Answers to problems in "Electrode Potentials"
Question 9
Cell diagram:
Pt | H_{2} (1 atm) | NH_{3}
(c), NH_{4}Cl (c) | AgCl | Ag
This is (again) a Harned Cell ie.
Pt | H_{2 }| HCl | AgCl |Ag |
RHE: | AgCl(s) + e^{-} ⇌ Ag(s) + Cl^{-}(aq) |
LHE: | H^{+}(aq) + e^{- }⇌ ½H_{2}(g) |
Cell Reaction:
AgCl(s) + ½H_{2}(g) ⇌ Ag(s) + H^{+}(aq) + Cl^{-}(aq) |
Nernst Equation: E_{cell} = E^{0}_{cell} - {RT/F}ln{a_{H+} a_{Cl-}},
for P_{H2 }= 1 bar (which approximates to 1 atm., and where E^{0}_{cell} = E^{0}_{Ag|AgCl} (by definition of E^{0}_{H+|H2 }= 0V).
Consider the acid dissociation equilibrium for the ammonium ion:
The acid dissociation constant for this equilibrium is: K_{a} = a_{NH3} a_{H+} / a_{NH4+}
We notice from the cell diagram that [NH_{4}^{+}]
= [NH_{3}] = c.
Thus, K_{a} = a_{H+} f_{NH3} / f_{NH4+}. We can assume that f_{NH3} = 1. This is very reasonable, because ammonia is an uncharged species. Therefore, K_{a }= a_{H+} / f_{NH4+}, which implies that:
Thus, E = 0.2223 - {RT/F}ln{c K_{a} f_{NH4+} f_{Cl-}}
Now, c < 10^{-2}M, and so the DHL applies. (lg f_{i} = -Az_{i}^{2} I^{½}) Since z_{i}^{2} and I are the same for both NH_{4}^{+} and Cl^{-},
Hence, E + {RT/F}ln(c) - 0.2223 = +4.606RT A I^{½} / F - {RT/F}ln{K_{a}}
We now plot a graph of E + {RT/F}ln(c) - 0.2223 against I^{½} (= c^{½}). This should be a straight line with ordinate-intercept -{RT/F}ln{K_{a}}.
c / M | E / V | c / mol dm^{-3/2 } | E + {RT/F}ln(c) - 0.2223 / V |
0.00857 |
0.8965 | 0.09257 | 0.55198 |
0.01580 |
0.8824 | 0.12570 | 0.55359 |
0.02120 |
0.8757 | 0.14560 | 0.55464 |
0.03030 |
0.8661 | 0.18166 | 0.55620 |
From graph, ordinate-intercept = 5.5 x 10^{-1}V mol^{-½} dm^{3/2}
Therefore, Ka = 5.5 x 10^{-10} mol dm-3 (pKa = 9.26)
Kb refers to the forward reaction of the following equilibrium:
Now, Kw = 1.008 x 10^{-14} mol^{2} dm^{-6} = Ka Kb
Therefore, Kb = 1.008 x 10^{-14} / 5.5 x 10^{-10} = 1.7 x 10^{-5} mol dm^{-3}
Hence, using DG0 = -RT ln Kb, DG0 = +27.2 kJ mol^{-1}
Since DG0 = DH0 - TDS0, DS0 = (4.53 - 27.2) x 10^{-3} / 298
DS0 = - 76.00 J mol^{-1} K^{-1}
The protonation of ammonia by water is an entropically disfavoured reaction, a facet manifested in the negative sign of DS0. This is primarily due to the increased ordering of the solvent (water) around the ionic species formed. The unfavourable entropy of solvation combined with the positive (unfavourable) enthalpy of reaction, causes the Gibbs free energy for the (forward) reaction to be positive. Hence, ammonia is only a weak base in water at 298K.