reactions are and from there we can help.

Start with this and then we'll move on.