[TIGERHULL]

Hi, I've got 2 questions that I'm finding quite tough.
 
1) The EMF of the cell
Pt l H2 (g) P= 1atm l HBr (10^-4M) l CuBr (s) l Cu

is 0.559 at 298K. The standard electrode potential for the Cu/Cu+ couple is 0.522V. Calculate the solubility product of CuBr.

My attempt:

The first half cell reaction is Cu+ 1/2H2 -> Cu (s) + H+ = 0.522V

The second half cell reaction is:

CuBr + 1/2 H2 -> Cu(s) + Br- (aq) + H+ (aq)

So if we add turn the top equation around and them, we get:

CuBr -> Cu+ + Br-

So that would be 0.559 - 0.522 = 0.007V

Then RTln K = -FE so e^ -FE/RT = K

So K=0.761

Is that right and would you mind explaining why they've put HBr in the equation, they haven't put in a salt bridge and they've given us the number of moles of HBr? Thanks


2) The second question is

The standard electrode potential of the cell

PtlH2 (g) P=1atm l HCl a=1 l AgCl l AG

can be expressed in: 'formula for E standard that uses temperature, it's quite long but i.e. 3T + 300-T^2 etc.

Write down the cell reactiona nd calculate dG, dS and dH

I can do this using dG = -dnFE and then find S from dg/dt and then plug them in to find dH. However, I haven't got two temperatures because it says to work it out for 298K, does that mean find the EMF at 288K and 308K and the average dG because they seem to have done this in an example?
Thanks a lot

[DevaDevil]

The first half cell reaction is Cu+ 1/2H2 -> Cu (s) + H+ = 0.522V

I am sorry, but it isn't... The Pt/H₂ side of the cell does not have any copper ions for one. And a half-cell reaction always produces or uses electrons... (half cell, remember)

The second half cell reaction is:

CuBr + 1/2 H2 -> Cu(s) + Br- (aq) + H+ (aq)

again it isn't. For one, the copper bromide solid may not be AT the electrode. What happens here is copper dissolution from your copper electrode.


The reason the CuBr is mentioned is because you can calculate the concentration of copper ions in your copper half-cell by using the nernst equation. From this you can calculate solubility product of the copper bromide (assuming the source of all copper ions in solution is from the copper bromide)
The HBr is the carrying electrolyte. You need it to calculate the Br concentration to use in the solubility product.