Answers to problems in "Electrode Potentials"

Question 6

The cell diagram is:

Ag | AgCl | Cl2(g) | Pt

The cell is reversible in both solid and liquid phases.

RHE: ½Cl2(g) + e- ⇌ Cl-(s/l)

LHE: AgCl(s/l) + e- ⇌ Ag(s/l) + Cl-(s/l)

Formal Cell Reaction:

½Cl2(g) + Ag(s) ⇌ AgCl(s/l)

We know that G = H - TS and H = U +PV

So, G = U + PV - TS

Therefore, dG = dU + PdV + VdP - TdS - SdT

But, dU = dq + dw (where the bold derivatives remind us that these two quantities are non-state functions. [See Price, G, "Thermodynamics of Chemical Processes"; Oxford Chemistry Primer, OUP (in press) for a treatment of state functions.]

Under reversible conditions, dS = dqrev/T, and thus, considering only expansion work,

dU = TdS - PdV and so dG = VdP - SdT

Hence, (∂G/∂T)P = -S, or (∂DG/∂T)P = - DS

(where the ∂ refers to a partial derivative. See Sivia, ibid. for more about partial derivatives.)

Since DG = -FE for a one-electron process,

DS = F(∂E/∂T)P

Therefore, a graph of E against T will enable us to calculate DS.

We observe two lines, as the temperature range of the experiment spans the transition temperature for the AgCl(s)/ AgCl(l) phase change. The melting point of the compound will be the temperature of the intersection of these two lines. Tfus = 742.5K One line (viz. the one with steeper gradient) refers to the cell reaction: ½Cl2(g) + Ag(s) ⇌ AgCl(l). The entropy of this reaction will therefore be related to the gradient of this line. From the graph, the gradient is 5.0286 x 10-4V K-1 Similarly, the other line refers to the cell reaction:

½Cl2(g) + Ag(s) ⇌ AgCl(s).

The gradient of this line relates to the entropy of this cell reaction. From the graph, the gradient is 3.36 x 10-4V K-1 The entropy corresponding to the phase transition: AgCl(s) → AgCl(l) will be the difference of these two entropies.

Hence, since DGfus = 0 at Tfus (as melting [fusion] is an equilibrium process),

DHfus = Tfus DSfus

From graph, DSfus = F{5.0286 - 3.36} x 10-4

= +16.1Jmol-1K-1

This is positive as expected, as there is an increase in molecular randomness (ie. entropy) on going from the solid to liquid phase.

DHfus = Tfus DSfus = +11.95kJmol-1

Reference to a data book (eg. Lide, D R (ed.) "CRC Handbook of Chemistry and Physics", 76th edn., CRC Press, Boca Raton, Florida (1995)) shows that these results are accurate within the limits of experimental error, again illustrating the usefulness of electrode potentials in determining thermodynamic data.