Answers to problems in "Electrode Potentials"
The cell diagram is:
|Pt | H2(g) (P = 1atm) | HBr (10-4M) | CuBr(s) | Cu(s)||E298K = +0.559V|
|RHE:||CuBr(s) + e- ⇌ Cu(s) + Br-(aq)||E0Cu|CuBr = ?|
|LHE:||H+(aq) + e- ⇌ ½H2(g)||E0H+|H2 = 0.00V (by definition)|
Therefore, E0cell = E0Cu|CuBr
CuBr(s) + ½H2(g) ⇌
Cu(s) + Br-(aq) + H+(aq)
Hence, the Nernst Equation for this equilibrium is:
Therefore, E0cell = ECu¦CuBr
Now, consider the cell:
The cell reaction is: CuBr(s) ⇌ Cu+(aq) + Br-(aq)
We notice that the equilibrium constant for
this process is the solubility product (KSP) for CuBr.
Now, E0cell = E0Cu|CuBr
- E0Cu+|Cu = 0.0861 - 0.522
-RTlnKSP = -nFE0cell
where n is the number of electron involved in the cell equilibrium (= 1 in this case).
The reader will find that the quoted solubility product at 298K in data books is 6.27 x 10-9 mol2 dm-6 - an order of magnitude lower than the above result, which in fact corresponds to the value as measured at 293K. The interested reader will find it constructive to calculate a value for the enthalpy of dissolution of CuBr, using the van't Hoff isochore (assuming, to a good approximation, that this value does not change with temperature).