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