Answers to problems in "Electrode Potentials"
Question 4
The cell diagram is:
Pt | H_{2}(g) (P = 1atm) | HBr (10^{-4}M) | CuBr(s) | Cu(s) | E_{298K} = +0.559V |
RHE: | CuBr(s) + e^{-} ⇌ Cu(s) + Br^{-}(aq) | E^{0}_{Cu|CuBr} = ? |
LHE: | H^{+}(aq) + e^{-} ⇌ ½H_{2}(g) | E^{0}_{H+|H2} = 0.00V (by definition) |
Therefore, E^{0}_{cell} = E^{0}_{Cu|CuBr}
Cell Reaction:
CuBr(s) + ½H_{2}(g) ⇌
Cu(s) + Br^{-}(aq) + H^{+}(aq)
Hence, the Nernst Equation for this equilibrium is:
Therefore, E^{0}_{cell} = E_{Cu¦CuBr}
= +0.0861V
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 (K_{SP}) for CuBr.
Now, E^{0}_{cell} = E^{0}_{Cu|CuBr}
- E^{0}_{Cu+|Cu }= 0.0861 - 0.522
= -0.4359V
DG^{0} = -RTlnK_{SP} = -nFE^{0}_{cell }
where n is the number of electron involved in the cell equilibrium (= 1 in this case).
Therefore,
The reader will find that the quoted solubility product at 298K in data books is 6.27 x 10^{-9} mol^{2} 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).