Answers to problems in "Electrode Potentials"
Question 5
The cell diagram is:
| Pt | H2(g) (P = 1 atm) | HCl (a = 1) | AgCl(s) | Ag(s) |
This is the Harned Cell. We first write down the potential determining equilibria
| at the RHE: | AgCl(s) + e- ⇌ Ag(s) + Cl-(aq) |
| at the LHE: | H+(aq) + e- ⇌ ½H2(g) |
Hence, the cell reaction is:
| AgCl(s) + ½H2(g) ⇌ Ag(s) + H+(aq) + Cl-(aq) |
For one electron transfer processes, DG0 = -FE0cell
Therefore, at 298K,
E0cell = 0.2366 - 4.856 x 10-4(T - 273) - 3.421 x 10-6(T - 273)2
= 0.2366 - 4.856 x 10-4(298-273) - 3.421 x 10-6(298 - 273)2
= +0.225V
Therefore, DG0 = -(96 490)(0.225) = -21.45kJ mol-1
The reader is advised to review pp. 25-28. We quote the result DS0 = F(∂E0/ ∂T)P
Differentiating the expression given in the question with respect to T (at constant P) using the chain rule (see Sivia, D, "Fundamentals of Mathematics for Chemists"; Oxford Chemistry Primer, OUP (to be published)) gives,
Therefore, at 298K, (∂E0/∂T)P = -6.5665 x 10-4 V K-1
Hence, DS0
= -63.38J mol-1 K-1
Since, DG0
= DH0 - TDS0
DH0 = -40.33kJ mol-1