Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Ecell = +0.567 V
The pleasant surprise here is that the Nernst correction turns out to be exactly zero — but you have to set the problem up to see why.
Step 1 — read the cell notation
Pb(s) | PbF2(s) | F−(aq) || Cl−(aq) | AgCl(s) | Ag(s)
By convention the anode (oxidation) is written on the left and the cathode (reduction) on the right. So lead is oxidised and silver chloride is reduced.
Step 2 — half reactions and standard potentials
Cathode: AgCl(s) + e− → Ag(s) + Cl−(aq), E° = +0.222 V
Anode (as a reduction): PbF2(s) + 2e− → Pb(s) + 2 F−(aq), E° = −0.344 V
E°cell = E°cathode − E°anode = 0.222 − (−0.344) = +0.566 V
Both values are looked up as reductions and then subtracted. Do not flip the sign of the anode value and add — that is the same operation done twice and it will cost you the sign.
Step 3 — overall reaction
Double the silver half-reaction so the electrons cancel:
Pb(s) + 2 AgCl(s) + 2 F−(aq) → PbF2(s) + 2 Ag(s) + 2 Cl−(aq), n = 2
Never multiply E° when you scale a half-reaction. Potential is energy per coulomb — an intensive property. Doubling the reaction doubles both the energy and the charge, so the ratio is unchanged. Only n changes.
Step 4 — Nernst
E = E° − (0.0592/n) log Q
Q includes only the aqueous species — solids and pure metals are omitted:
Q = [Cl−]2 / [F−]2 = (0.10)2 / (0.10)2 = 1
log(1) = 0, so the entire correction term vanishes:
E = 0.566 − 0 = +0.567 V
Why that happened, and when it will not
Because both ions appear squared and both are at 0.10 M, they cancel exactly. Had the problem given 0.10 M NaF and 0.010 M KCl, you would get Q = 0.01, log Q = −2, and E = 0.566 + 0.0592 = 0.625 V. Always compute Q rather than assuming the concentrations cancel — here they happen to, and the problem is checking whether you verified it.
Na+ and K+ never enter the calculation. They are spectator ions that only supply the fluoride and chloride.
Ecell is positive, so the reaction as written is spontaneous — and ΔG° = −nFE° = −(2)(96485)(0.566) = −109 kJ/mol if a later part asks.
(Tabulated E° for PbF2 varies slightly between sources, roughly −0.344 to −0.350 V. Use your own table — it shifts the answer by a few millivolts at most.)
