Answers to problems in "Electrode Potentials"

Question 3

The cell diagram is:

Pt | Fe2+(aq), (a=1), Fe3+(aq), (a=1) || Ce3+(aq), (a=1), Ce4+(aq), (a= 1) | Pt E0cell = +0.84V
At 298K

Potential determining equilibria:


RHE Ce4+(aq) + e- ⇌ Ce3+(aq)
LHE Fe3+(aq) + e- ⇌ Fe2+(aq)

Formal cell reaction

Ce4+(aq) + Fe2+(aq) ⇌ Ce3+(aq) + Fe3+(aq)

This is a 1 electron transfer.

DG0 = -FE0cell

DG0 = -RTln K

-FE0cell = -RTln K

ln K = FE0cell/RT

ln K = (96 490 x 0.84) / (8.313 x 298) = 32.72

K = exp(32.72) = 1.6 x 1014

Notice that in this case K is a dimensionless quantity.