Answers to problems in "Electrode Potentials"

Question 4

The cell diagram is:

Pt | H2(g) (P = 1atm) | HBr (10-4M) | CuBr(s) | Cu(s) E298K = +0.559V
Potential Determining Equilibria:
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

Cell Reaction: CuBr(s) + ½H2(g) ⇌ Cu(s) + Br-(aq) + H+(aq)

Hence, the Nernst Equation for this equilibrium is:

E = E0cell - {RT/F}ln{aBr- aH+ / PH2½} = 0.559V

Therefore, E0cell = ECu¦CuBr = +0.0861V

Now, consider the cell:

Cu(s) | Cu+(aq) || Br-(aq) | CuBr(s) | Cu(s)

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 = -0.4359V

DG0 = -RTlnKSP = -nFE0cell

where n is the number of electron involved in the cell equilibrium (= 1 in this case).


KSP = exp{ FE0cell / RT}

KSP = 4.225 x 10-8 mol2 dm-6

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).